TABLE OF CONTENTS
1 Functions 1
1.1
1.2
1.3
1.4
1.5
1.6
Functions and Their Graphs 1
Combining Functions; Shifting and Scaling Graphs 8
Trigonometric Functions 18
Graphing with Software 26
Exponential Functions 31
Inverse Functions and Logarithms 34
Practice Exercises 43
Additional and Advanced Exercises 52
2 Limits and Continuity 59
2.1
2.2
2.3
2.4
2.5
2.6
Rates of Change and Tangents to Curves 59
Limit of a Function and Limit Laws 62
The Precise Definition of a Limit 73
One-Sided Limits 81
Continuity 86
Limits Involving Infinity; Asymptotes of Graphs 92
Practice Exercises 102
Additional and Advanced Exercises 108
3 Differentiation 115
3.1 Tangents and the Derivative at a Point 115
3.2 The Derivative as a Function 121
3.3 Differentiation Rules 131
3.4 The Derivative as a Rate of Change 138
3.5 Derivatives of Trigonometric Functions 144
3.6 The Chain Rule 152
3.7 Implicit Differentiation 162
3.8 Derivatives of Inverse Functions and Logarithms 170
3.9 Inverse Trigonometric Functions 180
3.10 Related Rates 186
3.11 Linearization and Differentials 192
Practice Exercises 199
Additional and Advanced Exercises 214
4 Applications of Derivatives 219
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Extreme Values of Functions 219
The Mean Value Theorem 233
Monotonic Functions and the First Derivative Test 239
Concavity and Curve Sketching 253
Indeterminate Forms and L Hôpital s Rule 280
Applied Optimization 290
Newton's Method 304
Antiderivatives 309
Practice Exercises 318
Additional and Advanced Exercises 336
5 Integration 343
5.1
5.2
5.3
5.4
5.5
5.6
Area and Estimating with Finite Sums 343
Sigma Notation and Limits of Finite Sums 348
The Definite Integral 354
The Fundamental Theorem of Calculus 369
Indefinite Integrals and the Substitution Method 379
Substitution and Area Between Curves 387
Practice Exercises 407
Additional and Advanced Exercises 422
6 Applications of Definite Integrals 431
6.1
6.2
6.3
6.4
6.5
6.6
Volumes Using Cross-Sections 431
Volumes Using Cylindrical Shells 443
Arc Length 454
Areas of Surfaces of Revolution 462
Work and Fluid Forces 468
Moments and Centers of Mass 479
Practice Exercises 492
Additional and Advanced Exercises 501
7 Integrals and Transcendental Functions 507
7.1
7.2
7.3
7.4
The Logarithm Defined as an Integral 507
Exponential Change and Separable Differential Equations 515
Hyperbolic Functions 521
Relative Rates of Growth 529
Practice Exercises 535
Additional and Advanced Exercises 540
8 Techniques of Integration 543
8.1 Using Basic Integration Formulas 543
8.2 Integration by Parts 555
8.3
8.4
8.5
8.6
8.7
8.8
8.9
Trigonometric Integrals 569
Trigonometric Substitutions 577
Integration of Rational Functions by Partial Fractions 585
Integral Tables and Computer Algebra Systems 594
Numerical Integration 607
Improper Integrals 617
Probability 629
Practice Exercises 637
Additional and Advanced Exercises 650
9 First-Order Differential Equations 661
9.1
9.2
9.3
9.4
9.5
Solutions, Slope Fields, and Euler's Method 661
First-Order Linear Equations 670
Applications 674
Graphical Solutions of Autonomous Equations 678
Systems of Equations and Phase Planes 986
Practice Exercises 692
Additional and Advanced Exercises 698
10 Infinite Sequences and Series 701
10.1 Sequences 701
10.2 Infinite Series 712
10.3 The Integral Test 720
10.4 Comparison Tests 728
10.5 Absolute Convergence; The Ratio and Root Tests 738
10.6 Alternating Series and Conditional Convergence 744
10.7 Power Series 752
10.8 Taylor and Maclaurin Series 764
10.9 Convergence of Taylor Series 769
10.10 The Binomial Series and Applications of Taylor Series 777
Practice Exercises 786
Additional and Advanced Exercises 795
11 Parametric Equations and Polar Coordinates 801
11.1
11.2
11.3
11.4
11.5
11.6
11.7
Parametrizations of Plane Curves 801
Calculus with Parametric Curves 809
Polar Coordinates 819
Graphing Polar Coordinate Equations 825
Areas and Lengths in Polar Coordinates 832
Conic Sections 838
Conics in Polar Coordinates 849
Practice Exercises 860
Additional and Advanced Exercises 871
CHAPTER 1 FUNCTIONS
1.1
FUNCTIONS AND THEIR GRAPHS
(
1. domain
, ); range [1, )
2. domain [0, ); range
3. domain [ 2, ); y in range and y
5 x 10
5. domain
3 t
(
4
3 t
0
6. domain
4 t
, 3)
x2
(
(3, ); y in range and y
0
( 4, 4)
16 t 2
real number
4
t
3x
0
16
range
(
2
16
0
1]
8
,
2
t 2 16
, or if t
3
3 t
0
range
(
, 0)
2 ,
t 2 16
2
(4, ); y in range and y
range [0, ).
y can be any positive real number
, now if t
y can be any nonzero real number
, 4)
4
3
, 1]
y can be any positive real number
0
4. domain ( , 0] [3, ); y in range and y
range [0, ).
(
4
now if t
16
t
0
3
4
0, or if t
t
3
(0, ).
t 2 16
4
2
t 2 16
2
t 2 16
0
0
0, or if
y can be any nonzero
(0, ).
7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.
8. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Not the graph of a function of x since it fails the vertical line test.
x; (height) 2
9. base
x
2
perimeter is p ( x)
10. s
D2
x2
x x x
s2
side length
11. Let D
2
s2
3
2
height
3 2
(base)(height)
d2
d ; and
2
s
d2
d . The
3
s2
area is a
a
1
m2
13. 2 x 4 y
5
5 x2
4
5
4
14. y
x 3
the length of an edge. Then 2
6d 2
3
surface area is 6 2
D2
y4
d 2 and
2d 2 and the volume is 3
x
x
d2
3
1
x
(x
3/2
d3 .
3 3
0).
, m1 .
1
2
y
x
3
2 y2 1 y2
( x 0)2
5;L
4
20 x 2
25
x 16
y2
3 2
x ;
4
x
1 d2
2
12. The coordinates of P are x, x so the slope of the line joining P to the origin is m
Thus, x, x
3
2
1 ( x)
2
3x.
diagonal length of a face of the cube and
2 2
1
2
x; area is a ( x)
x; L
y4
20 x
16
25
( x 4)2
( y 0)2
20 x 2
20 x
4
( y 0)2
x2
( 12 x
5 )2
4
4)2
y2
x2
1 x2
4
5
4
25
x 16
25
( y2
3
( y 2 1) 2
y2
y2 1
Copyright
2014 Pearson Education, Inc.
1
2
Chapter 1 Functions
15. The domain is (
, ).
16. The domain is (
, ).
17. The domain is (
, ).
18. The domain is (
, 0].
19. The domain is (
, 0)
20. The domain is (
, 0)
21. The domain is (
, 5)
(0, ).
( 5, 3] [3, 5)
23. Neither graph passes the vertical line test
(a)
Copyright
(5, ) 22. The range is [2, 3).
(b)
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(0, ).
Section 1.1 Functions and Their Graphs
24. Neither graph passes the vertical line test
(a)
(b)
x
x
y
1
26.
x 0 1 2
y 0 1 0
27. F ( x)
4 x2 , x 1
x
2
29. (a) Line through (0, 0) and (1, 1): y
x, 0 x 1
f ( x)
x 2, 1 x 2
(b) f ( x)
2,
0,
2,
0,
0
1
2
3
x
y
1,
x
x
0
x, 0
x
x; Line through (1, 1) and (2, 0): y
x 2
x
x
2
3
4
x 2, 0
x
2
5,
3
x
5
1x
3
1
x 1
30. (a) Line through (0, 2) and (2, 0): y
Line through (2, 1) and (5, 0): m
f ( x)
y
2
x 2
0
5
Copyright
1
2
1
3
1 , so
3
y
1 (x
3
2) 1
2014 Pearson Education, Inc.
x
or
x 0 1 2
y 1 0 0
28. G ( x)
2 x, x 1
y 1
or
x
25.
y 1
1x
3
5
3
1 x
3
4
Chapter 1 Functions
(b) Line through ( 1, 0) and (0, 3): m
Line through (0, 3) and (2, 1) : m
3 0
0 ( 1)
1 3
4
2 0
2
3, so y
3x 3
2, so y
2x 3
3 x 3, 1 x 0
2 x 3, 0 x 2
f ( x)
31. (a) Line through ( 1, 1) and (0, 0): y
x
Line through (0, 1) and (1, 1): y 1
0 1
Line through (1, 1) and (3, 0): m 3 1
1
2
x
1 , so
2
1 (x
2
y
1 x 0
0 x 1
1 x 3
x
1
f ( x)
1
2
3
2
1
2
1
2
(b) Line through ( 2, 1) and (0, 0): y
x
Line through (0, 2) and (1, 0): y
2x 2
Line through (1, 1) and (3, 1): y
1
32. (a) Line through
f ( x)
33. (a)
34.
x
x
and (T, 1): m
0, 0
2
T
x
x 1, T2
A,
(b) f ( x)
T ,0
2
0
0
(T /2)
x
2
x
2x 2
0
x 1
1
2 , so
T
y
(b)
x
2
T
x T2
3
2
x
0
2
T
1 x
0
3
x 1
T
2
x T
x
T
2
A, T2
x T
A,
T
x
3T
2
A, 32T
x
2T
0 for x
1
T
f ( x)
1
2
1) 1
[0, 1)
0 for x
( 1, 0]
x only when x is an integer.
(n 1)
35. For any real number x, n x n 1, where n is an integer. Now: n x n 1
By definition: x
n and x
n
x
n. So x
x for all real x.
36. To find f(x) you delete the decimal or
fractional portion of x, leaving only
the integer part.
Copyright
2014 Pearson Education, Inc.
x
n.
Section 1.1 Functions and Their Graphs
37. Symmetric about the origin
Dec:
x
Inc: nowhere
38. Symmetric about the y-axis
x 0
Dec:
Inc: 0 x
39. Symmetric about the origin
Dec: nowhere
x 0
Inc:
0 x
40. Symmetric about the y-axis
Dec: 0 x
Inc:
x 0
41. Symmetric about the y-axis
x 0
Dec:
Inc: 0 x
42. No symmetry
x 0
Dec:
Inc: nowhere
Copyright
2014 Pearson Education, Inc.
5
6
Chapter 1 Functions
43. Symmetric about the origin
Dec: nowhere
Inc:
x
44. No symmetry
Dec: 0 x
Inc: nowhere
45. No symmetry
Dec: 0 x
Inc: nowhere
46. Symmetric about the y-axis
Dec:
x 0
Inc: 0 x
47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the
origin, the function is even.
x 5
48. f ( x)
1
x5
and f ( x)
x2
49. Since f ( x)
1 ( x)2
50. Since [ f ( x) x 2
even nor odd.
3x 2
1 ( x) 4
3( x)2
53. g ( x)
1
x2 1
54. g ( x)
x ;
x2 1
1
1
( x)2
x] [ f ( x)
x3
x4
1
( x )2 1
; h( t )
t
1
1
x5
f ( x). Thus the function is odd.
x ] and [ f ( x)
( x3
x
1
x)
x2
x] [ f ( x)
( x)2
x] the function is neither
g ( x ). So the function is odd.
g ( x), thus the function is even.
g ( x). Thus the function is even.
x
x2 1
g ( x)
1
( x )5
f ( x). The function is even.
x, g ( x )
52. g ( x)
t
1
x3
51. Since g ( x)
55. h(t )
( x) 5
1
g ( x). So the function is odd.
; h (t )
1
1
t
. Since h(t )
Copyright
h(t ) and h(t )
h( t ), the function is neither even nor odd.
2014 Pearson Education, Inc.
Section 1.1 Functions and Their Graphs
56. Since |t 3 |
|( t )3 |, h(t )
57. h(t ) 2t 1, h( t )
nor odd.
58. h(t )
59. s
60. K
h( t ) and the function is even.
2t
2| t | 1 and h( t )
kt
25
c v2
k (75)
12960
61. r
k
s
6
k
4
62. P
k
v
14.7
63. v
f ( x)
1. So h(t )
2| t |
k
1
3
k
24
k
1000
k
c
r
40
K
24 ; 10
s
24
s
2t 1, so h(t )
1t
3
t
180
40v 2 ; K
s
72 x 2
h(t ). The function is neither even
h( t ) and the function is even.
40(10) 2
4000 joules
12
5
14700 ; 23.4
v
P
4 x3
h( t )
1. So h(t )
1 t ; 60
3
14700
x(14 2 x )(22 2 x)
64. (a) Let h
1 2| t |
s
c(18)2
h( t ).
14700
v
308 x; 0
x
v
24500
39
628.2 in 3
7.
height of the triangle. Since the triangle is isosceles, AB
2
h 2 12
2
h 1 B is at (0, 1) slope of AB
y f ( x)
x 1; x [0, 1].
(b) A( x) 2 xy 2 x( x 1)
2 x 2 2 x; x [0, 1].
1
2
AB
2
22
The equation of AB is
65. (a) Graph h because it is an even function and rises less rapidly than does Graph g.
(b) Graph f because it is an odd function.
(c) Graph g because it is an even function and rises more rapidly than does Graph h.
66. (a) Graph f because it is linear.
(b) Graph g because it contains (0, 1).
(c) Graph h because it is a nonlinear odd function.
67. (a) From the graph, 2x
(b)
1 4x
x 1 4 0
x
2
x
( 2, 0)
x
2
1
x
0:
x
x
2x 8
0: 2x 1 4x 0
0
2x
x
2 since x is negative;
sign of ( x 4)( x 2)
4
x
x
2
x2 2 x
8
1 4x 0
0
2x
x 4 since x is positive;
2
Solution interval: ( 2, 0)
(4, )
(x
4)( x
2x
2)
0
( x 4)( x
2x
2)
0
(4, )
Copyright
AB
2014 Pearson Education, Inc.
2. So,
7
8
Chapter 1 Functions
68. (a) From the graph, x 3 1 x 2 1
x ( , 5)
3( x 1)
(b) Case x
1: x 3 1 x 2 1
2
x 1
( 1, 1)
3x 3 2 x 2 x
5.
Thus, x ( , 5) solves the inequality.
3( x
1)
x 1: x 3 1 x 2 1
2
x 1
3x 3 2 x 2 x
5 which
is true if x
1. Thus, x ( 1, 1)
solves the inequality.
5
Case 1 x : x 3 1 x 2 1 3 x 3 2 x 2 x
Case 1
which is never true if 1 x,
so no solution here.
In conclusion, x ( , 5) ( 1, 1).
69. A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and ( x, y ) lie
on the same vertical line. The graph of the function y f ( x) 0 is the x-axis, a horizontal line for which there
is a single y-value, 0, for any x.
70. price
71. x 2
40 5 x, quantity
x2
h2
x
300 25x
2h
;
2
h
2
cost
R ( x)
(40 5 x)(300 25 x)
5(2 x) 10h
72. (a) Note that 2 mi 10,560 ft, so there are 8002
C ( h) 10
2h
2
10h
2 2
x 2 feet of river cable at $180 per foot and (10,560 x)
feet of land cable at $100 per foot. The cost is C ( x ) 180 8002
(b) C (0) $1, 200, 000
C (500)
5h
x2
100(10,560
x).
$1,175,812
C (1000)
$1,186,512
C (1500)
$1, 212, 000
C (2000)
$1, 243, 732
C (2500)
$1, 278, 479
C (3000) $1,314,870
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet
from the point P.
1.2
COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. D f :
x
2. D f : x 1 0
Rf
3. D f :
Rg / f :
4. D f :
Rg : y
x
1
2
, Dg : x 1
x
0, R f
Df
g
D fg : x 1. R f :
1, Dg : x 1 0
g:
, Dg :
y
x 1. Therefore D f
2, R fg : y
x
y
g
, Rg : y
0, R f
g:
y 1, R fg : y
0
D fg : x 1.
0
, D f /g :
x
, D g /f :
x
, Rf : y
2, Rg : y 1, R f /g : 0
y
x
, Dg : x
0, D f /g : x
0, Dg /f : x
Copyright
0; R f : y 1, Rg : y 1, R f /g : 0
2014 Pearson Education, Inc.
y 1, Rg /f : 1
y
y
2,
Section 1.2 Combining Functions; Shifting and Scaling Graphs
5. (a) 2
(d) ( x 5) 2 3
(g) x 10
6. (a)
(d)
1
x
x 2 10 x 22
1
3
(g) x 2
6
(c)
(e) 0
(f )
(h)
1
1
x
x
1
1
1
f ( g (h( x)))
f ( g (4 x))
8. ( f g h)( x)
f ( g ( h( x)))
f ( g ( x 2 ))
f (2( x 2 ) 1)
9. ( f g h)( x)
f ( g ( h( x)))
f g 1x
f
f g
x
x
2
1
7. ( f g h)( x)
f ( g ( h( x)))
x4 6 x2
(b) 2
x
10. ( f g h)( x)
(c) x 2 2
(f ) 2
(b) 22
(e) 5
(h) ( x 2 3)2 3
f (3(4 x ))
1
x
2 x
1
2
f (12 3 x)
4
x
2
f
2
x
2
x
4x
x
2
3
2
3
x
x
3
x
x
2
3
2
x
x
8
7
(b) ( j g )( x)
(e) ( g h f )( x)
(c) ( g g )( x)
(f ) (h j f )( x)
12. (a) ( f j )( x)
(d) ( f f )( x )
(b) ( g h)( x)
(e) ( j g f )( x)
(c) ( h h)( x )
(f ) ( g f h)( x)
f (x)
( f g )( x)
(a) x 7
x
x 7
(b) x 2
3x
g(x)
(c) x 2
(d)
x
x
3( x 2)
x
x
1
x
(e)
x
1
1
(f ) 1
x
1 1x
x
1
x
x
1
14. (a) ( f g )( x) |g ( x )|
(b) ( f g )( x )
x
g ( x) 1
g ( x)
(c) Since ( f g )( x)
(d) Since ( f g )( x)
3x
2x
3x 6
x2 5
x 5
1
1
6x2 1
4
11. (a) ( f g )( x)
(d) ( j j )( x)
13.
x
5x 1
1 4x
1
2
1
1
1
3(2 x 2 1)
1
f
1
(12 3x) 1 13 3x
f (2 x 2 1)
f 1 x4 x
1
x
3
4
x
1
x
g ( x)
f x
x
x
x
1
1
1
x
x
(x
1)
x
.
1
1
1 , so g ( x ) x 1.
1 g (1x ) x x 1 1 x x 1 g (1x)
x 1 g ( x)
| x |, g ( x) x 2 .
| x |, f ( x) x 2 . (Note that the domain of the composite is [0, ).)
Copyright
2014 Pearson Education, Inc.
9
10
Chapter 1 Functions
The completed table is shown. Note that the absolute value sign in part (d) is optional.
g(x)
f(x)
1
| x|
x
1
x
x 1
x
( f g )(x)
1
x
1
x
x
x
1
2
x
| x|
x
2
| x|
x
15. (a) f ( g ( 1)) f (1) 1
(d) g ( g (2)) g (0) 0
16. (a)
(b)
(c)
(d)
(e)
1
f ( g (0))
g ( f (3))
g ( g ( 1))
f ( f (2))
g ( f (0))
(b) g ( f (0)) g ( 2)
(e) g ( f ( 2)) g (1)
2
1
(c) f ( f ( 1)) f (0)
2
(f) f ( g (1)) f ( 1) 0
f ( 1) 2 ( 1) 3, where g (0) 0 1
1
g ( 1)
( 1) 1, where f (3) 2 3
1
g (1) 1 1 0, where g ( 1)
( 1) 1
f (0) 2 0 2, where f (2) 2 2 0
g (2) 2 1 1, where f (0) 2 0 2
(f ) f g 12
1
2
f
17. (a) ( f g )( x)
f ( g ( x))
( g f )( x)
g ( f ( x))
5 , where
2
1
2
2
1
1 1
x
1
x 1
g 12
1
2
1
2
1
x
x
(b) Domain ( f g ): ( , 1] (0, ), domain ( g f ): ( 1, )
(c) Range ( f g ): (1, ), range ( g f ): (0, )
18. (a) ( f g )( x ) f ( g ( x)) 1 2 x x
( g f )( x ) g ( f ( x)) 1 | x |
(b) Domain ( f g ): [0, ), domain ( g f ): ( , )
(c) Range ( f g ): (0, ), range ( g f ): ( , 1]
19. ( f g )( x)
x
f ( g ( x))
g ( x) x g ( x)
20. ( f g )( x )
21. (a) y
22. (a) y
x 2
2x
g ( x)
f ( g ( x ))
g ( x)
g ( x) 2
2x
1 x
x 2
x
g ( x)
2( g ( x))3
x2 3
( x 1) 2
( g ( x) 2) x
4
x 2
( g ( x))3
(b) y
( x 4) 2
(b) y
(b) Position 1
4
(b) y
x
(c) y
6
2
g ( x)
3 x
6
2
x2 5
(c) Position 2
( x 2)2 3
Copyright
x g ( x) 2 x
2x
x 1
( x 7)2
23. (a) Position 4
24. (a) y
x
( x 4)2 1
2014 Pearson Education, Inc.
(d) Position 3
(d) y
( x 2) 2
Section 1.2 Combining Functions; Shifting and Scaling Graphs
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
Copyright
2014 Pearson Education, Inc.
11
12
Chapter 1 Functions
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
Copyright
2014 Pearson Education, Inc.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
47.
48.
49.
50.
51.
52.
53.
54.
55. (a) domain: [0, 2]; range: [2, 3]
(b) domain: [0, 2]; range: [ 1, 0]
Copyright
2014 Pearson Education, Inc.
13
14
Chapter 1 Functions
(c) domain: [0, 2]; range: [0, 2]
(d) domain: [0, 2]; range: [ 1, 0]
(e) domain: [ 2, 0]; range: [0, 1]
(f ) domain: [1, 3]; range: [0,1]
(g) domain: [ 2, 0]; range: [0, 1]
(h) domain: [ 1, 1]; range: [0, 1]
56. (a) domain: [0, 4]; range: [ 3, 0]
(b) domain: [ 4, 0]; range: [0, 3]
(c) domain: [ 4, 0]; range: [0, 3]
(d) domain: [ 4, 0]; range: [1, 4]
Copyright
2014 Pearson Education, Inc.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
(e) domain: [2, 4]; range: [ 3, 0]
(f ) domain: [ 2, 2]; range: [ 3, 0]
(g) domain: [1, 5]; range: [ 3, 0]
(h) domain: [0, 4]; range: [0, 3]
57. y
3x 2 3
59. y
1
2
1
58. y
1
x2
61. y
4x 1
63. y
4
1
2
x 2
2
65. y 1 (3 x)3
67. Let y
h( x )
j ( x)
60. y 1
1
2 x2
1
2
16 x 2
1 27 x3
2x 1
f ( x) and let g ( x)
1 1/2
2
1
( x /3)2
62. y
3 x 1
64. y
1
3
66. y 1
1/2
x 12
, i( x)
2 x
(2 x )2 1 4 x 2 1
1
4 x2
x
2
3
3
1 x8
x1/2 ,
1/2
2 x 12
, and
f ( x). The graph of h( x)
is the graph of g ( x) shifted left 12 unit; the graph
of i ( x) is the graph of h( x) stretched vertically by
a factor of 2; and the graph of j ( x) f ( x ) is the
graph of i ( x) reflected across the x-axis.
68. Let y
h( x )
1 2x
1 2x
f ( x). Let g ( x) ( x )1/2 ,
( x 2)1/2 , and i ( x )
1
2
( x 2)1/2
f ( x ). The graph of g ( x) is the graph
of y
x reflected across the x-axis. The graph
of h( x) is the graph of g ( x) shifted right two units.
And the graph of i ( x) is the graph of h( x)
compressed vertically by a factor of 2.
Copyright
9
x2
2014 Pearson Education, Inc.
15
16
Chapter 1 Functions
69. y f ( x) x3 . Shift f ( x) one unit right followed by
a shift two units up to get g ( x) ( x 1)3 2 .
70.
y
(1 x)3 2
[( x 1)3 ( 2)]
3
3
Let g ( x) x , h( x) ( x 1) ,
f ( x).
i( x) ( x 1)3 ( 2),
and j ( x)
[( x 1)3 ( 2)]. The graph of h( x) is the
graph of g ( x) shifted right one unit; the graph of i ( x)
is the graph of h( x) shifted down two units; and the
graph of f ( x) is the graph of i ( x) reflected across
the x-axis.
71. Compress the graph of f ( x)
1
x
horizontally by a
1 . Then
2x
factor of 2 to get g ( x)
vertically down 1 unit to get h( x )
72. Let f ( x)
1
x/ 2
2
1
x2
1
2
x2
and g ( x)
1
1/ 2 x
2
1
shift g ( x)
1
2x
1.
1
1
x2
2
1. Since 2
1.4, we see
that the graph of f ( x) stretched horizontally by
a factor of 1.4 and shifted up 1 unit is the graph
of g ( x).
73. Reflect the graph of y
3
x.
to get g ( x)
f ( x)
3
x across the x-axis
Copyright
2014 Pearson Education, Inc.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
74.
y
( 2 x) 2/3 [( 1)(2) x]2/3
f ( x)
( 1)2/3 (2 x) 2/3
(2 x)2/3 . So the graph of f ( x) is the graph of
g ( x) x 2/3 compressed horizontally by a factor of 2.
75.
76.
77. (a) ( fg )( x)
f ( x) g ( x )
f ( x)( g ( x))
( fg )( x), odd
(b)
f
g
( x)
f ( x)
g ( x)
f ( x)
g ( x)
f
g
( x), odd
(c)
g
f
( x)
g ( x)
f ( x)
g ( x)
f ( x)
g
f
( x), odd
(d) f 2 ( x)
f ( x) f ( x)
(e) g 2 ( x)
( g ( x))2
f 2 ( x), even
f ( x) f ( x)
( g ( x)) 2
g 2 ( x), even
(f ) ( f
g )( x)
f ( g ( x))
f ( g ( x))
(g) ( g
f )( x)
g ( f ( x))
g ( f ( x )) ( g
(h) ( f
f )( x)
f ( f ( x))
(i) ( g g )( x)
78. Yes, f ( x)
g ( g ( x))
f ( f ( x))
f ( g ( x))
g )( x), even
f )( x), even
(f
g ( g ( x))
(f
f )( x ), even
g ( g ( x))
0 is both even and odd since f ( x )
0
( g g )( x ), odd
f ( x ) and f ( x)
0
(b)
79. (a)
Copyright
2014 Pearson Education, Inc.
f ( x).
17
18
Chapter 1 Functions
(d)
(c)
80.
1.3
TRIGONOMETRIC FUNCTIONS
1. (a)
s
r
10
8
(10) 45
5
4
8 m
2.
s
r
3.
80
80 180
4. d
1 meter
r
radians and 54 180
4
9
30
50
0
2
3
2
0
1
0.6 rad or 0.6 180
3
4
1
2
cos
1
0
1
2
1
0
1
2
3
0
und.
1
cot
und.
1
3
und.
0
sec
1
2
1
und.
csc
und.
und.
(10)(110 ) 180
110
18
55
9
m
1
Copyright
1
2
2
radius
6 in.)
34
3
2
3
6
4
5
6
sin
1
3
2
1
2
1
2
1
2
cos
0
3
2
1
2
3
2
tan
und.
3
1
3
1
1
3
cot
0
1
3
3
1
6.
0
2
3
r
8.4 in. (since the diameter 12 in.
sin
tan
s
225
(6) 49
s
s
r
50 cm
2
3
5.
(b)
sec
und.
csc
1
2014 Pearson Education, Inc.
1
2
2
2
3
2
3
2
2
2
3
2
3
2
Section 1.3 Trigonometric Functions
7. cos x
4,
5
9. sin x
8
,
3
11. sin x
tan x
3
4
8. sin x
tan x
8
10. sin x
1 , cos x
5
2
5
12. cos x
13.
2 ,
5
12 ,
13
cos x
tan x
3
,
2
tan x
14.
period
15.
period
4
period
4
16.
period
2
17.
18.
period
period 1
6
19.
20.
period
2
period
Copyright
2
2014 Pearson Education, Inc.
1
5
12
5
1
3
19
20
Chapter 1 Functions
21.
22.
period
23. period
2
2
period
2
24. period 1, symmetric about the origin
, symmetric about the origin
s
3
2
s=
tan t
1
2
1
1
0
2
t
1
2
3
25. period
4, symmetric about the s-axis
26. period
4 , symmetric about the origin
27. (a) Cos x and sec x are positive for x in the interval
, ; and cos x and sec x are negative for x
2 2
, 2 and 2 , 32 . Sec x is
undefined when cos x is 0. The range of sec x is
( , 1] [1, ); the range of cos x is [ 1, 1].
in the intervals
3
2
Copyright
2014 Pearson Education, Inc.
Section 1.3 Trigonometric Functions
21
(b) Sin x and csc x are positive for x in the intervals
3 ,
and (0, ); and sin x and csc x are
2
negative for x in the intervals ( , 0) and
, 32 . Csc x is undefined when sin x is 0. The
range of csc x is (
sin x is [ 1, 1].
, 1] [1, ); the range of
28. Since cot x tan1 x , cot x is undefined when tan x 0
and is zero when tan x is undefined. As tan x
approaches zero through positive values, cot x
approaches infinity. Also, cot x approaches negative
infinity as tan x approaches zero through negative
values.
29. D :
31. cos x
32. cos x
33. sin x
34. sin x
x
2
; R: y
1, 0, 1
cos x cos
2
30. D :
sin x sin
2
(cos x)(0) (sin x)( 1)
2
cos x cos 2
sin x sin 2
(cos x)(0) (sin x)(1)
2
sin x cos 2
cos x sin 2
(sin x)(0) (cos x)(1)
2
35. cos( A B)
sin x cos
2
cos( A ( B))
cos x sin
2
x
1, 0, 1
sin x
sin x
cos x
(sin x)(0) (cos x)( 1)
cos A cos( B) sin A sin( B )
; R: y
cos x
cos A cos B sin A( sin B )
cos A cos B sin A sin B
36. sin( A B )
sin( A ( B ))
sin A cos( B ) cos A sin( B)
sin A cos B cos A( sin B )
sin A cos B cos A sin B
37. If B A, A B 0 cos( A B) cos 0 1. Also cos( A B)
cos 2 A sin 2 A. Therefore, cos 2 A sin 2 A 1.
cos( A A)
cos A cos A
sin A sin A
38. If B 2 , then cos( A 2 ) cos A cos 2
sin A sin 2
(cos A)(1) (sin A)(0) cos A and
sin( A 2 ) sin A cos 2
cos A sin 2
(sin A)(1) (cos A)(0) sin A . The result agrees with the fact that the
cosine and sine functions have period 2 .
39. cos(
x)
cos cos x sin sin x
( 1)(cos x) (0)(sin x)
Copyright
cos x
2014 Pearson Education, Inc.
22
Chapter 1 Functions
40. sin(2
x)
sin 2 cos( x ) cos(2 ) sin( x )
41. sin 32
x
sin 32 cos( x) cos 32 sin( x)
42. cos 32
x
cos 32 cos x sin 32 sin x
43. sin 712
sin 4
44. cos 11
12
2
3
cos 4
45. cos 12
cos 3
46. sin 512
sin 23
1
47. cos2 8
cos
sin 4 cos 3
3
cos 3 cos
2
8
1
2
1
2
49. sin 12
cos
2
2
12
1
2
51. sin 2
3
4
52. sin 2
cos 2
2
2
2
2
3
2
2
3
2
sin
sin 2
cos 2
1
2
2
2
2
2
1
2
4
3
2
3
50.
4
2
3
2
cos
1
3
2 2
1
cos
3
2
1
10
12
tan 2
54. cos 2 cos
0 2 cos 2
1 cos
cos
1 0 or 2cos 1 0 cos
6
8
2
1
0
tan
1
4
2
4
, 34 , 54 , 74
cos (2sin
1) 0 cos
5
,
,
, 5 , 32
6 6
6 2 6
2 cos 2
1 or cos
cos
1
2
1 0
or
0 or 2sin
sin A cos B cos A cos B
cos A cos B sin A sin B
sin A cos B
cos A cos B
cos A cos B
cos A cos B
cos A sin B
cos A cos B
sin A sin B
cos A cos B
tan A tan B
1 tan A tan B
56. tan( A B)
sin( A
cos( A
B)
B)
sin A cos B cos A cos B
cos A cos B sin A sin B
sin A cos B
cos A cos B
cos A cos B
cos A cos B
cos A sin B
cos A cos B
sin A sin B
cos A cos B
tan A tan B
1 tan A tan B
57. According to the figure in the text, we have the following: By the law of cosines, c 2
2 2cos( A B) . By distance formula, c 2
cos 2 A 2 cos A cos B cos 2 B sin 2 A 2sin A sin B sin 2 B
2 2(cos A cos B sin A sin B)
Copyright
1 0
(cos
1)(2 cos 1) 0
5
,
, , 53
3 3
3
B)
B)
2 2 cos( A B )
2
2
sin( A
cos( A
c2
3
4
2
2
1
55. tan( A B )
2cos( A B )
2
2
, 23 , 43 , 53
53. sin 2 cos
0 2sin cos
cos
0
3 , or
1
cos
0 or sin
,
2
2 2
12 12
1
3
2 2
2
2
1
2
2
sin 2 38
6
4
2
2
2
2
1
2
4
3
2
48. cos2 512
4
6
2
2
4
cos x
sin x
3
2
2
2
1
2
cos 23 sin
4
2
3
cos 2
cos 2
2
2
sin 3 sin
sin 23 cos
4
(0)(cos x) ( 1)(sin x)
sin 4 sin 23
4
sin x
( 1)(cos x) (0)(sin( x ))
cos 4 sin 3
cos 4 cos 23
4
(0)(cos( x )) (1)(sin( x))
a2
b2
2ab cos
(cos A cos B )2 (sin A sin B )2
2 2(cos A cos B sin A sin B ) . Thus
cos( A B)
cos A cos B sin A sin B .
2014 Pearson Education, Inc.
Section 1.3 Trigonometric Functions
23
58. (a) cos( A B) cos A cos B sin A sin B
sin
cos 2
and cos
sin 2
Let
A B
sin( A B )
cos 2 ( A B)
cos 2
sin A cos B cos A sin B
(b) cos( A B ) cos A cos B sin A sin B
cos( A ( B))
A
B
cos 2
A cos B sin 2
cos A cos( B ) sin A sin( B )
cos( A B) cos A cos( B) sin A sin( B) cos A cos B sin A( sin B )
Because the cosine function is even and the sine functions is odd.
59. c 2
a 2 b2
Thus, c
60. c 2
2ab cos C
7
a2
22
32
2(2)(3) cos(60 )
22
32
2(2)(3) cos(40 ) 13 12 cos(40 ). Thus, c
b2
4 9 12 cos(60 ) 13 12 12
2ab cos C
h . If
c
c sin B
ah
ab sin C
a
By the law of cosines, cos C
2
h
c
a
2
b
c
2 ab
2
a
2
2
b
c2
2ab
c b
2ac
2
a2
and cos B
sin(
h
b
Combining our results we have ah
sin A sin C
sin B
h
.
bc
a
c
b
13 12 cos 40°
1.951.
h . On the other hand,
b
h . Thus, in either case,
b
sin(
C)
ac sin B.
2
angles of triangle is , we have sin A
2
7.
C is an acute angle, then sin C
if C is obtuse (as in the figure on the right in the text), then sin C
b sin C
cos A cos B sin A sin B
2.65.
61. From the figures in the text, we see that sin B
h
A sin B
( B C ))
h
2 abc
(2a 2
ab sin C, ah
c2 b2
. Moreover,
2 ac
sin( B C )
b2
c2
c2
ac sin B, and ah
since the sum of the interior
sin B cos C
b2 )
ah
bc
cos B sin C
ah bc sin A.
bc sin A. Dividing by abc gives
law of sines
62. By the law of sines, sin2 A
sin B
3
3/2
. By Exercise
c
59 we know that c
7. Thus sin B
3 3
2 7
0.982.
63. From the figure at the right and the law of cosines,
b 2 a 2 22 2(2a ) cos B
a2
4 4a 12
a2
2a 4.
sin A
sin B
Applying the law of sines to the figure, a
b
2/2
3/2
3 a. Thus, combining results,
b
a
b
2
a2
a
2a 4
b2
0, we have a
3 a2
2
4
0
4
2
2
1 a2
2
4(1)( 8)
2a 4
0
4 3 4
2
a2
4a 8 . From the quadratic formula and the fact that
1.464.
64. (a) The graphs of y sin x and y x nearly coincide when x is near the origin (when the calculator is in
radians mode).
(b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves
look like intersecting straight lines near the origin when the calculator is in degree mode.
Copyright
2014 Pearson Education, Inc.
24
Chapter 1 Functions
65. A
2, B
2 ,C
66. A
1,
2
2, C
67. A
68.
A
69 72.
B
2,
L
2
,D
1, D
1
1
2
B
4, C
0, D
1
,B
L, C
0, D
0
Example CAS commands:
Maple:
f : x - A*sin((2*Pi/B)*(x-C)) D1;
A: 3; C: 0; D1: 0;
f_list : [seq(f(x), B [1,3,2*Pi,5*Pi])];
plot(f_list, x -4*Pi..4*Pi, scaling constrained,
color [red,blue,green,cyan], linestyle [1,3,4,7],
legend ["B 1", "B 3","B 2*Pi","B 3*Pi"],
title "#69 (Section 1.3)");
Mathematica:
Clear[a, b, c, d, f, x]
f[x_]: a Sin[2 /b (x c)] d
Plot[f[x]/.{a
3, b
1, c
0, d
0}, {x, 4 , 4 }]
69. (a) The graph stretches horizontally.
Copyright
2014 Pearson Education, Inc.
Section 1.3 Trigonometric Functions
(b) The period remains the same: period
| B |. The graph has a horizontal shift of 12 period.
70. (a) The graph is shifted right C units.
(b) The graph is shifted left C units.
(c) A shift of one period will produce no apparent shift. | C |
6
71. (a) The graph shifts upwards | D | units for D 0
(b) The graph shifts down | D | units for D 0.
72. (a) The graph stretches | A| units.
Copyright
(b) For A
0, the graph is inverted.
2014 Pearson Education, Inc.
25
26
1.4
Chapter 1 Functions
GRAPHING WITH SOFTWARE
1 4.
The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the
graphs and has little unused space.
1. d.
2. c.
3. d.
4. b.
5 30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5 30
are not unique in appearance.
5. [ 2, 5] by [ 15, 40]
6. [ 4, 4] by [ 4, 4]
7. [ 2, 6] by [ 250, 50]
8. [ 1, 5] by [ 5, 30]
Copyright
2014 Pearson Education, Inc.
Section 1.4 Graphing with Software
9. [ 4, 4] by [ 5, 5]
10. [ 2, 2] by [ 2, 8]
11. [ 2, 6] by [ 5, 4]
12. [ 4, 4] by [ 8, 8]
13. [ 1, 6] by [ 1, 4]
14. [ 1, 6] by [ 1, 5]
15. [ 3, 3] by [0, 10]
16. [ 1, 2] by [0, 1]
Copyright
2014 Pearson Education, Inc.
27
28
Chapter 1 Functions
17. [ 5, 1] by [ 5, 5]
18. [ 5, 1] by [ 2, 4]
19. [ 4, 4] by [0, 3]
20. [ 5, 5] by [ 2, 2]
21. [ 10, 10] by [ 6, 6]
22. [ 5, 5] by [ 2, 2]
23. [ 6, 10] by [ 6, 6]
24. [ 3, 5] by [ 2, 10]
25. [ 0.03, 0.03] by [ 1.25, 1.25]
26. [ 0.1, 0.1] by [ 3, 3]
Copyright
2014 Pearson Education, Inc.
Section 1.4 Graphing with Software
27. [ 300, 300] by [ 1.25, 1.25]
28. [ 50, 50] by [ 0.1, 0.1]
29. [ 0.25, 0.25] by[ 0.3, 0.3]
30. [ 0.15, 0.15] by [ 0.02, 0.05]
31. x 2 2 x 4 4 y y 2
y 2
x2 2 x
The lower half is produced by graphing
y 2
32. y 2 16 x 2
x2
1
8.
2 x 8.
y
1 16 x 2 . The upper branch
is produced by graphing y
1 16 x 2 .
33.
34.
Copyright
2014 Pearson Education, Inc.
29
30
Chapter 1 Functions
35.
36.
37.
38.
200
8
150
6
100
4
50
2
60
64
68
72
76
0
1970 1980 1990 2000 2010 2020
80
39.
40.
300
26
225
22
R
18
T
150
14
75
10
6
1972 1980 1988 1996 2004 2012
41.
2000 2002 2004 2006 2008
42.
1
600
450
0.5
300
1955
1935
1975
1995
2015
150
0
0.5
Copyright
0
2
2014 Pearson Education, Inc.
4
6
8
10
Section 1.5 Exponential Functions
1.5 EXPONENTIAL FUNCTIONS
1.
2.
3.
4.
5.
6.
7.
8.
Copyright
2014 Pearson Education, Inc.
31
32
Chapter 1 Functions
10.
9.
11. 162 16 1.75
1
12. 91/3 91/6
35/3
32/3
16.
13 2
18.
3
19.
2
2
1
6
93
5
14.
162 ( 1.75)
2
3
33
31
2 /2
1/2
1/2
24
1/ 2 4
(2
21. Domain: (
)
161/4
2
3
13.
44.2 3.7
12
1/2
36
1/2
6
3
20.
y
1
2 ex
41/2
251/2
(2 7) 3
2
5
14 3
61/2
4
, ); y in range
40.5
254/8
17. 2 3 7 3
13
3
16
22
44.2
43.7
15. (251/8 ) 4
3
132/2
12
4
91/2
160.25
2
(61/ 2 )2
32
6
9
2
3
. As x increases, e x becomes infinitely large and y becomes a smaller
and smaller positive real number. As x decreases, e x becomes a smaller and smaller positive real number,
y
1,
2
and y gets arbitrarily close to 12
22. Domain: (
1
cos x
23. Domain: (
24. If e2 x
, ); y in range
1
y
Range: 0, 12 .
cos(e t ). Since the values of e t are (0, ) and
Range: [ 1, 1].
, ); y in range
1, then x = 0
y
Domain: (
< y < 0. If x < 0, then 0
e2 x
1 3 t . Since the values of 3 t are (0, )
, 0)
1
Copyright
(0, ); y in range
3<y<
Range: (
y
3
1 e2 x
, 0)
2014 Pearson Education, Inc.
Range: (1, ).
. If x > 0, then 1 e2 x
(3, ).
Section 1.5 Exponential Functions
25.
33
26.
x
2.3219
x
1.3863
28.
27.
x
0.6309
x
29. Let t be the number of years. Solving 500,000(1.0375)t
population will reach 1 million in about 19 years.
1.5850
1, 000, 000 graphically, we find that t
18.828. The
30. (a) The population is given by P (t ) 6250(1.0275)t , where t is the number of years after 1890.
Population in 1915: P(25) 12,315
Population in 1940: P(50) 24,265
(b) Solving P(t) = 50,000 graphically, we find that t 76.651. The population reached 50,000 about 77 years
after 1890, in 1967.
31. (a)
A(t )
t /14
6.6 12
(b) Solving A(t) = 1 graphically, we find that t
38. There will be 1 gram remaining after about 38.1145 days.
32. Let t be the number of years. Solving 2300(1.60)t 4150 graphically, we find that t 10.129. It will take
about 10.129 years. (If the interest is not credited to the account until the end of each year, it will take
11 years.)
33. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve
A(1.0625)t 2 A, which is equivalent to 1.0625t 2. Solving graphically, we find that t 11.433. It will take
about 11.433 years. (If the interest is credited at the end of each year, it will take 12 years.)
34. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve
Ae0.0575t 3 A, which is equivalent to e0.0575t
about 19.106 years.
Copyright
3. Solving graphically, we find that t
2014 Pearson Education, Inc.
19.106. It will take
34
Chapter 1 Functions
35. After t hours, the population is P(t )
P(24)
248
2t /0.5 , or equivalently, P(t )
22t . After 24 hours, the population is
2.815 1014 bacteria.
36. (a) Each year, the number of cases is 100%
20% = 80% of the previous year s number of cases. After
t years, the number of cases will be C (t ) 10, 000(0.8)t . Solving C(t) = 1000 graphically, we find that
t 10.319. It will take 10.319 years.
(b) Solving C(t) = 1 graphically, we find that t 41.275. It will take about 41.275 years.
1.6 INVERSE FUNCTIONS AND LOGARITHMS
1. Yes one-to-one, the graph passes the horizontal line test.
2. Not one-to-one, the graph fails the horizontal line test.
3. Not one-to-one since (for example) the horizontal line y = 2 intersects the graph twice.
4. Not one-to-one, the graph fails the horizontal line test.
5. Yes one-to-one, the graph passes the horizontal line test.
6. Yes one-to-one, the graph passes the horizontal line test.
7. Not one-to-one since the horizontal line y = 3 intersects the graph an infinite number of times.
8. Yes one-to-one, the graph passes the horizontal line test.
9. Yes one-to one, graph passes the horizontal line test.
10. Not one-to-one since (for example) the horizontal line y = 1 intersects the graph twice.
11. Domain: 0 < x
13. Domain: 1
1, Range: 0
x
1, Range:
y
2
12. Domain: x < 1, Range: y > 0
y
2
Copyright
14. Domain:
< x < , Range:
2014 Pearson Education, Inc.
2
y
2
Section 1.6 Inverse Functions and Logarithms
15. Domain: 0
x
6, Range: 0
y
3
16. Domain: 2
x
1, Range: 1
17. The graph is symmetric about y = x.
(b)
y
1 x2
y2
1 x2
x2
1 y2
18. (a) The graph is symmetric about y = x.
19. Step 1: y
x2 1
Step 2: y
20. Step 1: y
x 1
x2
Step 2: y
21. Step 1: y
Step 2: y
22. Step 1: y
Step 2: y
f
y 1
1
x
y
1
x
x
1 x2
f 1 ( x)
1
y
1
x
y
y 1
( x)
y , since x
x3 1
x3
3
f 1 ( x)
x 1
x2
x
(b)
y
0.
f 1 ( x)
x
Step 2: y 1
23. Step 1: y
x2
1 y2
x
y 1
2x 1
x
y
( y 1)1/3
( x 1)2
y
x 1, since x
1
x 1
f 1 ( x)
( x 1)2
x 1
x
y
f
1
x 1, since x
1
x
y 1
( x)
Copyright
2014 Pearson Education, Inc.
y
f 1 ( x)
y<3
35
36
Chapter 1 Functions
24. Step 1: y
x 2/3
x
Step 2: y
x3/2
f 1 ( x)
25. Step 1: y
Step 2: y
y 3/2
x5
x
5
f 1 ( x);
x
y1/5
Domain and Range of f 1: all reals;
f ( f 1 ( x))
26. Step 1: y
Step 2: y
( x1/5 )5
x4
x
4
f 1 ( x);
x
f(f
27. Step 1: y
Step 2: y
0, Range of f 1: y
1/4 4
( x))
(x
( x5 )1/5
x
y1/4
Domain of f 1: x
1
x and f 1 ( f ( x))
)
x and f
x3 1
x3
3
f 1 ( x );
x 1
1
y 1
( f ( x))
0;
( x 4 )1/4
x
( y 1)1/3
x
Domain and Range of f 1: all reals;
f ( f 1 ( x))
28. Step 1: y
Step 2: y
(( x 1)1/3 )3 1 ( x 1) 1
x 72
1
2
1
2
x
y
7
2
x
x and f 1 ( f ( x))
(( x3 1) 1)1/3
( x3 )1/3
x
2y 7
f 1 ( x);
2x 7
Domain and Range of f 1: all reals;
f ( f 1 ( x))
7) 72
1 (2 x
2
29. Step 1: y
1
x2
Step 2: y
1
x
x2
x 72
1
y
7
2
x and f 1 ( f ( x))
2 12 x 72
7
1
y
x
f 1 ( x)
Domain of f 1: x > 0, Range of f 1: y > 0;
1
f ( f 1 ( x))
1
2
1
x
x3
1
y
1
x
30. Step 1: y
Step 2: y
1
x3
1
3 1
x
1/3
x
Domain of f 1: x
f ( f 1 ( x))
)
1
1
x2
1
x
x since x > 0.
1/3
y
f 1 ( x );
1
1/3 3
1
1
x
0, Range of f 1: y
1
(x
x and f 1 ( f ( x))
x
1
0;
x and f 1 ( f ( x))
Copyright
1
x3
1/3
1
x
1
x
2014 Pearson Education, Inc.
( x 7) 7
x
Section 1.6 Inverse Functions and Logarithms
31. Step 1: y
Step 2: y
x 3
x 2
y(x
2x 3
x 1
1
f 1 ( x);
Domain of f
2x 3
x 1
2x 3
x 1
Domain of f
f(f
(2 x 3) 3( x 1)
(2 x 3) 2( x 1)
2
y
x 3
Step 2: y
1
3
x
32. Step 1: y
3x 2
x 1
1
2y = x + 3
x 3
x = 2y + 3
2y 3
y 1
x
2;
y x 3y
x
x
x
x
x
2
x and f 1 ( f ( x))
5x
5
x
xy
y x
3
2
3
2
x
3
1
2( x 3) 3( x 2)
( x 3) ( x 2)
3y
5x
5
x
3y 2
y 1
x
f 1 ( x);
: (
(1, ), Range of f 1: [0, 9)
, 0]
3x 2
x 1
( x))
xy
1, Range of f 1: y
: x
f ( f 1 ( x ))
2) = x + 3
; If x > 1 or x
3x 2
x 1
3x
x 1
0
3
(9, );
3x 2
x 1
0
3x 2
x 1
3x
x 1
3x 3
x 1
3
3x
3 x 3( x 1)
3x
3
x and
2
x
3
f 1 ( f ( x))
9x
x 3
x
x
1
x 3
33. Step 1: y
x2
Step 2: y 1
x
x 3
2
9x
9
y 1 ( x 1)2 , x
2 x, x
1
x 1
f 1 ( x);
Domain of f 1: [ 1, ), Range of f 1: (
f ( f 1 ( x))
1
f 1 ( f ( x)) 1
34. Step 1: y
Step 2: y
f
1
( f ( x))
35. (a) y = mx
(x2
2 1
5
3 x 1
2
1
y5
y 1
x 1, x
1
x 1
1 2 x 1 x 1 2 2 x 1
( x 1)2 , x
1=1
|x
1| = 1
y 5 1 2 x3
y5 1
2
x3
x
x and
(1
x) = x
5
3 y 1
2
f 1 ( x);
, ), Range of f 1: (
: (
5
2 3 x2 1
3
1/5
3 (2 x 1)
2
x
x 1
2 x3 1
1
, 1];
2 x ) 1, x 1 1
(2 x3 1)1/5
Domain of f
f ( f 1 ( x))
2
x 1
x
1
m
(b) The graph of y
y
3
5
2 x2 1
1
5
, );
1/5
1
3
3 (2 x 1) 1
2
f 1 ( x)
1
m
1
1/5
3
3 2x
2
(( x5 1) 1)1/5
( x5 )1/5
x
x
f 1 ( x) is a line through the origin with slope m1 .
Copyright
2014 Pearson Education, Inc.
x and
y 1
37
38
Chapter 1 Functions
y
m
36. y = mx + b
x
y-intercept
b.
m
37. (a) y = x + 1
f 1 ( x)
b
m
x=y
1
m
f 1 ( x)
1
b;
m
x
the graph of f 1 ( x) is a line with slope m1 and
x 1
(b) y = x + b x = y b
f 1 ( x) x b
(c) Their graphs will be parallel to one another
and lie on opposite sides of the line y = x
equidistant from that line.
38. (a) y = x + 1
x= y+1
f 1 ( x ) 1 x;
the lines intersect at a right angle
(b) y = x + b
x= y+b
f 1 ( x)
the lines intersect at a right angle
(c) Such a function is its own inverse
39. (a)
ln 34
ln 0.75
ln 4 ln 9
(c)
ln 12
ln1 ln 2
(e)
ln 3 2
(f)
ln 13.5
40. (a)
1
ln 125
ln 73/2
(e)
7
ln 0.056 ln 125
(c)
ln sin5
1 ln(4t 4 )
2
3
ln 2)
ln 7 ln 53
2 ln 3
3
ln 49
5
ln 7 2 ln 5
ln 352
2ln 35
2 ln 7 ln 5
2 ln 5 2 ln 7
ln 7 3ln 5
1
2
ln 4t 4
2
1 ln 32
3
1 ln 9
3
ln 2)
(d) ln1225
(b) ln(3 x 2 9 x ) ln 31x
ln 5
5
+ ln cos
1 (3ln 3
2
(b) ln 9.8
3 ln 7
2
ln sin
sin
ln 2
(b) ln(8 x 4) ln 2
1 (ln 33
2
3ln 5
ln 5 ln 7 ln 7
2 ln 5
ln sin
42. (a) ln sec
(d) ln 3 9
1 ln 27
2
2
1 ln13.5
2
ln 7 7
41. (a)
2 ln 2 2 ln 3
ln 3 12 ln 2
ln1 3ln 5
ln 35
ln 25
ln 3 2 ln 2
ln 2
(c)
(f)
ln 22 ln 32
ln 3 ln 21/2
ln 17
ln 3 ln 22
ln 3 ln 4
(b) ln 94
b x;
ln 2
ln 2t 2 ln 2
2
ln 22t
2
ln 3 x 3x 9 x
ln( x 3)
ln(t 2 )
= ln[(sec )(cos )] = ln 1 = 0
ln(8 x 4) ln 4
(c) 3ln t 2 1 ln(t 1)
ln 8 x4 4
3ln(t 2 1)1/3 ln(t 1)
Copyright
ln(2 x 1)
3 13 ln(t 2 1) ln(t 1)
2014 Pearson Education, Inc.
ln
(t 1)(t 1)
(t 1)
ln(t 1)
Section 1.6 Inverse Functions and Logarithms
eln 7.2
43. (a)
7.2
2
e ln x
(b)
1
e
2
y2 )
44. (a)
eln( x
45. (a)
2 ln e
x2
y2
2 ln e1/2
2
2
ln e( x
( x2
46. (a)
ln esec
(sec )(ln e)
ln(e2ln x )
ln eln x
eln y
47. ln y = 2t + 4
49. ln(y
40) = 5t
50. ln(1
2y) = t
51. ln(y
1)
ln 2 = x + ln x
y
1 = sin x
53. (a) e2 k
200
(c) ek /1000
a
(b) 80ek
(c) e
55. (a)
(b)
(c)
1
e
2
(ln 0.2)t
e
e 2t 4
e5t
eln x ln 2
e5t
e5t
y
et
1 2y
1)
eln( x /2)
ln e 1
(e x )(ln e)
ln 2
e t 5
e t 5
y
40
et 1
2y
ex
eln y
48. ln y = t + 5
y 40
ln(y
ln 4
y2 1
y 1
ln
2k ln e
10k
e
et 1
2
y
y 1
ln x = x
ln 2 x
ln(sin x )
ln(y
80 1
0.8
27
ln e
0.4
(e
ln 0.8 k
)
ln 2
0.8
ln 33
1
(eln 0.2 )t
2k = 2 ln 2
ln 2
x
e
ln
y 1
2x
ex
y 1
2x
eln( y 1)
1) = ln(sin x)
ex
eln(sin x )
k ln e
1000
ln 80 1
ln a
0.4
k
1000
k
5k = ln 4
0.8
0.2t
Copyright
t
0.4
k
k
k
ln 2
10
1000 ln a
ln 4
5
k = ln 80
k=1
( 0.3t)ln e = 3 ln 3
ln 2
10k = ln 2
ln a
k ln e = ln 80
(0.8)
kt ln e
k = ln 2
10k ln e = ln 2
5k ln e = ln 4
ln e k
ln e 0.3t
kt
ln e
ln a
ln 4 1
ln 22
10 k
2
ln ek /1000
ek
1
(ln 0.8) k
kt
y
ln(sin x)
ln e5k
1
4
e 0.3t
(c)
2 xe x 1
ln e2 k
(b) 100e
e5 k
1
0.3
ln(e ln e)
e ln( x / y )
y = sin x + 1
4
10k
eln x ln y
x
(b) ln e(e )
2 ln x
et
(c)
y2
ln x 2
e 2t 4
eln(1 2 y )
52. ln( y 2 1) ln( y 1)
54. (a)
2
x2
sec
eln( y 40)
y 1 2 xe x
y
y 2 ) ln e
1
eln 0.3
1
x2
(b) ln(ln ee )
(2) 12 ln e 1
(c)
(c)
y )
e ln 0.3
(b)
ln x 2
0.3t = 3 ln 3
t = 10 ln 3
ln 2
k
ln 0.2t
ln 0.4
t ln 0.2 = ln 0.4
2014 Pearson Education, Inc.
t
ln 0.4
ln 0.2
39
x
y
x
2
40
Chapter 1 Functions
56. (a)
(b)
(c)
57. e t
e 0.01t
ln e 0.01t
1000
kt
1
e
10
e(ln 2)t 12
x2
kt ln e
(eln 2 )t
2 1
2t
ln x 2
t
ln e t
2
58. e x e2 x 1
59. (a) 5log5 7
ln10
2
ex 2x 1
et
2
(e)
log 3 3
log3 31/2
(f)
log 4 14
log 4 4 1
2 log 4 4
1
2
1log 4 4
(f)
log 3 19
log3 3 2
1
2
2 log3 3
4z
x
22 z
(b) Let z
log3 x
3z
x
(3 z )2
log 2 2sin x
log5 (3 x 2 )
log 4 (2e sin x )
ln x
ln 2
x
÷ ln
ln 3
(c)
ln x a
ln 2 a
ln a
ln x
÷ ln a2
log9 x
log3 x
ln x
ln 9
x
÷ ln
ln 3
(c)
65. (a)
log
log
10
x
2
x
log a b
logb a
6
5z
x
2
(c)
1.3log3 75
(c)
log 7
75
1
10log10 (1/2)
1
2
7
1
2
1
1
2
(2 z ) 2
x
2
x2
32 z
x2
2z
9z
x
x2
sin x
3x 2
25 z
x
log 4 4(e sin x )/2
log 2 x
log3 x
(b)
2x 1
9x4
x
x
63. (a)
64. (a)
t
0.5
21
log 4 x
x
1
2
1
log121 121
61. (a) Let z
(b) log e (e x )
x2
t
t = 100 ln 1000
ln10
k
2 log11 11 2 1 2
log121 11 log121 121
log 2 (e(ln 2) sin x )
ln et
8log8 2
(b)
(e)
(c)
4(ln x) 2
11
3
1/2
62. (a) Let z
t
0.01t = ln 1000
21 2
1 log 3
3
2
(d) log11 121 log11 112
(c)
t= 1
2
ln e x 2 x 1
et
kt = ln 10
2 1
(b)
log 4 4
2log 2 3
( 0.01t)ln e = ln 1000
ln10
2ln x
7
(d) log 4 16
60. (a)
ln1000
1
ln e
kt
ln x
ln x
ln 10
ln b
ln a
ln x ln 3
ln 2 ln x
÷ ln x
a
÷ ln
ln b
ln 3
ln 2
ln a ln x 2
ln x ln a
ln x ln 3
2 ln 3 ln x
ln 2
1
2
(b)
2 ln x
ln x
log 2 x
log8 x
ln x
ln 2
x
÷ ln
ln 8
ln x ln 8
ln 2 ln x
2
1
2
ln x
ln10
ln b ln b
ln a ln a
e x sin x
2
1
2
ln 2
ln 2
ln10
ln x
ln b 2
ln a
(b)
Copyright
4
2014 Pearson Education, Inc.
(c)
3
3ln 2
ln 2
3
Section 1.6 Inverse Functions and Logarithms
66. (a)
67. (a) arccos( 1) =
(b) arccos(0)
68. (a)
3
4
(b)
3
since cos( ) = 1 and 0
2
arcsin( 1)
since cos 2
2
1
2
(b) arcsin
(c)
4
since sin
.
2
1 and
2
6
.
0 and 0
since sin
1
2
4
2
2
and
2
2
.
4
2
.
69. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x1
f ( x1 )
f ( x2 ), so
f ( x1 )
f ( x2 ) and therefore g ( x1 )
x2 then
g ( x2 ). Therefore g(x) is one-to-one as well.
70. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x1
f ( x1 )
f ( x2 ), so f (1x )
1
1 ,
f ( x2 )
and therefore h( x1 )
g ( x1 )
x2
g ( x2 ), we also have f ( g ( x1 ))
f ( g ( x1 ))
x2 then g ( x1 )
g ( x2 ) because g is one-to-one. Since
f ( g ( x2 )) because f is one-to-one; thus, f g is one-to-one because
f ( g ( x2 )).
72. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x1
with g ( x1 ) g ( x2 ). For these numbers we would also have f ( g ( x1 ))
assumption that f g is one-to-one.
73. (a)
y
x
1 2 x
100
1 2 x
x2 then
h( x2 ).
71. The composite is one-to-one also. The reasoning: If x1
x1
log 2 100
1
y
100
y
2 x
log 2
100 y
y
100
y
log 2 (2 x )
1
x2 in the domain of g
f ( g ( x2 )), contradicting the
log 2 100
1
y
log 2 100
1
y
x
y
log 2 100 y .
f 1 ( x)
log 2 100x x
Interchange x and y: y
log 2 100x x
Verify.
(f
f 1 )( x)
( f 1 f )( x)
(b)
y
x
50
1 1.1
x
100
f log 2 100x x
f 1 100 x
1 2
1 1.1 x
log1.1 50
1
y
Interchange x and y: y
1 2
log 2
log 2
100
x
100 x
100
1 2 x
100 100
1
log 2 100 x
x
2
log 2
1 2 x
50
y
log1.1
41
1.1 x
50
y
50 y
y
log1.1 50 y .
log1.1 50x x
1
100
1
100 x
x
100
100(1 2 x ) 100
log1.1 (1.1 x )
log1.1 50x x
Verify.
Copyright
log 2
log1.1 50
1
y
y
f 1 ( x)
100
x 100 x
2014 Pearson Education, Inc.
100 x
100
x
log 2 (2 x )
1
2 x
x
x
log1.1 50
1
y
42
Chapter 1 Functions
f 1 )( x)
(f
f log1.1 50x x
( f 1 f )( x)
f 1
74. sin 1 (1) cos 1 (1)
If x
(f)
50
log1.1 x
50 x
1.1
log1.1 50 x
x
1.1
x
0
; sin 1 (0) cos 1 (0)
2
log1.1
1
a cos
1
a)
1
50
1 1.1 x
50
50
50
1 1.1
1 1.1 x
2
2
0
1 50x x
50
50(1 1.1 x ) 50
log1.1
( 1, 0) and x = a, then sin 1 ( x) cos 1 ( x )
(sin
75. (a)
(b)
(c)
(d)
(e)
2
1
50
50
2
2
1
1.1
x
x
log1.1 (1.1x )
sin 1 a (
cos
from Equations (3) and (4) in the text.
Begin with y = ln x and reduce the y-value by 3
y = ln x 3.
y = ln(x 1).
Begin with y = ln x and replace x with x 1
Begin with y = ln x, replace x with x + 1, and increase the y-value by 3 y = ln(x + 1) + 3.
y = ln(x 2) 4.
Begin with y = ln x, reduce the y-value by 4, and replace x with x 2
Begin with y = ln x and replace x with x y = ln( x).
Begin with y = ln x and switch x and y
(b) Begin with y = ln x and replace x with
x = ln y or y
x
3
y
(d) Begin with y = ln x and replace x with 2x
ex .
y = 2 ln x.
ln 3x .
(c) Begin with y = ln x and multiply the y-value by 14
y
1 ln x.
4
y = ln 2x.
77. From zooming in on the graph at the right, we
0.76666.
estimate the third root to be x
78. The functions f ( x) x ln 2 and g ( x) 2ln x
appear to have identical graphs for x > 0. This is
no accident, because
79. (a)
log1.1
50 x
50
; and sin 1 ( 1) cos 1 ( 1)
sin 1 ( a) cos 1 ( a )
76. (a) Begin with y = ln x and multiply the y-value by 2
x ln 2
50 x
x 50 x
eln 2 ln x
Amount
(eln 2 )ln x
2ln x.
t /12
8 12
t /12
t
1 t /12 1
1 t /12
1 3
(b) 8 12
1
2
8
2
2
12
There will be 1 gram remaining after 36 hours.
Copyright
3
t
36
2014 Pearson Education, Inc.
2
1
2
a)
x
.
Chapter 1 Practice Exercises
80. 500(1.0475)t
1.0475t
1000
ln(1.0475t )
2
ln(2)
t ln(1.0475)
ln(2)
ln(2)
ln(1.0475)
t
43
14.936
It will take about 14.936 years. (If the interest is paid at the end of each year, it will take 15 years.)
81. 375, 000(1.0225)t
ln
t
8
3
ln(1.0225)
1.0225t
1, 000, 000
8
3
ln(1.0225t )
ln 83
t ln(1.0225)
ln 83
44.081
It will take about 44.081 years.
82.
y
y0 e 0.18t represents the decay equation; solving (0.9) y0
y0 e 0.18t
t
ln(0.9)
0.18
0.585 days
CHAPTER 1 PRACTICE EXERCISES
r 2 and the circumference is C
1. The area is A
2. The surface area is S
4 r2
for surface area gives S
4 r2
1/2
S
4
r
4
C
2
2 r. Thus, r
. The volume is V
4
3
r3
A
r
C
2
3 3V
4
2
C2 .
4
. Substitution into the formula
3V 2/3.
4
3. The coordinates of a point on the parabola are (x, x2). The angle of inclination joining this point to the origin
x2
x. Thus the point has coordinates ( x, x 2 ) (tan , tan 2 ).
satisfies the equation tan
x
4. tan
rise
run
h
500
h 500 tan ft.
5.
6.
Symmetric about the origin.
Symmetric about the y-axis.
7.
8.
Neither
9. y ( x)
Symmetric about the y-axis.
( x)2 1 x 2 1
y ( x). Even.
Copyright
2014 Pearson Education, Inc.
44
Chapter 1 Functions
10. y ( x)
( x )5
( x )3
11. y ( x ) 1 cos( x ) 1 cos x
12. y ( x)
sec( x) tan( x )
x
13. y ( x)
x
3
4
2
1
x
14. y ( x)
( x) sin( x)
15. y ( x)
x cos( x)
16. y ( x)
( x) cos( x )
x5
( x)
sin
cos2
x3
x
y ( x). Odd.
y ( x ). Even.
x
sin x
cos2 x
x
x4 1
x4 1
x3 2 x
x3 2 x
( x) sin x
sec x tan x
y ( x ). Odd.
y ( x). Odd.
( x sin x )
y ( x). Odd.
x cos x. Neither even nor odd.
x cos x
y ( x). Odd.
17. Since f and g are odd
f ( x)
f ( x) and g ( x)
g ( x).
(a) ( f g )( x ) f ( x) g ( x ) [ f ( x)] [ g ( x )] f ( x) g ( x) ( f g )( x)
f g is even.
3
f ( x) f ( x) f ( x)
f 3 ( x)
(b) f ( x) f ( x) f ( x) f ( x) [ f ( x)] [ f x ] [ f ( x)]
(c) f (sin( x )) f ( sin( x ))
f (sin( x))
f (sin( x )) is odd.
g (sec( x)) is even.
(d) g (sec( x)) g (sec( x))
(e) | g ( x )| | g ( x )| | g ( x ) |
| g | is even.
18. Let f (a x ) f ( a x) and define g ( x) f ( x a). Then g ( x)
f ( x a ) g ( x ) g ( x ) f ( x a ) is even.
f (( x) a )
19. (a) The function is defined for all values of x, so the domain is (
(b) Since | x | attains all nonnegative values, the range is [ 2, ).
f (a
x)
f 3 is odd.
f (a
x)
, ).
20. (a) Since the square root requires 1 x 0, the domain is ( ,1].
(b) Since 1 x attains all nonnegative values, the range is [ 2, ).
21. (a) Since the square root requires 16 x 2
(b) For values of x in the domain, 0 16
0, the domain is [ 4, 4].
x2
16, so 0
16 x 2
4. The range is [0, 4].
22. (a) The function is defined for all values of x, so the domain is (
(b) Since 32 x attains all positive values, the range is (1, ).
,
).
23. (a) The function is defined for all values of x, so the domain is (
(b) Since 2e x attains all positive values, the range is ( 3, ).
,
).
24. (a) The function is equivalent to y
tan 2 x, so we require 2 x
k
4
x
for odd integers k.
(b) Since the tangent function attains all values, the range is (
k
2
for odd integers k. The domain is given by
, ).
25. (a) The function is defined for all values of x, so the domain is ( , ).
(b) The sine function attains values from 1 to 1, so 2 2sin (3x
) 2 and hence 3 2 sin (3x
The range is [ 3, 1].
Copyright
2014 Pearson Education, Inc.
) 1 1.
Chapter 1 Practice Exercises
26. (a) The function is defined for all values of x, so the domain is (
,
).
5 2
(b) The function is equivalent to y
x , which attains all nonnegative values. The range is [0, ).
27. (a) The logarithm requires x 3 0, so the domain is (3, ).
(b) The logarithm attains all real values, so the range is ( , ).
28. (a) The function is defined for all values of x, so the domain is (
(b) The cube root attains all real values, so the range is ( , ).
29. (a)
(b)
(c)
(d)
, ).
Increasing because volume increases as radius increases.
Neither, since the greatest integer function is composed of horizontal (constant) line segments.
Decreasing because as the height increases, the atmospheric pressure decreases.
Increasing because the kinetic (motion) energy increases as the particles velocity increases.
30. (a) Increasing on [2, )
(c) Increasing on ( , )
(b) Increasing on [ 1, )
(d) Increasing on 12 ,
31. (a) The function is defined for 4 x 4, so the domain is [ 4, 4].
(b) The function is equivalent to y
| x |, 4 x 4, which attains values from 0 to 2 for x in the domain.
The range is [0, 2].
32. (a) The function is defined for 2
(b) The range is [ 1, 1].
2, so the domain is [ 2, 2].
x
0 1
1
1 0
1
0 1
0). m 2 1 11
33. First piece: Line through (0, 1) and (1, 0). m
Second piece: Line through (1, 1) and (2,
f ( x)
1 x,
0
2 x,
1 x
10
35. (a) ( f g )( 1)
0
x
2
2
x
4
f ( g ( 1))
(Note: x
g ( f (2))
g 12
(c) ( f f )( x)
f ( f ( x ))
f 1x
36. (a) ( f g )( 1)
y
( x 1) 1
5
2
g ( g ( x))
g
f ( g ( 1))
0
5
0
2
0 5
4 2
5x
2
5
2
y
5
2
y
2 can be included on either piece.)
1
1 2
1
1 2
2
f
(b) ( g f )(2)
(d) ( g g )( x)
1
x 1 1 x
x
2
2 x
2
Second piece: Line through (2, 5) and (4, 0). m
f ( x)
y
x 1
34. First piece: Line through (0, 0) and (2, 5). m
5 x,
2
5x ,
2
1
1
1/ x
1
1
1
or
2
5
f (1)
1
2.5
x, x
0
4
1
1
x 2
1
x 2
f 3 1 1
2
f (0)
x
2
1 2 x
2 0
2
2
3
(b) ( g f )(2)
(c) ( f f )( x)
f ( g (2)) g (2 2) g (0)
0 1 1
f ( f ( x)) f (2 x) 2 (2 x) x
(d) ( g g )( x)
g ( g ( x))
g 3x 1
33
Copyright
x 1 1
2014 Pearson Education, Inc.
5 (x
2
2) 5
5
2
x 10 10 52x
45
46
Chapter 1 Functions
37. (a) ( f g )( x)
f ( g ( x))
f
x 2
( g f )( x)
g ( f ( x))
g (2 x 2 )
2
x 2
2 x2
2
x, x
2.
4 x2
2
(b) Domain of f g : [ 2, ).
Domain of g f : [ 2, 2].
(c) Range of f g : ( , 2].
Range of g f : [0, 2].
38. (a) ( f g )( x)
f ( g ( x))
f
1 x
( g f )( x)
g ( f ( x))
g
x
1 x
1
41
x.
x
(b) Domain of f g : ( , 1].
Domain of g f : [0, 1].
39.
y f ( x)
(c) Range of f g : [0, ).
Range of g f : [0, 1].
y ( f f )( x)
40.
41.
42.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
Copyright
It does not change the graph.
2014 Pearson Education, Inc.
Chapter 1 Practice Exercises
43.
47
44.
Whenever g1 ( x) is positive, the graph of y
g 2 ( x) | g1 ( x)| is the same as the graph of y g1 ( x).
When g1 ( x) is negative, the graph of y g 2 ( x) is
the reflection of the graph of y g1 ( x ) across the
x-axis.
Whenever g1 ( x) is positive, the graph of y
g 2 ( x) g1 ( x ) is the same as the graph of y
g1 ( x). When g1 ( x) is negative, the graph of y
g 2 ( x) is the reflection of the graph of y g1 ( x )
across the x-axis.
45.
46.
Whenever g1 ( x) is positive, the graph of
y g 2 ( x) | g1 ( x)| is the same as graph of
y g1 ( x). When g1 ( x) is negative, the graph of
y g 2 ( x) is the reflection of the graph of
y g1 ( x ) across the x-axis.
47.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
48.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
49. (a) y g ( x 3) 12
(c) y g ( x )
(e) y 5 g ( x)
Copyright
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
(b) y
g x
(d) y
(f ) y
g ( x)
g (5 x)
2
3
2
2014 Pearson Education, Inc.
48
Chapter 1 Functions
50. (a)
(c)
(d)
(e)
(f )
Shift the graph of f right 5 units
(b) Horizontally compress the graph of f by a factor of 4
Horizontally compress the graph of f by a factor of 3 and then reflect the graph about the y-axis
Horizontally compress the graph of f by a factor of 2 and then shift the graph left 12 unit.
Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units.
Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the
graph up 14 unit.
x about the x-axis
51. Reflection of the graph of y
followed by a horizontal compression by a factor of
1 then a shift left 2 units.
2
52. Reflect the graph of y x about the x-axis, followed
by a vertical compression of the graph by a factor
of 3, then shift the graph up 1 unit.
53. Vertical compression of the graph of y
factor of 2, then shift the graph up 1 unit.
1
x2
by a
54. Reflect the graph of y x1/3about the y-axis, then
compress the graph horizontally by a factor of 5.
Copyright
2014 Pearson Education, Inc.
Chapter 1 Practice Exercises
55.
56.
period
57.
period
4
period
4
period
2
58.
period
2
59.
60.
period
2
61. (a) sin B
a2
b
c
sin 3
b2
(b) sin B
c2
b
2
c2
a
b
c
sin 3
b
2
c
2sin 3
b2
c
2
sin 3
2 23
3. By the theorem of Pythagoras,
4 3 1.
2
3
2
4 . Thus,
3
a
c2
b2
4
3
2
62. (a) sin A
a
c
a
c sin A
(b) tan A
a
b
a
b tan A
63. (a) tan B
b
a
a
b
tan B
(b) sin A
a
c
c
a
sin A
64. (a) sin A
a
c
(b) sin A
a
c
c 2 b2
c
Copyright
2014 Pearson Education, Inc.
(2)2
4
3
2 .
3
49
50
Chapter 1 Functions
65. Let h height of vertical pole, and let b and c denote
the distances of points B and C from the base of the
pole, measured along the flat ground, respectively.
h , tan 35
h , and b c 10.
Then, tan 50
c
b
Thus, h c tan 50 and h b tan 35 (c 10) tan 35
c tan 50 (c 10) tan 35
c(tan 50 tan 35 ) 10 tan 35
10 tan 35
c
h c tan 50
tan 50 tan 35
10 tan 35 tan 50
16.98
tan 50 tan 35
m.
66. Let h height of balloon above ground. From the
h , tan 70
h , and
figure at the right, tan 40
a
b
a b 2. Thus, h b tan 70
h (2 a) tan 70
and h a tan 40
(2 a ) tan 70 a tan 40
a (tan 40 tan 70 ) 2 tan 70
2 tan 70
a tan 40
h a tan 40
tan 70
2 tan 70 tan 40
tan 40 tan 70
1.3 km.
67. (a)
(b) The period appears to be 4 .
x
4
sin( x 2 ) cos 2x 2
sin x cos 2x
since the period of sine and cosine is 2 . Thus, f(x) has period 4 .
(c) f ( x 4 )
sin( x 4 ) cos
2
68. (a)
(b) D
(
,0)
(0, ); R
[ 1, 1]
(c) f is not periodic. For suppose f has period p. Then f 21
Choose k so large that 21
kp
1
0
1
1/(2 )
kp
kp
f 21
. But then f 21
which is a contradiction. Thus f has no period, as claimed.
69. (a) D:
(b) D: x
x
0
Copyright
2014 Pearson Education, Inc.
sin 2
kp
0 for all integers k.
sin (1/(2 1)) kp
0
Chapter 1 Practice Exercises
70. (a)
D
(
, 0)
71. (a) D:
3
x
(b)
(0, )
3
D
(b) D: 0
72. (a) D [ 1, 1)
(b)
73. ( f g )( x ) ln(4 x 2 ) and domain: 2
x
(
, 2)
x
( 2, 2)
(2, )
4
D [ 1, 1]
2;
( g f )( x ) 4 (ln x )2 and domain: x 0;
( f f )( x ) ln(ln x ) and domain: x 1;
x 4 8 x 2 12 and domain:
( g g )( x )
74. (a) Even
(b) Neither even nor odd
76. For c
0, D
(
, )
For c
0, D
(
,
c)
( c, )
y
ln( x 2 c )
x
.
(c) Neither even nor odd
(d) Even
y
4
2
4
2
0
2
x
4
2
4
78. For large values of x, y
79. (a) D : (
, )
80. (a) D :
x
R:
2
; R: 0
a x has the largest values; y
log a x has the smallest.
,2
(b)
D : [ 1, 1]
y
(b)
D : 1 x 1; R : 1
R : [ 1, 1]
y 1
y
y
y
cos 1 (cos x )
y
1
cos(cos 1 x )
0.5
2
0
2
x
1
0.5
0
0.5
1
Copyright
2014 Pearson Education, Inc.
0.5
1
x
51
52
Chapter 1 Functions
81. (a) No
(b) Yes
82. Answers depend on the view screen used. For [15, 17] [5 106 , 107 ] it appears that e x
83. (a)
3
f ( g ( x ))
x
3
x, g ( f ( x ))
3 3
x
107 for x 16.128.
x
(b)
y
y
2
x3
1
2
1
0
y
x
y
x1/ 3
1
x
2
1
2
84. (a) h (k ( x ))
1
4
3
(4 x )1/3
3
4 x4
x, k (h ( x ))
1/3
x
(b)
y
4
y
2
x
y (4 x )1/ 3
2
4
y
x3
4
0
2
4
x
2
4
CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
1. There are (infinitely) many such function pairs. For example, f ( x )
f ( g ( x )) f (4 x ) 3(4 x ) 12 x 4(3 x) g (3 x) g ( f ( x )).
2. Yes, there are many such function pairs. For example, if g ( x)
( f g )( x ) f ( g ( x)) f ((2 x 3)3 ) ((2 x 3)3 )1/3 2 x 3.
3x and g ( x )
(2 x 3)3 and f ( x)
4 x satisfy
x1/3, then
3. If f is odd and defined at x, then f ( x )
f ( x ). Thus g ( x) f ( x) 2
f ( x) 2 whereas
g ( x)
f ( x ) 2. Then g cannot be odd because g ( x)
g ( x)
f ( x) 2
f ( x) 2
( f ( x ) 2)
4 0, which is a contradiction. Also, g ( x) is not even unless f ( x ) 0 for all x. On the other hand, if f is
even, then g ( x) f ( x) 2 is also even: g ( x) f ( x ) 2 f ( x) 2 g ( x).
4. If g is odd and g(0) is defined, then g (0)
Copyright
g ( 0)
g (0). Therefore, 2 g (0)
2014 Pearson Education, Inc.
0
g (0)
0.
Chapter 1 Additional and Advanced Exercises
53
5. For (x, y) in the 1st quadrant, | x | | y | 1 x
x y 1 x
y 1. For (x, y) in the 2nd
x y x 1
quadrant, | x | | y | x 1
y 2 x 1. In the 3rd quadrant, | x | | y | x 1
x y x 1
y
2 x 1. In the 4th
x ( y) x 1
quadrant, | x | | y | x 1
y
1. The graph is given at the right.
6. We use reasoning similar to Exercise 5.
(1) 1st quadrant: y | y | x | x |
2 y 2x
y x.
(2) 2nd quadrant: y | y | x | x |
2 y x ( x) 0
y 0.
(3) 3rd quadrant: y | y | x | x |
y ( y ) x ( x)
0 0
all points in the 3rd quadrant
satisfy the equation.
(4) 4th quadrant: y | y | x | x |
y ( y) 2 x
0 x. Combining
these results we have the graph given at the right:
7. (a) sin 2 x cos2 x 1
sin 2 x 1 cos 2 x
(1 cos x) (1 cos x)
1 cos x
sin 2 x
1 cos x
1
cos x
sin x
(b) Using the definition of the tangent function and the double angle formulas, we have
cos 2 x
2
2
cos 2 x
2
2
1
sin 2
x
2
cos 2 2x
tan 2 2x
1
1
1
cos x
.
cos x
8. The angles labeled in the accompanying figure are
equal since both angles subtend arc CD. Similarly, the
two angles labeled are equal since they both subtend
arc AB. Thus, triangles AED and BEC are similar which
a c
2a cos
b
implies b
a c
(a c)(a
a2 c2
c2 a2
c) b(2a cos
2ab cos
b2
2
b 2ab cos .
b)
9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah bc sin A ab sin C
the area of ABC 12 (base)(height) 12 ah 12 bc sin A 12 ab sin C 12 ac sin B.
10. As in Section 1.3, Exercise 61, (Area of ABC )2
1 a 2b 2 (1
4
cos 2 C ) . By the law of cosines, c
(area of ABC ) 2
1
16
1
16
4a 2b 2
[(( a b) 2
a
b
2
c
1 a 2b 2 (1
4
(a 2
b2
c 2 )2
c 2 )(c 2
a
b
2
cos 2 C )
c
1
16
( a b)2 )]
a
b
2
c
1 a 2b 2
4
[(2ab
(a 2
1 [(( a
16
a b c
2
1 a 2 b 2 sin 2 C
4
a2 b2 c 2
b 2 ab cos C cos C
. Thus,
2 ab
2
2
2
b2 c 2 )
a 2 b2 c 2
a 2b 2 1 ( a
4
2 ab
4 a 2b 2
1 (base) 2 (height) 2
4
2
2
2
a
1
b2
c 2 )) (2ab (a 2
1 a2 h2
4
b2
c 2 ))]
b) c)(( a b) c)(c ( a b))(c ( a b))]
s ( s a )( s b)( s c), where s
Therefore, the area of ABC equals s ( s a )( s b)( s c) .
Copyright
ac sin B
2014 Pearson Education, Inc.
a
b
2
c
.
1
sin x
cos x
54
Chapter 1 Functions
11. If f is even and odd, then f ( x)
f ( x ) 0.
Thus 2 f ( x ) 0
f ( x)
12. (a) As suggested, let E ( x )
function. Define O ( x)
f ( x)
f ( x)
f ( x)
E ( x)
2
f ( x) E ( x)
f ( x)
2
f ( x) and f ( x )
f ( x)
O( x)
2
f ( x)
f ( x)
f ( x)
f ( x)
f ( x)
f ( ( x ))
2
f ( x)
f ( x)
2
f ( x ) for all x in the domain of f.
f ( x)
f ( x)
2
O is an odd function
f ( x)
2
. Then O( x )
f ( x)
E ( x)
f ( x)
E is an even
f
2
( x)
E ( x ) O( x) is the sum of an even
and an odd function.
(b) Part (a) shows that f ( x ) E ( x ) O ( x ) is the sum of an even and an odd function. If also
f ( x) E1 ( x) O1 ( x), where E1 is even and O1 is odd, then f ( x) f ( x) 0
( E1 ( x) O1 ( x )) ( E ( x) O ( x )) . Thus, E ( x) E1 ( x) O1 ( x) O ( x) for all x in the domain of f (which is
the same as the domain of E E1 and O O1). Now ( E E1 )( x) E ( x) E1 ( x) E ( x) E1 ( x) (since E
and E1 are even) ( E E1 )( x) E E1 is even. Likewise, (O1 O )( x) O1 ( x) O( x )
O1 ( x) ( O ( x)) (since O and O1 are odd)
(O1 ( x) O ( x ))
(O1 O ) ( x) O1 O is odd.
Therefore, E E1 and O1 O are both even and odd so they must be zero at each x in the domain of f by
Exercise 11. That is, E1 E and O1 O, so the decomposition of f found in part (a) is unique.
13. y
ax 2
bx c
a x2
b
a
x
b2
4a 2
b2
4a
c
a x
b
2a
2
b2
4a
c
(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of
the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward. Decreasing
a causes a vertical stretching and a shift of the vertex toward the y-axis and downward.
(b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the graph
downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the right.
If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward to the
right. If b 0, decreasing b shifts the graph upward to the left.
(c) Changing c (for fixed a and b) by c shifts the graph upward c units if c 0, and downward c units
if c 0.
14. (a) If a 0, the graph rises to the right of the vertical line x
b and falls to the left. If a < 0, the graph falls
to the right of the line x
b and rises to the left. If a 0, the graph reduces to the horizontal line y c.
As | a | increases, the slope at any given point x x0 increases in magnitude and the graph becomes steeper. As
| a | decreases, the slope at x0 decreases in magnitude and the graph rises or falls more gradually.
(b) Increasing b shifts the graph to the left; decreasing b shifts it to the right.
(c) Increasing c shifts the graph upward; decreasing c shifts it downward.
15. Each of the triangles pictured has the same base
b v t v (1 sec) . Moreover, the height of each
triangle is the same value h. Thus 12 (base)(height)
1 bh
2
A1 A2 A3
. In conclusion,
the object sweeps out equal areas in each one
second interval.
Copyright
2014 Pearson Education, Inc.
Chapter 1 Additional and Advanced Exercises
16. (a) Using the midpoint formula, the coordinates of P are
y
x
of OP
b /2
a /2
(b) The slope of AB
b
a
slopes is 1
b.
a
b 0
0 a
b
a
sin
EB, cos
area AEB
1 sin
2
1 (1)2
2
cos
2
and AB
CD
AD
AE , tan
area sector DB
, 2
0
a, b
2 2
. Thus the slope
of their
b.
17. From the figure we see that 0
AE
AB
0 b
2
b . The line segments AB and OP are perpendicular when the product
a
b 2 . Thus, b 2 a 2
a b (since both are positive). Therefore, AB is
a2
perpendicular to OP when a
EB
AB
a
55
area
1 (1)(tan
2
AD 1. From trigonometry we have the following:
CD, and tan
1 ( AE )( EB )
ADC
2
1 sin cos
1
)
2
2
EB
AE
2
1 ( AD )
2
1 sin
2 cos
sin . We can see
cos
1 ( AD ) (CD )
2
that:
18. ( f g )( x) f ( g ( x)) a (cx d ) b acx ad b and ( g f )( x) g ( f ( x )) c (ax b) d acx cb d
Thus ( f g )( x ) ( g f )( x ) acx ad b acx bc d
ad b bc d . Note that f (d ) ad b and
g (b) cb d , thus ( f g )( x ) ( g f )( x) if f (d ) g (b).
19. (a) The expression a (bc x ) d is defined for all values of x, so the domain is (
, ). Since bc x attains all
positive values, the range is (d, ) if a > 0 and the range is ( d, ) if a < 0.
(b) The expression a logb ( x c ) d is defined when x c > 0, so the domain is (c, ). Since
a logb ( x c ) d attains every real value for some value of x, the range is (
20. (a) Suppose f ( x1 )
f ( x2 ). Then:
ax1 b
cx1 d
(ax1 b)(cx2
acx1 x2
d)
adx1 bcx2 bd
adx1 bcx2
Since ad
(b)
y
cxy dy
(cy a ) x
x
ax2 b
cx2 d
(ax2
b)(cx1 d )
acx1 x2
adx2 bcx1 bd
adx2 bcx1
(ad bc) x1 (ad bc ) x2
bc 0, this means that x1 x2 .
ax b
cx d
ax b
dy b
dy b
cy a
Interchange x and y.
dx b
y
cx a
f 1 ( x)
dx b
cx a
Copyright
2014 Pearson Education, Inc.
, ).
56
Chapter 1 Functions
21. (a) y = 100,000
10,000x, 0
(b)
x
y
55, 000
100, 000 10, 000 x
55, 000
10, 000 x
55, 000
10
x 4.5
The value is $55,000 after 4.5 years.
22. (a) f(0) = 90 units
(b) f(2) = 90 52 ln 3
(c)
23. 1500(1.08)t
5000
32.8722 units
1.08t
5000
1500
10
3
ln(1.08)t
ln 10
3
t ln1.08
ln 10
3
t
ln(10/3)
ln1.08
15.6439
It will take about 15.6439 years. (If the bank only pays interest at the end of the year, it will take 16 years.)
24.
A(t )
A0 ert ; A(t )
x
25. ln x ( x )
2 A0
2 A0
x x ln x and ln( x x ) x
x
Therefore, x ( x )
A0 ert
x ln( x x )
ert
2
rt = ln 2
x 2 ln x; then, x x ln x
t
ln 2
r
x 2 ln x
( x x ) x when x = 2.
26. (a) No, there are two intersections: one at x = 2
and the other at x = 4.
(b) Yes, because there is only one intersection.
Copyright
2014 Pearson Education, Inc.
t
0.7
r
xx
x2
70
100 r
x ln x
70
( r %)
2 ln x
x
2.
Chapter 1 Additional and Advanced Exercises
27.
ln x
ln 4
ln x
ln 2
log 4 x
log 2 x
28. (a)
ln x ln 2
ln 4 ln x
ln 2 ,
ln x
f ( x)
ln 2
ln 4
ln 2
2 ln 2
1
2
ln x
ln 2
g ( x)
(b) f is negative when g is negative, positive
when g is positive, and undefined when
g = 0; the values of f decrease as those of g
increase.
(c)
ln 2
ln x
ln x
ln 2
(ln 2)2
(ln x) 2
(ln 2 ln x)(ln 2 + ln x) = 0
or ln x = ln 2
e
ln x
e
ln(1/2)
e
ln x
e
ln 2
x = 2 or x
ln x = ln 2
or
1.
2
Therefore, the two curves cross at the two
ln(1/2)
points 12 , ln 2
2
2, ln
ln 2
1,
2
1 and
(2, 1).
Copyright
2014 Pearson Education, Inc.
57
58
Chapter 1 Functions
Copyright
2014 Pearson Education, Inc.
CHAPTER 2
2.1
LIMITS AND CONTINUITY
RATES OF CHANGE AND TANGENTS TO CURVES
1. (a)
f
x
f (3) f (2)
3 2
28 9
1
2. (a)
g
x
g (3) g (1)
3 1
3 ( 1)
2
3. (a)
h
t
1 1
4. (a)
g
t
h
3
4
3
4
R
R (2) R (0)
2 0
6.
P
P (2) P (1)
2 1
(b)
8. (a)
(b)
9. (a)
y
x
4
f (1) f ( 1)
1 ( 1)
2
(b)
g
x
g (4) g ( 2)
4 ( 2)
4
(b)
h
t
(2 1) (2 1)
0
8 1
2
1
3 1
2
2
4( x 2)
2 2
y
x
11. (a)
y
x
4x
12. (a)
y
x
4h h 2
h
y 3
y
y
2x 2
3 3
(2 1) (2 1)
2
0
Copyright
4
at P(2, 1) the slope is 4.
4
0, 4 h
at P(2, 3) the slope
2 h h2
h
2 h. As h
0, 2 h
2
at
2 x 7.
y
h2 2 h
h
h 2. As h
0, h 2
2
at P (1, 3) the
2 x 1.
12 4h h 2 . As h
0, 12 4h h 2
12,
at P (2, 8)
y 12 x 16.
3h 3h 2 h3
h
2 1 3h 3h 2 h3 1
h
3x 3
0, 4 h
4 x 11
12h 4 h 2 h3
h
y 8 12 x 24
P (1, 1) the slope is 3.
(b) y 1 ( 3)( x 1)
y 1
3
3
g( ) g( )
( )
4 h. As h
1 2 h h 2 4 4 h ( 3)
h
8 12 h 4h 2 h3 8
h
2 (1 h)3 (2 13 )
h
4 h. As h
4 4h h 2 4 2h 3 ( 3)
h
2x 4
((1 h) 2 4(1 h)) (12 4(1))
h
the slope is 12.
(b) y 8 12( x 2)
6
0
4x 9
y
8
((2 h )2 2(2 h ) 3) (22 2(2) 3)
h
(2 h)3 23
h
4h h 2
h
7 4 4h h2 3
h
y 3
slope is 2.
(b) y ( 3) ( 2)( x 1)
8 8
6
0
6
1
0
4 4h h2 5 1
h
P (2, 3) the slope is 2.
(b) y ( 3) 2( x 2)
y 3
10. (a)
h
2
2
g
t
(b)
y 1 4x 8
(7 (2 h )2 ) (7 22 )
h
h
2 0
2
1
(8 16 10) (1 4 5)
1
is 4.
y 3 ( 4)( x 2)
y
x
f
x
2
((2 h )2 5) (22 5)
h
y ( 1)
y
x
4
g ( ) g (0)
0
5.
7. (a)
h
(b)
19
y
3 3h h 2 . As h
0, 3 3h h2
3,
at
3x 4.
2014 Pearson Education, Inc.
59
60
Chapter 2 Limits and Continuity
13. (a)
(1 h)3 12(1 h ) (13 12(1))
h
2
y
x
As h 0, 9 3h h
(b) y ( 11) ( 9)( x 1)
14. (a)
y
x
As h
(b) y 0
9 3h h 2 .
8 12 h 6 h 2 h3 12 12 h 3h 2 4 0
h
3 h 2 h3
h
3h h 2 .
0, 3h h
0 at P (2, 0) the slope is 0.
0( x 2)
y 0.
Q
Q1 (10, 225)
Q2 (14,375)
Q3 (16.5, 475)
Q4 (18,550)
16. (a)
9 h 3h 2 h 3
h
9 at P (1, 11) the slope is 9.
y 11
9x 9
y
9 x 2.
(2 h )3 3(2 h )2 4 (23 3(2) 2 4)
h
2
15. (a)
(b) At t
1 3h 3h 2 h3 12 12 h ( 11)
h
Slope of PQ
650 225
20 10
650 375
20 14
650 475
20 16.5
650 550
20 18
p
t
42.5 m/sec
45.83 m/sec
50.00 m/sec
50.00 m/sec
20, the sportscar was traveling approximately 50 m/sec or 180 km/h.
Slope of PQ
Q
Q1 (5, 20)
Q2 (7,39)
Q3 (8.5,58)
Q4 (9.5, 72)
80 20
10 5
80 39
10 7
80 58
10 8.5
80 72
10 9.5
p
t
12 m/sec
13.7 m/sec
14.7 m/sec
16 m/sec
(b) Approximately 16 m/sec
p
17. (a)
200
160
120
80
40
0
(b)
p
t
2010 2011 2012 2013 2014
Ye ar
174 62
2014 2012
112
2
t
56 thousand dollars per year
(c) The average rate of change from 2011 to 2012 is
The average rate of change from 2012 to 2013 is
p
t
p
t
62 27
2012 2011
35 thousand dollars per year.
111 62
2013 2012
49 thousand dollars per year.
So, the rate at which profits were changing in 2012 is approximately 12 (35 49)
per year.
18. (a) F ( x ) ( x 2)/( x 2)
x
1.2
1.1
F ( x)
4.0
3.4
F
x
F
x
F
x
4.0 ( 3)
1.2 1
3.04 ( 3)
1.01 1
3.0004 ( 3)
1.0001 1
1.01
3.04
1.001
3.004
F
x
F
x
5.0;
4.04;
3.4 ( 3)
1.1 1
3.004 ( 3)
1.001 1
1.0001
3.0004
4.4;
4.004;
4.0004;
(b) The rate of change of F ( x) at x 1 is 4.
Copyright
2014 Pearson Education, Inc.
42 thousand dollars
1
3
Section 2.1 Rates of Change and Tangents to Curves
g
x
g
x
19. (a)
g (2) g (1)
2 1
0.414213
2 1
2 1
g (1 h ) g (1)
1 h 1
h
(1 h ) 1
(b) g ( x)
g
x
g (1.5) g (1)
1.5 1
1.5 1
0.5
61
0.449489
x
1 h
1 h
1 h 1 /h
1.1
1.01
1.001
1.0001
1.00001
1.000001
1.04880
1.004987
1.0004998
1.0000499
1.000005
1.0000005
0.4880
0.4987
0.4998
0.499
0.5
0.5
(c) The rate of change of g ( x) at x 1 is 0.5.
(d) The calculator gives lim 1 hh 1
h 0
20. (a) i)
ii)
f (3)
3
f (T )
T
1
3
f (2)
2
f (2)
2
1
2
1
1
T
1
2
T 2
1
6
1
2
2T
T
2T
1.
2
1
6
2 T
2T (2 T )
2 T
2T (T 2)
T 2
1
2T
,T
2
(b) T
f (T )
( f (T )
2.1
2.01
2.001
0.476190
0.497512
0.499750
f (2))/(T 2)
0.2381
0.2488
0.2500
(c) The table indicates the rate of change is 0.25 at t 2.
1
(d) lim 21T
4
T
2.0001
0.4999750
0.2500
2.00001
0.499997
0.2500
2.000001
0.499999
0.2500
2
NOTE: Answers will vary in Exercises 21 and 22.
21. (a) [0, 1]:
(b) At P
s
t
15 0
1 0
1 , 7.5
2
15 mph; [1, 2.5]:
s
t
20 15
2.5 1
10
3
mph; [2.5, 3.5]:
: Since the portion of the graph from t
s
t
30 20
3.5 2.5
10 mph
0 to t 1 is nearly linear, the instantaneous rate of
change will be almost the same as the average rate of change, thus the instantaneous speed at t 12 is
15 7.5 15 mi/hr. At P (2, 20): Since the portion of the graph from t 2 to t 2.5 is nearly linear, the
1 0.5
20 0 mi/hr.
instantaneous rate of change will be nearly the same as the average rate of change, thus v 20
2.5 2
For values of t less than 2, we have
Q
Q1 (1, 15)
Q2 (1.5, 19)
Q3 (1.9, 19.9)
s
t
Slope of PQ
15 20
1 2
19 20
1.5 2
19.9 20
1.9 2
5 mi/hr
2 mi/hr
1 mi/hr
Thus, it appears that the instantaneous speed at t
At P (3, 22):
Q
Q1 (4, 35)
Q2 (3.5, 30)
Q3 (3.1, 23)
Slope of PQ
35 22
4 3
30 22
3.5 3
23 22
3.1 3
s
t
2 is 0 mi/hr.
Q
13 mi/hr
Q1 (2, 20)
16 mi/hr
Q2 (2.5, 20)
10 mi/hr
Q3 (2.9, 21.6)
Thus, it appears that the instantaneous speed at t
Copyright
3 is about 7 mi/hr.
2014 Pearson Education, Inc.
Slope of PQ
20 22
2 3
20 22
2.5 3
21.6 22
2.9 3
s
t
2 mi/hr
4 mi/hr
4 mi/hr
62
Chapter 2 Limits and Continuity
(c) It appears that the curve is increasing the fastest at t
s
Slope of PQ
Q
t
Q1 (4, 35)
Q2 (3.75, 34)
Q3 (3.6, 32)
35 30 10 mi/hr
4 3.5
34 30
16 mi/hr
3.75 3.5
32 30
20 mi/hr
3.6 3.5
A
t
10 15
3 0
gal
Q2 (3.25, 25)
Q3 (3.4, 28)
3.9 15
5 0
A
t
1.67 day ; [0, 5]:
22 30
3 3.5
25 30
3.25 3.5
28 30
3.4 3.5
Q1 (3, 22)
Thus, it appears that the instantaneous speed at t
22. (a) [0, 3]:
3.5. Thus for P (3.5, 30)
Slope of PQ
Q
s
t
16 mi/hr
20 mi/hr
20 mi/hr
3.5 is about 20 mi/hr.
gal
2.2 day ; [7, 10]:
gal
0 1.4
10 7
A
t
0.5 day
(b) At P (1, 14):
Q
Q1 (2, 12.2)
Q2 (1.5, 13.2)
Q3 (1.1, 13.85)
A
t
Slope of PQ
12.2 14
2 1
13.2 14
1.5 1
13.85 14
1.1 1
Slope of PQ
Q
1.8 gal/day
Q1 (0, 15)
1.6 gal/day
Q2 (0.5, 14.6)
1.5 gal/day
15 14
0 1
14.6 14
0.5 1
14.86 14
0.9 1
Q3 (0.9, 14.86)
A
t
1 gal/day
1.2 gal/day
1.4 gal/day
Thus, it appears that the instantaneous rate of consumption at t 1 is about 1.45 gal/day.
At P (4, 6):
A
A
Slope of PQ
Q
Slope of PQ
t
Q
t
Q1 (5, 3.9)
Q2 (4.5, 4.8)
Q3 (4.1, 5.7)
3.9 6
5 4
4.8 6
4.5 4
5.7 6
4.1 4
2.1 gal/day
Q1 (3, 10)
2.4 gal/day
Q2 (3.5, 7.8)
3 gal/day
Q3 (3.9, 6.3)
10 6
3 4
7.8 6
3.5 4
6.3 6
3.9 4
4 gal/day
3.6 gal/day
3 gal/day
Thus, it appears that the instantaneous rate of consumption at t 1 is 3 gal/day.
At P (8, 1):
A
Slope of PQ
A
Q
t
Slope of PQ
Q
t
1.4
1
Q1 (7, 1.4)
0.6 gal/day
0.5 1
Q1 (9, 0.5)
7 8
0.5 gal/day
9 8
1.3
1
Q2 (7.5, 1.3)
0.6 gal/day
0.7 1
Q2 (8.5, 0.7)
7.5 8
0.6 gal/day
8.5 8
1.04 1
Q3 (7.9, 1.04)
0.6 gal/day
0.95 1
Q3 (8.1, 0.95)
7.9 8
0.5 gal/day
8.1 8
Thus, it appears that the instantaneous rate of consumption at t 1 is 0.55 gal/day.
(c) It appears that the curve (the consumption) is decreasing the fastest at t 3.5. Thus for P (3.5, 7.8)
s
Slope of PQ
A
Q
Slope of PQ
t
Q
t
11.2
7.8
Q1 (2.5, 11.2)
4.8 7.8
3.4 gal/day
Q1 (4.5, 4.8)
3 gal/day
2.5 3.5
Q2 (4, 6)
Q3 (3.6, 7.4)
4.5 3.5
6 7.8
4 3.5
7.4 7.8
3.6 3.5
Q2 (3, 10)
3.6 gal/day
Q3 (3.4, 8.2)
4 gal/day
Thus, it appears that the rate of consumption at t
2.2
10 7.8
3 3.5
8.2 7.8
3.4 3.5
4.4 gal/day
4 gal/day
3.5 is about 4 gal/day.
LIMIT OF A FUNCTION AND LIMIT LAWS
1. (a) Does not exist. As x approaches 1 from the right, g ( x ) approaches 0. As x approaches 1 from the left, g ( x)
approaches 1. There is no single number L that all the values g ( x) get arbitrarily close to as x 1.
(b) 1
(c) 0
(d) 0.5
2. (a) 0
(b) 1
Copyright
2014 Pearson Education, Inc.
Section 2.2 Limit of a Function and Limit Laws
63
(c) Does not exist. As t approaches 0 from the left, f (t ) approaches 1. As t approaches 0 from the right,
f (t ) approaches 1. There is no single number L that f (t ) gets arbitrarily close to as t
0.
(d) 1
3. (a) True
(d) False
(g) True
(b) True
(e) False
(c) False
(f) True
4. (a) False
(d) True
(b) False
(e) True
(c) True
x
0 | x|
5. lim
x
does not exist because | xx |
x
x
0 and | xx |
1 if x
x
x
1 if x
0. As x approaches 0 from the left, | xx |
approaches 1. As x approaches 0 from the right, | xx | approaches 1. There is no single number L that all the
function values get arbitrarily close to as x
0.
6. As x approaches 1 from the left, the values of x1 1 become increasingly large and negative. As x approaches 1
from the right, the values become increasingly large and positive. There is no number L that all the function
values get arbitrarily close to as x 1, so lim x1 1 does not exist.
x 1
7. Nothing can be said about f ( x) because the existence of a limit as x x0 does not depend on how the function
is defined at x0 . In order for a limit to exist, f ( x ) must be arbitrarily close to a single real number L when x is
close enough to x0 . That is, the existence of a limit depends on the values of f ( x ) for x near x0 , not on the
definition of f ( x) at x0 itself.
8. Nothing can be said. In order for lim f ( x) to exist, f ( x) must close to a single value for x near 0 regardless of
x
the value f (0) itself.
0
9. No, the definition does not require that f be defined at x 1 in order for a limiting value to exist there. If f (1) is
defined, it can be any real number, so we can conclude nothing about f (1) from lim f ( x) 5.
x 1
10. No, because the existence of a limit depends on the values of f ( x) when x is near 1, not on f (1) itself. If
lim f ( x) exists, its value may be some number other than f (1) 5. We can conclude nothing about lim f ( x),
x 1
whether it exists or what its value is if it does exist, from knowing the value of f (1) alone.
lim ( x 2 13) ( 3)2 13 9 13
11.
12. lim ( x 2 5 x 2)
x
2
(2) 2 5(2) 2
13. lim 8(t 5)(t 7) 8(6 5)(6 7)
t
14.
15.
6
lim ( x3 2 x 2
x
2
lim 2 x 53
x
4
3
x
2 11 x
4 x 8)
2(2) 5
11 (2)3
9
3
4 10 2
4
8
( 2)3 2( 2)2
4( 2) 8
8 8 8 8
16
3
Copyright
2014 Pearson Education, Inc.
x 1
64
16.
Chapter 2 Limits and Continuity
t
17.
x
2/3
lim 4 x (3x 4)2
y
y 2
y
z 2 10
4
21. lim
3
3h 1 1
22. lim
5h 4 2
h
h
h
0
0
x
24.
25.
5 x
lim
x
x 3
2
3 x 4x 3
lim
x
2
27. lim t 2 t 2
t 1 t
29.
t
1
y
5 y3 8 y 2
31.
1
lim xx 11
x 1
32.
lim
x
lim
4
2
0 3 y 16 y
0
1
x 1
x
1
x 1
1 x
lim x
x 1x 1
lim
0
2
25
2
y
lim 1 xx x1 1
1 2
1 2
5y 8
lim
x 1
lim
h
7
3
1
3
8
16
1
x
2x
1
lim ( x 1)(
x 1) x
0
Copyright
5h
5h 4 2
1
2
2
0 3 y 16
x
0 h
1
2
2 5
2
4
2
2
2 x
lim
h
3
2
1 2
1 1
lim
x 1
x
2
16
5 2
5
lim
y 2 (5 y 8)
( x 1) ( x 1)
( x 1)( x 1)
1
3 1
lim tt 22
1
x
2
2
0 y (3 y 16)
y
x
t
2
2 x ( x 2)
24
5h 4 2
lim ( x 2)
x
lim t 2
t 1t 1
2( x 2)
lim
( 2) 25
1
10
lim ( x 5)
(t 2)(t 1)
x
1
5 5
lim x1 1
3
x
lim (t 2)(t 1)
t
1
t 2
0 h
x
( x 5)( x 2)
x 2
(t 2)(t 1)
2x 4
3
2
2 x 2x
30. lim
2
lim (t 1)(t 1)
1
lim
x
lim
x
4
(5h 4) 4
lim
h
x
( x 5)( x 2)
x 5
t
2
lim t 2 3t 2
1 t
5
2
6
5h 4 2
5h 4 2
x 3
lim ( x 3)(
x 1)
3
2
lim x x3 x5 10
5
4
2
3
2
x
x
(6) 13
1
1
5
(8)1/3
lim x 1 5
5
x
2
26. lim x x7 x2 10
x 2
28.
0
3
2
( 2)
4
20
16 10
5h 4 2
h
lim
h
2
(8)4/3
3
1 1
(8 2) 43
1
4
4
4 10 6
x 5
lim ( x 5)(
x 5)
5
25
1
2
3
4 2 10
3
3(0) 1 1
23. lim 2x 5
2 23
[5 ( 3)]4/3
3
lim
z
1
2
2 2
(2)2 5(2) 6
2
2 y 5y 6
lim (5 y ) 4/3
20.
4
1/2
18. lim
19.
8 5 23
lim (8 3s )(2 s 1)
1
2
1
lim ( x 1)(2 x 1)
0
x
2
1
2
2014 Pearson Education, Inc.
0
5
5h 4 2
5
4 2
5
4
Section 2.2 Limit of a Function and Limit Laws
4
33. lim u 3 1
u 1u
1
u 1 (u
3
34. lim v4 8
2v
35.
lim xx 93
9
x
2
36. lim 4 x x
x
x 1
38.
lim
x
x (4 x )
4 2 x
x
1
( x 1)
lim
lim
x
1
lim
x 2 12 4
x 2
x
2
x
2
40.
x 2
lim
x
x2 5 3
2
x
41.
lim
x
3
x2 5
x 3
2
lim
x
lim
x
42. lim
x
45
4 x
x2 9
lim
x
x
43. lim (2sin x 1)
0
45. lim sec x
x
0
4
16 4
( x 2)
x2 5 3
x2 5 3
4
x
3 ( x 3) 2
x2 5
(4 x ) 5
x2 9
x2 9 5
2sin 0 1 0 1
1
1
lim
x
8 3
( x 2 12) 16
2 ( x 2)
x
2
2
4
2
4
( x 1)( x 1)
x2 8 3
1 ( x 1)
( x 2)( x 2)
lim
x
12 4
2
x
x2 5 3
( x 2)
lim
2 ( x 2)
lim
( x 2 5) 9
x
x
5
lim
x
3 ( x 3) 2
3 x
32
lim
x
4 ( x 2 5)
4
1
1
Copyright
x
2
6
2
x2 5
( x 2)
x 2 12 4
x2 5 3
( x 2)( x 2)
2
4
lim
x
5
lim
25 ( x 2 9)
5
25
8
44.
46.
9 x2
x2 5
3 ( x 3) 2
3
2
x2 9
(4 x ) 5
2
lim 5 4 x x 9
x 4
(4 x)(4 x )
x
2
3
3
2
x2 9
x2 9
1 ( x 1)
lim
x2 5
2
x
1
2
9 3
4
( x 3) 2
1
cos 0
x
x2 5 3
x2 5 2
lim
x 1
( x 2 8) 9
lim
x 2 12 4
x 2 12 4
(4 x ) 5
lim cos1 x
0
x
x 2 12 4
4(2 2) 16
1
3
(3 x )(3 x )
4 5
lim
x
x 2 12 4
2
3
x
8 3
x2 5 3
x 2
lim
2
x
x
4
x 3 2
lim
2
3 3
2
x
x2 8 3
2
1
6
( x 3) 4
x 1
x2 8 3
lim
( x 1)
3
8
12
32
lim x 2
x
lim
x 3 2
x 2
2
x)
4
3
4 4 4
(4)(8)
1
9 3
x )(2
x
x 3 2
( x 2)
lim
x
2
x 1
lim
9
x (2
( x 1)
1
1
x 3
lim
x2 8 3
x
39. lim
v
4
x 3 2
x 1
x2 8 3
x 1
x
v 2 2v 4
2
2 ( v 2)(v 4)
lim
x
lim
(1 1)(1 1)
1 1 1
u2 u 1
u 1
x 3
9 ( x 3)( x 3)
lim
x
x 1
x 3 2
37. lim
u 1)(u 1)
lim
x
(u 2 1)(u 1)
lim
2
2 ( v 2)( v 2)(v 4)
v
4 2
2
(v 2)(v 2 2v 4)
lim
16
v
(u 2 1)(u 1)(u 1)
lim
65
x
(4 x ) 5
4
x2 9
16 x 2
5
4
2
lim sin 2 x
x
0
lim tan x
x
lim sin x
0
x
0
sin x
lim cos
x
0
x
2014 Pearson Education, Inc.
sin 0
cos 0
(sin 0)2
0
1
0
02
0
66
Chapter 2 Limits and Continuity
x sin x
47. lim 1 3cos
x
x 0
1 0 sin 0
3cos 0
48. lim ( x 2 1)(2 cos x)
49.
x
lim
x 4 cos( x
)
50. lim 7 sec2 x
x
1
3
(02 1)(2 cos 0)
0
x
1 0 0
3
x
lim
x
4
x
lim (7 sec2 x)
0
( 1)(1)
lim cos( x
4 cos 0
7 sec 2 0
0
x
1
)
lim sec2 x
7
0
x
( 1)(2 1)
4
7 (1)2
1
4
2 2
51. (a) quotient rule
(c) sum and constant multiple rules
(b) difference and power rules
52. (a) quotient rule
(c) difference and constant multiple rules
(b)
53. (a) lim f ( x) g ( x)
x
lim f ( x )
c
x
(b) lim 2 f ( x) g ( x)
x
c
x
c
x
x
4
x
4
x
x
x
x
x
4
(d) lim f ( x)/g ( x)
x
56. (a)
lim f ( x)
b
x
b
(c)
2
2
lim 1 2 h h h 1
h 0
lim r ( x)
x
lim
h
0
lim r ( x)
Copyright
x
x
5 lim r ( x)
x
2
0
4 0 ( 3) 1
(4)(0)( 3)
2
lim (2 h)
h
2
lim s ( x)
2
2
h (2 h )
h
lim s ( x)
2
4 lim p( x)
2
(1 h )2 12
h
0
57. lim
7
3
x
x
21
12
7
3
2
lim [ 4 p ( x) 5r ( x)]/s ( x )
x
(4)( 3)
lim p ( x)
x
(7)( 3)
b
b
x
4
b
lim p( x)
2
2
3
x
x
x
1
9
b
b
5 3( 2)
0
lim f ( x )/ lim g ( x)
x
lim p ( x) r ( x) s ( x)
x
(4)(0)
lim g ( x)
x
20
5
7
lim g ( x)
b
2(5)( 2)
lim g ( x) 7 ( 3)
b
lim [ p ( x ) r ( x ) s( x)]
x
(b)
h
lim f ( x)
x
lim 4
x
4
10
3 3 0
3
0 1
4
b
b
5
5 ( 2)
[ 3]2
lim f ( x ) lim 1
(b) lim f ( x) g ( x)
x
4
2
4
lim g ( x )
b
(c) lim 4 g ( x)
x
lim g ( x)
55. (a) lim [ f ( x) g ( x)]
x
4
c
lim 3
4
lim x lim f ( x)
x
g ( x)
x
x
c
lim g ( x)
4
(d) lim f ( x ) 1
x 4
c
c
c
x
(c) lim [ g ( x)]2
x
x
lim f ( x) 3 lim g ( x)
x
54. (a) lim [ g ( x) 3]
(b) lim xf ( x)
lim g ( x)
c
lim f ( x ) lim g ( x )
x
(5)( 2)
c
x c
lim f ( x )
f ( x)
(d) lim f ( x ) g ( x)
x c
4
x
2 lim f ( x)
(c) lim [ f ( x) 3 g ( x)]
x
lim g ( x )
c
power and product rules
0
lim s( x) [ 4(4)
x
2
2
2014 Pearson Education, Inc.
5(0)]/ 3
16
3
Section 2.2 Limit of a Function and Limit Laws
( 2 h )2 ( 2) 2
h
0
2
lim 4 4h h h 4
h 0
58. lim
h
[3(2 h) 4] [3(2) 4]
h
0
lim 3hh
59. lim
h
1
2 h
60. lim
1
2
0
h
61. lim
7 h
h
h
h
0
7
h
3(0 h ) 1
h
0
x
0
64. lim (2 x 2 )
x
0
2 ( 2 h)
7 h
7
h
0
h
2
65. (a) lim 1 x6
0h
h
3h 1 1
x
0
2 and lim 2 cos x
0
7 h
lim
3h 1 1
5 and lim
1
4
(7 h ) 7
lim
7
h
0
x
4
0
h
7
3h 1 1
5 2(0)2
2 0
7 h
7 h
lim
lim ( h 4)
h
lim h(4 h2 h)
0
lim 2h( 2 h)
0
h
3(0) 1
5 2x2
63. lim
1
0
3
0
2h
0
lim
62. lim
h
2
2 h
lim
h
h
h( h 4)
h
lim
h
h
5 x2
2(1)
lim
7
0h
h
(3h 1) 1
0h
h
7 h
lim
3h 1 1
h
5 (0)2
0h
0
66. (a) lim 12
x 0
x2
24
h
3h
3h 1 1
x
0
x2
lim 24
x
0
1
2
0
1
2
2; by the sandwich theorem, lim g ( x)
x
1
2
and lim 12
x 0
1;
2
0
3
3h 1 1
3
2
0
0
2
1
x
by the sandwich theorem, lim 1 cos
2
x
0
x
x 2 /24, and the graphs converge as
0.
x
67. (a)
lim
h
1
2 7
7
x
(b) For all x 0, the graph of f ( x) (1 cos x)/x 2
lies between the line y 12 and the parabola
y
0
5; by the sandwich theorem, lim f ( x )
1 60
lim 12
1
7 h
lim
7
x
1 and lim 1 1; by the sandwich theorem, lim 2 x2sin
cos x
x 0
x 0
(b) For x 0, y ( x sin x)/(2 2 cos x) lies
between the other two graphs in the figure,
and the graphs converge as x 0.
x
f ( x)
( x 2 9)/( x 3)
x
3.1
3.01
3.001
3.0001
3.00001
3.000001
f ( x)
6.1
6.01
6.001
6.0001
6.00001
6.000001
x
2.9
2.99
2.999
2.9999
2.99999
2.999999
5.999
5.9999
5.99999
5.999999
f ( x)
5.99
5.9
The estimate is lim f ( x)
6.
x
67
3
Copyright
2014 Pearson Education, Inc.
1.
2
5
68
Chapter 2 Limits and Continuity
(b)
(c) f ( x)
68. (a) g ( x)
x
g ( x)
x2 9
x 3
( x2
( x 3)( x 3)
x 3
2)/ x
1.4
2.81421
x 3 if x
3, and lim ( x 3)
x
3 3
3
6.
2
1.41
2.82421
1.414
2.82821
1.4142
2.828413
1.41421
2.828423
1.414213
2.828426
(b)
(c) g ( x)
69. (a) G ( x)
x
G ( x)
x
G ( x)
x2 2
x 2
x
2 x
x
2
x
2
2 if x
2, and lim
x
2
x
2
2
2
2 2.
( x 6)/( x 2
5.9
.126582
4 x 12)
5.99
.1251564
5.999
.1250156
5.9999
.1250015
5.99999
.1250001
5.999999
.1250000
6.1
.123456
6.01
.124843
6.001
.124984
6.0001
.124998
6.00001
.124999
6.000001
.124999
x 6
( x 2 4 x 12)
x 6
( x 6)( x 2)
(b)
(c) G ( x)
1
x 2
if x
Copyright
6, and lim x 1 2
x
6
1
6 2
2014 Pearson Education, Inc.
1
8
0.125.
Section 2.2 Limit of a Function and Limit Laws
70. (a) h( x)
x
h( x )
( x2
2 x 3)/( x 2
4 x 3)
2.9
2.052631
2.99
2.005025
2.999
2.000500
2.9999
2.000050
2.99999
2.000005
2.999999
2.0000005
x
3.1
h( x) 1.952380
3.01
1.995024
3.001
1.999500
3.0001
1.999950
3.00001
1.999995
3.000001
1.999999
(b)
(c) h( x)
71. (a) f ( x)
x
f ( x)
x2 2 x 3
x2 4 x 3
( x 3)( x 1)
( x 3)( x 1)
x 1 if
x 1
x
3, and lim xx 11
x 3
3 1
3 1
4
2
2.
( x 2 1)/(| x | 1)
1.1
2.1
1.01
2.01
1.001
2.001
x
f ( x)
.9
1.9
.99
1.99
(c) f ( x)
x2 1
x 1
( x 1)( x 1)
x 1
( x 1)( x 1)
( x 1)
.999
1.999
1.0001
2.0001
.9999
1.9999
1.00001
2.00001
.99999
1.99999
1.000001
2.000001
.999999
1.999999
(b)
72. (a) F ( x)
x 1, x
0 and x 1
1 x, x
0 and x
1
, and lim (1 x) 1 ( 1)
x
1
( x 2 3 x 2)/(2 | x |)
x
F ( x)
2.1
1.1
2.01
1.01
2.001
1.001
2.0001
1.0001
2.00001
1.00001
2.000001
1.000001
x
F ( x)
1.9
.9
1.99
.99
1.999
.999
1.9999
.9999
1.99999
.99999
1.999999
.999999
Copyright
2014 Pearson Education, Inc.
2.
69
70
Chapter 2 Limits and Continuity
(b)
(c) F ( x)
73. (a) g ( )
g( )
( x 2)( x 1)
,
2 x
( x 2)( x 1)
2 x
x2 3x 2
2 x
x
0
x 1, x
, and lim ( x 1)
0 and x
x
2
2
(sin )/
.1
.998334
.01
.999983
.1
g( )
.998334
lim g( ) 1
.001
.999999
.01
.999983
.0001
.999999
.001
.999999
.0001
.999999
.00001
.999999
.00001
.999999
.000001
.999999
.000001
.999999
0
(b)
74. (a) G (t )
t
G (t )
(1 cos t )/t 2
.1
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
t
.1
G (t ) .499583
lim G (t ) 0.5
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
t
2 1
0
(b)
Copyright
2014 Pearson Education, Inc.
1.
Section 2.2 Limit of a Function and Limit Laws
f ( x) x1/(1 x)
x
.9
.99
f(x) .348678 .366032
75. (a)
x
1.1
1.01
f(x) .385543 .369711
lim f ( x) 0.36788
.999
.367695
.9999
.367861
.99999
.367877
.999999
.367879
1.001
.368063
1.0001
.367897
1.00001
.367881
1.000001
.367878
71
x 1
(b)
Graph is NOT TO SCALE. Also, the intersection of the axes is not the origin: the axes intersect at the
point (1, 2.71820).
f ( x) (3 x 1)/x
x
.1
.01
f(x) 1.161231 1.104669
76. (a)
x
.1
.01
f(x) 1.040415 1.092599
lim f ( x) 1.0986
.001
1.099215
.0001
1.098672
.00001
1.098618
.000001
1.098612
.001
1.098009
.0001
1.098551
.00001
1.098606
.000001
1.098611
x 1
(b)
77. lim f ( x) exists at those points c where lim x 4
x
c
lim x 2
Moreover, lim f ( x )
0
x
0
x
x
lim x 2 . Thus, c 4
c
x
0 and lim f ( x)
x
c
79. 1
lim f ( x ) lim 5
f ( x) 5
x
lim
x 4 x 2
80. (a) 1
lim
x
4
x
x
4
lim x lim 2
x
f ( x)
2 x
2
x
4
lim f ( x )
x
2
lim x
x
2
lim f ( x)
x
c
2
4
2
Copyright
x
4
lim f ( x )
x
2
0, 1, or 1.
x 1
lim f ( x) 5
4 2
4
0
2. Yes, f (2) could be 0. Since the conditions
2
lim f ( x ) 5
4
c 2 (1 c 2 )
lim f ( x) 1.
1
78. Nothing can be concluded about the values of f , g , and h at x
5 0.
of the sandwich theorem are satisfied, lim f ( x)
x
c2
2(1)
lim f ( x)
x
4
4.
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7.
72
Chapter 2 Limits and Continuity
(b) 1
81. (a) 0
f ( x)
lim
lim
2
2 x
x
3 0
x
x
f ( x) 5
2 x 2
x
2
4 0
x
lim
2
0 x
x
0
(b) 0 1 0
lim
83. (a) lim x sin 1x
0.
f ( x)
x
lim x
x
1 sin 1x 1 for x 0:
x 0
x x sin 1x x
x sin 1x
x
84. (a) lim x 2 cos 13
x
x
0
1 cos 13
(b)
lim x 2
x
0
x
x
( x 2)
lim f ( x )
f ( x)
lim x 2
f ( x)
0
x2
x
lim
0
x
lim
x
2
2
0
x
0
f ( x)
. That
x
f ( x)
x2
x2
is, lim
x
x
lim x sin 1x
0 by the sandwich theorem;
lim x sin 1x
0 by the sandwich theorem.
x
0
x
0
0
lim f ( x ).
x
f ( x)
x
0
0.
0
1 for x
0
x2
x 2 cos 13
x
x2
lim x 2 cos 13
x
0
x
85-90. Example CAS commands:
Maple:
f : x - (x^4 16)/(x 2);
x0-1..x0 1, color
black,
title "Section 2.2, #85(a)" );
limit( f (x), x
2.
lim [ f ( x) 5]
x
0.
x0 : 2;
plot( f (x), x
f ( x)
x
5 as in part (a).
2
2
0 x
lim
x
0
2
x
lim
0
f ( x) 5
x 2
lim
x
0
0
0
2)
lim x
2
0 x
x
x
2
f ( x)
1
2
2
lim
5.
x 2
f ( x) 5
lim ( x
x 2
2 x 2
x
x
x
2
f ( x)
x
lim
x
2
lim ( x 2)
lim f ( x)
That is, lim f ( x)
(b)
lim 1x
x
lim
82. (a) 0 1 0
x
f ( x)
x
lim
lim f ( x ) 5
(b) 0
2
x 0 );
Copyright
2014 Pearson Education, Inc.
0 by the sandwich theorem since
Section 2.3 The Precise Definition of a Limit
In Exercise 87, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be
overcome in Maple by entering the function as f : x - (surd(x 1, 3) 1)/x.
Mathematica: (assigned function and values for x0 and h may vary)
Clear[f , x]
f[x _]: (x 3 x 2 5x 3)/(x 1)2
x0 1; h 0.1;
Plot[f[x],{x, x0 h, x0 h}]
Limit[f[x], x
x0]
2.3
THE PRECISE DEFINITION OF A LIMIT
1.
Step 1:
Step 2:
x 5
x 5
5 x
5
5 7
2,or
5 1
4.
The value of which assures x 5
1 x 7 is the smaller value,
2.
2.
Step 1:
Step 2:
x 2
2 1
The value of
x 2
2 x
2
1, or
2 7
5.
which assures x 2
1 x 7 is the smaller value,
1.
3.
Step 1:
Step 2:
x ( 3)
3
7
2
The value of
x 3
or
1,
2
3
x
3
5.
2
1
2
3
which assures x ( 3)
7
2
x
1
2
is the smaller value,
1.
2
4.
Step 1:
x
3
2
Step 2:
3
2
x
7
2
The value of
2, or
3
2
3
2
1
2
which assures x
3
2
3
2
x
1.
3
2
7
2
x
1
2
is the smaller value,
5.
Step 1:
Step 2:
x 12
1
2
4
9
The value of
x 12
1 , or
18
1
2
4
7
which assures x 12
Copyright
1
2
x
1.
14
4
9
1
2
x
4
7
is the smaller value,
2014 Pearson Education, Inc.
1.
18
1.
73
74
Chapter 2 Limits and Continuity
6.
Step 1:
Step 2:
7. Step 1:
x 3
x 3
3 x
3
3 2.7591
0.2409, or
3 3.2391
0.2391.
The value of which assures x 3
2.7591 x 3.2391 is the smaller value,
Step 2:
x 5
From the graph,
x 5
5 4.9
8. Step 1:
Step 2:
x ( 3)
From the graph,
9. Step 1:
Step 2:
x 1
x 1
9
From the graph,
1 16
5 x
0.1, or
x 3
3
3.1
3 x
0.1, or
1
10. Step 1:
Step 2:
x 3
From the graph,
x 3
3 2.61
3
11. Step 1:
Step 2:
x 2
From the graph,
5 2.
thus
x 2
2
2
2
12. Step 1:
x ( 1)
Step 2:
thus
5 2
.
2
13. Step 1:
Step 2:
x ( 1)
From the graph,
14. Step 1:
x 12
Step 2:
x 12
( x 1) 5
x 4
16. Step 1:
(2 x 2) ( 6)
4.02 2 x
x ( 2)
17. Step 1:
Step 2:
1
2
From the graph,
thus
0.00248.
15. Step 1:
Step 2:
Step 2:
x 1
16
1
9
0.01
x 1 1
0.19
x 0
x 4
x 4
1 x
1
1 x
1
0.77, or
1
2
1
2
0.01
4
0.41; thus
3
2
16
25
9
25
0.00248, or
0.01 x 4
x
4
0.01
5
1
1
2
x
1
2.01
1
7.
16
thus
2
0.118 or
0.1.
1
2
3.99
0.01.
0.39.
5 2
2
3
0.2361;
0.1340;
2
0.36; thus
1
1.99
x
1
1.99
4.01
0.02
2 x 4 0.02
0.02 2 x 4 0.02
3.98
2.01 x
1.99
x 2
2 x
2
0.01.
0.1
x
1
2.01
9 ;
16
2
0.2679, or
5 2
2
7
9
0.1 in either case.
0.1; thus
1
25
1 16
x
3
5
2
1
2.9
x
3
0.39, or
3 3.41
x 1
From the graph,
0.1; thus
3
3
x
or
7 ,
16
3
5
5 5.1
0.21
x
0.1
x 1 1 0.1
. Then,
Copyright
0.19
0.2391.
0.9
x 1 1.1
0.19 or
0.81 x 1 1.21
0.21; thus,
2014 Pearson Education, Inc.
0.19.
9
25
0.36.
1
2
0.00251;
Section 2.3 The Precise Definition of a Limit
18. Step 1:
Step 2:
x 14
19. Step 1:
Step 2:
21. Step 1:
Step 2:
22. Step 1:
Step 2:
Step 2:
24. Step 1:
Step 2:
25. Step 1:
Step 2:
26. Step 1:
Step 2:
27. Step 1:
Step 2:
0.1
1
2
x
x 14
1
4
0.16
1
0.1
x 7 4
0.09 or
1
1
x 23
Then
1
x
1
4
0.05
x 4
Then
4
x2 3
x
10
3
x 23
7, or
x
3
0.0286
4
1
4
x2
( 1)
x ( 1)
Then
0.1
4
1
10
9
11
10
1
x 2 16
x 4
Then
x 4
4
15
17 4 0.12.
120
x
5
1
4
15
1 120
5 1
x
1
4
20
4, or
mx 2m 0.03
0.03
x 2
x 2
Then
2 2 0.03
m
3.5
x
or 10
3
x
5.
32
10
11
1.
x2
15
1
6
x
120
24.
6; thus
4.5
4.5
2
10
9
x
3
3.5
or 10
9
0.0286;
x
17
15
17 4
30
x
0.1292;
10 .
11
x
x
17.
0.1231; thus
20 or 20
x
30.
4.
2m 0.03
0.03 2m mx 0.03 2m 2 0.03
x
m
2 x
2.
0.03 , or
0.03 . In either case,
0.03 .
2 2 0.03
m
m
m
m
mx
Copyright
3.5,
1.
11
thus
120
x
1
4
3.1
3.5
9
10
1;
11
24 x
30
x
3.1
3.1
17
24
x
3
4 x
4.
0.1270, or
4
x 24
24
4.5
1
x
6
16
3.
x 2 16 1
1
10
3
x
2.9
x
10
11
1
25
2.
3
3.1
x2
1 x
1 , or
9
10
2
2 x
2.
0.1213, or
2
1 0.1
x 7
7.
0.0291, or
3.5
x 1
( x 2 5) 11
x 24
Then
0.5
0.09.
4 19 x 16
9
0.3
x2
2.9
0.36
5.
5
4.
1; thus
2.9
4.5 2
1
x
0.1
4
23.
9; thus
1
x
3
for x near 2.
x ( 2)
x 2
Then
2
4.5
thus
4.5 2 0.12.
1
x
x 7
4 x
4 5 or
3
0.5
19 x
0.2
3
x
0.11; thus
23 x
32
0.05
2.9
0.5
0.36
3
23
x 4
2 , or
or
3
3
x2
1
x
0.16
1.
4
2
4 1
0.1 x 2 3 0.1
0.1
1
4
0.6
3 or 3 x 15
10 x
10.
10 15
5; thus
x 7
0.05
x
x
19 x 3 1
1
23 16
0.4
1
4
4 x 19
16 15 x
x 10
x 10
Then
10 3
7, or
thus
23. Step 1:
0.1
19 x 3
20. Step 1:
Step 2:
1
2
x
75
2014 Pearson Education, Inc.
2
0.03 .
m
76
Chapter 2 Limits and Continuity
28. Step 1:
Step 2:
mx 3m c
c mx 3m c
x 3
x 3
c , or
Then
3 3 mc
m
29. Step 1:
m
2
(mx b)
x 12
Step 2:
1
2
Then
30. Step 1:
1
Step 2:
x 1
Then
31. lim (3 2 x)
x
3
Step 1:
Step 1:
Step 2:
34.
35.
1
2
c
m
mx m
1
x
0.05 , or
m
1.
2
1.95
x 2
Then
x
either case,
c.
m
mx
c
0.05
m
2
1.
1 1 0.05
m
c
m
1
2
x
0.05 . In
m
mx
c.
m
1
2
c.
m
either case,
0.05 m
6.02
0.05 m
0.05 .
m
either case,
2x
5.98
3.01
x
2.99 or
3.
0.01; thus
lim
x
2)
2 2
4, x
0.01.
( x 2)( x 2)
( x 2)
5
(x 5)(x 1)
(x 5)
( 4)
4 0.05
0.05
4, x
5
( x 5)( x 1)
( x 5)
0.05
5.
4.95, x
x 5
5.05
0.05, or
1 5( 3)
1 5x 4
lim ( x 1)
x
0.5
16
0.5
x 1
1.01 x
0.01
0.01; thus
0.99.
0.01.
2
2.05, x 2.
x 2
2 x
2.
2 1.95
0.05, or
2 2.05
5.05 x
x ( 5)
Then
5
1 5x
c . In
m
c . In
m
0.03
3x 3 0.03 0.01
x 1
1 x
1.
1.01
0.01, or
1
0.99
Step 2:
Step 2:
x
1
2
m
2
c
0.03
x2 6 x 5
x 5
Step 1:
c
3 mc
3
Step 1:
3
0.05
( x 2)( x 2)
lim ( x
( x 2)
x 2
x 2
x2 4
4 0.05
0.05
x 2
2
lim x x 6x5 5
5
lim
0.05
or
0.05 .
m
lim
x
x
c,
m
x 1
1 1 0.05
m
( 3x 2) 1
x ( 1)
Then
1
2
33. lim xx 24
x 2
Step 2:
b)
m
2
1
2
mx
x
( 3)( 1) 2 1
1
Step 1:
c
m
x 1
3 2(3)
lim ( 3x 2)
x
c
3 mc
c 3m
(3 2 x) ( 3) 0.02
0.02 6 2 x 0.02
2.99 x 3.01.
0 x 3
x 3
3 x
Then
3 2.99
0.01, or
3 3.01
Step 2:
32.
c
x 12
1
2
(mx b) ( m
0.05
m
b
c 3m mx
3 x
3.
3 3 mc
3.95
x 2 4.05, x
0.05; thus
2
0.05.
5.
4
5 x
5.
5
4.95
0.05
4.05
0.05; thus
x 1
3.95, x
5
0.05.
4
1 5x 4
11.25
5 x 19.25
3.85 x
x ( 3)
x 3
Then
3
3.85
0.85, or
Copyright
0.5
3.5
1 5x
4.5
2.25.
3 x
3.
3
2.25 0.75; thus
2014 Pearson Education, Inc.
12.25 1 5 x
0.75.
20.25
Section 2.3 The Precise Definition of a Limit
36. lim 4x
x 2
Step 1:
Step 2:
37. Step 1:
Step 2:
38. Step 1:
Step 2:
4
2
2
4
x
41. Step 1:
Step 2:
0.4
Step 2:
Step 2:
2
1.6
2
1,
3
or
4 x
x 4
4
, or
x 5 2
x
4
x
2.
4
4 x
4
x
2
(2 )2 x
x 0
Then
(2
smaller distance,
4
x
)2
2
4
4
2
x
For x
x2 1
1
, 1
x2
4
near x
2.
x ( 2)
x 2
Then
2
4
1
x
2, 2
1
x
1
x 1
Then 1
1
x2
1
3
1 3
1
3
1 3
4
x
4
.
2
x
.
3 3.
. Thus choose
2
2
9
2
x
(2
, or
3
.
)2
(2
4
x 5 (2
2
4
)2
(2
)2
)2
. Thus choose the smaller
4 x
(2
)2
4.
)2
(2
x2
1
4
1
2
4
. Thus choose the
x
1
1.
1
1
1
x2 4
4
2 x
2, or
1. Choose
4
x
4
2.
2
4
1
1
1
1
1
x 1
, or 1
.
1
1
1
x
2
1
1
1
3
3
1 3
1
3
1
x2
x
3
1 3
Copyright
1 3
3
1
3
, or
3
1 3
1
1
, the smaller of the two distances.
1
x2
x2
4
1
x
1
1
1
3
4
4
. Choose
.
x 1
1
3
4 x
4
1
4
4
1
5.
2
x
1 , that is, the smaller of the two distances.
2, x 2 4
4
x
3 3
9.
9
4
1
near x 1.
x 1
1 x
1
1 1 , or
1
or 53
.
For x 1, x 2 1
1
x 1
Then
4
10
6
x
. Thus choose
x 5
)2
(2
4
4
2
)2
(2
.
10
4
1.
3
4
2
4 x 2
10
24
4.
9
3x 9
3 x
3.
3 3 3
x 5 2
x
4
1 ; thus
2
(2 ) 2 5 x (2 ) 2 5.
x 9
x 9
9 x
2
2
Then
9
4 9
4
, or
2
4
.
distance,
4 x
10
16
2.4
5
2
2
(3 x 7) 2
3x 9
x 3
x 3
Then
3 3 3
, or
3
Choose
44. Step 1:
0.4
x 2
min
43. Step 1:
2
5
3
min 1
42. Step 1:
4
x
0.4
(9 x) 5
x 4
Then
4
40. Step 1:
Step 2:
2
x 2
Then
39. Step 1:
Step 2:
77
.
1
1
1
x2
x
1
.
1 3
3
3
1 3
2014 Pearson Education, Inc.
for x near 3.
4
x
78
Chapter 2 Limits and Continuity
Step 2:
x
3
x
Then 3
Choose
Step 2:
x2 1
x 1
47. Step 1:
Step 2:
48. Step 1:
3
1 3
3
,x
3
, or
(x 1) 2
x 1
1
, or 1
1
0
0:
x 0
Then
x
2
0
0
x
x
.
, or
2
2
x sin 1x
49. By the figure, x
.
0
x 1
3
x
.
.
since x 1. Thus, 1
2
x
0;
since x 1. Thus, 1 x 1 6 .
x 1 .
1 6
. Choose
.
6
6
x
2
0;
2
2 . Choose
0 and x
0
x sin 1x
2
.
x for x
0. Since lim ( x)
x
x
c
f (h c) L
53. Let f ( x)
whenever 0
0
0, then by
0.
0. Since lim ( x 2 )
x
0
lim x 2
x
0
0, then by the
0.
x c
0, there exists
h 0
,x
0
(h c) c
0 such that f ( x) L
lim f (h c)
h
c
c
whenever 0
h
x c
L.
x
2
x never gets arbitrarily close to 1 for x near 0.
Copyright
0, there exists a
,h c
x 2 . The function values do get closer to 1 as x approaches 0, but lim f ( x)
function f ( x)
lim x
x
0
51. As x approaches the value 0, the values of g ( x) approach k. Thus for every number
such that 0 x 0
g ( x) k
.
52. Write x h c. Then 0 x c
h 0 0 h 0
.
Thus, lim f ( x) L
for any
3.
.
. Choose
x 2 for all x except possibly at x
sandwich theorem, lim x 2 sin 1x
3.
2.
x
x
x 3
. Choose
x 1
1
0
x for all x
x 2 sin 1x
3
1 3
3.
1
the sandwich theorem, in either case, lim x sin 1x
50. By the figure, x 2
3
,x 1
2 2x
2x
3
1 3
x
3
3
x 1: (6 x 4) 2
0 6x 6
x 1
x 1
1
Then 1
1 2
, or 1
2
0: 2 x 0
.
3 .
x 3
2
3
, or 3
3
1 3
,
3
x 1: (4 2 x ) 2
x
x
3
1 3
( x 3) 6
3
x 1
Then 1
x
Step 2:
3
3
( 6)
x ( 3)
Then
46. Step 1:
Step 2:
min
x2 9
x 3
45. Step 1:
3
3
1 3
2014 Pearson Education, Inc.
0
0, not 1. The
0
,
Section 2.3 The Precise Definition of a Limit
54. Let f ( x)
given
sin x, L
1 , and x
0
2
0. There exists a value of x (namely x
0
sin 1x , L
As another example, let g ( x)
1
2
for any
1,
2
and x0
0. We can choose infinitely many values of x near 0 such
as you can see from the accompanying figure. However, lim sin 1x fails to exist. The wrong
x 0
0 of L 12 . Again you can
statement does not require all values of x arbitrarily close to x0 0 to lie within
that sin
1
x
1
2
) for which sin x
0, not 12 . The wrong statement does not require x to be arbitrarily close to x0 .
0. However, lim sin x
x
6
79
see from the figure that there are also infinitely many values of x near 0 such that sin 1x
we cannot satisfy the inequality sin 1x
55.
A 9
2
0.01
8.99
x
9.01
2
2
x
2
0.01
9
or 3.384
1
2
for all values of x sufficiently near x0
0.01
x
x2
4
8.99
x2
4 (8.99)
9.01
0. If we choose
1
4
0.
4 (9.01)
3.387. To be safe, the left endpoint was rounded up and
the right endpoint was rounded down.
56. V
V
R
(120)(10)
51
RI
V 5
R
(120)(10)
49
I
R
0.1
0.1
23.53
R
120
R
5 0.1
4.9
120
R
10
49
5.1
10
51
R
120
24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
57. (a)
x 1 0
f ( x) 2
1
1
1
2
x 1
f ( x)
x. Then f ( x) 2
no matter how small
x 2
is taken when 1
2 x
x 1
lim f ( x)
(b) 0 x 1
1 x 1
f ( x) x 1. Then f ( x) 1 ( x 1) 1
no matter how small is taken when 1 x 1
lim f ( x) 1.
(c)
2 1 1. That is,
x 1
x
2.
x 1. That is, f ( x) 1
1
x 1
x 1 0 1
x 1
f ( x) x. Then f ( x) 1.5 x 1.5 1.5 x 1.5 1 0.5.
Also, 0 x 1
1 x 1
f ( x) x 1. Then f ( x) 1.5 ( x 1) 1.5 x 0.5
x 0.5 1 0.5 0.5. Thus, no matter how small is taken, there exists a value of x such that
but f ( x) 1.5 12
lim f ( x) 1.5.
x 1
x 1
58. (a) For 2 x 2
h( x ) 2
how small we choose
0
(b) For 2 x 2
h( x) 2
how small we choose
0
(c) For 2
x
2
h( x )
h( x ) 4 2. Thus for
lim h( x) 4.
x
2
x
2
h( x) 3 1. Thus for
lim h( x) 3.
x 2 so h( x) 2
x2
when x is near 2 and to the left on the real line
whenever 2
x
1, h( x) 3
whenever 2
whenever 2
x2
2 will be close to 2. Thus if
0
lim h( x)
x
2
2014 Pearson Education, Inc.
x
x
2
2
no matter
no matter
0 is chosen, x 2 is close to 4
2 . No matter how small
2 no matter how small we choose
Copyright
2, h( x) 4
2.
1, h( x) 2
80
Chapter 2 Limits and Continuity
x 3
f ( x ) 4.8
59. (a) For 3
matter how small we choose
0
f ( x ) 4 0.8. Thus for
lim f ( x) 4.
x
3
(b) For 3 x 3
f ( x) 3
f ( x) 4.8 1.8. Thus for
no matter how small we choose
0
lim f ( x) 4.8.
(c) For 3
x 3
f ( x ) 4.8
matter how small we choose
x
0.8, f ( x) 4
x
1.8, f ( x) 4.8
whenever 3
1.8, f ( x) 3
whenever 3
3
f ( x) 3 1.8. Again, for
0
lim f ( x) 3.
x
whenever 3
x
3 no
3
x
3 no
3
60. (a) No matter how small we choose
0, for x near 1 satisfying 1
x
1 , the values of g ( x) are
1
1
near 1
g ( x) 2 is near 1. Then, for
we have g ( x) 2 2 for some x satisfying 1
x
1 ,
2
or 0 x 1
lim g ( x) 2.
x
1
(b) Yes, lim g ( x) 1 because from the graph we can find a
x
1
0 such that g ( x) 1
if 0
x ( 1)
61-66. Example CAS commands (values of del may vary for a specified eps):
Maple:
f : x - (x^4-81)/(x-3); x0 : 3;
.
plot( f (x), x x0-1..x0 1, color black,
title "Section 2.3, #61(a)" );
# (a)
L : limit( f (x), x x0 );
# (b)
epsilon : 0.2;
# (c)
plot( [f (x), L-epsilon,L epsilon], x x0-0.01..x0 0.01,
color black, linestyle [1,3,3], title "Section 2.3, #61(c)" );
q : fsolve( abs( f (x)-L ) epsilon, x x0-1..x0 1 );
# (d)
delta : abs(x0-q);
plot( [f (x), L-epsilon, L epsilon], x x0-delta..x0 delta, color black, title "Section 2.3, #61(d)" );
for eps in [0.1, 0.005, 0.001 ] do
# (e)
q : fsolve( abs( f (x)-L ) eps, x x0-1..x0 1 );
delta : abs(x0-q);
head : sprintf ("Section 2.3, #61(e)\n epsilon %5f , delta %5f \n", eps, delta );
print(plot( [f (x), L-eps, L eps], x x0-delta..x0 delta,
color black, linestyle [1,3,3], title head ));
end do:
Mathematica (assigned function and values for x0, eps and del may vary):
Clear[f , x]
y1: L eps; y2: L eps; x0 1;
f[x _]: (3x 2 (7x 1)Sqrt[x] 5)/(x 1)
Plot[f [x], {x, x0 0.2, x0 0.2}]
L: Limit[f [x], x
x0]
eps 0.1; del 0.2;
Plot[{f [x], y1, y2}, {x, x0 del, x0 del}, PlotRange
Copyright
{L 2eps, L 2eps}]
2014 Pearson Education, Inc.
.
Section 2.4 One-Sided Limits
2.4
81
ONE-SIDED LIMITS
1. (a) True
(e) True
(i) False
(b) True
(f) True
(j) False
(c) False
(g) False
(k) True
(d) True
(h) False
(l) False
2. (a) True
(e) True
(i) True
(b) False
(f) True
(j) False
(c) False
(g) True
(k) True
(d) True
(h) True
3. (a)
lim f ( x)
x
2
2
2
1 2, lim f ( x)
x
3 2 1
2
(b) No, lim f ( x) does not exist because lim f ( x)
x
(c)
2
lim f ( x)
x
4
4
2
x
4
(d) Yes, lim f ( x) 3 because 3
x
4. (a)
4
lim f ( x)
x
2
2
2
x
2
(c)
x
2
lim f ( x)
1
3 ( 1)
(d) Yes, lim f ( x)
x
4 because 4
1
lim f ( x)
2
1
x
(b)
0
lim f ( x)
x
0
x
0
lim 0
x
0
4
2
lim f ( x)
x
2
3 ( 1)
lim f ( x)
1
5. (a) No, lim f ( x) does not exist since sin
x
2
lim f ( x)
x
3 2 1, f (2)
4, lim f ( x)
x
x
1 3
4
x
lim f ( x)
2
lim f ( x)
x
1, lim f ( x)
(b) Yes, lim f ( x ) 1 because 1
x
4
2
1 3, lim f ( x)
x
1
x
x
4
lim
1
f ( x)
does not approach any single value as x approaches 0
0
(c) lim f ( x) does not exist because lim f ( x) does not exist
x
6. (a) Yes, lim g ( x)
x
0
0
0 by the sandwich theorem since
x
g ( x)
(b) No, lim g ( x) does not exist since x is not defined for x
x
0
x
0
x when x
0
0
(c) No, lim g ( x) does not exist since lim g ( x) does not exist
x
7. (a)
0
(b)
lim f ( x) 1
x 1
lim f ( x)
x 1
(c) Yes, lim f ( x) 1 since the right-hand and left-hand
x 1
limits exist and equal 1
Copyright
2014 Pearson Education, Inc.
82
Chapter 2 Limits and Continuity
8. (a)
(b)
lim f ( x)
0
x 1
lim f ( x)
x 1
(c) Yes, lim f ( x)
0 since the right-hand and left-hand
x 1
limits exist and equal 0
9. (a) domain: 0 x 2
range: 0 y 1 and y 2
(b) lim f ( x) exists for c belonging to (0, 1)
x
(1, 2)
c
(c) x
(d) x
2
0
10. (a) domain:
x
range: 1 y 1
(b) lim f ( x) exists for c belonging to
x
c
( , 1)
(c) none
(d) none
11.
13.
14.
15.
x
x
0.5
2
lim
x 1
lim
h
x 2
x 1
lim
lim
0
( 1, 1)
x
x 1
1
x 1
(1, )
0.5 2
0.5 1
2x 5
x2 x
x 6
x
h2 4h 5
h
3/2
1/2
5
( 2)2 ( 2)
h
h2 4 h 5
h
0
lim
h
0
6
5h 2 11h 6
h
0 h
lim
h
h2 4h 5
6
0
lim
h
(2) 12
3 1
7
5
0 h
6 (5h
6
2
7
h 2 4h 5
5
h 2 4h 5
5
5h 2 11h 6
h
2
11h 6)
5h 2 11h 6
Copyright
0
0
1
lim
h
( h2 4 h 5) 5
0 h
h 2 4h 5
5
2
5
6
5h 2 11h 6
6
5h 2 11h 6
h(5h 11)
lim
h
1 1
1 2
1
0 4
5 5
5
x 1
x 2
lim
x 1
7
1
1
2
h( h 4)
lim
h
16.
1 6
1
1
1 1
lim
12.
2( 2) 5
2
2 1
3 x
7
3
0 h
6
5h 2 11h 6
2014 Pearson Education, Inc.
(0 11)
6 6
11
2 6
Section 2.4 One-Sided Limits
x 2
17. (a)
x
(b)
x
( x 2)
lim ( x 3) x 2
2
x
x
x 2
x 2
lim ( x 3)
2
lim ( x 3) ( x 2)
(|x 2|
2
lim ( x 3) (( 2) 3) 1
2 x ( x 1)
x 1
lim
x 1
x
x
(b)
2 x ( x 1)
x 1
lim
x 1
( x 2)
lim ( x 3) ( x 2)
(|x 2|
2
x
lim ( x 3)( 1)
( 2 3)
1
lim
2 x ( x 1)
( x 1)
lim
2x
1
1
3
20. (a)
t
lim (t
lim sin 2
21.
0
2 x ( x 1)
( x 1)
lim
2x
x
22. lim sint kt
t
kt
lim k sin
0 kt
sin 3 y
0 4y
24.
lim sinh3h
0
h
25. lim tanx2 x
lim
x
0
2t
26. lim tan
t
t 0
0
t
csc 2 x
27. lim xcos
5x
x 0
0
cos t
2 lim tsin
t
sin t
cos t
t
lim sinx2 x cos15 x
0
0
x x cos x
29. lim sin
x cos x
x 0
lim sin xxcos x
0
lim
x
0
1
sin x
x
1
3
1
lim sin
lim (t
t )
4
4 3 1
1
3
3y )
1
3
1
1
lim sint t
(where
lim cos15 x
0
1
2
lim 3cos x sinx x sin2 x2 x
0
lim sinx x cos1 x
0
lim
0
1
sin x
x
2
211 2
x
x
1 2
0
x
Copyright
(where
2x
lim 2 sin
0 2x
t
1 lim 2 x
2 x 0 sin 2 x
x
kt )
x
0
x cos x
sin x cos x
(where
0
lim cos12 x
0
t
lim cos1 x
0
x
3
4
x
6 x 2 cos x
lim sin
x sin 2 x
x 0
x
0
2 lim cos t
0
x
28. lim 6 x 2 (cot x )(csc 2 x )
x
sin 3 h
3h
0
x
t
2 lim
lim sin
1
lim
h
3
4
2x
lim xsin
0 cos 2 x
x
0
1
3
t
k 1 k
0
3 lim sin 3 y
4 y 0 3y
sin 2 x
cos 2 x
3
2 )
k lim sin
0
1 3h
3 sin 3h
lim
h
x
lim k sin
2
3
lim
(b)
(where x
1 lim 3sin 3 y
4 y 0 3y
23. lim
( x 1) for x 1)
2
0
1
0
t
0
y
4 4
x 1 for x 1)
(|x 1|
(b)
lim sinx x
2
(|x 1|
1
t )
4
2)
2
lim
x 1
3
3
lim
( x 2) for x
2
x 1
19. (a)
2)
2
x
18. (a)
( x 2) for x
1 (1)
1
2
311 3
lim sinx x
0
x
(1)(1) 1 2
2014 Pearson Education, Inc.
3h)
83
84
Chapter 2 Limits and Continuity
2
30. lim x x2 xsin x
x 0
lim 2x
x
1 sin x
2
x
1
2
0
lim (2sin cos )(1 cos )
0
lim (2 cos sin
)(1 cos )
0
32. lim x x 2cos x
lim
0 sin 3 x
34. lim
h
0
sin 3 x
lim sin
33. lim
t
2
0
x
sin(1 cos t )
0 1 cos t
x (1 cos x )
lim sin
sin
35. lim sin
2
0
sin
lim sin
2
0
sin 5 x
36. lim sin
4x
x 0
x
37. lim
0
2
2
sin h
tan 3 x
39. lim sin
8x
x 0
lim
x
0 as t
1 (0)
9
2
1
3x
0
0
0
0
2
sin 2
1
2
1
2
11
5 lim sin 5 x
4x
4 x 0 5 x sin 4 x
5
4
5
4
11
cos 2
lim sin 2 sin
cos
0
sin 3 x
1
lim cos
3 x sin 8 x
0
sin 3 y cot 5 y
0 y cot 4 y
40. lim
tan
2
cot 3
0
41. lim
sin 3 x
3x
8x
sin 8 x
sin 3 y sin 4 y cos 5 y
0 y cos 4 y sin 5 y
lim
y
lim
cot 4
2
2
0 sin cot 2
42. lim
0
0
sin 3 y
3y
sin
cos
2 cos 3
sin 3
lim
cos 4
sin 4
0 sin
lim
0
a
L
4 cos 4 cos 2
y
5y
sin 5 y
cos 5 y
cos 4 y
sin sin 3
cos cos 3
a
Copyright
a
1 1 1 1 12
5
lim
2
x
3 4 5y
3 4 5y
sin 3
3
0
lim sin4 4
0
cos 5 y
sin 5 y
34
5
lim sin
2
lim f ( x), then lim f ( x)
x
sin 4 y
cos 4 y
cos 4 sin 2 2
2
0 sin cos 2 sin 4
0 cos 2 sin 4
43. Yes. If lim f ( x)
3
8
111
sin 3 y
y
0
lim
cos 2 2
sin 2 2
2
3
8
lim
sin 4 y
4y
lim
2
1
2
0
x
lim
y
2
lim 2cos
cos
sin 3 x
8x 3
1
lim cos
3 x sin 8 x 3 x 8
0
x
3 lim
1
8 x 0 cos 3 x
exist.
0
0 as h
0
2
lim sin cos
sin 2
0
0
x
sin 3 x 2
3x
01 0
38. lim sin cot 2
y
1 cos t
lim sin
1
2
sin 5 x 4 x 5
lim sin
4 x 5x 4
0
cos
x
9 x2
1 since
0
sin 2
lim (2sin cos
)(1 cos )
0
1 lim 1 cos x
9x 0
x
sin 3 x 2
1 cos x
9x
lim
sin 2 3 x
0
x
0
0
x (1 cos x )
9 x2
1 since
0
sin(sin h)
sin h
0
(2)(2)
lim
1 (1)
2
2
lim (2sin 1coscos)(1 cos )
0
(1 cos )(1 cos )
31. lim 1sincos
2
0
x
0 12
0
cos 4 cos 2
3
cos cos 3
cos 2
sin
2
2
cos 2 sin 4
0
L. If lim f ( x)
x
a
3
(1)(1) 11
cos 4 (2sin cos )2
lim
2
12
5
3
cos 4 (4sin 2 cos 2 )
lim
0
sin 2 cos2 2 sin 4
cos 4 cos2
1
sin 4
4
1
1
2
cos 2
112
12
1
lim f ( x), then lim f ( x) does not
x
2014 Pearson Education, Inc.
a
x
a
Section 2.4 One-Sided Limits
44. Since lim f ( x )
x
L if and only if lim f ( x )
c
x
calculating lim f ( x).
x
L and lim f ( x)
c
x
L, then lim f ( x) can be found by
x
c
c
c
45. If f is an odd function of x, then f ( x )
f ( x). Given lim f ( x)
3, then lim f ( x)
46. If f is an even function of x, then f ( x )
f ( x). Given lim f ( x)
7 then lim
can be said about lim
x
47. I
(5, 5
48. I
(4
)
5
, 4)
4
0
x
2
x
2
2
5
. Also, x 5
x 5
x
4. Also, 4 x
4 x
x
2
x
long as x
51. (a)
(b)
x
lim
x
400
400. Just observe that if 400
any number
0 that 400 x 400
lim
x 399. Just observe that if 399
x
400
0 that 400
x
lim
any number
(c) Since lim
x
52. (a)
400
lim f ( x)
x
0
0
lim f ( x)
x
x
lim
x
lim f ( x)
x
(b)
0, and thus 2
0
Since x 2 0
0
401, then x
x
0
1
which is always true so
. Thus, lim
x
2
400. Thus if we choose
x 2
x 2
x2 0
1.
1, we have for
1, we have for
0
0;
x 0
400
x
0
x
2
2
for x positive. Choose
0.
0
0.
1.
x
x 400 400 400 0 .
x 400 then x 399. Thus if we choose
x
lim x 2 sin 1x
x
x 2
x 2
4 x
4
which is always true
0
2
lim
x
0.
x 400
x 399 399 399 0 .
x we conclude that lim x does not exist.
400
x
x
x
2
x 5
5
lim x
0
1
lim
x
0
x
x 2
x 2
x 2
2. Hence we can choose any
1
7. However, nothing
2
. Choose
0 with
x 2. Then, x 2 1
2 and x 2
f ( x)
2
. Choose
2
4
x
x
independent of the value of x. Hence we can choose any
2 we have x
2
5
3.
0
x
( 1)
x
50. Since x
x
f ( x) because we don t know lim f ( x).
2
x
x
0 the number x is always negative. Thus, x
49. As x
85
x2
if
x 0.
(c) The function f has limit 0 at x0
0 by the sandwich theorem since x 2
whenever x
, we choose
x 2 sin 1x
x 2 for all x
and obtain x 2 sin 1x
0.
0
0 since both the right-hand and left-hand limits exist and equal 0.
Copyright
2014 Pearson Education, Inc.
86
2.5
Chapter 2 Limits and Continuity
CONTINUITY
1. No, discontinuous at x
2, not defined at x
2. No, discontinuous at x
3, 1
lim g ( x)
x
2
g (3) 1.5
3
3. Continuous on [ 1, 3]
4. No, discontinuous at x 1, 1.5
lim k ( x)
x 1
lim k ( x )
x 1
5. (a) Yes
0
(b) Yes, lim
(c) Yes
(d) Yes
6. (a) Yes, f (1) 1
(d) No
7. (a) No
(b) No
9. f (2)
(0, 1)
(1, 2)
f ( x)
x 1
0
2
(2, 3)
0, since lim f ( x)
x
1
(b) Yes, lim f ( x)
(c) No
8. [ 1, 0)
x
2
10. f (1) should be changed to 2
2(2) 4
0
lim f ( x )
x
2
lim f ( x)
x 1
11. Nonremovable discontinuity at x 1 because lim f ( x) fails to exist ( lim f ( x) 1 and lim f ( x)
x 1
Removable discontinuity at x
0 by assigning the number lim f ( x)
x
than f (0) 1.
0
x 1
0 to be the value of f (0) rather
12. Nonremovable discontinuity at x 1 because lim f ( x) fails to exist ( lim f ( x)
x 1
Removable discontinuity at x
f (2)
15. Discontinuous only when x 2
2 and lim f ( x) 1).
x 1
x 1
2 by assigning the number lim f ( x ) 1 to be the value of f (2) rather than
x
2.
13. Discontinuous only when x 2
0).
x 1
0
4x 3
16. Discontinuous only when x 2 3 x 10
x
2
0
0
2
14. Discontinuous only when ( x 2)2
( x 3)( x 1)
( x 5)( x 2)
0
x
0
0
x
3 or x 1
x
5 or x
2
17. Continuous everywhere. (|x 1| sin x defined for all x; limits exist and are equal to function values.)
18. Continuous everywhere. (|x| 1 0 for all x; limits exist and are equal to function values.)
Copyright
2014 Pearson Education, Inc.
2
Section 2.5 Continuity
19. Discontinuous only at x
87
0
20. Discontinuous at odd integer multiples of
2
, i.e., x
(2n 1) 2 , n an integer, but continuous at all other x.
21. Discontinuous when 2x is an integer multiple of , i.e., 2 x
continuous at all other x.
n , n an integer
22. Discontinuous when 2x is an odd integer multiple of 2 , i.e., 2x
integer (i.e., x is an odd integer). Continuous everywhere else.
23. Discontinuous at odd integer multiples of
2
, i.e., x
3
2
1
3
26. Discontinuous when 3 x 1 0 or x
n
2
, n an integer, but
(2n 1) 2 , n an integer
x
2n 1, n an
(2n 1) 2 , n an integer, but continuous at all other x.
24. Continuous everywhere since x 4 1 1 and 1 sin x 1
equal to the function values.
25. Discontinuous when 2 x 3 0 or x
x
0 sin 2 x 1
1 sin 2 x 1; limits exist and are
3,
2
continuous on the interval
1,
3
continuous on the interval
.
.
27. Continuous everywhere: (2 x 1)1/3 is defined for all x; limits exist and are equal to function values.
28. Continuous everywhere: (2 x)1/5 is defined for all x; limits exist and are equal to function values.
2
29. Continuous everywhere since lim x x x3 6
x 3
30. Discontinuous at x
31. lim sin( x sin x )
x
32. lim sin( 2 cos(tan t ))
t
0
lim
x
3
( x 3)( x 2)
x 3
lim ( x 2)
x
2 since lim f ( x) does not exist while f ( 2)
x
sin(
sin(
0)
sin( 2 cos(tan(0)))
sin
33. lim sec ( y sec2 y tan 2 y 1)
sin
2
0, and function continuous at x
cos(0)
sin
lim sec ( y sec2 y sec2 y )
y 1
g (3)
4.
2
sin )
5
3
y 1
.
1, and function continuous at t
2
lim sec (( y 1) sec2 y )
y 1
0.
sec ((1 1)sec 2 1)
sec 0 1, and function continuous at y 1.
34. lim tan
x
4
0
35. lim cos
t
x
19 3 sec 2t
0
36. lim
tan
cos
csc 2 x 5 3 tan x
6
at x
cos(sin x1/3 )
6
4
cos(sin(0))
cos
19 3 sec 0
csc 2
6
tan
5 3 tan
16
6
4
cos(0)
cos 4
tan
2
2
4 5 3
4
1, and function continuous at x
, and function continuous at t
1
3
9
.
Copyright
2014 Pearson Education, Inc.
0.
3, and function continuous
0.
88
37.
38.
Chapter 2 Limits and Continuity
lim sin
x
2
0
1
lim cos
e x
ln x
x 1
x2 9
x 3
39. g ( x)
sin
1
cos
42. g ( x)
x 2 16
x 3x 4
x 3, x
3
43. As defined, lim f ( x)
x
a
s2 s 1 ,
s 1
x 4,
x 1
s 1
x
12
2a 2
4
g (4)
b
b 1
x
b
b
0 or b
47. As defined, lim
x
lim f ( x)
x 1
1
4
1
8
5
2
a 2 (2) 2a
2
2a 2
b
b 1
and lim g ( x)
x
0
(0)2 b
b. For g ( x) to be continuous we must have
f ( x)
2 and lim
x
1
f ( x)
a( 1) b
a b, and lim f ( x)
x 1
3. For f ( x) to be continuous we must have 2
x
0
x
and b
0
3.
2
2a. For f ( x) to be continuous we must have
2.
4 3a b and lim g ( x)
3
2
4
2.
0 b
b 1
48. As defined, lim g ( x)
a
lim xx
x
4b. For g ( x) to be continuous we must have
3 or a
0
3
2
b( 2)2
x
46. As defined, lim g ( x)
2
lim s s s1 1
s 1
3
2 and lim g ( x)
x
a
7
2
6a. For f ( x) to be continuous we must have
2
2a
lim (t 5)
t
f (1)
45. As defined, lim f ( x) 12 and lim f ( x)
x
6
3
(3)2 1 8 and lim (2a)(3)
2
1.
2
b
lim ( x 3)
x
h(2)
4.
3
x
2
2
, and the function is continuous at x = 1.
2
g (3)
x
3
44. As defined, lim g ( x)
4b
t 5, t
( x 4)( x 4)
( x 4)( x 1)
2
1, and the function is continuous at x = 0.
2
cos 1 (0)
( s 2 s 1)( s 1)
( s 1)( s 1)
s3 1
s3 1
8
sin
ln 1
(t 5)(t 2)
t 2
41. f ( s )
6a
e0
( x 3)( x 3)
( x 3)
t 2 3t 10
t 2
40. h(t )
2
a (0) 2b
2b and lim g ( x)
x
0
(0) 2
a b and a b
3a b
3
5
2
a
2014 Pearson Education, Inc.
x
2
a b and
and b
3a b, and lim g ( x)
3(2) 5 1. For g ( x) to be continuous we must have 2b
Copyright
a(1) b
1.
2
(2) 2 3a b
3a b and 4 3a b 1
Section 2.5 Continuity
49. The function can be extended: f (0)
89
50. The function cannot be extended to be continuous at
x 0. If f (0) 2.3, it will be continuous from the
right. Or if f (0)
2.3, it will be continuous from
the left.
2.3.
51. The function cannot be extended to be continuous
at x 0. If f (0) 1, it will be continuous from the
right. Or if f (0)
1, it will be continuous from
the left.
52. The function can be extended: f (0)
7.39.
53. f ( x) is continuous on [0, 1] and f (0) 0, f (1) 0
by the Intermediate Value Theorem f ( x ) takes on
every value between f (0) and f (1) the equation
f ( x) 0 has at least one solution between x 0
and x 1.
54. cos x
x
(cos x ) x
some x between
continuous.
2
and
0. If x
2
2
, cos
2
2
0. If x
2
, cos
2
2
0. Thus cos x x
0 for
according to the Intermediate Value Theorem, since the function cos x x is
3, f ( 1) 15, f (1)
13, and f (4) 5.
55. Let f ( x) x3 15 x 1, which is continuous on [ 4, 4]. Then f ( 4)
By the Intermediate Value Theorem, f ( x) 0 for some x in each of the intervals 4 x
1, 1 x 1, and
1 x 4. That is, x3 15 x 1 0 has three solutions in [ 4, 4]. Since a polynomial of degree 3 can have at
most 3 solutions, these are the only solutions.
56. Without loss of generality, assume that a b. Then F ( x) ( x a) 2 ( x b) 2 x is continuous for all values of x,
so it is continuous on the interval [a, b]. Moreover F (a ) a and F (b) b. By the Intermediate Value Theorem,
since a a 2 b b, there is a number c between a and b such that F ( x) a 2 b .
Copyright
2014 Pearson Education, Inc.
90
Chapter 2 Limits and Continuity
57. Answers may vary. Note that f is continuous for every value of x.
(a) f (0) 10, f (1) 13 8(1) 10 3. Since 3
10, by the Intermediate Value Theorem, there exists a c so
.
that 0 c 1 and f (c)
(b) f (0) 10, f ( 4) ( 4)3 8( 4) 10
22. Since 22
3 10, by the Intermediate Value Theorem,
there exists a c so that 4 c 0 and f (c)
3.
(c) f (0) 10, f (1000) (1000)3 8(1000) 10 999,992, 010. Since 10 5, 000, 000 999,992, 010, by the
Intermediate Value Theorem, there exists a c so that 0 c 1000 and f (c) 5, 000, 000.
58. All five statements ask for the same information because of the intermediate value property of continuous
functions.
(a) A root of f ( x) x3 3 x 1 is a point c where f (c) 0.
(b) The point where y x3 crosses y 3x 1 have the same y-coordinate, or y x3 3x 1
f ( x)
x3 3 x 1 0.
(c) x3 3 x 1 x3 3 x 1 0. The solutions to the equation are the roots of f ( x) x3 3 x 1.
(d) The points where y x3 3 x crosses y 1 have common y-coordinates, or y x3 3 x 1
f ( x)
x3 3 x 1 0.
(e) The solutions of x3 3 x 1 0 are those points where f ( x) x3 3 x 1 has value 0.
sin( x 2)
is discontinuous at x 2 because it is not defined there.
59. Answers may vary. For example, f ( x)
x 2
However, the discontinuity can be removed because f has a limit (namely 1) as x 2.
1
x 1
60. Answers may vary. For example, g ( x)
x
lim g ( x)
and lim g ( x)
x
1
has a discontinuity at x
1 because lim g ( x ) does not exist.
x
1
.
1
1 . For any
61. (a) Suppose x0 is rational
f ( x0 ) 1. Choose
0 there is an irrational number x (actually
2
, x0
)
f ( x) 0. Then 0 |x x0 |
but | f ( x) f ( x0 )|
infinitely many) in the interval ( x0
1 12 , so lim f ( x) fails to exist
f is discontinuous at x0 rational.
x
x0
On the other hand, x0 irrational
Again lim f ( x) fails to exist
x
x0
f ( x0 ) 0 and there is a rational number x in ( x0
, x0
)
f ( x) 1.
f is discontinuous at x0 irrational. That is, f is discontinuous at every point.
(b) f is neither right-continuous nor left-continuous at any point x0 because in every interval ( x0
, x0 ) or
( x0 , x0
) there exist both rational and irrational real numbers. Thus neither limits lim f ( x) and
x
x0
lim f ( x ) exist by the same arguments used in part (a).
x
x0
62. Yes. Both f ( x)
g
1
2
63. No. For instance, if f ( x)
64. Let f ( x)
1
( x 1) 1
1
x
1
x 1
and g ( x )
x
f ( x)
1
2
are continuous on [0, 1]. However g ( x) is undefined at x
f ( x)
is discontinuous at x 12 .
g ( x)
0
x and g ( x)
0, g ( x)
x , then h( x)
0 x
0 is continuous at x
x 1. Both functions are continuous at x
is discontinuous at x
1
2
since
0 and g ( x ) is not.
0. The composition f g
f ( g ( x))
0, since it is not defined there. Theorem 10 requires that f ( x) be continuous
at g (0), which is not the case here since g (0) 1 and f is undefined at 1.
Copyright
2014 Pearson Education, Inc.
Section 2.5 Continuity
91
65. Yes, because of the Intermediate Value Theorem. If f (a ) and f (b) did have different signs then f would have
to equal zero at some point between a and b since f is continuous on [a, b].
66. Let f ( x) be the new position of point x and let d ( x) f ( x) x. The displacement function d is negative if x is
the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the
Intermediate Value Theorem, d ( x ) 0 for some point in between. That is, f ( x ) x for some point x,
which is then in its original position.
67. If f (0) 0 or f (1) 1, we are done (i.e., c 0 or c 1in those cases). Then let f (0) a 0 and f (1) b 1
because 0 f ( x ) 1. Define g ( x) f ( x ) x
g is continuous on [0, 1]. Moreover, g (0) f (0) 0 a 0 and
g (1) f (1) 1 b 1 0 by the Intermediate Value Theorem there is a number c in (0, 1) such that
g (c ) 0
f (c) c 0 or f (c) c.
f ( c)
2
68. Let
0. Since f is continuous at x c there is a
0 such that x c
f (c )
f ( x ) f (c ) .
1 f (c)
1 f (c)
If f (c) 0, then
f ( x) 23 f (c)
f ( x ) 0 on the interval (c
2
2
If f (c)
1
2
0, then
3
2
f (c)
Thus, f ( x) is continuous at x
f (c )
c
1
2
f ( x)
lim f ( x )
x
f (c )
f (c )
c
Now lim sin(c h)
h
h
lim (sin c)(cos h) (cos c)(sin h)
h
0
0
By Example 11 Section 2.2, lim cos h 1 and lim sin h
continuous at x
lim cos(c h)
h
0
g ( x)
h
c. Similarly,
h
0
0
lim (cos c )(cos h) (sin c )(sin h)
h
0
cos x is continuous at x
0
,c
).
,c
cos c.
0
(sin c ) lim cos h
(cos c) lim sin h .
0. So lim sin(c h)
sin c and thus f ( x)
h
h
0
0
(cos c) lim cos h
h
c.
0
h
h
0
72. x 1.4516, 0.8547, 0.4030
73. x 1.7549
74. x 1.5596
75. x
3.5156
76. x
3.9058, 3.8392, 0.0667
77. x
0.7391
78. x
1.8955, 0, 1.8955
2014 Pearson Education, Inc.
0
(sin c) lim sin h
71. x 1.8794, 1.5321, 0.3473
Copyright
).
f (c).
sin c and lim cos(c h)
0
f (c )
0 on the interval (c
lim f (c h)
h
70. By Exercise 67, it suffices to show that lim sin(c h)
h
f ( x)
f ( x)
sin x is
cos c. Thus,
92
2.6
Chapter 2 Limits and Continuity
LIMITS INVOLVING INFINITY; ASYMPTOTES OF GRAPHS
1. (a) lim f ( x)
(c)
(e)
x
2
lim
x
3
0
f ( x)
lim f ( x)
x
0
(g) lim f ( x)
(i)
x
0
lim f ( x)
x
(e)
(g)
(i)
x
4
x
2
lim
x
3
(d) lim f ( x)
1
(f)
(h)
0
2
(b)
x
3
x
0
x
lim f ( x)
x
2
x
2
lim
(f)
3
f ( x)
lim f ( x)
x
(h)
3
lim f ( x)
x
(j)
0
0
x
(l)
Note: In these exercises we use the result lim
x
1
xm / n
Theorem 8 and the power rule in Theorem 1: lim
xm / n
x
3. (a)
lim
x
x
3
3
x
does not exist
f ( x)
0
lim f ( x)
x
x
0
lim f ( x)
1
x
lim
(b)
4. (a)
3
lim f ( x)
0 whenever mn
1
does not exist
lim f ( x) 1
(d) lim f ( x)
x
2
lim f ( x)
lim f ( x) 1
(k) lim f ( x)
does not exist
1
0. This result follows immediately from
m/ n
m/ n
x
3
(b)
5. (a)
1
2
(b)
1
2
6. (a)
1
8
(b)
1
8
5
3
7. (a)
8. (a)
f ( x)
2
does not exist
2. (a) lim f ( x)
(c)
(b)
3
4
(b)
9.
1
x
sin 2 x
x
1
x
10.
1
3
cos
3
1
3
sin t
11. lim 2t t cos
t
t
5
3
(b)
lim
x
sin 2 x
x
lim
2
t
t
1
r
12. lim 2 r r 7 sin5sin
r
r
lim
r
1
sin t
t
cos t
t
0 by the Sandwich Theorem
0 1 0
1 0
sin r
r
7
sin r
5
r
r
1
2
0 by the Sandwich Theorem
cos
3
lim
3
4
1
lim 2 1 00 0
r
Copyright
1
2
2014 Pearson Education, Inc.
lim
1
x
0m / n
0.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
13. (a)
lim 52xx 73
x
14. (a)
lim
x
3
x
7
x
2
lim
5
x
(b) 52 (same process as part (a))
2
5
7
x3
7
1
x 2 x3
2
2 x3 7
3
x x2 x 7
lim
1
x
1
x
(b) 2 (same process as part (a))
15. (a)
16. (a)
17. (a)
18. (a)
19. (a)
20. (a)
(b)
21. (a)
(b)
22. (a)
(b)
23.
24.
25.
lim x2 1
x
x
3
x
lim
2
x
lim
9 x4 x
2 x4 5x2 x 6
lim
10 x5 x 4 31
x6
x
x
x
x
x
x
x
7x 3
3 x 4x
x
lim
x2 x 1
8 x2 3
lim
1 x3
x2 7 x
x
8
lim
2
x
1/3
x
x
1
x 3
3x 3
2
x
9x
4
5
3
8
3x
x2
1
x
lim
x
1
x
1
8
x2
1
x
7
x
1
2
5
2
, since x n
0, 5x 3
.
, and the denominator
0, 5x 3
, and the denominator
3
x2
1
x
8 0
2 0
1/3
x2
3
x2
x
5
lim
1
x
.
0 and 3x 4
, since x n
4
.
0 and 3x 4
, since x n
4
.
0 and x 7
, since x n
9x
x
4
lim
lim
x 3
3x 3
0 and x 7
x n
,
2
3
lim 5 x 25 x
1
5
1
6 7x
3x
x n
1
4
lim 3 x 5 x2
x
8
3
lim 5 x 2 x 5 9
x
x
(b) 0 (same process as part (a))
,
2
1 x
3
lim 5 x 25 x
3 x 4x
0
1
6 7x
x
(b) 92 (same process as part (a))
31
1
x
9
2
6
x4
x6
lim x 7 1 2 x
x
7
2
lim 3 x 3 5 x 1
8 x2 3
2 x2 x
1
x2
4
lim 3 x 5 x2
7x 3
8
3
lim 5 x 2 x 59
x
10
x
1 x
x
(b) 7 (same process as part (a))
1
x3
1
x3
5
x2
2
x
x
x 1
6x
7
9
x2
9
lim
lim
7
2
lim 3 x 3 5 x 1
x
7
3
x
x
3
2
lim x 2 7 x 2
6x
(b) 0 (same process as part (a))
lim x 7 1 2 x
x 1
x
0
1
x
3
2
lim x 2 7 x 2
(b) 0 (same process as part (a))
7
lim
x
0
x2
2
x2
1
x
7 x3
3x 2 6 x
3
lim
x
3
x
lim
2
1
x2
3
x2
1
x
lim 3 2x 7
x
1
x
lim
93
lim
Copyright
x2
1
x
7
x
4
1
x
1
8
1
2
1/3
x2
3
x2
5
0
1 0
1 0 0
8 0
1/3
5
2014 Pearson Education, Inc.
1
8
1/3
1
2
4.
4.
94
26.
27.
29.
30.
31.
32.
33.
34.
35.
36.
Chapter 2 Limits and Continuity
x2 5 x
x3 x 2
lim
x
x
3
lim x
x
x
x
5
x
1
x
x
2
x
4
3
5/3
1/3
lim 2 x8/5 x 7
x
x
x
3x
x2 1
x 1
x
lim
x
lim
x2 1
x 1
lim
x 3
x
x
4x
2
x
9
(1/3)
1
3
x3/5
1
5
3
x
1
x1/3
4
x
x 2/3
2
( x 1)/ x
x
lim
x
lim 2
28.
lim
1
1
2
x
( x 1)/ x
2
6
x
lim
(4 3 x3 )/( x3 )
(x
6
9)/ x
6
lim
( 4/ x3 3)
x
lim x2 x8
8
x
4
2
7 ( x 7)
1 0
( 1 0)
1
1
1 9/ x
6
1
(1 0)
4 0
4 25/ x 2
x
41.
(0 3)
1 0
1
2
3
lim 25x
positive
negative
40.
lim x1 3
x 3
positive
positive
negative
positive
42.
lim 2 x3 x10
5
x
negative
negative
positive
positive
44. lim
2
1/3
0 3x
lim
x
2
x1/ 2
1
(1 3/ x )
lim
(4 x 2 25)/ x 2
x
(4 3 x3 )/ x6
9/ x
2
( x 3)/ x
lim
4 x 2 25/ x 2
1 0
(1 0)
lim ( 11 1/1/xx)
x
lim ( x 1)/( x )
x
( x 3)/ x 2
x
1/ x 2
lim (11 1/
x)
x
( x 2 1)/ x 2
x 2 1/ x 2
6
( x 2 1)/ x 2
( x 1)/ x
lim
positive
negative
45. (a)
1
5
2
lim x 3 2
x 2
x
x
2
x1/ 2
1
x11/10
39.
43. lim
lim
1
1
x 2 /15
38.
0
x
x
1
x 2 /15
positive
positive
x
2
x
lim 31x
37.
0
7
x8/5
1
x19/15
x 2 1/ x 2
lim
x
(1/3)
2 x1/15
lim
lim
25
3
lim 4 63x
x
x
0
x2
1
x
x
x 5x 3
2 x x 2/3 4
lim
x
1
3
lim
0 0
1 0 0
2
x3
1
x
lim
x
1
x2
1
5
x2
0
7
x
1 x
x
1
x
1
(1/5)
lim 1 x(1/5)
lim
x
x2
3
x
lim
2
x3
2
x1/ 2
lim
5
x
3
x
lim
5
x2
1
x2
1
x
1
lim 2 3xx 7x
x
1
x
lim
x
(b)
Copyright
x
0
1
2
0 x ( x 1)
2
1/3
0 3x
lim
x
2014 Pearson Education, Inc.
negative
positive positive
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
2
x1/5
lim
46. (a)
0
x
4
47. lim 2/5
49.
4
1/5 2
0 (x )
x
x
lim tan x
x
2
1/5
0 x
lim
x
1
48. lim 2/3
lim
0 x
x
(b)
50.
0 x
lim
x
2
51.
lim (1 csc )
52.
lim (2 cot )
1
1/3 2
0 (x )
lim
x
sec x
2
0
and lim (2 cot )
0
53. (a)
(b)
(c)
(d)
54. (a)
(b)
(c)
(d)
55. (a)
(b)
(c)
(d)
56. (a)
(c)
(d)
1
x2 4
x
1
2
2 x 4
x
lim
x
2
lim
x
x
x
lim ( x 2)(1 x 2)
2
1
positive positive
lim ( x 2)(1 x 2)
2
1
positive negative
lim
1
x2 4
x
lim ( x 2)(1 x 2)
2
1
positive negative
lim
1
x2 4
x
lim ( x 2)(1 x 2)
2
1
negative negative
2
2
lim 2x
x 1 x 1
lim 2x
x 1 x 1
lim ( x 1)(x x 1)
positive
positive positive
lim ( x 1)(x x 1)
positive
positive negative
x 1
x 1
x
2
x
1 x 1
lim 2x
x
1 x 1
x2
2
1
x
0
lim
x2
2
1
x
0
lim
2
x2
2
1
x
22/3
2
1
x2
2
1
x
1
2
x
x
0
3
lim
x
2
lim 2xx 14
2
x
2
lim 2xx 14
x 1
2
lim 2xx 14
x 0
negative
negative negative
lim ( x 1)(x x 1)
x
1
lim
0
negative
positive negative
lim ( x 1)(x x 1)
x
1
lim
x
, so the limit does not exist
0
lim 1x
1
negative
lim 1x
1
positive
x
x
0
0
2 1/3 2 1/3
1
21/3
3
2
1
1
positive
positive
lim
x 1
1
4
0
( x 1)( x 1)
2x 4
20
2 4
(b)
2
lim 2xx 14
2
x
0
Copyright
2014 Pearson Education, Inc.
positive
negative
95
96
Chapter 2 Limits and Continuity
57. (a)
(b)
(c)
2
lim x 3 3 x 22
x
x
0
2x
2
lim x 3 3 x 22
x
x
2
2x
x
2
2x
2
(d) lim x 3 3 x 22
2 x
x
0 x
58. (a)
(b)
(c)
(d)
(e)
x
2
lim
( x 2)( x 1)
2
0
x
0
x
4x
2
lim x 3 3 x 2
x 1
x
4x
x 2 ( x 2)
lim x 21
1,x
4
2
lim x 21
1,
4
x
2
x
2
x
2
x
2
x
x
lim x 21
x
1,
4
2 x
( x 2)( x 1)
negative negative
positive negative
x2 ( x 2)
( x 2)( x 1)
( x 1)
( x 2)( x 1)
( x 2)( x 1)
lim x ( x 2)( x 2)
0
x
( x 2)( x 1)
lim x ( x 2)( x 2)
x 1
1
2(4)
lim x ( x 2)
x 2
1
8
( x 1)
lim x ( x 2)( x 2)
x
2
4x
2
lim x 3 3x 2
x 2 ( x 2)
lim x( x 2)( x 2)
x 2
2
lim x 3 3 x 2
2
negative negative
positive negative
x 2 ( x 2)
x 2 ( x 2)
2
( x 2)( x 1)
lim
4x
x
2
x
x
2
lim x 3 3 x 2
x
( x 2)( x 1)
x
2x
x
lim
lim
2x
2
(e) lim x 3 3 x 22
x
( x 2)( x 1)
0
x
2
lim x 3 3 x 22
x
lim
x
lim x ( x 2)
x
2
negative
negative positive
( x 1)
negative
negative positive
lim x( x 2)
0
x
( x 1)
lim x ( x 2)
x 1
0
(1)(3)
0
negative
positive positive
lim x (xx 12)
x 0
negative
and lim x (xx 12)
negative positive
x 0
so the function has no limit as x 0.
59. (a)
60. (a)
61. (a)
(c)
62. (a)
(c)
63. y
t
t
3
t1/3
(b)
7
(b)
lim
2
lim
1
t 3/5
lim
1
x 2/3
2
( x 1) 2/3
(b)
lim
1
(d)
0
0
x
0
x 1
x 2/3
2
( x 1) 2/3
lim
1
x1/3
1
( x 1)4/3
(b)
lim
1
1
( x 1) 4/3
(d)
x
0
x 1
x1/3
1
x 1
t
t
2
lim
1
t 3/5
lim
1
x 2/3
2
( x 1)2/3
lim
1
x 2/3
2
( x 1)2/3
lim
1
x1/3
1
( x 1) 4/3
lim
1
x1/3
1
( x 1)4/3
0
0
x
0
x 1
x
0
x 1
64. y
Copyright
3
t1/3
lim
7
1
x 1
2014 Pearson Education, Inc.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
65. y
1
2x 4
67. y
x 3
x 2
1 x1 2
66. y
3
x 3
68. y
2x
x 1
2
2
x 1
69. Here is one possibility.
70. Here is one possibility.
71. Here is one possibility.
72. Here is one possibility.
Copyright
2014 Pearson Education, Inc.
97
98
Chapter 2 Limits and Continuity
73. Here is one possibility.
74. Here is one possibility.
75. Here is one possibility.
76. Here is one possibility.
f ( x)
77. Yes. If lim g ( x )
x
f ( x)
2 then the ratio the polynomials leading coefficients is 2, so lim g ( x)
x
2 as well.
78. Yes, it can have a horizontal or oblique asymptote.
f ( x)
79. At most 1 horizontal asymptote: If lim g ( x )
x
f ( x)
80.
so lim g ( x )
x
L as well.
lim
x 4
x
x 9
lim
x 9
lim
5
x
x 9
x
81.
lim
x
x2
25
x2 1
x2
x
x
lim
x2
3 x
x
x2 1
1
x
3
x2
x
x2
Copyright
x
lim
x
lim
x
1
1
1
( x 9) ( x 4)
x 9 x 4
0
x 2 25
x2 1
x 2 25
x2 1
26
x
x
3
x
3
x2
0
1 1
4
x
1
x2 3 x
x2
1
9
x
x2 3 x
x
lim
5
x
3
lim
x 4
x 4
x2 1
25
x 2 25
x2 3
lim
lim
x
26
lim
x
82.
x 9
x 9
x 4
x 4
lim
L, then the ratio of the polynomials leading coefficients is L,
25
x2
lim
0
1 1
1
1
x2
0
1 1
lim
x
0
2014 Pearson Education, Inc.
x 2 25
x2 1
0
( x 2 3) ( x 2 )
x2 3 x
( x 2 25) ( x 2 1)
x
lim
3
x2 3 x
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
83.
x
lim
4 x2 3x 2
2x
x
x
x
84.
9x2
lim
x
x 3x
x2
x
x2
3x
2
x2
x
x2
x
x
x
9 x2
2x
x
x
1
x
1
0, take N
1. Then for all x
88. For any
0, take N
1. Then for all y
Now,
1
x2
1
x2
B
x
0
90. For every real number B
1
x
B
0
x
B
0
B
0
x 3
92. For every real number B
Now,
1
( x 5)2
x 5
B
1
B
0
1
( x 5)2
x2 3 x
2x
x
x
x
1
2
1 .
B
x
2x
x
2
x2
2
5
3
x
x2 2 x
3x
1
x
9 x2 x 3x
x
2
( x 2 x) ( x 2 x)
x2 x
x
k k
0
k k
1
B
Choose
x 2 3x
x
2x
5
5
1 1 2
lim
x
( x2 3 x) ( x 2 2 x )
lim
2
2
x
x2 x
x
x
lim
x
1
6
0 such that for all x, 0
0, we must find a
1 . Choose
B
2
( x 3)2
2
( x 3)2
1
3 3
N we have that f ( x) k
1
B
3x 2
x
4 3x
x2 2 x
lim
x2 x
x
2x
x2 x
.
0
.
1
x2
x 0
, then 0
B.
x
1
B
x
.
91. For every real number B
2
( x 3)2
1
x
9 x2 x 3 x
N we have that f ( x) k
x2
0
1
x2
B so that lim
2
1 1
0, we must find a
1
x2
0
1
x
1
87. For any
89. For every real number B
x
2
x2
lim
4 x2 3x 2
2x
(9 x 2 x ) (9 x 2 )
x
x2 x
x2
x
3
x
4
lim
lim
x2 2 x
2
lim
x
2x
x2
(4 x 2 ) (4 x 2 3 x 2)
lim
3
4
3
x2
5x
x2 3x
x2
lim
lim
1
x 2 3x
lim
x
3x 2
x2
9 x2 x 3x
1
x
4 x2 3 x 2
2x
3 0
2 2
9
4 x2 3 x 2
2x
9 x2 x 3x
x
3x
x
x
2
x2
lim
x
x2
x2
x
4
x 3x
lim
lim
2
x
3
x
3
x
86.
4 x2 3x 2
2x
9x2
x
lim
3x 2
lim
x
4 x2 3 x 2
2x
lim
lim
lim
85.
lim
99
0 such that for all x, 0
1 . Then 0
B
x 0
0, we must find a
B
B
( x 3)
2
0
2
x
( x 3)2
1
B
2
2
3 ( x 3)
0, we must find a
( x 5)2
1
B
x
1
2
5 ( x 5)
Copyright
2
B
0
B so that
1
B
1
x
B. Now,
lim 1
x 0 x
.
2
( x 3) 2
x 3
3
X
2 . Choose
B
B. Now,
2 , then
B
.
0 such that for all x, 0
x 5
B so that lim
1
x
0 such that for all x, 0
0 so that lim
x
1
B
x 0
. Choose
x ( 5)
1 . Then 0
B
.
2014 Pearson Education, Inc.
x ( 5)
1
( x 5)2
B.
x2 x
100
Chapter 2 Limits and Continuity
93. (a) We say that f ( x) approaches infinity as x approaches x0 from the left, and write lim f ( x)
x
,
x0
if for every positive number B, there exists a corresponding number
0 such that for all x,
x0
x x0
f ( x ) B.
(b) We say that f ( x) approaches minus infinity as x approaches x0 from the right, and write lim f ( x)
x
,
x0
if for every positive number B (or negative number B) there exists a corresponding number
0 such
f ( x)
B.
that for all x, x0 x x0
(c) We say that f ( x) approaches minus infinity as x approaches x0 from the left, and write lim f ( x)
, if
x
for every positive number B (or negative number B) there exists a corresponding number
x x0
f ( x)
B.
for all x, x0
94. For B
0, 1x
95. For B
0, 1x
1
x
B
0
B
x
0, x 1 2
Then 2
x
1
x
0
B so that lim
96. For B
1 . Choose
B
x
0
1
x
1
x 2
B
2
B
1 . Then 0
B
0
1
B
x
.
B
x 2
( x 2)
0
1
B
1 x
Then 1
x near 1
99. y
x2
x 1
x 1
lim 1 2
x 1 1 x
x 1 0
1 x
1
B
1
B
x 2
0, x 1 2 B
0 x 2 B1 . Choose
1
B 0 so that lim x 1 2
.
x 2
x 2
98. For B 0 and 0 x 1, 1 2 B 1 x 2 B1
97. For B
x
0
1
B
x 2
1
x 2
0
2
x
2
0
Now 1 2x
(1 x)(1 x )
1 1 x
B 2
x 2
.
x 1 x1 1
100. y
Copyright
x2 1
x 1
x 1 x2 1
2014 Pearson Education, Inc.
0
1
1 x2
x
.
x 2
1 since x 1. Choose
1
B
.
1.
B
2
x
1
B
0
2 B1 . Choose
0 so that lim x 1 2
1.
B
(1 x)(1 x)
x
x
B
0 such that
B so that lim 1x
x 0
1
x
1 . Then
B
x. Choose
1 . Then
B
1
2B
1
B
x
x0
B for 0
1
B
1 .
2B
x 1 and
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
101. y
x2 4
x 1
x 1 x3 1
102. y
x2 1
2x 4
1
2
103. y
x2 1
x
x 1x
104. y
x3 1
x2
x
105. y
x
106. y
4 x2
Copyright
x 1 2 x3 4
1
x2
1
4 x2
2014 Pearson Education, Inc.
101
102
Chapter 2 Limits and Continuity
107. y
x 2/3
1
x1/3
108. y
sin
x2 1
109. (a) y
(see accompanying graph)
(see accompanying graph)
(b) y
(c) cusps at x
1 (see accompanying graph)
110. (a) y 0 and a cusp at x 0 (see the
accompanying graph)
3 (see accompanying graph)
(b) y
2
(c) a vertical asymptote at x 1 and contains the
CHAPTER 2
1. At x
3
23 4
1,
point
(see accompanying graph)
PRACTICE EXERCISES
1:
x
lim f ( x)
1
lim f ( x) 1
x
lim f ( x) 1
x
f ( 1)
1
1
f is continuous at x
At x
0:
lim f ( x)
x
0
lim f ( x)
x
0
But f (0) 1
1.
lim f ( x)
x
0.
0
0
lim f ( x)
x
0
f is discontinuous at x 0.
If we define f (0) 0, then the discontinuity at
x 0 is removable.
At x 1:
lim f ( x)
x 1
1 and lim f ( x) 1
x 1
lim f ( x) does not exist
x 1
f is discontinuous at x 1.
Copyright
2014 Pearson Education, Inc.
Chapter 2 Practice Exercises
2. At x
1:
x
x
lim f ( x)
0 and lim f ( x)
1
x
lim f ( x) does not exist
1
f is discontinuous at x
At x
0:
lim f ( x)
x
and lim f ( x)
0
x
0
f is discontinuous at x
At x 1:
lim f ( x)
x 1
But f (1)
0
1
1
1.
lim f ( x) does not exist
x
103
0
0.
lim f ( x) 1
lim f ( x) 1.
x 1
x 1
lim f ( x )
x 1
f is discontinuous at x 1.
If we define f (1) 1, then the discontinuity at
x 1 is removable.
3. (a) lim (3 f (t ))
t
3 lim f (t )
t0
t
(b) lim ( f (t )) 2
t
t0
t
t0
f (t )
t
t
0
t
t0
lim | f (t )|
t
t
1
f (t )
(h) lim
t
t0
4. (a) lim
x
t
(d) lim f 1( x )
x 0
(e) lim ( x
x
(f)
0
x
f ( x ) cos x
x 1
0
5. Since lim x
x
0
7 0
t0
2
x
1
0
lim x
0
0 12
lim f ( x)
x
1
2
0
lim x lim 1
x
0
x
(1)
0
we would have lim
x
0
x
0
and lim
x
0
1
2
1
2
0 1
0 we must have that lim (4 g ( x))
4 g ( x)
x
2
2
1
2
x 0
x 0
lim f ( x ) lim cos x
x
1
2
lim g ( x)
x
0
2
2
1
2
2
0
lim f ( x )
x
7
1
7
0
0
f ( x ))
1
lim g ( x) lim f ( x)
x
1
lim f ( x )
lim
x
1
7
0
(c) lim ( f ( x) g ( x))
0
7
0 7
t0
cos 0 1
t
lim g ( x)
x
0
x
t
lim g (t )
t0
t0
(b) lim ( g ( x ) f ( x))
x
t0
lim f (t )
t
0
| 7| 7
1
lim f (t )
g ( x)
0
t
t0
t0
t0
( 7)(0)
t0
cos lim g (t )
(g) lim ( f (t ) g (t ))
t
t
lim f (t )
t0
t0
lim f (t )
lim g ( t ) lim 7
t0
t0
t
t
lim ( g (t ) 7)
(e) lim cos ( g (t ))
(f)
49
lim f (t ) lim g (t )
t0
lim f (t )
(d) lim g (t ) 7
t t
21
( 7) 2
t0
(c) lim ( f (t ) g (t ))
t
2
lim f (t )
t0
t
3( 7)
0. Otherwise, if lim (4 g ( x)) is a finite positive number,
4 g ( x)
x
x
0
so the limit could not equal 1 as x
reasoning holds if lim (4 g ( x)) is a finite negative number. We conclude that lim g ( x )
x
0
x
Copyright
2014 Pearson Education, Inc.
0
4.
0. Similar
104
Chapter 2 Limits and Continuity
6. 2
x
lim
x lim g ( x)
4
x
x
0
7. (a) lim f ( x )
c
(b) lim g ( x)
x
c
x
n
I
n
I
(b)
0 x
0
4 lim g ( x) (since lim g ( x) is a
0
x
lim x3/4
c3/4
g (c) for every nonnegative real number c
x
c
0
x
0
f is continuous on (
, ).
g is continuous on [0, ).
1
c 2/3
h(c) for every nonzero real number c
h is continuous on (
lim x
1/6
1
c1/ 6
k (c) for every positive real number c
k is continuous on (0, )
c
x
c
, n 12
3
5 x 2 14 x
x 2
x ( x 7)
x2 4 x 4
(b) lim
x
lim g ( x)
4 x
f (c) for every real number c
x2 4 x 4
lim
x
x
, where I
, 0) and
the set of all integers.
the set of all integers.
, ) ( , )
, 0) (0, )
9. (a) lim
x
4 lim
0
1.
2
(n , (n 1) ), where I
(c) (
(d) (
x
c1/3
c
n 12
8. (a)
lim g ( x)
4
lim x1/3
x
x
c
x
2/3
c
( , ).
(d) lim k ( x)
4
lim x
(c) lim h( x )
x
x
2
4
lim g ( x)
constant)
x
lim x lim
0
2 x
3
( x 2)( x 2)
lim x( x 7)( x 2) lim x (xx 27) , x
x 0
0
and lim x (xx 27)
x 0
x
( x 2)( x 2)
5 x 2 14 x
lim x(xx 27) , x
2
lim x( x 7)( x 2)
2
x
x
2; the limit does not exist because
2, and lim x(xx 27)
x 2
0
2(9)
0
x ( x 1)
x 1
x2 x
1
lim 3 2
lim
lim
, x 0 and x
1.
5
4
3
x
2
x
x
x
(
x 2 x 1) x 0 x 2 ( x 1)( x 1) x 0 x 2 ( x 1)
x 0
x 0
2
Now lim 2 1
and lim 2 1
lim 5 x 4x 3
.
x 0 x 2x x
x 0 x ( x 1)
x 0 x ( x 1)
2
x ( x 1)
lim 5 x 4x 3
lim 3 2
lim 2 1 , x 0 and x
1. The limit does not
1 x 2x x
1 x ( x 2 x 1) x
1 x ( x 1)
x
x
lim 2 1
and lim 2 1
.
1 x ( x 1)
1 x ( x 1)
x
x
10. (a) lim
(b)
11. lim 11 xx
x 1
2
2
12. lim x 4 a 4
x
a x
a
x
a (x
( x h )2 x 2
h
0
lim
h
( x h )2 x 2
h
0
14. lim
x
15.
lim
x
0
1
2 x
x
1
2
(2 x )3 8
x
0
16. lim
x
2
1
lim
x)
a 2 )( x 2 a 2 )
1
2
2
a x a
( x 2 2hx h2 ) x 2
h
0
h
( x 2 2hx h2 ) x 2
h
0
x
2 (2 x )
lim 2 x (2 x )
0
x
x
lim (2 x h)
2x
lim (2 x h)
h
0
0
lim 4 21 x
0
1
4
x
( x3 6 x 2 12 x 8) 8
x
0
lim
1
2a 2
lim
x
lim
x
1
2
x
x 11
( x2 a 2 )
lim
13. lim
h
1 x
x )(1
lim
x 1 (1
lim ( x 2
x
Copyright
0
6 x 12) 12
2014 Pearson Education, Inc.
exist because
Chapter 2 Practice Exercises
( x1/3 1) ( x 2/3 x1/3 1)( x 1)
2/3
x1/3 1)
x 1 ( x 1) ( x 1)( x
1/3
17. lim x 1
x 1
x 1
18.
lim
tan 2 x
sin 2 x cos x
lim cos
2 x sin x
0
19. lim tan x
x 0
20.
x
x
sin 2
22. lim cos 2 ( x tan x)
23. lim 3sin8 xx x
x 0
8
sin x
03 x 1
lim
x
cos 2 x 1
sin x
4sin x cos 2 x
0 cos 2 x 1
25. Let x = t
3
t
8
3(1) 1
lim ln(t 3)
2x
x
111 2
2
ln1 0
27.
1
e 1
( 1) 2
1
4
2
sin 2 2 x
lim sin x (cos
2 x 1)
x 0
lim cos 2 x 1
x 0 sin x (cos 2 x 1)
0
lim ln x
3
t
1 cos
x
sin x
1
cos 2 ( )
tan )
4(0)(1)2
1 1
26. lim t 2 ln 2
t 1
sin 2
2x 1
lim cossin2xx 1 cos
cos 2 x 1
x 0
lim
x
cos x
cos 2 x
x
sin
cos 2 (
x
0
2
3
lim sin1 x
21. lim sin 2x sin x
x
x
lim sin2 x2 x
0
x
lim csc x
24. lim
1 1
1 1 1
lim
x 8
64
1
lim 2/3 x 1/3
x
1
x 1x
( x1/3 4)( x1/3 4)
( x1/3 4)( x1/3 4) ( x 2/3 4 x1/3 16)( x 8)
lim
x 8
x 8
( x 8)( x 2/3 4 x1/3 16)
x 64
x 64
1/3
1/3
( x 64) ( x
4) ( x 8)
(x
4) ( x 8) (4 4) (8 8) 8
lim
lim
2/3
16 16 16
3
4 x1/3 16) x 64 x 2/3 4 x1/3 16
x 64 ( x 64) ( x
2/3
lim x 16
x
( x 1)( x 1)
lim
2/3
x1/3 1)
x 1 ( x 1)( x
x
0
ecos( / )
e1
e 1
e cos( / )
e
ecos( / )
lim
0
0 by the
Sandwich Theorem
28.
29.
30.
lim
z
0
2e1/ z
1/ z
e
1
z
lim [4 g ( x)]1/3
x
x
0
lim x g1( x )
5
2
32.
2
lim 5 x
x
2
g ( x)
2
lim 4 g ( x)
x
x
5
lim g ( x)
2
1
2
x
5
x
0
lim g ( x)
0 since lim (3 x 2 1)
lim g ( x )
x
lim 4 g ( x) 8, since 23
2
0
lim ( x g ( x))
x 1
0
2
1 0
1/3
2
2
31. lim 3gx( x)1
x 1
2
1 e1/ z
lim
x 1
5
1
2
x
lim g ( x )
5
4
since lim (5 x 2 ) 1
x
Copyright
8. Then lim g ( x)
2
2014 Pearson Education, Inc.
x
1
2
0
5
2.
105
106
Chapter 2 Limits and Continuity
33. At x
1:
x
x
x
lim
lim
x2 1
f ( x)
1
lim ( x )
x
x
1
lim
f ( x)
1
x
x
x ( x 2 1)
lim
x
1
x ( x 2 1)
1
lim
f ( x)
| x 2 1|
1
lim x
1, and
1
x ( x 2 1)
lim
|x
1
2
1|
x
x ( x 2 1)
lim
( 1) 1. Since lim
x
1
1
( x 2 1)
f ( x)
lim f ( x) does not exist, the
x
1
function f cannot be extended to a continuous
function at x
1.
At x 1:
lim f ( x )
lim
x 1
x 1
lim
x 1
x ( x 2 1)
x ( x 2 1)
lim
2
| x 1|
x ( x 2 1)
x 1
x2 1
x 1
(x
2
1)
lim ( x )
x 1
1, and lim f ( x)
x 1
lim
x 1
x ( x 2 1)
| x 2 1|
lim x 1.
Again lim f ( x) does not exist so f cannot be extended to a continuous function at x 1 either.
x 1
34. The discontinuity at x
0 of f ( x)
1
x
sin
is nonremovable because lim sin 1x does not exist.
x 0
35. Yes, f does have a continuous extension at a 1:
4.
define f (1) lim x 41
3
x 1x
x
36. Yes, g does have a continuous extension at a
g 2
lim
2
5 cos
4 2
5.
4
37. From the graph we see that lim h(t )
t
0
t
2
:
lim h(t )
0
so h cannot be extended to a continuous function
at a 0.
Copyright
2014 Pearson Education, Inc.
Chapter 2 Practice Exercises
38. From the graph we see that lim k ( x)
x
lim k ( x)
0
x
0
so k cannot be extended to a continuous function at
a 0.
39. (a) f ( 1)
1 and f (2) 5
(b), (c) root is 1.32471795724
f has a root between 1 and 2 by the Intermediate Value Theorem.
40. (a) f ( 2)
f has a root between 2 and 0 by the Intermediate Value Theorem.
2 and f (0)
(b), (c) root is
41.
43.
lim 52xx 73
x
x
2
1.76929235424
lim
x
2
lim x 43x 8
3x
3
x
7
x
2
5
x
2 0
5 0
2
5
1
3x
4
3 x2
lim
42.
8
3 x3
0 0 0
x
2
lim 2 x2 3
5x
7
x
lim
2
5
3
2 0
5 0
x2
7
x2
0
1
44.
45.
47.
lim
x
1
7x 1
x2
lim
2
lim x x 71 x
x
lim sin x
x
lim cos
49.
x 2 x
lim x sin
x
x
sin
x
50.
x 2/3 x 1
lim 2/3
2
51.
52.
53.
x
lim e1/ x cos 1x
x
lim ln 1 1t
x
x
lim tan 1 x
46.
0 since
lim 2
cos x
0
x
x
48.
x
1
x
1
0
1 0 0
lim x 17
lim 1
x
x
1
x
x2
7 1
x x2
lim
x
sin x 2
x
x
1 sinx x
lim 1 x
x
1
as x
lim cos 1
0
1
x
5/3
cos2 x
x 2/3
1 0 0
1 0
1 0
1 0
lim
x
x 4 x3
12 x3 128
lim sin x
x
x
x
lim
0.
0.
1
1
e0 cos(0) 1 1 1
ln1 0
2
Copyright
2014 Pearson Education, Inc.
x 1
12 128
3
x
2
5
107
108
54.
Chapter 2 Limits and Continuity
t
lim e3t sin 1 1t
x2 4
x 3
55. (a) y
0 sin 1 (0)
x 1 x
2x 8
2
x
lim 2 x 6
x
2x 8
4
x
x
lim xx 34
4
. Thus x
1
1 x 2 : lim 1 x 2
x2 1 x
x2 1
1
x2
lim
1
1
1
x2
1
x
x
, thus x
2
and lim x2 x 2
2x 1
x 1 x
2
4: lim x2 x 6
2 and
x
y
2
and lim xx 34
x 3
2
undefined at x 1: lim x2 x 2
asymptote.
2
(c) y x2 x 6 is undefined at x
56. (a)
0
2
3 : lim xx 34
x 3
is undefined at x
x 2 x 2 is
x2 2x 1
(b) y
0 0
2x
2x 8
, thus x 1 is a vertical
2x 1
lim x 3
x 2x 4
3 is a vertical asymptote.
5;
6
x
x2 x 6
2
4 x 2x 8
lim
x
lim xx 34
4
4 is a vertical asymptote.
1
1 x2
x2 1
1 and lim
x
x
lim
x2
1
1
1
1
1
x2
1, thus y
1 is a
horizontal asymptote.
(b)
x 4
:
x 4
y
x2 4
x
(c) y
lim
x
:
lim
x2 9 :
9 x2 1
1
3
thus y
CHAPTER 2
1. (a) x
x
x2
1 0
1
1
x
x2
x
x
x
x2 4
x
1 and lim
x
4
1
lim
1, thus y 1 is a horizontal asymptote.
4
1
lim
1 0
1 0
4
x
lim
1
x
x
lim
x2
x
x2
4
x2
1
1 0
1
1
1
1,
x2 9
9 x2 1
lim
x
1
9
9
x2
1
x2
1 0
9 0
1
3
and lim
x
x2 9
9 x2 1
is a horizontal asymptote.
ADDITIONAL AND ADVANCED EXERCISES
0.01
0.001
0.0001 0.00001
0.7943 0.9550 0.9931 0.9991 0.9999
x
4
1
1 are horizontal asymptotes.
Apparently, lim x x
(b)
1
x
x
lim
0.1
x
4
x
1
lim
x2 4
x
x
thus y 1 and y
(d) y
x 4
x 4
0
1
Copyright
2014 Pearson Education, Inc.
x
lim
1
9
9
x2
1
x2
1 0
9 0
1,
3
Chapter 2 Additional and Advanced Exercises
2. (a) x
10
1
x
(b)
3.
100
c
1/(ln x )
c
1
e
0.3678
lim v 2
v2
c2
lim L0 1
v
1000
0.3679 0.3679 0.3679
Apparently, lim 1x
x
lim L
v
1/(ln x )
109
v
L0 1
2
L0 1 c 2
c
c
2
0
c
The left-hand limit was needed because the function L is undefined if v
the speed of light).
4. (a)
x
2
1
0.2
0.2
x
2
1 0.2
0.8
(b)
x
2
1
0.1
0.1
x
2
1 0.1
0.9
5. |10 (t 70) 10 4 10| 0.0005
5 t 70 5 65 t 75
x
2
1.2
x
2
1.1
1.6
1.8
|(t 70) 10 4 | 0.0005
Within 5 F.
x
x
c (the rocket cannot move faster than
2.4
2.56
2.2
x
3.24
x
0.0005 (t 70) 10
5.76.
4.84.
4
0.0005
6. We want to know in what interval to hold values of h to make V satisfy the inequality
|V 1000| |36 h 1000| 10. To find out, we solve the inequality:
|36 h 1000| 10
10
36 h 1000 10
990
36 h 1010
990
36
h
1010
36
8.8
h 8.9
where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe.
The interval in which we should hold h is about 8.9 8.8 0.1 cm wide (1 mm). With stripes 1 mm wide, we can
expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.
7. Show lim f ( x)
x 1
2
lim ( x 2 7)
Step 1: |( x 7) 6|
Step 2: | x 1|
Then
1
1
or
|( x
2
7) 6|
8. Show lim g ( x)
x
1
4
Step 1: 21x 2
6
x 1
x
2
x 1
1
1
and lim f ( x)
x
1
4
1
2x
2
1
2x
1
x2
1
1 x
1.
. Choose
min 1
1
1
1
, 1
x
1
.
1 , then 0 | x 1|
6. By the continuity text, f ( x ) is continuous at x 1.
1
lim
x
f (1).
g
2
1
4
.
2
Copyright
1
2x
2
1
4 2
x
1
4 2
2014 Pearson Education, Inc.
.
110
Chapter 2 Limits and Continuity
Step 2:
1
4
1
4
X
Then
Choose
1
4
x
1
4 2
4(2
)
1
4
1
4
1
4 2
4(2
)
x
Step 1:
2
1
4
x
2x 3 1
x 2
(1 ) 2 3
2
. Choose
or
2
2
(1 )2 3
2
2
2
1
10. Show lim F ( x )
Step 1:
x
2.
1 (1 ) 2
2
2
, or
2
2
5
9 x 2
2x 3 1
)2
9 (2
By the continuity test, F ( x ) is continuous at x
x
)2 .
9 (2
)2
2
4 (2 )2
2.
9 x 2
, so lim 9 x 2.
x
5.
L1. Let
f ( x) L1
x0
1 (L
3 2
L1 ) L1
4 L1 L2
3 f ( x)
x
any
c
L. If k
0, there is a
|(kf ( x)) ( kL)|
13. (a) Since x
0 ,0
0, then lim k f ( x)
x
lim 0
c
x
0 so that 0 | x c |
. Thus lim k f ( x)
x
x3
( x3
x 1
x
x)
x
L2 f ( x ) 13 ( L2 L1 ) L2
2 } both inequalities must
L2 . That is, L1 L2
c
| k || f ( x) L |
|k |
L2 so
x0
0 and
0, then given
| k ( f ( x) L)|
x3
0
( x3
x)
(c) Since x
0 ,0
x4
x2
1
( x2
x4 )
(d) Since x
0 , 1 x
0
0
x2
c
lim f ( x3
0
0 , 1 x
Copyright
x
L1 ). Since
k lim f ( x) .
(b) Since x
x4
1 (L
3 2
0 lim f ( x) and we are done. If k
| f ( x) L |
kL
c
0
c
5
2 L1 L2 . Likewise, lim f ( x)
there is a 2
1 (L
such that 0 | x x0 |
| f ( x) L2 |
f ( x) L2
L1 )
2
3 2
2 L2 L1 3 f ( x) 4 L2 L1
L1 4 L2
3 f ( x)
2 L2 L1. If
min{ 1 ,
4 L L 3 f ( x) 2 L1 L2
: 1 2
5( L1 L2 ) 0 L1
hold for 0 | x x0 |
L1 4 L2
3 f ( x)
2 L2 L1
L1 L2 0, a contradiction.
12. Suppose lim f ( x)
,
2.
11. Suppose L1 and L2 are two different limits. Without loss of generality assume L2
lim f ( x) L1 there is a 1 0 such that 0 | x x0 | 1 | f ( x) L1 |
f ( x)
(1 )2 1
2
2
F (5).
9 x 2
L1 ) L1
(1 ) 2 3
.
2
x
(1 ) 2 3
2
, the smaller of the two values. Then, 0 | x 2|
lim 9 x
x
5
1 (L
3 2
2.
4
(1 )2 3
2
2
Step 2: 0 | x 5|
x 5
5 x
5.
Then
5 9 (2 )2
(2 )2 4 2 2 , or
5 9 (2
2
Choose
2 , the smaller of the two values. Then, 0 | x 5|
x
.
and lim 21x
x 1
2
(1 ) 2 3
2
2x 3 1
2
x
)
1.
4
so lim 2 x 3 1. By the continuity test, h( x) is continuous at x
x
4(2
1
2x
2
Step 2: | x 2|
2
1
4
lim 2 x 3 1 h(2).
2x 3 1
2
1
4 2
, the smaller of the two values. Then 0
x
2
Then
1
4
, or
By the continuity test, g ( x ) is continuous at x
9. Show lim h( x )
1.
4
1
4 2
x
x
0
1
0
lim f ( x 3
x
0
( x2
x)
0
lim f ( x 2
x
0
x4 )
0
lim f ( y )
y
B where y
0
x)
lim f ( y )
x4 )
y
0
x
0
2014 Pearson Education, Inc.
A where y
lim f ( y )
y
0
lim f ( x 2
x3
x4 )
A where y
x3
x.
x.
x2
A as in part (c).
x4 .
Chapter 2 Additional and Advanced Exercises
14. (a) True, because if lim ( f ( x) g ( x)) exists then lim ( f ( x) g ( x))
x
a
x
lim g ( x) exists, contrary to assumption.
x
a
1
x
(b) False; for example take f ( x)
lim ( f ( x) g ( x))
x
0
lim 1x
x
1
x
0
1 . Then
x
and g ( x)
lim 0
x
lim f ( x )
a
x
a
lim [( f ( x) g ( x))
x
a
111
f ( x)]
neither lim f ( x) nor lim g ( x) exists, but
x
0
x
0
0 exists.
0
(c) True, because g ( x) | x | is continuous g ( f ( x )) | f ( x)| is continuous (it is the composite of
continuous functions).
1, x 0
(d) False; for example let f ( x)
f ( x) is discontinuous at x 0. However | f ( x)| 1 is
1, x 0
continuous at x 0.
15. Show lim f ( x)
x
1
2
lim xx 11
x
1
lim
x
( x 1)( x 1)
1 ( x 1)
2, x
1.
x2 1 ,
x 1
Define the continuous extension of f ( x) as F ( x)
x
1
. We now prove the limit of f ( x) as x
1
2,x
1
exists and has the correct value.
2
Step 1: xx 11 ( 2)
Step 2: | x ( 1)|
Then
1
1
x2 1
x 1
( 2)
( x 1)( x 1)
( x 1)
x 1
, or
1
lim F ( x)
x
x
1 x
2
1.
. Choose
,x
1
1 x
1.
. Then 0 | x ( 1)|
2. Since the conditions of the continuity test are met by F ( x ), then f ( x) has
1
2
lim x 2 x2 x6 3
x 3
3
( x 1)
1
a continuous extension to F ( x) at x
16. Show lim g ( x)
2
1.
lim
x
3
( x 3)( x 1)
2( x 3)
2, x
Define the continuous extension of g ( x) as G ( x)
3.
x2 2x 3 ,
2x 6
x
3
2
x
3
,
. We now prove the limit of g ( x) as x
3
exists and has the correct value.
2
Step 1: x 2 x2 x6 3 2
Step 2: | x 3|
Then, 3
3 2
x2 2 x 3
2x 6
( x 3)( x 1)
2( x 3)
x 3
2 , or
3
x
3 3 2
( x 3)( x 1)
lim 2( x 3)
x 3
2
x 1
2
2
continuously extended to G ( x) at x
2
,x
3.
2 . Choose
3
3 2
x
3 2.
2 . Then 0 | x 3|
2. Since the conditions of the continuity test hold for G ( x ), g ( x) can be
3.
17. (a) Let
0 be given. If x is rational, then f ( x ) x | f ( x) 0| | x 0|
| x 0| ; i.e., choose
.
Then | x 0|
| f ( x) 0|
for x rational. If x is irrational, then f ( x) 0 | f ( x) 0|
0
which is true no matter how close irrational x is to 0, so again we can choose
. In either case, given
0 such that 0 | x 0|
| f ( x ) 0| . Therefore, f is continuous at x 0.
0 there is a
(b) Choose x c 0. Then within any interval (c , c
) there are both rational and irrational numbers. If c
c
is rational, pick
. No matter how small we choose
0 there is an irrational number x in
2
(c
,c
)
| f ( x)
f (c)| |0 c |
f (c )
other hand, suppose c is irrational
there is a rational number x in (c
| x|
If x c
value x
c
2
c
,c
c
2
. That is, f is not continuous at any rational c
0. Again pick
) with | x c |
f is not continuous at any irrational c
0, repeat the argument picking
c.
Copyright
|c|
2
c
2
c . No
2
c
2
0. On the
matter how small we choose
x
3c . Then
2
| f ( x)
f (c )| | x 0|
0.
c . Therefore
2
f fails to be continuous at any nonzero
2014 Pearson Education, Inc.
0
112
Chapter 2 Limits and Continuity
m
n
18. (a) Let c
be a rational number in [0, 1] reduced to lowest terms
0 is taken, there is an irrational number x in the interval (c
how small
0 1n
f (c)
1
n
1
2n
. Therefore f is discontinuous at x
f (c )
(b) Now suppose c is an irrational number
1 . Pick
n
,c
)
1 . No
2n
matter
| f ( x)
f (c)|
c, a rational number.
0. Let
0 be given. Notice that 12 is the only rational
number reduced to lowest terms with denominator 2 and belonging to [0, 1]; 13 and 23 the only rationals with
denominator 3 belonging to [0, 1]; 14 and 34 with denominator 4 in [0, 1]; 15 , 25 , 53 and 54 with denominator 5
in [0, 1]; etc. In general, choose N so that N1
there exist only finitely many rationals in [0, 1] having
denominator N , say r1 , r2 , , rp . Let
min {| c ri |: i 1, , p}. Then the interval (c , c
)
contains no rational numbers with denominator N . Thus, 0 | x c |
| f ( x) f (c)| | f ( x) 0|
| f ( x)| N1
f is continuous at x c irrational.
(c) The graph looks like the markings on a typical
ruler when the points ( x, f ( x)) on the graph of
f ( x) are connected to the x-axis with vertical
lines.
19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero
R represents the midnight point (at the same exact time). Suppose x1 is a point
point, 0, on the equator 0
x1
R is simultaneously just after midnight. It seems reasonable that the
on the equator just after noon
temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after
R) 0. At exactly the same moment in time pick x2 to be a point just before
midnight: That is, T ( x1 ) T ( x1
x2
R is just before noon. Then T ( x2 ) T ( x2
R) 0. Assuming the temperature function T is
midnight
continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c
R) 0; i.e., there is always a pair
between 0 (noon) and R (simultaneously midnight) such that T (c) T (c
of antipodal points on the earth s equator where the temperatures are the same.
lim 14 ( f ( x) g ( x)) 2 ( f ( x) g ( x ))2
20. lim f ( x) g ( x)
x
c
21. (a) At x
At x
x c
1 (32
4
( 1) 2 )
0: lim r ( a )
a
0
a
1
2
lim ( f ( x) g ( x))
x
c
2
lim ( f ( x) g ( x))
x
c
2.
lim 1 a1 a
a 0
1: lim r (a )
1
4
lim
a 0
1 (1 a )
lim
a
1 a( 1 1 a )
Copyright
1
1 a
1 a
1 a
a
lim
a
1 a( 1 1 a )
a
1
1
1 (1 a )
a
(
1 1 a)
a 0
1
1
1 0
lim
2014 Pearson Education, Inc.
1
1
1 0
1
2
Chapter 2 Additional and Advanced Exercises
(b) At x
0: lim r (a )
a
1
1 a
a
1
1 1 a
1
1 1 a
lim
0
a
0
lim
a
0
lim
a
0
1
lim
a
1 a
1
1
a
0
1 a
1 a
1 (1 a )
0 a( 1 1 a )
lim
a
lim
113
0 a( 1
a
a
1 a)
(because the denominator is always negative); lim r (a )
a
0
(because the denominator is always positive).
Therefore, lim r ( a ) does not exist.
a
At x
(c)
0
1: lim r (a )
a
1
a
lim
1
1 a
a
1
a
lim
1
1
1
1 a
1
(d)
22. f ( x) x 2 cos x
f (0) 0 2 cos 0 2 0 and f ( )
2 cos( )
2 0. Since f ( x) is
2, 2].
continuous on [ , 0], by the Intermediate Value Theorem, f ( x ) must take on every value between [
Thus there is some number c in [ , 0] such that f (c) 0; i.e., c is a solution to x 2 cos x 0.
23. (a) The function f is bounded on D if f ( x) M and f ( x) N for all x in D. This means M f ( x) N for
all x in D. Choose B to be max {| M |, | N |}. Then | f ( x)| B. On the other hand, if | f ( x )| B, then
B f ( x) B
f ( x)
B and f ( x) B
f ( x ) is bounded on D with N B an upper bound and
M
B a lower bound.
L N . Since lim f ( x ) L there is a
0 such that
(b) Assume f ( x) N for all x and that L N . Let
2
0 | x x0 |
| f ( x) L |
3L N
2
L
L N
2
. But L N
N
N
contradiction proves L N .
(c) Assume M f ( x) for all x and that L
f ( x)
L
M L
2
3L M
2
f ( x)
Copyright
f ( x)
L
L
x x0
L N
2
f ( x)
L
L N
2
L N
2
f ( x) contrary to the boundedness assumption f ( x)
M . Let
M L
2
M L . As
2
in part (b), 0 | x x0 |
M , a contradiction.
2014 Pearson Education, Inc.
L
f ( x)
N . This
M L
2
114
Chapter 2 Limits and Continuity
24. (a) If a
b, then a b
0
|a b|
If a
b, then a b
0
|a b|
25.
lim
x
0
sin(1 cos x )
x
lim
x
2
1 lim x (1sincosx x )
x 0
26.
lim sin x
x
x
0
sin(sin x )
x
sin( x 2 x )
x
0
28. lim
x
x
x
x
sin( x 3)
x 9
x
30. lim
9
0
lim
sin( x 2 4)
x 2
2
29. lim
x
lim
x
0
lim
2
lim
9
x
sin
x
x
x
sin(sin x) sin x
sin x
x
sin( x 2 x )
( x 1)
x2 x
sin( x 2 4)
x
2
b a
sin(1 cos x ) 1 cos x 1 cos x
1 cos x
x
1 cos x
x
lim sinx x 1 sin
cos x
0
0
x
( a b)
1 20
x
lim sinx x
0 sin x
27. lim
0
| a b| a b a b
2 a a.
2
2
2
2
|a b| a b b a
max {a, b} a 2 b
2
2
2
( x 2)
4
sin( x 3)
x 3
1
x 3
lim
x
0
1
0
sin( x 2 x )
x2 x
0
x
lim
x
x
0
x2 4
9
11 0
0.
1 1 1.
lim ( x 1) 1 1 1.
x
sin( x 2 4)
2
lim
b.
2
sin(1 cos x )
lim 1 cos x
1 cos x
x 0 x (1 cos x )
sin(sin x )
lim sin x
sin x
x 0 x
lim
x
sin x
x
0
lim
x
lim
x
2b
2
0.
1 lim
x
a b
2
max {a, b}
|a b|
.
2
a b
2
(b) Let min {a, b}
a b
0
lim ( x 2) 1 4
x
2
sin( x 3)
lim
x 3
x 9
1
x 3
1 16
4.
1.
6
31. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is
3/ 2
3 , thus the oblique asymptote is y 2 x.
an oblique asymptote. y 2 x 2 x 3 2 x
x 1
32. As x
, 1x
0
sin 1x
the oblique asymptote is y
33. As x
, x2 1 x2
asymptotes are y x and y
0
x 1
1 sin 1x
1, thus as x
,y
x x sin 1x
x 1 sin 1x
x; thus
x.
x2 1
x.
34. As x
,x 2 x
x2 2 x
asymptotes are y x and y
x.
x 2 ; as x
x( x 2)
Copyright
, x2
x 2 ; as x
x, and as x
, x2
2014 Pearson Education, Inc.
, x2
x, and as x
x; thus the oblique
, x2
x;
CHAPTER 3
3.1
DERIVATIVES
TANGENTS AND THE DERIVATIVE AT A POINT
1. P1: m1 1, P2 : m2
5, P :m
2
2 2
3. P1: m1
5. m
5
2. P1: m1
1
2
4. P1: m1
[4 ( 1 h ) 2 ] (4 ( 1) 2 )
h
h 0
h(2 h)
lim
2; at ( 1,3): y
h 0 h
lim
y
6. m
lim 2 1 hh 2 1
0
h
lim
h
1
lim
h
lim
0
y
1
lim
h
y 1 2( x ( 1))
0
0;
0 1 h 1
( 1)2
( 2h h 2 )
lim h
h
lim 2 1 hh 2 2 1 h 2
2 1 h 2
0
2
lim
1;
h
2
0 h( 1 h )
h
y 1, tangent line
2 1( x 1)
( 1 h )2
h2
0 h
lim
h
4(1 h) 4
0 2h 1 h 1
at (1, 2): y
h
(1 2h h 2 ) 1
h
0
3 2( x ( 1))
[(1 h 1)2 1] [(1 1) 2 1]
h
0
at (1,1) : y 1 0( x 1)
8. m
3
2 x 5, tangent line
h
h
3, P2 : m2
0
lim
lim
7. m
2, P2 : m2
h
1 ( 1 h )2
2
0 h( 1 h)
2 h
2
0 ( 1 h)
lim
h
x 1, tangent line
y
2x
2; at ( 1,1):
3, tangent line
Copyright
2014 Pearson Education, Inc.
115
116
Chapter 3 Derivatives
9. m
( 2 h)3 ( 2)3
h
0
2
2
3
lim 8 12h 6h h h 8
h 0
lim
h
lim (12 6h h ) 12;
0
h
at ( 2, 8): y
tangent line
8 12( x ( 2))
1
10. m
lim
h
lim
y
12. m
13. m
[(2 h )2 1] 5
h
0
lim
h
15. m
16. m
17. m
3 h
(3 h ) 2
h
h
h
(2 h)3 8
h
0
lim
h
h
lim 4 hh 2
0
lim
h
0
h
8 2(4 4 h h 2 )
1 (x
6
m
h ( 3 2h)
h
4 h 2
4 h 2
h (12 6 h h2 )
h
0
lim
h
h (6 3h h 2 )
h
0
4 h 2
2( x 3), tangent line
8
4
2; at (2, 2): y 2
h
4 h 2
lim
0h
h
6; at (1, 4): y 4
1
4 2
1;
4
4), tangent line
lim 9 hh 3
0
h
9 h 3
9 h 3
lim
h
0h
(9 h ) 9
lim
9 h 3
0h
h
h
9 h 3
1
9 3
8), tangent line
5( 1 h ) 2 5
h
0
lim
h
5(1 2h h 2 ) 5
h
0
lim
h
Copyright
3( x 1), tangent line
2( x 2)
12; at (2,8): y 8 12(t 2), tangent line
lim
(4 h ) 4
0h
h
3; at (1, 1) : y 1
2
0 h (2 h )
h
lim
h
4( x 2), tangent line
2 h(4 h)
lim
h (2 h ) 2
0
(1 3h 3h2 h3 3 3h) 4
h
0
lim 4 hh 2
0
1 (x
4
5
lim
2
0
2; at (3,3): y 3
lim
h
(8 h ) 1 3
h
1, y
lim h ( h2h1)
0
(8 12 h 6 h2 h3 ) 8
h
0
h
lim
h
h
lim
[(1 h)3 3(1 h)] 4
h
0
h
h
0 h(2 h)
lim
h
(1 h 2 4 h 2h 2 ) 1
h
0
8 2(2 h )2
lim
h
0
0
4; at (2,5): y 5
lim
(3 h ) 3( h 1)
h( h 1)
lim
2
at (8,3): y 3
19. At x
h
h
(2 h )2
lim
3
h
0
at (4, 2): y 2
18. m
(5 4 h h 2 ) 5
h (4 h )
lim
h
0
h 0 h
[(1 h ) 2(1 h )2 ] ( 1)
h
0
lim
2))
lim
lim
8
14. m
3
0 8h ( 2 h )
2
lim 12 6h h3
h 0 8( 2 h)
8 h ( 2 h )3
3;
16
3 (x (
1 :y
1
8
8 16
3 x 1 , tangent line
16
2
2,
h
8 ( 2 h )3
lim
h
0
(12 h 6h 2 h3 )
0
12
8( 8)
11. m
( 2)3
h
h
at
1
( 2 h )3
y 12 x 16,
lim
h
0
5h ( 2 h )
h
2014 Pearson Education, Inc.
10, slope
1;
6
6(t 1), tangent line
Section 3.1 Tangents and the Derivative at a Point
20. At x
2, y
21. At x
3, y
22. At x
0, y
3
1
2
m
h
m
1
[1 (2 h )2 ] ( 3)
h
0
lim
1
1
(3 h ) 1 2
lim
0
h
m
h 1
h 1
lim
0
h
(1 4 4 h h 2 ) 3
h
0
lim
h
2 (2 h )
h
( 1)
h
h
lim
0
0
lim 2 h(2h h )
0
lim 2h (2 h )
0
h
4, slope
1 , slope
4
h
( h 1) ( h 1)
h ( h 1)
h (4 h )
h
lim
h
117
lim h (2hh 1)
0
2
h
23. (a) It is the rate of change of the number of cells when t 5. The units are the number of cells per hour.
(b) P (3) because the slope of the curve is greater there.
6.10(5 h ) 2 9.28(5 h ) 16.43 [6.10(5) 2 9.28(5) 16.43]
h 0
h
lim 51.72 6.10h 51.72 52 cells/hr.
(c)
P (5)
h
lim
25. At a horizontal tangent the slope m
0
( x 2 2 xh h 2 4 x 4h 1) ( x 2 4 x 1)
lim
h
h 0
x
2. Then f ( 2)
a horizontal tangent.
m
lim (3x
h
h
0
24. (a) From t 0 to t 3, the derivative is positive.
(b) At t 3, the derivative appears to be 0. From t
26. 0
61.0h 6.10h 2 9.28h
0
h
lim
4 8 1
[( x h)3 3( x h )] ( x3 3 x)
h
0
3xh h
0
2
3)
3x
2
m
3, the derivative is positive but decreasing.
[( x h ) 2 4( x h) 1] ( x 2 4 x 1)
h
0
lim
h
(2 xh h 2 4 h )
lim
h
h 0
5
lim
h
2
0
2 to t
2 x 4; 2 x 4
lim (2 x h 4)
h
0
( 2, 5) is the point on the graph where there is
( x3 3x 2 h 3 xh2 h3 3 x 3h ) ( x3 3 x)
h
0
lim
h
2
0
3; 3 x
3 0
3 x 2 h 3 xh 2 h 3 3h
h
0
lim
h
1 or x 1. Then f ( 1)
x
2 and f (1)
2
( 1, 2)
and (1, 2) are the points on the graph where a horizontal tangent exists.
1 m
27.
lim
h
1
( x h) 1
1
x 1
( x 1) ( x h 1)
0 h( x 1)( x h 1)
lim
h
0
h
h
0 h ( x 1)( x h 1)
x( x 2) 0
x 0 or x 2. If x 0, then y
1 and m
then y 1 and m
1 y 1 ( x 2)
( x 3).
28.
1
4
m
Thus, 14
lim
h
0
1
x h
h
2 x
f (2 h ) f (2)
h
0
29. lim
h
x
lim
h
2
x
0
x h
h
x
x
4
x h
x h
x
x
0h
x h
(100 4.9(2 h )2 ) (100 4.9(2)2 )
h
0
h 0 h
1 (x
4
f (10 h ) f (10)
h
0
h
f (3 h ) f (3)
h
0
31. lim
h
3(10 h ) 2 3(10) 2
h
0
lim
h
(3 h )2
h
0
lim
h
(3) 2 )
3(20 h h 2 )
h
0
lim
h
[9 6 h h 2 9]
h
0
lim
h
Copyright
60 ft/sec.
lim (6 h)
h
0
2014 Pearson Education, Inc.
6
x2
2x
0
( x 1). If x
2,
1 .
2 x
1.
lim ( 19.6 4.9h)
h
The minus sign indicates the object is falling downward at a speed of 19.6 m/sec.
30. lim
x
x
4
4)
4.9(4 4 h h2 ) 4.9(4)
h
0
lim
h
h
x h
lim
2
1
1 ( x 0)
y
x
2. The tangent line is y
y
lim
h
1
( x h) x
lim
h
( x 1) 2
1
( x 1) 2
lim
h
0
19.6.
118
Chapter 3 Derivatives
f (2 h ) f (2)
h
0
32. lim
h
33. At ( x0 , mx0
lim
h
4
3
(2 h )3
4
3
(2)3
h
0
4
3
lim
h
[12 h 6 h2 h3 ]
b) the slope of the tangent line is lim
h
The equation of the tangent line is y (mx0 b)
34. At x
1
2
0
2h 4 h 2
35. Slope at origin
h
g (0 h) g (0)
h
0
38.
0
h
h sin
h
0
lim
h
1
h
0
h
0
lim h sin 1h
h
0
U (0 h ) U (0)
h
lim 0h1
h
0
0
0
h
, and lim
h
0
, and lim
h
0
39. (a) The graph appears to have a cusp at x
lim
h
0
of y
f (0 h ) f (0)
h
2/5
x
2/5
lim h h 0
h 0
U (0 h ) U (0)
h
m.
lim 2
h
4 h 2
0 2h 4 h 2
1
2 4 2
4
4 h
4 h
1
16
yes, f ( x) does have a tangent at the
lim 1 h0
0
1
3/5
0 h
does not have a vertical tangent at x
f (0 h ) f (0)
h
Copyright
. Therefore, lim
h
h
lim 1h1
h
0
0
0
f (0 h ) f (0)
h
no, the graph of f does not have a
0.
1
3/5
0 h
limit does not exist
the graph
1
1/5
0 h
limit does not exist
y
and lim
h
0.
0.
4/5
1
lim h h 0 lim 1/5
h 0
h 0
h 0 h
does not have a vertical tangent at x 0.
lim
0
0
f (0 h ) f (0)
h
lim
h
40. (a) The graph appears to have a cusp at x
(b)
0
lim m
h
lim sin h1 . Since lim sin 1h does not exist, f ( x) has no tangent at the origin.
vertical tangent at (0, 1) because the limit does not exist.
(b)
4 h
0 2h 4 h
1
2 4 h 2 4 h
lim
1
h
lim 2
h
mh
0 h
lim
h
mx b.
y
2 4 h
2 4 h
h
h
0
1
2
h
4 h
h2 sin
lim
m( x x0 )
0
h
2h 4 h 2
0
h
1 0
h
h
[12 6h h 2 ] 16
( m( x0 h ) b ) ( mx0 b)
( x0 h ) x0
1
4 h
lim
h
yes, the graph of f has a vertical tangent at the origin.
lim
h
h
1
2
h
0
f (0 h) f (0)
h
0
lim
f (0 h ) f (0)
h
lim
h
h
lim
36. lim
37.
1
4 h
lim
lim
4 h
origin with slope 0.
h
and m
4 (4 h )
lim
h
1
4
4, y
0
4
0 3
lim
h
0
and lim
h
2014 Pearson Education, Inc.
x 4/5
Section 3.1 Tangents and the Derivative at a Point
119
41. (a) The graph appears to have a vertical tangent
at x 0.
1/5
1
lim h h 0 lim 4/5
h 0
h 0 h
h
42. (a) The graph appears to have a vertical tangent
at x 0.
f (0 h ) f (0)
h
0
(b) lim
f (0 h ) f (0)
h
0
(b) lim
h
h3/5 0
h
0
lim
h
lim 21 5
h
(b)
lim
h
0
2/5
lim 4h h 2 h
h 0
f (0 h ) f (0)
h
the graph of y
lim
f (0 h ) f (0)
h
5/3
2/3
lim h h5h
0
lim
0
4
h3/5
2
0
5
h1/3
h
0
y
x5/3 5 x 2/3 does not have a vertical tangent at x
Copyright
4
h3/5
and lim
h
0
2
0.
limit does not exist
0.
0.
lim h 2/3
h
0.
x3/5 has a vertical tangent at x
4 x 2/5 2 x does not have a vertical tangent at x
h
x
0.
h
44. (a) The graph appears to have a cusp at x
(b)
x1/5 has a vertical tangent at
the graph of y
0 h
43. (a) The graph appears to have a cusp at x
y
0
5 does not exist
lim 1/3
h
0 h
0.
2014 Pearson Education, Inc.
the graph of
120
Chapter 3 Derivatives
45. (a) The graph appears to have a vertical tangent
at x 1 and a cusp at x 0.
(b) x 1:
(1 h )2/3 (1 h 1)1/3 1
h
0
lim
h
at x 1;
x
0:
f (0 h ) f (0)
h
0
2/3
lim
h
y
x
(1 h )2/3 h1/3 1
h
0
lim
h
h 2/3 ( h 1)1/3 ( 1)1/3
h
h 0
1/3
lim
( x 1)
y
( h 1)1/3
h
1
1/3
0 h
lim
h
x 2/3 ( x 1)1/3 has a vertical tangent
does not have a vertical tangent at x
1
h
does not exist
0.
46. (a) The graph appears to have vertical tangents
at x 0 and x 1.
(b) x
0:
x 1:
f (0 h ) f (0)
h1/3 ( h 1)1/3 ( 1)1/3
lim
h
h
h 0
h 0
f (1 h ) f (1)
(1 h )1/3 (1 h 1)1/3 1
lim
lim
h
h
h 0
h 0
lim
y
x1/3 ( x 1)1/3 has a vertical tangent at x
y
x1/3 ( x 1)1/3 has a vertical tangent at x 1.
47. (a) The graph appears to have a vertical tangent
at x 0.
(b)
f (0 h ) f (0)
h
h 0
1
lim
h 0 |h|
lim
lim
x
0
h 0
h
lim 1
h
0 h
; lim
y has a vertical tangent at x
Copyright
h
0
f (0 h ) f (0)
h
lim
h
0
0.
2014 Pearson Education, Inc.
|h| 0
h
lim
h
0
|h|
|h|
0;
Section 3.2 The Derivative as a Function
48. (a) The graph appears to have a cusp at x
(b)
f (4 h ) f (4)
h
lim
h
0
lim
h
49-52.
0
| h|
| h|
lim
h
lim
h
0
0
1
|h|
4.
|4 (4 h)| 0
h
y
lim
h
0
|h|
h
lim 1
h
0
; lim
h
h
0
f (4 h ) f (4)
h
4 x does not have a vertical tangent at x
lim
h
0
4.
Example CAS commands:
Maple:
f : x - x^3 2*x;x0 : 0;
plot( f (x), x x0-1/2..x0 3, color black,
# part (a)
title "Section 3.1, #49(a)" );
q : unapply( (f (x0 h)-f (x0))/h, h );
# part (b)
L : limit( q(h), h 0 );
sec_lines : seq( f(x0) q(h)*(x-x0), h 1..3 );
tan_ line : f(x0) L*(x-x0);
plot( [f(x),tan_line,sec_lines], x x0-1/2..x0 3, color black,
# part (c)
# part (d)
linestyle [1,2,5,6,7], title "Section 3.1, #49(d)",
legend ["y f(x)","Tangent line at x 0","Secant line (h 1)",
"Secant line (h 2)","Secant line (h 3)"] );
Mathematica: (function and value for x0 may change)
Clear[f , m, x, h]
x0 p;
f[x_ ]: Cos[x] 4Sin[2x]
Plot[f [x],{x, x0 1, x0 3}]
dq[h_ ]: (f [x0 h] f [x0])/h
m Limit[dq[h], h 0]
ytan: f [x0] m(x x0)
y1: f [x0] dq[1](x x0)
y2: f [x0] dq[2](x x0)
y3: f [x0] dq[3](x x0)
Plot[{f [x], ytan, y1, y2, y3}, {x, x0 1, x0 3}]
3.2
THE DERIVATIVE AS A FUNCTION
1. Step 1:
Step 2:
Step 3:
f ( x)
4 x 2 and f ( x h)
f ( x h) f ( x)
h
f ( x)
[4 ( x h ) ] (4 x 2 )
h
lim ( 2 x h)
h
0
4 ( x h) 2
2
(4 x 2 2 xh h 2 ) 4 x 2
h
2 x; f ( 3)
Copyright
6, f (0)
0, f (1)
2 xh h2
h
2
2014 Pearson Education, Inc.
h( 2 x h)
h
2x h
|4 (4 h)|
h
121
122
Chapter 3 Derivatives
( x 1)2 1 and F ( x h)
2. F ( x)
( x h 1)2 1
( x 2 2 xh h 2 2 x 2 h 1 1) ( x 2 2 x 1 1)
lim
h
h 0
F ( 1)
3. Step 1:
4, F (0)
2, F (2)
1
t2
and g (t h)
g (t )
1
Step 2:
g (t h ) g (t )
h
Step 3:
g (t )
1 z
2z
4. k ( z )
p(
Step 2:
6. r ( s )
3
p( )
2s h 1
0
0h
0 3
2s 1
1 ,r
3
s3 2s 2 3
2
0
2;g
t3
2
dr
ds
3
(z)
( 1)
lim
3(
1 ( z h) 1 z
2( z h )
2z
1,
2
(1)
3h
3
3
2s 1
2s 2h 1
2s 1
h
k
2
zh z h z 2 zh
lim z z 2(
z h ) zh
h 0
3
3h
3
3
3h
3
(3
h
3h ) 3
3
3h
3
3
2 3
3
p (3)
lim 2 s 2h h1
0
r (s)
1,
2
p 32
3
2 2
2s 1
h
(2 s 2h 1) (2 s 1)
lim
0h
h
3 ,
2 3
; p (1)
2 s 2h 1
2s 1
2
2s 1
2s 1
2
2 2s 1
2s 1
1 ;
2s 1
1
2
2( x h)3
h (6 x 2 6 xh 2 h2 )
lim
h
h 0
dy
dx
2( x h )3 2 x3
h
0
lim
h
lim (6 x
h
2
0
h (3s 2 3sh h2 4 s 2h )
h
0
lim
h
Copyright
h
6 xh 2h )
(( s h )3 2( s h )2 3) ( s3 2 s 2 3)
h
0
2( x3 3 x 2h 3 xh 2 h3 ) 2 x3
h
0
lim
2
6 x2
3
2
2
3
2
2
3
2
lim s 3s h 3sh h 2 s h 4 sh 2 h 3 s 2 s 3
h 0
lim
h
(1 z h) z (1 z )( z h )
2( z h ) zh
0
1
2
4
3
3
3
2t h
(t h ) 2 t 2
lim
h
h
3
3h
h ( 2t h)
2th h 2
(t h ) 2 t 2 h
(t h ) 2 t 2 h
1,g
2
3
4
3 3
h)
3
2 s 2h 1
1
2
2( x 1);
0
2, g (2)
0
1 ,k
2
h
( 1)
2( s h) 1
2
lim 3s h 3sh hh 4 sh 2h
h
(t h ) 2 t 2 h
2
lim
h 0 2s 2h 1
2s 1
2
2
3
lim 6 x h 6hxh 2h
h 0
8. r
3
3h
2 x3 and f ( x h)
f ( x)
h
3
3
3
lim
h
2h
2 s 2h 1
r (0) 1, r (1)
7. y
3h
3h
h
lim
h
t 2 (t 2 2th h 2 )
h)
h)
h
2 s 1 and r ( s h)
lim
h
3(
lim (2 x h 2)
h
t 2 ( t h )2
( t h )2 t 2
2t
t2 t2
3 and p(
h
Step 3:
1
t2
1 ( z h)
k
2( z h)
1 ;k
1
lim
2
h 0 2( z h) z 2 z
h) p( )
h
h
1
(t h ) 2
and k ( z h)
p( )
[( x h 1)2 1] [( x 1)2 1]
h
0
lim
2
lim 2 xh hh 2 h
h 0
2
h
lim 2t 2h 2
h 0 (t h ) t
h
lim
h 0 2( z h ) zh
5. Step 1:
( t h )2
F ( x)
lim (3s 2 3sh h 2
h
0
2014 Pearson Education, Inc.
4 s 2h)
3s 2
2s
Section 3.2 The Derivative as a Function
9. s
10. dv
dt
11.
(t h )
lim
h
dp
dq
h
1
t h
h
( q h )3/2 q 3/2
h
0
h
0
lim
h
0 h ( w h )2 1 w2 1
13. f ( x)
9
x
x
h
0
h
( q h )1/2
lim
f ( 3)
h
0
w2 1
( w h )2 1
[(t h )3 (t h )2 ] (t 3 t 2 )
h
0
h (3t 2 3th h 2 2t h )
lim
h
h 0
16.
dy
dx
lim
h
( x h) 3
1 ( x h)
0
x 3
1 x
h
lim (1 x h4)(1 x)
0
h
17. f ( x)
8
x 2
0
q [( q h )1/2 q1/2 ]
h
h
lim
w2 1
( w h )2 1
h
0
t2 1
t2
( x h)
( q h )1/2
w2 1
0 h ( w h ) 2 1 w2 1
f ( x h) f ( x)
h
2
lim t (t hth)t1
h 0
q
( q h )1/2 q1/2
( q h )1/2
lim
lim (2t 2 h 11)(2t 1)
0
h
1
t2
1
h ( q h )1/2
h
2w h
0 h ( w h ) 2 1 w2 1 w2 1
9
x
k ( x h) k ( x)
h
h 0
1 ; k (2)
1
16
x)2
lim
f ( x)
3
2
lim
h
and f ( x h)
1
2 x h
w
( w2 1)3/2
2
lim xx ( xxhh )9
h 0
h
2
3th h
0
2
2t h) 3t
( x h 3)(1 x ) ( x 3)(1 x h )
(1 x h )(1 x )
h
dy
; dx
x 2
8
( x h) 2
4
(3)2
2
x2 9
x2
1
9
x2
(2 x) (2 x h )
lim h(2 x )(2 x h)
0
h
2
2
3
2
lim 3t h 3th hh 2th h
h 0
ds
dt t 1
5
2
3 x x 3 x 2 3 x xh 3h
lim x h 3 x xh
h
x h)(1 x )
(1
h 0
lim h (1 x 4hh)(1 x )
0
h
4
9
f ( x h) f ( x)
h
8
( x h) 2
8h
h x h 2 x 2 x 2 x h 2
h x h 2 x 2 x 2 x h 2
8
4
4
1
; m f (6)
2
( x 2) x 2
4 4
x 2 x 2 x 2 x 2
1
1
1
4
( x 6)
y
x 3 4
y
x 7.
2
2
2
Copyright
2t ; m
1
2 x
h
0
(t 3 3t 2 h 3th2 h3 ) (t 2 2th h 2 ) t 3 t 2
h
0
0
4
(1 x )2
h
lim
lim (3t
h
lim
8
x 2
h
f ( x)
8
x 2
x h 2
x 2
x h 2
h x h 2 x 2
x 2
x h 2
8
lim
h 0 x h 2 x 2 x 2
x h 2
the equation of the tangent line at (6, 4) is
2014 Pearson Education, Inc.
q1/2
x ( x h )2 9 x x2 ( x h) 9( x h )
x( x h)h
h
x 2 xh 9 ;
x( x h)
q1/2
( w h)2 1
( w h )2 1
x
q1/2
2
( w h )2 1
w2 1
9
( x h)
h ( x 2 xh 9)
x ( x h) h
x 2 h xh 2 9h
x( x h) h
8[( x 2) ( x h 2)]
y
h
0
lim
h
h
lim
h
( x h) ( x 9 h )
and f ( x h)
lim
0 h ( w h ) 2 1 w2 1
h
and k ( x h) 2 ( 1x h )
k ( x)
2 x
h
1
lim
lim
(2
h 0 h (2 x )(2 x h) h 0 (2 x )(2 x h )
15. ds
dt
0
( w h )2 1
1
14. k ( x)
( t h )(2 t 1) t (2 t 2 h 1)
(2t 2 h 1)(2 t 1)
lim
2
2
lim hth(t h ht)t h
h 0
h
q [( q h ) q ]
h [( q h )1/2 q1/2 ]
lim
w2 1
x3 2 x 2 h xh2 9 x x3 x 2 h 9 x 9 h
x( x h) h
m
t
2t 1
h (t h) t t (t h)
(t h )t
lim
w2 1 ( w2 2 wh h 2 1)
lim
h
h
h
0
1
1
t h t
h
( q h )( q h )1/2 q q1/2
h
0
1
w2 1
( w h )2 1
t h
2( t h ) 1
lim
lim
q [( q h )1/2 q1/2 ][( q h )1/2 q1/2 ]
h [( q h )1/2 q1/2 ]
1
dz
dw
lim
h
0
ds
dt
h
h 0
h 0
2
2
2
t
t
2
ht
h
2
t
2
ht
t
h
lim
lim
(2t 2h 1)(2t 1) h
h 0
h 0 (2t 2 h 1)(2t 1) h
1)
t
(t
lim
lim
12.
t h
2(t h) 1
t and r (t h)
2t 1
(t h)(2t 1) t (2t 2 h 1)
lim (2t 2h 1)(2t 1) h
h 0
1
1
(2t 1)(2t 1) (2t 1) 2
r (t )
123
;
124
Chapter 3 Derivatives
18. g ( z )
lim
h
4 ( z h) ) 1
0h
4 z
h
0
h
4 z h
lim
h
(1
h
0
of the tangent line at (3, 2) is w
4 z
4 z h
h
h 0
1
4 z h 4 z
1 (z
2
2
lim
4 z
4 z h
lim
4 z h
1
2 4 z
3)
4 z
;m
g (3)
1
2
w
3
2
z
f (t ) 1 3t 2 and f (t h) 1 3(t h) 2 1 3t 2 6th 3h 2
19. s
(1 3t 2 6th 3h2 ) (1 3t 2 )
h
0
lim
h
lim ( 6t 3h)
6t
dy
dx
lim
h
0
f ( x) 1 1x and f ( x h) 1 x 1 h
20. y
h
lim 1
h 0 x( x h)
lim x ( x h ) h
0
h
21. r
2
4
f( )
lim 2 4
h
0 h 4
22. w
0 4
4
f ( z)
lim h
h
z
z h
h
0
h
2 4
2 4
h
4
z
0
z
f ( z ) f ( x)
z x
x
z h
z h
z
z h
z
lim
25. g ( x)
lim
26. g ( x)
g ( z ) g ( x)
z x
x
g ( z ) g ( x)
z x
x
lim
z
0
lim
4
dr
d
4
h
0
1
8
f ( z h) f ( z )
h
0
( z h) z
1 lim
0h
h
z h
z
h
0
0 h 4
h
z h
lim
z h
2 4
4
(z
z)
1
1
2 z
h
h
h
0
1
0 z h
1 lim
h
1
x h
h
lim 2 4
h
0
lim
h
2
4
h
4
1
x
lim
h
2
h)
h
1
x
1
h
) 4(4
1
) 4
dw
dz
h
1
x h
1
lim
h
4
(4
lim
z
1
z 2
1
x 2
( x 2) ( z 2)
x
lim
z
x
x
x 1
z x
x
lim
z
z
z 1
(1
z ) (1
z x
z
lim ( z x)( zx 2)(
x 2)
x
lim
z x ( z x )( z 2)( x 2)
z x
f ( z) f ( x)
( z 2 3 z 4) ( x 2 3 x 4)
lim
z
x
z x
z x
z x
( z x) ( z x) 3
lim
lim ( z x) 3 2 x 3
z x
z x
z x
24. f ( x)
z
0
h
h) f ( )
h
4(4
2
) 2 4
( z h)
z h
h
lim 1
h
h
0 2h 4
h
1
f(
lim
f (t h ) f (t )
h
0
lim
6
f ( x h) f ( x )
h
0
lim
h
(4
h
z and f ( z h)
lim
z
4
dr
d
ds
dt
z
5
4
dw
dz z 4
23. f ( x)
h
2
lim
h
2 4
h)
h
1
3
3
2
4 ( h)
2 4
h
and f (
2 4
4
dy
dx x
1
x2
ds
dt t
(4 z h ) (4 z )
lim
4 z
h 0h 4 z h 4 z
1
1
the equation
2
2 4 3
1 z 7.
2 w
2
2
2
2
lim z 3zz xx 3 x
z x
z ( x 1) x ( z 1)
lim ( z x)( z 1)( x 1)
x
z
x)
lim zz x x
z x
lim ( z 2)(1x 2)
x
2
2
lim z x z 3x z 3x
z x
x
lim ( z x )( zz 1)(
x 1)
x
z
z
z
x
x
z x
x ( z x )( z
lim
z
1
( x 2)2
z
z
lim
z
x
( z x )( z x ) 3( z x )
z x
lim ( z 1)(1x 1)
x
1
( x 1)2
z
x)
lim
z
x
1
z
x
1
2 x
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x
the slope is always increasing which matches (b).
positive
0), then
28. Note that the slope of the tangent line is never negative. For x negative, f 2 ( x) is positive but decreasing as
x increases. When x 0, the slope of the tangent line to x is 0. For x 0, f 2 ( x) is positive and increasing. This
graph matches (a).
Copyright
2014 Pearson Education, Inc.
Section 3.2 The Derivative as a Function
125
29. f3 ( x) is an oscillating function like the cosine. Everywhere that the graph of f3 has a horizontal tangent we
expect f3 to be zero, and (d) matches this condition.
30. The graph matches with (c).
31. (a) f is not defined at x
example, lim
x
0
0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For
f ( x) f (0)
x 0
joining (0, 2) and (1, 2)
slope of line joining ( 4, 0) and (0, 2)
1
2
4. Since these values are not equal, f (0)
(b)
32. (a)
33.
f ( x ) f (0)
slope of line
x 0
0
f ( x ) f (0)
lim
does not exist.
x 0
x 0
but lim
x
(b) Shift the graph in (a) down 3 units
y
2
1
6
7
8
9
10
11
x
1
2
3
4
5
34. (a)
(b) The fastest is between the 20th and 30th days;
slowest is between the 40th and 50th days.
35. Answers may vary. In each case, draw a tangent line and estimate its slope.
dT 1.54 °F
dT
ii) slope 2.86
(a) i) slope 1.54
hr
dt
dt
iii) slope
0
dT
dt
0 °F
hr
iv) slope
Copyright
3.75
2014 Pearson Education, Inc.
dT
dt
2.86 °F
hr
3.75 °F
hr
126
Chapter 3 Derivatives
(b) The tangent with the steepest positive slope appears to occur at t
dT
dt
7.27
°F .
hr
12 p.m. and slope
The tangent with the steepest negative slope appears to occur at t
dT
dt
8.00
slope
6
12
7.27
6 p.m. and
8.00 °F
hr
(c)
36. Answers may vary. In each case, draw a tangent line and estimate the slope.
lb
20.83 dW
20.83 month
ii) slope
35.00
(a) i) slope
dt
dW
lb
iii) slope
6.25
6.25 month
dt
(b) The tangent with the steepest positive slope appears to occur at t
dW
lb
53.13 month
dt
dW
dt
lb
35.00 month
2.7 months. and slope
7.27
(c)
37. Left-hand derivative: For h
2
lim h h 0
h 0
lim h
h
0, f (0 h)
0
0, f (0 h)
lim 1 1; Then lim
h
x 2 curve)
h
0
0
f ( h)
f (0 h ) f (0)
h
h (using y
lim
h
0
x curve)
f (0 h ) f (0)
h
38. Left-hand derivative: When h
0, 1 h 1
f (1 h)
2
Right-hand derivative: When h
0, 1 h 1
f (1 h)
2(1 h)
lim 2hh
h
0
Then lim
h
0
lim 2
h 0
f (1 h ) f (1)
h
h
0
1 h 1
h
f (0 h ) f (0)
h
lim
f (0 h) f (0)
h
h
0
0
f (1 h ) f (1)
h
lim
h
h
the derivative f (0) does not exist.
0
2 2h
lim 2 h 2
0
h
lim
h
0
lim 0
h
f (1 h) f (1)
h
0
lim
h
2;
lim
h
0
f (1 h) f (1)
h
39. Left-hand derivative: When h
lim
lim
0;
Right-hand derivative: For h
lim h h 0
h 0
h 2 (using y
f ( h)
1 h 1
1 h 1
0,1 h 1
lim
h
(1 h) 1
0 h 1 h 1
Copyright
the derivative f (1) does not exist.
f (1 h)
lim
h
0
1 h
1
1 h 1
lim
h
0
f (1 h ) f (1)
h
1;
2
2014 Pearson Education, Inc.
lim
h
0
1 h 1
h
0
0;
(2 2 h) 2
h
Section 3.2 The Derivative as a Function
Right-hand derivative: When h
(2 h 1) 1
lim
h
h 0
h 0
f (1 h ) f (1)
Then lim
h
h 0
lim
2
h
f (1 h ) f (1)
h
0
Right-hand derivative: lim
h
h
0
f (1 h) f (1)
h
0
40. Left-hand derivative: lim
Then lim
f (1 h ) f (1)
h
41. f is not continuous at x
2(1 h) 1 2h 1
f (1 h ) f (1)
h
lim
h
0
2;
lim
h
f (1 h)
0,1 h 1
0
0
(1 h ) 1
h
lim
h
0
1
f (1 h) f (1)
h
lim
h
the derivative f (1) does not exist.
lim
h
x
0
0
1
lim
h
0
f (1 h ) f (1)
h
0 since lim f ( x)
1 h
lim 1 1;
h
h
0
1 (1 h )
1 h
lim h (1 hh)
0
h
lim 1 1h
0
1;
h
h
the derivative f (1) does not exist.
does not exist and f (0)
1
1/3
g ( h ) g (0)
1
lim h h 0 lim 2/3
;
h
h 0
h 0
h 0 h
2/3
g ( h) g (0)
1
Right-hand derivative: lim
lim h h 0 lim 1/3
;
h
h 0
h 0
h 0 h
g ( h ) g (0)
g ( h ) g (0)
Then lim
lim
the derivative g (0) does
h
h
h 0
h 0
42. Left-hand derivative: lim
not exist.
43. (a) The function is differentiable on its domain 3
(b) none
(c) none
x
2 (it is smooth)
44. (a) The function is differentiable on its domain 2
(b) none
(c) none
x
3 (it is smooth)
45. (a) The function is differentiable on 3 x 0 and 0 x 3
(b) none
(c) The function is neither continuous nor differentiable at x
0 since lim f ( x)
h
0
lim f ( x)
h
0
46. (a) f is differentiable on 2 x
1, 1 x 0, 0 x 2, and 2 x 3
(b) f is continuous but not differentiable at x
1: lim f ( x) 0 exists but there is a corner at x
lim
h
0
f ( 1 h ) f ( 1)
h
3 and lim
h
0
x
f ( 1 h) f ( 1)
h
1
3
f ( 1) does not exist
(c) f is neither continuous nor differentiable at x 0 and x 2:
at x 0, lim f ( x) 3 but lim f ( x ) 0
lim f ( x) does not exist;
at x
x
0
x
2
x
x
0
2, lim f ( x) exists but lim f ( x)
x
2
47. (a) f is differentiable on 1 x 0 and 0 x
(b) f is continuous but not differentiable at x
so f (0)
(c) none
f (0 h ) f (0)
lim
h
h 0
0
f (2)
2
0: lim f ( x)
x
0
0 exists but there is a cusp at x
does not exist
Copyright
2014 Pearson Education, Inc.
0,
1 since
127
128
Chapter 3 Derivatives
48. (a) f is differentiable on 3 x
2, 2 x 2, and 2 x 3
(b) f is continuous but not differentiable at x
2 and x 2: there are corners at those points
(c) none
49. (a) f ( x)
(b)
f ( x h) f ( x )
h
0
lim
h
( x h) 2 ( x 2 )
h
0
lim
h
2
2
2
lim x 2 xhh h x
h 0
lim ( 2 x h)
h
2x
0
(c) y
2 x is positive for x 0, y is zero when x 0, y is negative when x 0
(d) y
x 2 is increasing for
x 0 and decreasing for 0 x
; the function is increasing on intervals
where y 0 and decreasing on intervals where y 0
50. (a) f ( x)
(b)
f ( x h) f ( x)
h
0
lim
h
lim
h
0
1
x h
1
x
h
x ( x h)
lim x ( x h ) h
0
lim x( x1 h )
0
h
h
1
x2
(c) y is positive for all x 0, y is never 0, y is never negative
1 is increasing for
(d) y
x 0 and 0 x
x
51. (a) Using the alternate formula for calculating derivatives: f ( x )
( z x )( z 2 zx x 2 )
3( z x )
x
lim
(b)
z
(c) y is positive for all x
x3
3
2
x2
lim z zx
3
z
x
0, and y
is increasing for all x
(d) y
never decreasing
x2
0 when x
f ( x)
lim
z
x
z3
3
x3
3
z x
3
3
lim 3(z z xx )
x
z
x2
0; y is never negative
0 (the graph is horizontal at x
Copyright
f ( z ) f ( x)
lim
z x
z x
0 ) because y is increasing where y
2014 Pearson Education, Inc.
0; y is
Section 3.2 The Derivative as a Function
52. (a) Using the alternate form for calculating derivatives: f
4
4
z x
lim 4(
x z x)
lim
z
(b)
z
( z x )( z
x
(c) y is positive for x
x4
4
(d) y
53. y
2
3
3
xz x z x )
4( z x )
x
2
lim z xz 4
x
z
0, y is negative for x
and decreasing on
x
z4
4
lim
z
x4
4
z x
x
x3
0
0
(2( x h ) 2 13( x h ) 5) (2 x 2 13 x 5)
h
0
2
2
2
2
lim 2 x 4 xh 2 h 13 xh13h 5 2 x 13 x 5 lim 4 xh 2hh 13h
h 0
h 0
4 x 13, slope at x. The slope is 1 when 4 x 13
1 4 x 12
x 3
lim (4 x 2h 13)
h
2
0, y is zero for x
is increasing on 0
lim
h
3
f ( z ) f ( x)
( x) lim
z x
z x
2
3
x z x
x3
f ( x)
129
0
y 2 32 13 3 5
tangency is (3, 16).
54. For the curve y
16. Thus the tangent line is y 16
x , we have y
lim
h
x h
0
x
h
( 1)( x 3)
x h
x
x h
x
x 13 and the point of
( x h) x
lim
h
y
x h
0
x h
lim
h
0
1
x h
1 .
2 x
x
Suppose a, a is the point of tangency of such a line and ( 1, 0) is the point on the line where it crosses the
0
x-axis. Then the slope of the line is a a( 1)
x
a
a 1
a
1
2 a
1;
2
1
2a
2 a
a
a 1
f is
2 a
a 1. Thus such a line does exist: its point of tangency is (1, 1), its slope is
a 1
1 (x
2
and an equation of the line is y 1
55. Yes; the derivative of
which must also equal 1 ; using the derivative formula at
1)
y
f so that f ( x0 ) exists
56. Yes; the derivative of 3g is 3g so that g (7) exists
1
2
x
1.
2
f ( x0 ) exists as well.
3 g (7) exists as well.
g (t )
57. Yes, lim h(t ) can exist but it need not equal zero. For example, let g (t )
t 0
g (t )
but lim h(t )
t 0
lim mt
t
t
lim m
0
t
x 2 for 1
58. (a) Suppose | f ( x)|
lim
h
0
f (h) 0
h
0
mt and h(t )
t. Then g (0)
h(0)
0,
m, which need not be zero.
x 1. Then | f (0)| 02
f (h)
lim h . For | h | 1, h 2
0
h
f ( h)
h2
f (0)
h
0. Then f (0)
f ( h)
h
h
lim
h
f (0)
0
f (0 h ) f (0)
h
f (h)
lim h
0
h
0 by the
Sandwich Theorem for limits.
(b) Note that for x 0,
| f ( x)| | x 2 sin 1x | | x 2 ||sin 1x | | x 2 | 1
differentiable at x
0 and f (0)
x 2 (since 1 sin x 1).
x 2 (since 1 sin x 1). By part (a), f is
0.
Copyright
2014 Pearson Education, Inc.
130
Chapter 3 Derivatives
59. The graphs are shown below for h 1, 0.5, 0.1 The function y
lim x hh
0
that 1
2 x
h
smaller.
x
60. The graphs are shown below for h
3x 2
( x h)
h
0
lim
h
3
x
3
x h
h
. The graphs reveal that y
x
2,1,0.5. The function y
. The graphs reveal that y
3
( x h)
h
x
3
1
2 x
is the derivative of the function y
gets closer to y
as h gets smaller and
3x 2 is the derivative of the function y
gets closer to y
61. The graphs are the same. So we know that for
| x|
f ( x) | x |, we have f ( x) x .
62. Weierstrass s nowhere differentiable continuous function.
Copyright
1
2 x
2014 Pearson Education, Inc.
x so
x3 so that
3x 2 as h gets smaller and smaller.
Section 3.3 Differentiation Rules
63-68.
Example CAS commands:
Maple:
f : x -> x^3 x^2 - x;
x0 : 1;
plot( f(x), x x0-5..x0 2, color black,
title "Section 3.2, #63(a)" );
q : unapply( f(x h)-f(x))/h, (x,h) );
# (b)
L : limit( q(x,h), h 0 );
# (c)
m : eval( L, x x0 );
tan_line : f(x0) m*(x-x0);
plot( [f(x),tan_line], x x0-2..x0+3, color black,
linestyle [1, 7], title "Section 3.2 #63(d)",
legend ["y f(x)","Tangent line at x 1"] );
Xvals : sort( [x0 2^(-k) $ k 0..5, x0-2^(-k) $ k 0..5 ] ):
# (e)
Yvals : map( f, Xvals ):
evalf[4]( convert(Xvals,Matrix) , convert(Yvals,Matrix) >);
plot( L, x x0-5..x0 3, color black, title "Section 3.2 #63(f )" );
Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ):
Miscellaneous`RealOnly`
Clear[f, m, x, y, h]
x0 /4;
f[x_ ]: x 2 Cos[x]
Plot[f[x], {x, x0 3, x0 3}]
q[x_,h_ ]: (f[x h] f[x])/h
m[x_ ]: Limit[q[x, h], h 0]
ytan: f[x0] m[x0] (x x0)
Plot[{f[x], ytan},{x, x0 3, x0 3}]
m[x0 1]//N
m[x0 1]//N
Plot[{f[x], m[x]},{x, x0 3, x0 3}]
3.3
DIFFERENTIATION RULES
1. y
x2 3
dy
dx
dy
dx
2. y
x2
3. s
5t 3 3t 5
ds
dt
4. w
3z 7
21z 2
5.
4 x3
3
y
x 8
7 z3
x2 )
d (
dx
x 2e x
d (3)
dx
2x 1 0
d
dt
2x 0
2x 1
d2y
dx 2
dy
dx
21z 6
21z 2
4 x 2 1 2e x
Copyright
42 z
d2y
dx 2
2
dx 2
2
d (3t 5 ) 15t 2 15t 4
(5t 3 ) dt
dw
dz
d2y
2x
d 2s
dt 2
d 2w
dz 2
d
dt
d (15t 4 )
(15t 2 ) dt
126 z 5 42 z 42
8 x 2e x
2014 Pearson Education, Inc.
30t 60t 3
131
132
6.
Chapter 3 Derivatives
x3
3
y
x2
2
7. w 3 z 2
dy
dx
e x
z 1
dy
dx
10. y
4 2x x 3
dy
dx
11. r
1s 2
3
dr
ds
5s 1
2
1
12. r 12
48
24
3
5
4
3
dy
2 x 1 e x NOTE: dx e x
z 2
1
z2
d 2w
dz 2
6
z3
18 z 4
4t 3
2z 3
dr
d
d2y
3
x4
5s 2
2
2
3s3
5
2s2
2
12
4
12
d 2r
ds 2
5
4
x 1)
d ( x3
(3 x 2 ) dx
y
y
(2 x 3)(5 x 2
4 x)
(b)
y
(2 x 3)(5 x 2
4 x) 10 x3 7 x 2 12 x
d 2r
d 2
3
24
48
5
x 1) ( x3
20
6
(2 x 3)(10 x 4) (5 x 2
y
( x 2 1) ( x 5 1x )
d (3 x 2 )
x 1) dx
4 x)(2)
30 x 2 14 x 12
30 x 2 14 x 12
y
d ( x 5 1 ) ( x 5 1 ) d ( x 2 1)
( x 2 1) dx
x
x dx
( x 2 1) (1 x 2 ) ( x 5 x 1 ) (2 x) ( x 2 1 1 x 2 ) (2 x 2 10 x 2) 3 x 2 10 x 2
(b) y
x3 5 x 2
(1 x 2 )( x3/4
y
2 x 5 1x
1
x2
10 x 2
1
x2
x 3)
(1 x 2 ) 34 x 1/4 3 x 4
(b) y
x3/4
x 3
3x2
y
(a) y
x11/4
x 1
5 ; use the quotient rule:
2
6 x 15
19
2) 2
(3 x 2)2
( x3/4
y
x 3 )(2 x)
3
4x1/ 4
3
x4
3
4 x1/ 4
11 x 7/4
4
17. y
2x
3x
6x 4
(3 x
18. y
4 3 x ; use the quotient rule: u 4 3 x and v 3 x 2 x
3 x2 x
(3 x 2 x )( 3) (4 3 x )(6 x 1)
9 x 2 3 x 18 x 2 21x 4
9 x 2 24 x 4
(3 x 2 x) 2
(3 x 2 x )2
(3 x 2 x )2
u
x 2 4 ; use the quotient rule:
x 0.5
( x 0.5)(2 x ) ( x 2 4)(1)
2 x 2 x x2 4
( x 0.5)2
( x 0.5) 2
19. g ( x)
5
5
s3
2
s4
6
14. (a)
16. y
4
4
(3 x 2 ) (3x 2 1) ( x3 x 1) ( 2 x)
5 x 4 12 x 2 2 x 3
3
2
5
4
(b) y
x 4 x x 3x 3 y
5 x 12 x 2 2 x 3
15. (a) y
24
t4
12
x5
12
2
2
z3
x
2 s 4 5s 3
12
e x from Example 8(b).
12 0 30x 4 12 304
dx 2
0 12 x 5
dx 2
4
t3
d2y
x
2
18
z4
24t 4
12 x 10 10 x 3 12 x 10 103
2s 3
3
20
d 2s
dt 2
8
t3
2
t2
2 3x 4
4
(3 x 2 ) ( x3
13. (a) y
d2y
dx 2
2t 2 8t 3
ds
dt
6 x 2 10 x 5 x 2
9. y
x e x
6z 3
dw
dz
2t 1 4t 2
8. s
x2
2 x 5 and v
u
x2
3x 2
4 and v
u
x 0.5
11 x7/4
4
3
x4
1
x2
2 and v
u
1
x2
3
3 and v
u
2 x and v
x2 x 4
( x 0.5) 2
Copyright
2014 Pearson Education, Inc.
vu uv
v2
y
6x 1
y
1
g ( x)
(3 x 2)(2) (2 x 5)(3)
(3 x 2)2
vu uv
v2
vu uv
v2
Section 3.3 Differentiation Rules
t2 1
t2 t 2
20. f (t )
21. v
22. w
(t 1)(t 1)
(t 2)(t 1)
(1 t ) (1 t 2 ) 1
x 5
2x 7
(2 x )(5) (5 x 1)
5x 1
2 x
25. v
1 x 4 x
x
26. r
2 1
27. y
1
; use the
( x 2 1)( x 2 x 1)
2
2
(x
du
dx
y
x 2 3e x
2e x x
x
2
1
1
1
3/ 2
2 x3
x2
1
1/ 2
( x 2 1) ( x 2
2 x 1 2 x3
( x 2 3 x 2)(2 x 3) ( x 2 3 x 2)(2 x 3)
2
( x 1) ( x 2)
yt
2
2( e x ) e 2 x e x
x 1)
2 x2
u
0 and
4 x3 3x 2 1
2x
(2e x x )(2 x 3e x ) ( x 2 3e x )(2e x 1)
(2e
x 2 x e 3e
(2e x x ) 2
x
x)
2
6 x 2 12
( x 1)2 ( x 2)2
e x (2e2 x )
d (e x )
2e2 x from part b of Example 6 and dx
dy
dx
xe
y
x 3e x
y
x3 e x
32. w
re r
w
r e r ( 1) (1) e r
3x 2 e x
y
( x 1)2 ( x 2)2
2e x
3e3 x
(4 xe x 6e2 x 2 x 2 3 xe x ) (2 x 2 e x x 2 6e2 x 3e x )
(2e x x )2
( x 3 3 x 2 )e x
(1 r )e r
Copyright
6( x 2 2)
e x from part b of Example 8
x
y
31.
2 x 1
x2
2
x 1)(2 x)
e2 x e x
2 x
1
s ( s 1)2
2 s ( s 1)2
quotient rule: u 1 and v
2e x
y
( s 1) ( s 1)
2
y
d (e 2 x )
NOTE: dx
30.
(0) 1
x2 3x 2
x2 3x 2
e3 x
17
(2 x 7) 2
4 x3 3 x 2 1
( x 2 1)2 ( x 2 x 1)2
( x 2 1)2 ( x 2 x 1)2
2e x
1
2 s
t 2 2t 1
(1 t 2 )2
5x 1
4 x3/ 2
(1 x 4 x )
2
1)(2 x 1) ( x
29.
1
x
x2
0 1(4 x3 3 x 2 1)
y
2
x
x1
r
( x 1)( x 2)
( x 1)( x 2)
( s 1)
4x
v
28. y
1
2 s
1
(t 2)2
from Example 2 in Section 3.2
24. u
v
(1 t )
2 x 7 2 x 10
(2 x 7)2
t 2 t 1
(t 2)2
1 t 2 2t 2t 2
(1 t 2 ) 2
2 2
( s 1)2
1
2 s
s)
(t 2)2
(1 t 2 )( 1) (1 t )(2t )
dv
dt
( s 1)
(t 2)(1) (t 1)(1)
f (t )
(2 x 7) 2
f (s)
d (
ds
NOTE:
1 t
1 t2
1
(2 x 7)(1) ( x 5)(2)
w
s 1
s 1
23. f ( s )
t 1 ,t
t 2
2014 Pearson Education, Inc.
vu uv
v2
133
134
33.
Chapter 3 Derivatives
x9/4
y
e 2x
x9/4
d (e 2 x )
NOTE: dx
34.
x 3/5
y
35. s
1
z1/ 4
3/2
37.
y
7 2
38.
y
3 9.6
39. r
es
s
40. r
e
xe
/2
e
2
r
e
3
3
1 x4
2
42.
y
1 x5
120
43.
y
( x 1)( x 2)( x 3)
3 x2
2
x
1 x4
24
y
(4 x 3 3x )(2 x ) x
y
48 x 2 48 x 6
46. s
47. r
x2
(
1)(
2
3
/2
)
2
r
/2
5
t
6t
1)
1 x3
6
y
y
3.2 x 2.2
0
3
/2 1
3.2 x 2.2
y
e
2
r
e
2
3
/2 1
2
2
(
2
/2
)e
1
1
y (4)
12
y ( n)
1
y ( n)
0 for all n
2
/2
y
6x 8
10
t3
3
1
y
6
y iv
96 x 48
2x 7x 2
1 5t 1 t 2
3
1 x2
2
y
4 x3 8 x 2 3x 6 x
dy
dx
1
t2
4
6 x2 3
y
y
12 x
y (4)
y (5)
x
ds
dt
0 for n
y (n )
d2y
dx 2
0 5t 2
0 for n
2 14 x 3
2t 3
1
3
1
3
Copyright
dr
d
0 3 4
3
5
2 143
x
5t 2
4
16 x 3 24 x 2 6 x 6
y
2t 3
6
t4
1
6
4.
4 x 4 8 x3 3x 2 6 x
96
7
x2
2x
y (n )
0 for all n
x2 5x 6 x2 2 x 3
( x 2)( x 3) ( x 1)( x 3) ( x 1)( x 2)
y
7x 1
t 2 5t 1 1
t2
d 2 s 10t 3
dt 2
y
2 z 3/ 2
ex e 1
2
7 x5/ 7
y
2e1.3
2 x3 3 x 1
3x 2 8 x 1
x3 7
x
x3.2
2
y
y
45. y
ex e 1
1.4
z 2.4
w
/2 1
y
44.
x 5/7
z 3/2
2
s2
/2
2
2
x 8/5
1.4 z 2.4
2
2
x 2
3
5
e s ( s 1)
e (
41.
x2
2
7
r
2
2
r
2e 2 x
3t1/2
w
x9.6/3 2e1.3
s
1
9 x5/4
4
y
( e2 x ) 2
y
s
y
s e s e s (1)
r
0
0
z 1/2
x 2/7
2e1.3
x
3t1/2
s
xe
x
3 x 8/5
5
y
z 1.4
z
e 2 x 0 1(2e2 x )
9 x5/4
4
y
2e2 x from part b of Example 6
2t 3/2 3e2
36. w
1
e2 x
3
4
d 2r
d 2
2014 Pearson Education, Inc.
5
t2
2
t3
12
5
12
5
5
Section 3.3 Differentiation Rules
48. u
( x 2 x)( x2 x 1)
x ( x 1)( x 2 x 1)
x ( x3 1)
4
4
4
x
49. w
50. p
1 3 z (3
3z
d 2w 2 z 3
dz 2
x
1z 1
3
3
z)
0
2z
d 2u
dx 2
3
4
x
12 x
3 z 2 (2e2 z ) 6 ze 2 z
and w
dw
dz
z 2
0 1
q2 3
q2 3
2 q3 6 q
2q ( q 2 3)
1
2q
1q 1
2
dp
dq
e z ( z3
2z2
8
3
e z ( z3
z2
z 1)
z ) (3z 2
d 2w
dz 2
6 ze2 z (1 z )
dw
dz
d (e 2 z )
6e 2 z (1 4 z 2 z 2 ), NOTE: dz
e z ( z 1)( z 2 1)
z 1
z 2 1
1
z2
1q 2
2
1
2q 2
2
z3
( q 1)3 ( q 1)3
( q3 3q 2 3q 1) ( q3 3q 2 3q 1)
d2p
3
1
q
dq 2
q3
d 2w
dz 2
z
q2 3
dw
dz
1 x 3
x
x4
1
z 1 13 3 z
1 (3 z )
q2 3
51. w 3 z 2 e2 z
52. w
3x 4
0 3x 4
du
dx
x
x4 x
x4
5 12
x5
w
4 z 1) e z
135
1
6 ze2 z (1) (1 z )(6 z (2e2 z ) 6e2 z )
2e2 z from part b of Example 6
e z ( z3
z2
z 1) (3z 2
2 z 1) e z
e z ( z3 2 z 2
z)
e z ( z 3 5 z 2 3z 1)
53. u (0) 5, u (0)
3, v(0)
1, v (0) 2
d
d
(a) dx (uv) uv vu
(uv)
u (0)v (0) v (0)u (0) 5 2 ( 1)( 3) 13
dx
x 0
v (0)u (0) u (0)v (0) ( 1)( 3) (5)(2)
d
u
vu
uv
d
u
(b) dx v
7
2
2
2
dx v
(c)
(d)
d v
dx u
d (7v
dx
v
uv vu
u2
2u )
7v
(v (0))
( 1)
x 0
u (0)v (0) v (0)u (0) (5)(2) ( 1)( 3)
d v
dx u x 0
(u (0))2
(5)2
d
2u
(7v 2u ) | x 0 7v (0) 2u (0) 7
dx
54. u (1) 2, u (1) 0, v(1) 5, v (1)
1
d (uv ) |
(a) dx
u
(1)
v
(1)
v
(1)
u
(1)
x 1
v (1)u (1) u (1) v (1)
5 0 2 ( 1)
(v (1))2
(5)2
(b)
d u
dx v x 1
(c)
u (1)v (1) v (1)u (1)
d v
dx u x 1
(u (1))2
d (7v 2u ) |
x 1 7v (1) 2u
dx
(d)
2 ( 1) 5 0
2 ( 1) 5 0
(2)2
(1)
7
25
2 2( 3)
20
2
2
25
1
2
7 ( 1) 2 0
7
55. y x3 4 x 1. Note that (2, 1) is on the curve: 1 23 4(2) 1
(a) Slope of the tangent at ( x, y ) is y 3x 2 4 slope of the tangent at (2, 1) is y (2) 3(2) 2 4 8. Thus
the slope of the line perpendicular to the tangent at (2, 1) is 18
the equation of the line perpendicular to
the tangent line at (2, 1) is y 1
1 (x
8
2
2) or y
x
8
5.
4
(b) The slope of the curve at x is m 3 x 4 and the smallest value for m is 4 when x 0 and y 1.
(c) We want the slope of the curve to be 8
y 8 3 x 2 4 8 3 x 2 12 x 2 4 x
2. When
2,
x 2, y 1 and the tangent line has equation y 1 8( x 2) or y 8 x 15; When x
y
( 2)3 4( 2) 1 1, and the tangent line has equation y 1 8( x 2) or y
Copyright
2014 Pearson Education, Inc.
8 x 17.
136
Chapter 3 Derivatives
x3 3 x 2
56. (a) y
3x 2 3. For the tangent to be horizontal, we need m
y
y
0
0
3x 2 3
3x2 3 x
1. When x
1, y 0 the tangent line has equation y 0. The line perpendicular to
1. When x 1, y
4 the tangent line has equation y
4. The line
this line at ( 1, 0) is x
perpendicular to this line at (1, 4) is x 1.
(b) The smallest value of y is 3, and this occurs when x 0 and y
2. The tangent to the curve at (0, 2)
has slope 3 the line perpendicular to the tangent at (0, 2) has slope 13
y 2 13 ( x 0) or y 13 x 2
is an equation of the perpendicular line.
( x 2 1)(4) (4 x )(2 x )
dy
dx
4x
x2 1
57. y
(x
2
1)
to the curve at (0, 0) is the line y
line y 2.
8
58. y
x
2
4
4( x 2 1)
( x2 1)2
4 x. When x 1, y
( x 2 4)(0) 8(2 x )
y
4 x2 4 8 x2
( x 2 1)2
2
16 x . When x
( x 2 4) 2
1 ( x 2), or
1
2
( x 2 4)2
curve at (2, 1) has the equation y
. When x
2
y
0, y
x
2
4, so the tangent
0, so the tangent to the curve at (1, 2) is the
16(2)
2, y 1 and y
y
4(0 1)
1
0 and y
1 , so
2
(22 4)2
the tangent line to the
2.
ax 2 bx c passes through (0, 0) 0 a (0) b(0) c c 0; y ax 2 bx passes through (1, 2)
2 a b; y 2ax b and since the curve is tangent to y x at the origin, its slope is 1 at x 0
y 1
when x 0 1 2a (0) b b 1. Then a b 2 a 1. In summary a b 1 and c 0 so the curve is
y x 2 x.
59. y
60. y cx x 2 passes through (1, 0) 0 c (1) 1 c 1 the curve is y x x 2 . For this curve, y 1 2 x
and x 1 y
1. Since y x x 2 and y x 2 ax b have common tangents at x 1, y x 2 ax b must
1 21 a a
3
y x 2 3 x b. Since this last curve
also have slope 1 at x 1. Thus y 2 x a
3, b 2 and c 1 so the curves are
passes through (1, 0), we have 0 1 3 b b 2. In summary, a
2
2
y x 3 x 2 and y x x .
61. y
8x 5
m
8; f ( x)
3x 2
y
4 x 12
m
62. 8 x 2 y 1
1 (4)3
3
g (4)
63. y
2x 3
2
x 2
y 8
;
x 3
64. m
x
65. (a) y
y
m
2
x
x2
f ( x)
4 or x
3 (4) 2
2
2
4 or x
f (4)
5,
3
g ( 1)
1;y
2
x
x 2
0
f ( x)
4
f ( x)
if x
2 x; m
2
6 x 4; 6 x 4
1 x3 3 x 2 1
3
2
1 ( 1)3 3 ( 1) 2
3
2
4; g ( x)
1
m
4x
y
f ( x)
16, f (2)
x3 x
y 3x 2 1. When x
2( x 1) or y 2 x 2.
Copyright
2
2
1, y
g ( x)
y 8
x 3
4
( x 2)2
2, and if
2x
x
2
4,
2 ;
2
( x 2) 2 ( x 2)2
x 0, y 0 0 2
x2 8
x 3
3(2)2
f (2)
x 2 3 x; x 2 3 x
5
6
1
( x 2)(1) x (1)
4
4 2
4, y
8
2x
5
3
or
1,
1
2
4
0
x2 8
4
4(2)
x
4
4 or x
(2, 4)
1
5
6
( x 2) 2
(4, 2) or (0, 0).
2 x2 6 x
x2 6x 8
(4, 16) or (2, 4).
0 and y
2
the tangent line to the curve at ( 1, 0) is
2014 Pearson Education, Inc.
0
Section 3.3 Differentiation Rules
137
(b)
3
(c) yy 2xx 2x
x3 x 2 x 2
x3 3 x 2
other intersection point is (2, 6)
66. (a) y x3 6 x 2 5 x
(0, 0) is y 5 x.
(b)
( x 2)( x 1)2
3 x 2 12 x 5. When x
y
3
2
(c) yy 5x x 6 x 5 x
x3 6 x 2 5 x 5 x
x3 6 x 2
the other intersection point is (6, 30).
50
67. lim xx 11
x 1
68.
50 x 49
2/9
lim x x 1 1
1
2
9
x
50(1) 49
x 1
x 7/9
x
1
69. g ( x)
2x 3 x 0
a
x 0 , since
70. f ( x)
a
x
2bx x
1
1,
a
3
71. P ( x)
72. R
3
an x n
M 2 C2
2b
b
an 1 x n 1
M
3
2
9( 1)7 /9
C
2
M2
3.
2
2 or x
1. Since y
2(2) 2
6; the
0 and y
5
the tangent line to the curve at
x 2 ( x 6)
0
x
0 or x 6. Since y
5(6)
30,
2
9
g is differentiable at x
1
0
x
50
since f is differentiable at x
since f is continuous at x
0, y
0
lim (ax b)
x
1
a2 x 2
1 M 3,
3
a1 x a0
0
lim (2 x 3)
x
1
0
x
lim a
1
3 and lim a
x
a and lim (2bx)
x
a b and lim (bx 2 3)
x
P ( x)
where C is a constant
nan x n 1
dR
dM
0
1
1
b 3
(n 1)an 1 x n 2
CM
a
a
2b
a b
3
a
2b, and
b 3
2a2 x a1
M2
dc 0
d (u c) u dc c du u 0 c du c du . Thus when one of the functions is a
73. Let c be a constant
dx
dx
dx
dx
dx
dx
constant, the Product Rule is just the Constant Multiple Rule the Constant Multiple Rule is a special case of
the Product Rule.
Copyright
2014 Pearson Education, Inc.
138
Chapter 3 Derivatives
74. (a) We use the Quotient rule to derive the Reciprocal Rule (with u 1):
v 0 1 dv
dx
d 1
dx v
v
1 dv
dx
2
(b) Now, using the Reciprocal Rule and the Product Rule, we ll derive the Quotient Rule:
d u
dx v
d
dx
d u
dx v
75. (a)
u 1v
1 du
v dx
u dv
v du
dx
dx
v du
u dv
dx
dx
2
v2
v
d (uvw)
dx
d ((uv )
dx
uvw
uv w u vw
d (u u u u )
dx 1 2 3 4
(b)
d 1
dx v
u
d
dx
w)
(Product Rule)
(uv) dw
dx
d (uv)
dx
w
u1u2u3 u4
u1u2 u3
d
dx
u1u2u3u4
u1u2u3
du4
dx
u4 u1u2
d
dx
u1u2u3u4
u1u2u3
du4
dx
u1u2u4
d (x m )
dx
77. P
nRT
V nb
dP
dV
78.
3.4
A( q )
x m 0 1( m x m 1 )
d 1
dx x m
m xm
x2m
( x m )2
an 2 . We are holding T constant,
V2
(V nb ) 0 ( nRT )(1) V 2 (0) ( an 2 )(2V )
(V nb )
km
q
cm
2
du3
dx
du3
dx
1
(Reciprocal Rule)
uv dw
dx
du4
dx
w u dv
dx
d
dx
du
u3u1 dx2
u4
u1u3u4
h
2
uv dw
dx
wu dv
dx
wv du
dx
u1u2u3
u3u2
du2
dx
du1
dx
u2u3u4
un 1un
m x m 1 2m
(using (a) above)
du1
dx
u1u2
m x
un 2un 1un
u1u2
un
m 1
2an 2
V3
nRT
(V nb )2
(V )
(km )q 1 cm
v du
dx
and a, b, n, R are also constant so their derivatives are zero
2 2
hq
2
1 du
v dx
, the Quotient Rule.
u1u2u3u4 u1u2 u3u4 u1u2 u3u4 u1u2u3u4
d (u
(c) Generalizing (a) and (b) above, dx
1 un ) u1u2
76.
1 dv
v 2 dx
u
1 dv .
v 2 dx
v2
(km) q 2
dA
dq
q
h
2
km
q2
d 2A
dt 2
h
2
2( km) q 3
2 km
q3
THE DERIVATIVE AS A RATE OF CHANGE
1. s t 2 3t 2, 0 t
(a) displacement
(b) v
ds
dt
2t 3
2
s
s (2) s (0)
t
3. v
2
body changes direction at t
(b) v
ds
dt
6 2t
s
t
2 m, vav
2
2
d 2s
dt 2
| v(0)| | 3| 3 m/sec and | v(2)| 1 m/sec; a
a (2) 2 m/sec 2
(c) v 0 2t 3 0
2. s 6t t 2 , 0 t 6
(a) displacement
0m 2m
s
is negative in the interval 0
3.
2
s (6) s (0)
0 m, vav
s
t
0
6
t
3
2
2
a(0) 2 m/sec2 and
and v is positive when 32
t
2
the
0 m/ sec
| v(0)| |6| 6 m/ sec and | v(6)| | 6| 6 m/ sec; a
a (6)
2 m/ sec 2
(c) v 0 6 2t 0 t 3. v is positive in the interval 0 t
body changes direction at t 3.
Copyright
1 m/sec
d 2s
dt 2
2
a(0)
3 and v is negative when 3 t
2014 Pearson Education, Inc.
2 m/ sec 2 and
6
the
Section 3.4 The Derivative as a Rate of Change
3.
s
t 3 3t 2 3t , 0
(a) displacement
ds
dt
(b) v
3t
2
t 3
s s (3) s (0)
6t 3
s
t
9 m, vav
9
3
3 m/ sec
| v(0)| | 3| 3 m/ sec and | v(3)| | 12| 12 m/ sec; a
2
139
2
d 2s
dt 2
6t 6
a (0) 6 m/ sec and a (3)
12 m/ sec
2
2
(c) v 0
3t
6t 3 0 t 2t 1 0 (t 1) 2 0 t 1. For all other values of t in the interval
the velocity v is negative (the graph of v
3t 2 6t 3 is a parabola with vertex at t 1 which opens
the body never changes direction).
downward
4. s
t4
4
(a)
t3 t 2 , 0 t
s
3
9 m,
4
s (3) s (0)
(a)
25
t2
t
3
4
3
m/ sec
5
s (5) s (1)
50
t3
(c) v
(a)
5,1
t
s
(b) v
6. s
2
9
4
s
t
(b) v t 3t
2t | v (0)| 0 m/ sec and | v(3)| 6 m/sec; a 3t 2 6t 2 a(0) 2 m/ sec2 and
a (3) 11 m/ sec2
(c) v 0 t 3 3t 2 2t 0 t (t 2)(t 1) 0 t 0, 1, 2 v t (t 2)(t 1) is positive in the interval for
0 t 1 and v is negative for 1 t 2 and v is positive for 2 t 3 the body changes direction at t 1
and at t 2.
5. s
3
vav
5
| v(1)|
t2
50 5t
0
t3
0
25 ,
t 5
4 t
s
20 m, vav
(b) v
(c) v
0
5 m/ sec
45 m/sec and | v(5)|
50 5t
0
t
1
5
10
m/ sec; a
150
t4
10
t3
a(1) 140 m/ sec2 and a(5)
4
25
m/sec2
the body does not change direction in the interval
0
s (0) s ( 4)
25
( t 5) 2
20
4
| v( 4)|
25
( t 5) 2
0
20 m, vav
20
4
5 m/sec
25 m/ sec and | v(0)| 1 m/ sec; a
v is never 0
50
(t 5)3
a( 4)
50 m/ sec2 and a(0)
2
5
m/ sec 2
the body never changes direction
7. s t 3 6t 2 9t and let the positive direction be to the right on the s -axis.
(a) v 3t 2 12t 9 so that v 0 t 2 4t 3 (t 3)(t 1) 0 t 1 or 3; a 6t 12 a(1)
6 m/ sec2
and a(3) 6 m/ sec 2 . Thus the body is motionless but being accelerated left when t 1, and motionless
but being accelerated right when t 3.
(b) a 0 6t 12 0 t 2 with speed | v(2)| |12 24 9| 3 m/sec
(c) The body moves to the right or forward on 0 t 1, and to the left or backward on 1 t 2. The positions
are s (0) 0, s (1) 4 and s(2) 2 total distance | s (1) s (0)| | s (2) s (1)| |4| | 2| 6 m.
8. v t 2 4t 3 a 2t 4
(a) v 0 t 2 4t 3 0 t 1 or 3 a (1)
2 m/sec2 and a(3) 2 m/sec2
(b) v 0 (t 3) (t 1) 0 0 t 1 or t 3 and the body is moving forward; v
1 t 3 and the body is moving backward
(c) velocity increasing a 0 2t 4 0 t 2; velocity decreasing a 0
9. sm 1.86t 2 vm 3.72t and solving 3.72t 27.8
solving 22.88t 27.8 t 1.2 sec on Jupiter.
Copyright
t
7.5 sec on Mars; s j
2014 Pearson Education, Inc.
0
11.44t 2
(t 3)(t 1)
2t 4
vj
0
0
0 t
22.88t and
2
140
Chapter 3 Derivatives
10 . (a) v(t ) s (t ) 24 1.6t m/sec, and a (t ) v (t )
(b) Solve v(t ) 0 24 1.6t 0 t 15sec
(c) s (15) 24(15) .8(15)2 180 m
1.6 m/sec2
s (t )
2
(d) Solve s (t ) 90 24t .8t 2 90 t 30 15
4.39 sec going up and 25.6 sec going down
2
(e) Twice the time it took to reach its highest point or 30 sec
1
2
11. s 15t
gst 2
v 15 g s t so that v
0
15 g s t
0
gs
15 . Therefore
t
gs
15
20
3
4
0.75 m/sec2
12. Solving sm 832t 2.6t 2 0 t (832 2.6t ) 0 t 0 or 320 320 sec on the moon;
solving se 832t 16t 2 0 t (832 16t ) 0 t 0 or 52 52 sec on the earth. Also, vm 832 5.2t 0
t 160 and sm (160) 66,560 ft, the height it reaches above the moon s surface; ve 832 32t 0
t 26 and se (26) 10,816 ft, the height it reaches above the earth s surface.
13. (a) s 179 16t 2
(b) s
0
(c) When t
14. (a)
lim v
2
(b) a
dv
dt
v
179 16t 2
179 , v
16
32t
speed | v | 32t ft/sec and a
179
16
0
t
32
179
16
lim 9.8(sin )t
3.3 sec
8 179
107.0 ft/sec
9.8t so we expect v
9.8t m/sec in free fall
2
9.8 m/sec 2
15. (a) at 2 and 7 seconds
(c)
(b) between 3 and 6 seconds: 3 t
(d)
16. (a) P is moving to the left when 2 t
still when 1 t 2 or 3 t 5
(b)
17. (a)
(c)
(e)
(f)
(g)
32 ft/sec 2
3 or 5 t
6; P is moving to the right when 0 t 1; P is standing
190 ft/sec
(b) 2 sec
at 8 sec, 0 ft/sec
(d) 10.8 sec, 90 ft/sec
From t 8 until t 10.8 sec, a total of 2.8 sec
Greatest acceleration happens 2 sec after launch
v (10.8) v (2)
32 ft/sec2
From t 2 to t 10.8 sec; during this period, a
10.8 2
Copyright
6
2014 Pearson Education, Inc.
Section 3.4 The Derivative as a Rate of Change
18. (a) Forward: 0 t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t
Slows down: 0 t 1, 3 t 5, and 6 t 7
(b) Positive: 3 t 6; negative: 0 t 2 and 6 t 7; zero: 2 t 3 and 7 t 9
(c) t 0 and 2 t 3
(d) 7 t 9
19.
s
490t 2
v
980t
(a) Solving 160
490t
a
2
141
6;
980
t
4 sec. The
7
average velocity was
s (4/7) s (0)
4/7
280 cm/sec.
(b) At the 160 cm mark the balls are falling at v (4/7) 560 cm/sec. The acceleration at the 160 cm mark
was 980 cm/sec2.
17
29.75 flashes per second.
(c) The light was flashing at a rate of 4/7
20. (a)
(b)
21. C position, A velocity, and B acceleration. Neither A nor C can be the derivative of B because B s
derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes
while C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position.
Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves
B for acceleration.
22. C position, B velocity, and A acceleration. Curve C cannot be the derivative of either A or B because C
has only negative values while both A and B have some positive slopes. So, C represents position. Curve C
has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is
negative where B has negative slopes and positive where B has positive slopes.
23. (a) c (100) 11, 000
cav
2
11,000
100
$110
(b) c( x) 2000 100 x .1x
c ( x) 100 .2 x. Marginal cost c ( x) the marginal cost of producing
100 machines is c (100) $80
(c) The cost of producing the 101st machine is c (101) c(100) 100 201
$79.90
10
20000 1 1x
r ( x) 20000
, which is marginal revenue. r (100) 20000
$2.
100 2
x2
(b) r (101) $1.96.
(c) lim r ( x) lim 20000
0. The increase in revenue as the number of items increases without bound will
2
24. (a) r ( x)
x
x
x
approach zero.
Copyright
2014 Pearson Education, Inc.
142
Chapter 3 Derivatives
25. b(t ) 106 104 t 103 t 2 b (t ) 104 (2)(103 t ) 103 (10 2t )
(a) b (0) 104 bacteria/hr
(b) b (5) 0 bacteria/hr
104 bacteria/hr
(c) b (10)
26. S ( w)
1
120
27. (a) y
61
180
w
t
12
1
80 w
2
; S increases more rapidly at lower weights where the derivative is greater.
6 1 6t
t2
144
dy
dt
t
12
1
dy
(b) The largest value of dt is 0 m/h when t 12 and the fluid level is falling the slowest at that time.
dy
The smallest value of dt is 1 m/h, when t 0, and the fluid level is falling the fastest at that time.
dy
the graph of y is
(c) In this situation, dt 0
dy
always decreasing. As dt increases in value,
the slope of the graph of y increases from 1
to 0 over the interval 0 t 12.
200(30 t )2
28. Q (t )
200(900 60t t 2 )
Q (t )
200( 60 2t )
Q (10)
8, 000 gallons/min is the rate
Q (10) Q (0)
10
10, 000 gallons/min is the average rate the water
the water is running at the end of 10 min. Then
flows during the first 10 min. The negative signs indicate water is leaving the tank.
29. s ( v ) 1.1 0.108v; s (35) 4.88, s (70) 8.66. The units of ds / dv are ft/mph; ds / dv gives, roughly, the
number of additional feet required to stop the car if its speed increases by 1 mph.
r3
4
3
30. (a) V
(b) When r
dV
dr
2, dV
dr
4 r2
4 (2) 2 16 ft 3 /ft
dV
dr r 2
16 so that when r changes by 1 unit, we expect V to change by approximately 16 .
Therefore when r changes by 0.2 units V changes by approximately (16 )(0.2)
Note that V (2.2) V (2) 11.09 ft 3 .
31. 200 km/hr
t
32. s
25, D
55 95 m/sec
10 (25)
9
v0 t 16t 2
v0
v
2
500
9
6250 m
9
m/sec, and D
v0 32t ; v
0
t
(64)(1900) 80 19 ft/sec and,
10 t 2
9
V
20 t. Thus V
9
500
9
v0
v
; 1900 v0 t 16t 2 so that t 320
32
80 19 ft 60 sec 60 min 1 mi
finally, sec
1 hr
5280 ft
1 min
20 t
9
1900
3.2
10.05 ft 3 .
500
9
t
v02
32
25sec. When
v02
64
238 mph.
33.
(a) v
(b) v
0 when t 6.25sec
0 when 0 t 6.25
body moves right (up); v
Copyright
0 when 6.25 t 12.5
2014 Pearson Education, Inc.
body moves left (down)
Section 3.4 The Derivative as a Rate of Change
143
(c) body changes direction at t 6.25 sec
(d) body speeds up on (6.25, 12.5] and slows down on [0, 6.25)
(e) The body is moving fastest at the endpoints t 0 and t 12.5 when it is traveling 200 ft/sec. It s moving
slowest at t 6.25 when the speed is 0.
(f ) When t 6.25 the body is s 625 m from the origin and farthest away.
34.
(a) v 0 when t 32 sec
(b) v 0 when 0 t 1.5 body moves left (down); v
(c) body changes direction at t 32 sec
(d) body speeds up on
3,5
2
and slows down on 0,
0 when 1.5 t
5
body moves right (up)
3
2
3
2
(e) body is moving fastest at t 5 when the speed | v(5)| 7 units/sec; it is moving slowest at t
the speed is 0
(f ) When t 5 the body is s 12 units from the origin and farthest away.
when
35.
6
(a) v
0 when t
(b) v
6
15
3
sec
15
0 when 3
t 6 3 15
body moves right (up)
6
(c) body changes direction at t
(d) body speeds up on
6
15
3
body moves left (down); v
,2
(e) The body is moving fastest at t
(f ) When t
6
15
3
15
3
0 when 0
t
6
15
3
or 6
15
t
3
4
sec
6
15
3
, 4 and slows down on 0,
0 and t
the body is at position s
Copyright
6
15
3
2,
6
15
3
.
4 when it is moving 7 units/sec and slowest at t
6.303 units and farthest from the origin.
2014 Pearson Education, Inc.
6
15
3
sec
144
Chapter 3 Derivatives
36.
6
15
(a) v
0 when t
(b) v
0 when 0 t 6 3 15 or 6
body is moving right (up)
3
6
(c) body changes direction at t
6
(d) body speeds up on
15
3
15
t
3
15
3
,2
4
body is moving left (down); v
at t
t
6
15
3
6
15
3
, 4 and slows down on 0,
0 and t
6
15
3
2,
6
15
3
4; it is moving slowest and stationary
3
6
15
the position is s 10.303 units and the body is farthest from the origin.
3
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
1. y
3
x
3. y
x 2 cos x
4. y
y
6. y
dy
dx
10 x 3cos x
2. y
5.
15
3
15
(f ) When t
3.5
6
sec
(e) The body is moving fastest at 7 units/sec when t
6
0 when
dy
dx
5sin x
dy
dx
x sec x tan x
dy
dx
7
ex
dy
dx
1
x2
sin x tan x
8.
cos x
sin 2 x
g ( x)
sec x
2 x
csc x cot x
10 3sin x
5cos x
x 2 sin x 2 x cos x
0
4
2 x
x sec x tan x
2
x3
f ( x)
1 cos x
sin x sin x
2
2
sin x sec2 x cos x tan x
csc x cot x
g ( x)
sec x
2 x
7
ex
d (cot x ) cot x d ( x 2 )
x 2 dx
dx
x 2 csc2 x 2 x cot x
7. f ( x)
3
x2
x 2 ( sin x) 2 x cos x
x sec x 3
x 2 cot x
d (sin x)
5 dx
3
x2
dy
dx
csc x 4 x
d (cos x )
10 3 dx
x 2 csc2 x
2
x3
sin x sec2 x
(cot x)(2 x )
sin x
cos x cos
x
2
x3
sin x (sec2 x 1)
csc x( csc 2 x) ( csc x cot x) cot x
csc3 x csc x cot 2 x
csc x(csc x cot x )
9.
y
xe x sec x
dy
dx
d
dx
( x )e x sec x x dxd ( e x )sec x xe
x d
dx
(sec x )
e x sec x xe x sec x xe x sec x tan x
e x sec x (1 x x tan x )
Copyright
2014 Pearson Education, Inc.
Section 3.5 Derivatives of Trigonometric Functions
10. y
dy
dx
d (sec x) sec x d (sin x cos x )
(sin x cos x) dx
dx
(sin x cos x ) sin x cos x sin x
(sin x cos x)(sec x tan x) (sec x )(cos x sin x)
2
cos x
(sin x cos x) sec x
sin 2 x cos x sin x cos 2 x cos x sin x
cos 2 x
Note also that y
11. y
dy
dx
cot x
1 cot x
cos x
sec2 x
1
cos 2 x
sin x sec x cos x sec x
(1 cot x )
(1 cot x )2
csc x
(1 cot x )2
dy
dx
13. y
4
cos x
1
tan x
14. y
cos x
x
x
cos x
15. y
(sec x tan x) (sec x tan x)
4sec x cot x
dy
dx
(1 cot x )( csc2 x ) (cot x )( csc 2 x )
2
d
d
(1 sin x ) dx
(cos x ) (cos x ) dx
(1 sin x )
dy
2
dx
(1 sin x )
(1 sin x )
1
1 sin x
(1 sin x ) 2
sin x 1
(1 sin x )2
sec2 x.
2
(1 cot x )2
cos x
1 sin x
dy
dx
tan x 1
d
d
(1 cot x ) dx
(cot x ) (cot x ) dx
(1 cot x )
csc 2 x csc 2 x cot x csc 2 x cot x
12. y
sin x sin 2 x cos 2 x
(1 sin x )2
(1 sin x )( sin x ) (cos x )(cos x )
(1 sin x )
2
4sec x tan x csc 2 x
x ( sin x ) (cos x )(1)
(cos x )(1) x ( sin x )
x2
cos 2 x
dy
dx
x sin x cos x
x2
cos x x sin x
cos 2 x
d (sec x tan x) (sec x tan x ) d (sec x tan x )
(sec x tan x ) dx
dx
(sec x tan x)(sec x tan x sec2 x) (sec x tan x ) (sec x tan x sec2 x)
(sec 2 x tan x sec x tan 2 x sec3 x sec 2 x tan x) (sec 2 x tan x sec x tan 2 x sec3 x tan x sec2 x )
Note also that y
16. y
2
2
2
sec x tan x
dy
dx
2
(tan x 1) tan x 1
dy
dx
x 2 cos x 2 x sin x 2 cos x
x 2 sin x 2 x cos x 2 x cos x
0.
( x 2 ( sin x) (cos x)(2 x)) (2 x cos x (sin x)(2)) 2( sin x)
2sin x 2 sin x
x 2 sin x
17. f ( x) x3 sin x cos x
f ( x) x3 sin x( sin x ) x3 cos x(cos x) 3x 2 sin x cos x
3
2
3
2
x sin x x cos x 3 x 2 sin x cos x
18. g ( x) (2 x) tan 2 x
g ( x) (2 x) (2 tan x sec2 x) ( 1) tan 2 x
2
2(2 x) tan x(sec x tan x)
sec 2 t e t
19. s
tan t e t
ds
dt
21. s
1 csc t
1 csc t
ds
dt
(1 csc t )( csc t cot t ) (1 csc t )(csc t cot t )
22. s
sin t
1 cos t
ds
dt
23. r
4
24. r
145
2
sin
sin
cos
(1 csc t )
(1 cos t )(cos t ) (sin t )(sin t )
(1 cos t ) 2
2 d
d
dr
d
dr
d
(sin )(1)) sin
Copyright
ds
dt
2t sec t tan t 5et
csc t cot t csc 2 t cot t csc t cot t csc 2 t cot t
(1 csc t ) 2
cos t cos 2 t sin 2 t
(1 cos t ) 2
(sin ) (sin )(2 )
( cos
t 2 sec t 5et
20. s
2
2(2 x) tan x sec2 x tan 2 x
(
cos t 1
(1 cos t ) 2
2
cos
1
1 cos t
2 sin )
cos
2014 Pearson Education, Inc.
2 csc t cot t
(1 csc t ) 2
1
cos t 1
( cos
2sin )
0.
146
Chapter 3 Derivatives
25. r
dr
d
sec csc
1
sin 2
sec2
1
cos 2
(csc )(sec tan )
1
cos
(sin ) (sec tan )
(cos
(sec )( csc cot )
csc2
dr
d
26. r
(1 sec ) sin
27. p
5
28. p
(1 csc q) cos q
dp
dq
29. p
sin q cos q
cos q
(cos q )(cos q sin q ) (sin q cos q )( sin q )
30. p
tan q
1 tan q
31. p
q sin q
1
cot q
(1 sec ) cos
dp
dq
5 tan q
dp
dq
dp
dq
1) tan 2
(1 csc q )( sin q ) (cos q)( csc q cot q)
cos
( sin q 1) cot 2 q
cos 2 q cos q sin q sin 2 q cos q sin q
2
cos 2 q
cos q
(1 tan q )(sec 2 q ) (tan q )(sec2 q )
sec 2 q tan q sec 2 q tan q sec2 q
sec 2 q
(1 tan q )2
(1 tan q )2
(1 tan q ) 2
q 1
q3 cos q q 2 sin q q cos q sin q
1
sin
1
cos
sin
cos
sec 2
sec 2 q
( q 2 1)( q cos q sin q (1)) ( q sin q )(2 q )
dp
dq
2
cos
sin
1
sin
(q
2
1)
sin q csc2 q
1
cos 2 q
sec2 q
q3 cos q q 2 sin q q cos q sin q 2 q 2 sin q
2
( q 2 1) 2
( q 2 1) 2
3q tan q
q sec q
32. p
dp
dq
( q sec q )(3 sec2 q ) (3q
tan q )( q sec q tan q
sec q (1))
( q sec q ) 2
3q sec q q sec3 q (3q 2 sec q tan q 3q sec q q sec q tan 2 q sec q tan q )
3
q sec q
2
3q sec q tan q
( q sec q ) 2
q sec q tan 2 q
sec q tan q
( q sec q ) 2
csc x
y
csc x cot x
y
((csc x )( csc2 x ) (cot x)( csc x cot x)) csc3 x csc x cot 2 x
(csc x)(csc 2 x cot 2 x) (csc x)(csc 2 x csc2 x 1) 2 csc3 x csc x
(b) y sec x
y sec x tan x
y (sec x)(sec2 x ) (tan x )(sec x tan x) sec3 x sec x tan 2 x
(sec x )(sec2 x tan 2 x) (sec x)(sec2 x sec 2 x 1) 2sec3 x sec x
33. (a)
y
34. (a) y
(b) y
2 sin x
y
9 cos x
y
2 cos x
y
9sin x
y
2( sin x )
9 cos x
y
2sin x
y
9( sin x)
2 cos x
9sin x
35. y sin x
y cos x slope of tangent at x
is
y ( ) cos ( )
1; slope of tangent at x 0 is
y (0) cos (0) 1; and slope of tangent at x 32 is
y ( 32 )
cos 32
0. The tangent at (
, 0) is
y 0
1( x
), or y
x
; the tangent at (0, 0) is
y 0 1 ( x 0), or y x; and the tangent at
3 , 1 is y
1.
2
Copyright
2014 Pearson Education, Inc.
y (4)
y
(4)
2 sin x
9 cos x
Section 3.5 Derivatives of Trigonometric Functions
36. y
tan x
sec
sec2 x
y
2
slope of tangent at x
3
, tan
3
3
3
tangent at (0, 0) is y
, tan
3
37. y
x
3
3
sec x
y
is sec
3
at x
4
3
4
tan
38. y 1 cos x
sin
4
y
4
3
2
3
2
3
2
3
4 x
3
; the
3
.
slope of tangent at
2 3; slope of tangent
2. The tangent at the point
, sec
sin x
4
2 3 x
4
; the
3
, 2 is
slope of tangent at x
of tangent at x
3
, 1 cos
3
2
is sin
3
3
,
3
is
3
2
3
2
1.
; the tangent at the point
3
2
3
2
x sin x
y
,1 cos
39. Yes, y
x
4 x
, 2 is y 2
The tangent at the point
is y
3
.
3
; slope
2
3
4. The tangent
3
3
4
3
2 x
0 is sec (0) 1; and
2
3 is y
, 3 is y
3
2
is
x; and the tangent at
tangent at the point
y
,
sec x tan x
tan
3
3
is sec
, sec
is sec
3
2
4; slope of tangent at x
3
at
slope of tangent at x
, 1 is y 1
x
3
2
1 cos x; horizontal tangent occurs where 1 cos x
0
40. No, y 2 x sin x
y 2 cos x; horizontal tangent occurs where 2 cos x
2.
no x-values for which cos x
41. No, y x cot x
y 1 csc 2 x; horizontal tangent occurs where 1 csc2 x
x-values for which csc2 x
1.
42. Yes, y
1
2
cos x
0
0
x 2 cos x
y 1 2 sin x; horizontal tangent occurs where 1 2 sin x
sin x
x 6 or x 56
43. We want all points on the curve where the tangent
line has slope 2. Thus, y tan x
y sec 2 x so that
2
y 2 sec x 2 sec x
2
x
. Then the
4
tangent line at
tangent line at
4
, 1 has equation y 1 2 x
4
147
, 1 has equation y 1 2 x
Copyright
4
; the
4
.
2014 Pearson Education, Inc.
1
cos x
csc2 x
0
x
2. But there are
1. But there are no
1 2sin x
148
Chapter 3 Derivatives
44. We want all points on the curve y cot x where the tangent
y
csc2 x so that
line has slope 1. Thus y cot x
2
2
y
1
csc x
1 csc x 1 csc x
1 x 2.
The tangent line at
45. y
2
, 0 is y
4 cot x 2 csc x
(a) When x
2
x
2
.
csc2 x 2 csc x cot x
y
, then y
1; the tangent line is y
x
(b) To find the location of the horizontal tangent set y
then y
46. y 1
4
4
0
2 csc x cot x csc 2 x
y
, then y
4; the tangent line is y
3
4
When x
48.
x
49.
lim
sin
1
2
sin
50. lim
d
d
6
tan
4
d
d
1
4
lim sec e x
x
0
1
2
1
2
sin 0
3
radians. When x
4.
0
2 cos x 1 0
x
3
4
radians.
0
1 cos( csc(
0
lim cos
0
sin t
t
sin
cos
6
))
1 cos(
cos
6
6
sec2
(tan )
( 2))
2
1
sec 1
tan 0
tan 0 2 sec 0
sin
sin t
tan 1 lim t
t 0
cos
lim
0 sin
tan
4 sec0
sin
1
sec
cos
1
1
tan
1
2
tan (1 1)
cos
2
4
4
4 sec x
3
2
6
sec2
4
tan x
tan x 2 sec x
53. lim tan 1
t
x
2 cos x 1
sin x
2 is the horizontal tangent.
(sin )
tan
0
52. lim sin
54.
1
2
, then y
1 cos( csc x)
6
x
0
6
lim
51.
1 2 cos x
1
sin x
4x
(b) To find the location of the horizontal tangent set y
47. lim sin 1x
x 2
2.
2
3 is the horizontal tangent.
2 csc x cot x
(a) If x
1 2 cos x
sin x
1
sin x
0
1
lim
sin
1
0
Copyright
2014 Pearson Education, Inc.
4
sec ( )
1
3
,
Section 3.5 Derivatives of Trigonometric Functions
55. s
2 2 sin t
v
ds
dt
2 cos t
2 m/sec; speed | v
56. s
sin t cos t
v
58.
0
0
0
0
0
dv
dt
a
sin 3 x
3x
sin 3 x
3x
b and lim g ( x)
x
2 cos t. Therefore, velocity
2
a
2 m/sec ; jerk
4
sin t cos t
a
da
dt
j
j
0
j
0
0 m/sec3 .
4
lim f ( x)
x
f (0)
0
lim cos x 1 so that g is continuous at x
x
0
4
2 m/sec3 .
4
2 m/sec 2 ; jerk
4
v
cos t sin t. Therefore velocity
9 so that f is continuous at x
b 1. Now g is not differentiable at x
0
da
dt
j
0 m/sec; acceleration
4
lim 9
x
lim ( x b)
x
lim g ( x )
x
v
2 sin t
2 m/sec; acceleration
cos t sin t
sin 2 3 x
x2
lim
x
lim g ( x)
x
ds
dt
0 m/sec; speed
4
57. lim f ( x)
x
v
|
4
dv
dt
a
149
0
9
lim g ( x)
x
0
d ( x b)|
0, the left-hand derivative is dx
x
0: At x
c.
0
1,
d (cos x )|
but the right-hand derivative is dx
sin 0 0. The left- and right-hand derivatives can never agree
x 0
at x 0, so g is not differentiable at x 0 for any value of b (including b 1).
999
59. d 999 (cos x )
4
sin x because d 4 (cos x )
dx
cos x
dx
the derivative of cos x any number of times that is a
multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999
3
d
dx3
(cos x )
sec x
1
cos x
(b) y
csc x
1
sin x
cot x
cos x
sin x
dy
dx
dy
dx
dy
dx
(cos x )(0) (1)( sin x )
sin x
cos 2 x
cos x
sin 2 x
(sin x )2
(sin x )( sin x ) (cos x )(cos x )
x 10 cos(0) 10 cm; t
(b) t
0
v
62. (a) t
t
(b) t
0
x 3cos(0) 4sin(0) 3 ft; t 2
x 3cos( ) 4 sin( )
3 ft
ft ; t
v
3sin(0) 4 cos(0) 4 sec
t
v
10sin(0)
1
sin x
cm ; t
0 sec
x 10 cos
3
v
3
3 sin( ) 4 cos( )
ft
4 sec
2
2
sin x cos x
sin 2 x
5 cm; t
3
10sin
x
cos x
sin x
2
(sin x ) 2
0
sin x
cos x
1
cos x
(cos x ) 2
(sin x )(0) (1)(cos x)
61. (a) t
0
d 3 d 249 4
dx3 dx249 4
(cos x)
(cos x)
sin x.
60. (a) y
(c) y
d 999
dx999
249 4 3
d (sec x)
dx
csc x cot x
d (csc x )
dx
3
4
4sin
2
3sin
2
csc 2 x
1
sin 2 x
3
4
csc x cot x
csc 2 x
3
4
10sin
5 2 cm
3
4
cm
5 2 sec
4 ft;
2
4 cos
v
sec x tan x
d (cot x)
dx
x 10 cos
cm ; t
5 3 sec
3
3cos
v
sec x tan x
2
ft ;
3 sec
63.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y
and closer to the black curve y
cos x because
on the values of 1, 0.5, 0.3 and 0.1.
Copyright
d (sin x)
dx
sin( x h) sin x
lim
h
h 0
2014 Pearson Education, Inc.
sin( x h ) sin x
h
get closer
cos x. The same is true as h takes
150
Chapter 3 Derivatives
64.
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y
and closer to the black curve y
sin x because
d (cos x )
dx
takes on the values of 1, 0.5, 0.3, and 0.1.
cos( x h) cos x
lim
h
h 0
cos( x h) cos x
h
get closer
sin x. The same is true as h
65. (a)
sin( x h ) sin( x h)
The dashed curves of y
are closer to the black curve y cos x than the corresponding
2h
dashed curves in Exercise 63 illustrating that the centered difference quotient is a better approximation of
the derivative of this function.
(b)
cos( x h) cos( x h)
The dashed curves of y
are closer to the black curve y
sin x than the corresponding
2h
dashed curves in Exercise 64 illustrating that the centered difference quotient is a better approximation of
the derivative of this function.
|0 h| |0 h|
2h
0
66. lim
h
|h| |h|
0 2h
lim
x
lim 0
h
0
0
the limits of the centered difference quotient exists even though the
derivative of f ( x ) | x | does not exist at x
0.
67. y tan x
y sec2 x, so the smallest value
2
y sec x takes on is y 1 when x 0; y has no
maximum value since sec2 x has no largest value
on 2 , 2 ; y is never negative since sec2 x 1.
68. y cot x
y
csc 2 x so y has no smallest
2
value since csc x has no minimum value on
(0, ); the largest value of y is 1, when x 2 ;
the slope is never positive since the largest value
y
csc2 x takes on is 1.
Copyright
2014 Pearson Education, Inc.
Section 3.5 Derivatives of Trigonometric Functions
sin x appears to cross the y -axis at y 1, since
x
lim sinx x 1; y sinx2 x appears to cross the y -axis
x 0
y 2, since lim sinx2 x 2; y sinx4 x appears to
x 0
cross the y -axis at y 4, since lim sinx4 x 4.
x 0
69. y
at
However, none of these graphs actually cross the
y -axis since x 0 is not in the domain of the
x
sin kx k
x 0 x
sin( 3 x )
, and y
x
sin( 3 x )
3,
x
x 0
the graphs of y sinx5 x ,
sin kx
approach 5, 3, and k,
x
sin 5 x
0 x
functions. Also, lim
and lim
5, lim
y
respectively, as x 0. However, the graphs do not
actually cross the y -axis.
sin h
h
sin h
h
70. (a) h
.017452406
.017453292
.017453292
.017453292
1
0.01
0.001
0.0001
sin h
0 h
lim
h
lim
x
.99994923
1
1
1
sin h . 180
lim
h
0
180
h
180
sin h
0
180
(converting to radians)
180
lim
.h
180
sin
h
180
0
180
cos h 1
h
(b) h
1
0.01
0.001
0.0001
cos h 1
lim h
h 0
0.0001523
0.0000015
0.0000001
0
0, whether h is measured in degrees or radians.
d (sin x)
(c) In degrees, dx
lim sin x
h
0
cos h 1
h
(sin x)(0) (cos x)
sin( x h ) sin x
h
h 0
sin h
lim cos x h
h 0
lim
180
180
lim
h
(sin x cos h
0
(sin x )
cos x sin h ) sin x
h
cos h 1
sin h
lim
(cos x) lim h
h
h 0
h 0
cos x
(cos x cos h sin x sin h ) cos x
cos( x h ) cos x
lim
h
h
h 0
h 0
(cos x )(cos h 1) sin x sin h
cos h 1
sin h
lim
lim cos x
lim sin x h
h
h
h 0
h 0
h 0
cos h 1
sin h
(cos x) lim
(sin x) lim h
(cos x)(0) (sin x) 180
sin x
180
h
h 0
h 0
d (cos x )
(d) In degrees, dx
(e)
lim
d2
dx 2
(sin x)
d
cos x
dx 180
d2
dx 2
(cos x)
d
dx
180
sin x
2
180
3
sin x; d 3 (sin x)
dx
2
180
Copyright
3
cos x; d 3 (cos x)
dx
2
d
dx
180
d
dx
sin x
2
180
cos x
2014 Pearson Education, Inc.
3
180
cos x;
3
180
sin x
151
152
3.6
Chapter 3 Derivatives
THE CHAIN RULE
1. f (u )
6u 9
dy
therefore dx
f (u )
6
f ( g ( x))
6; g ( x )
6 2 x3
f ( g ( x)) g ( x)
2 x3 ;
g ( x)
2. f (u ) 2u 3
dy
therefore dx
f ( g ( x)) g ( x)
6(8 x 1)2 8
3. f (u ) sin u
dy
therefore dx
f (u ) cos u
f ( g ( x)) g ( x)
f ( g ( x)) cos(3x 1); g ( x) 3x 1
(cos(3x 1))(3) 3cos(3x 1)
4.
f (u ) cos u
dy
dx
5.
u
f (u )
sin u
therefore
f (u )
dy
dx
tan u
dy
therefore dx
8.
f (u )
f ( g ( x )) g ( x )
6. f (u )
7.
f (u )
f (u )
g ( x)
1
2 u
8x 1
g ( x)
8;
48(8 x 1)2
sin( e x ); g ( x )
f ( g ( x ))
sin( e x )( e x )
e
g ( x)
x
3;
e x ; therefore,
g ( x)
e x sin( e x )
1 ;
2 sin x
f ( g ( x))
g ( x ) sin x
cos x; therefore,
g ( x)
cos x
2 sin x
f (u )
cos u
f ( g ( x))
f g ( x)) g ( x)
cos( x cos x); g ( x )
f ( g ( x ))
sec 2 ( x 2 ); g ( x )
f ( g ( x )) g ( x ) sec2 ( x 2 )(2 x )
f (u )
x cos x
g ( x) 1 sin x;
(cos( x cos x))(1 sin x)
f (u ) sec2 u
sec u
1
x2
6(8 x 1) 2 ; g ( x)
f ( g ( x ))
sin u
f ( g ( x )) g ( x )
f (u)
dy
dx
6u 2
12 x3
1 x4
2
sec u tan u
dy
7; therefore, dx
sec 1x 7 x tan( 1x
f ( g ( x ))
1
x2
(2 x 1), y
u 5 : dx
dy
dy du
du dx
5u 4 2 10(2 x 1) 4
10. With u
(4 3x), y
u 9 : dx
dy
dy du
du dx
9u 8 ( 3)
11. With u
1 7x , y
dy
u 7 : dx
dy du
du dx
7u 8
dy
u 10: dx
dy du
du dx
10u 11
1, y
13. With u
x2
8
x 1x , y
u 4 : dx
14. With u
3x2
4 x 6, y
u1/2 : dx
dy
dy
dy du
du dx
dy du
du dx
Copyright
4u 3
2 x;
1
x
7 x ); g ( x )
7 sec( 1x 7 x ) tan( 1x 7 x )
9. With u
x
2
g ( x)
2 x sec 2 ( x 2 )
f ( g ( x )) g ( x )
12. With u
x2
27(4 3 x )8
8
1 7x
1
7
1
1
4 x
4 x
x
4
1 u 1/2
2
1
1
x2
(6 x 4)
x
2
2
4 x8
1
11
x 1x
3x 2
3x 2 4 x 6
2014 Pearson Education, Inc.
3
x
4
1
1
x2
7x
Section 3.6 The Chain Rule
15. With u
tan x, y
sec u: dx
dy
dy du
du dx
(sec u tan u )(sec2 x )
(sec(tan x) tan(tan x )) sec2 x
16. With u
1,y
x
cot u: dx
dy
dy du
du dx
( csc 2 u ) 12
csc 2
17. With u
tan x, y
u 3: dx
18. With u
cos x, y
dy
5u 4 : dx
dy
2x ,
3
21. With u = 5
22. With u
23. p
dy
dy
dy
24. q
3
2r r 2
25. s
4
3
sin 3t
26. s
sin 32 t
dy du
du dx
dy
dp
dt
(2r r 2 )1/3
4
5
cos 5t
cos 32 t
cos 32 t sin 32 t
cot ) 1
(csc
28. r
6(sec
29. y
x 2 sin 4 x x cos 2 x
tan )3/2
eu 4 12 x 1/2
dq
dr
1 (2r
3
ds
dt
cos 32 t
(csc
dr
d
7e5 7 x
d (3 t )
t ) 1/2 dt
4
3
dr
d
27. r
2 e 2 x /3
3
1 (3
2
ds
dt
20(cos 5 x )(sin x)
5e 5 x
eu ( 7)
dy du
du dx
eu : dx
(3 t )1/2
eu 32
1
x
3tan 2 x sec 2 x
( 20u 5 )( sin x )
eu ( 5)
dy du
du dx
eu : dx
x2 , y
4 x
dy du
du dx
eu : dx
7x, y
3 t
3
2
y
3u 2 sec2 x
dy du
du dx
eu : dx
19. With u = 5x, y
20. With u
dy du
du dx
1
x2
x
2
x
2x e
1 (3
2
t ) 1/2
1
2 3 t
d (2 r r 2 )
r 2 ) 2/3 dr
sin 32 t
cot ) 2 dd (csc
6 32 (sec
d 3 t
dt 2
cot )
tan )1/2 dd (sec
r 2 ) 2/3 (2 2r )
1 (2r
3
d (3t )
4 ( sin 5t ) d (5t )
cos 3t dt
5
dt
d 3 t
dt 2
4 x x2
2x
4
cos 3t
3
2
4 sin 5t
cos 32 t
csc cot csc2
(csc cot )2
tan ) 9 sec
2 2r
3(2 r r 2 )2/3
3
2
4
csc (cot
(csc
(cos 3t sin 5t )
sin 32 t
csc )
cot ) 2
tan (sec tan
dy
dx
d (sin 4 x ) sin 4 x d ( x 2 ) x d (cos 2 x ) cos 2 x d ( x )
x 2 dx
dx
dx
dx
d (sin x )) 2 x sin 4 x x ( 2 cos 3 x d (cos x )) cos 2 x
x 2 (4sin 3 x dx
dx
x 2 (4sin 3 x cos x ) 2 x sin 4 x x(( 2 cos 3 x) ( sin x )) cos 2 x
4 x 2 sin 3 x cos x 2 x sin 4 x 2 x sin x cos 3 x cos 2 x
30. y
153
1 sin 5
x
d (sin 5 x) sin 5 x d 1
x d (cos3 x ) cos3 x d ( x )
x 3x cos3 x
y 1x dx
3 dx
dx x
dx 3
x ((3cos 2 x )( sin x )) (cos3 x) 1
1 ( 5sin 6 x cos x) (sin 5 x)
1
x
3
3
x2
5 sin 6 x cos x 1 sin 5 x x cos 2 x sin x 1 cos3 x
2
x
3
x
Copyright
2014 Pearson Education, Inc.
csc
csc
cot
sec2 )
154
31.
Chapter 3 Derivatives
(5 2 x) 3
6(5 2 x)
33. y
1
1
2 x2
4
2)5 3 ( 1) 4
6 (3 x
18
32. y
2)6
1 (3 x
18
y
1 2
8 x
4
1
1
x2
2
x
2
1
2 x2
4
3
(3x 2)5
x3 (4
2
x
1
(4 x 3)
34. y
dy
dx
(2 x 5) 1 ( x 2 5 x)6
2( x 2 5 x )6
6 ( x 2 5 x )5
3
y
xe x
36.
y
(1 2 x)e 2 x
37.
y
( x2
38.
y
(9 x 2 6 x 2)e x
3
x tan 2 x
7
y
x 2 sec 1x
(4 x 3) (4 x 7)
( x 1)4
(2 x 5) 1 (6)( x 2 5 x )5 (2 x 5) ( x 2 5 x )6 ( 1)(2 x 5) 2 (2)
( x2
y
h ( x)
43. f ( )
sin
1 cos
1
x2
(2sin )(cos
2
f ( x)
cos
(1 cos )3
3
(1 x )e x 3 x 2e x
4 xe 2 x
2 x 2) e5 x /2 52
(2 x 2) e5 x /2
x sec 2 2 x
1
x
x sec 2 2 x
x sec x) 1/2 ( x (sec x tan x) (sec x) 1)
1 (7
2
2
)
d
d
(2 sin )(cos
(1 cos )3
Copyright
( x 7)
sin
1 cos
1)
d 1
dx x
2 sin
1 cos
tan 2 x
2 x sec 1x
(3( x 7)sec2 3 x 4 tan 3 x )
8
(1 cos )(cos ) (sin )( sin )
(1 cos )2
2 sin
(1 cos )2
2014 Pearson Education, Inc.
3
x sec x tan x sec x
2 7 x sec x
( x 7)3 (3( x 7) sec2 3 x 4 tan 3 x
[( x 7) ]
2 1 sin
cos
3 x 3 e5 x /2
(27 x 4 18 x3 6 x 2 18 x 6)e x
tan 2 x
4 2
sin
5 x2
2
d (tan (2 x1/2 )) tan (2 x1/2 ) d ( x ) 0
x dx
dx
( x 7) 4 (sec2 3 x 3) (tan 3 x )4( x 7)3 .1
f ( )
2
3
d sec 1
d ( x 2 ) x 2 sec 1 tan 1
x 2 dx
sec 1x dx
x
x
x
2 x sec 1x
2 x sec 1x sec 1x tan 1x
k ( x)
g ( x)
3 x 2e x
3
3
(9 x 2 6 x 2) e x (3x 2 ) (18 x 6) e x
y
7 x sec x
tan 3 x
( x 7) 4
3(4 x 3)4 ( x 1) 4 16(4 x 3)3 ( x 1) 3
(1 2 x ) e 2 x ( 2) (2) e 2 x
2 x 2)e5 x /2
42. g ( x)
2
x2
3
x e x ( 1) (1) e x
y
x 2 sec 1x tan 1x
41. f ( x)
1
2 x2
4
x2
d (2 x1/2 ) tan(2 x1/2 )
x sec 2 (2 x1/2 ) dx
40. k ( x)
d
dx
(2 x 5)2
35.
39. h( x)
ex
2
d ( x 1) ( x 1) 3 (4)(4 x 3)3 d (4 x 3)
(4 x 3) 4 ( 3)( x 1) 4 dx
dx
[ 3(4 x 3) 16( x 1)]
( x 1)4
1
2 x2
1 )2
2 x2
3
(4 x 3)4 ( 3)( x 1) 4 (1) ( x 1) 3 (4)(4 x 3)3 (4)
3
1
3
3(5 2 x) 4 ( 2) 84 2x 1
6
(5 2 x )4
dy
dx
d (3 x 2) ( 1) 4
2)5 dx
6 (3 x
18
1
x3
dy
dx
1
(4 x 3)4 ( x 1) 3
dy
dx
( x 7)5
Section 3.6 The Chain Rule
45. r
46. r
3 2t
1 sin 3t
sin( 2 ) cos(2 )
(1 sin 3t )( 2) (3 2t )(3cos 3t )
g (t )
48. q
cot( sint t )
49.
y
cos e
50.
y
3
2
cos
2(t 1) t
dq
dt
t
t 1
d
dt
dy
d
e 2 (3cos 5
t
t 1
2
d
d
t
t 1
cos
2
e
sin e
sec2 t
2
t 1
t 1
cos
2
1
t 1
t 1
t
2 t 1
t 1
t cos t sin t
t2
2
e
2
d
d
(
2
)
2 e
2
2
sin e
53. y
(1 cos 2t ) 4
54. y
1 cot 2t
d ( t 2)
2sin( t 2) cos( t 2) dt
d ( t)
(2sec t )(sec t tan t ) dt
d (sec t )
(2sec t ) dt
dy
dt
2
t 1(1) t . dtd
1
5 sin 5 )
52. y
(t tan t )10
sec2
1
2
csc2 sint t
dy
d (1 cos 2t )
4(1 cos 2t ) 5 dt
dy
dt
2
t
t 1
d sin( t 2)
sin 2 ( t 2)
2sin( t 2) dt
dt
2 sin( t 2) cos( t 2)
2
tan
1
)
(3 2 )(e 2 cos 5 ) ( 3 cos 5 )e 2 dd ( 2 ) 5(sin 5 )( 3e 2 )
2 cos 5
dy
dt
tan
tan
sec
51. y
dy
dt
tan 1 (sec
2
d sin t
dt
t
sin e
dy
d
1
tan
cos
csc2 sint t
e 2 cos 5
2
tan 1 sec
t 2
2(t 1)3/ 2
2(t 1)3/ 2
2
sec 2 1
sec
1
dq
dt
t
t 1
cos
dr
d
sec2 1
t
t 1
sin
dr
d
tan 1
sec
sec
2
55. y
2 2sin 3t 9 cos 3t 6t cos 3t
(1 sin 3t )
(1 sin 3t )2
sin( 2 )( sin 2 ) dd (2 ) cos(2 ) (cos( 2 )) dd ( 2 )
sin( 2 )( sin 2 )(2) (cos 2 )(cos ( 2 ))(2 )
2sin( 2 ) sin(2 ) 2 cos(2 ) cos( 2 )
1
47. q
1
1 sin 3t
3 2t
44. g (t )
155
3
2 1 cot 2t
d
dt
d (2t )
4(1 cos 2t ) 5 ( sin 2t ) dt
1 cot 2t
10 (t tan t )9 t sec2 t 1 tan t
2 sec 2 t tan t
2 1 cot 2t
3
csc 2 2t
8sin 2t
(1 cos 2t )5
d t
dt 2
csc 2
t
2
1 cot
t
2
10t 9 tan 9 t (t sec2 t tan t )
10t10 tan 9 t sec t 10t 9 tan10 t
56. y
(t 3/4 sin t )4/3 t 1 (sin t ) 4/3
dy
dt
t 1 43 (sin t )1/3 cos t t 2 (sin t ) 4/3
(sin t )1/3 (4t cos t 3cos t )
4(sin t )1/3 cos t
3t
(sin t )4/3
t2
3t 2
57.
y
2
ecos ( t 1)
dy
dt
2
ecos ( t 1) 2 cos( t 1) ( sin( t 1))
Copyright
2
2 sin( t 1) cos( t 1)ecos t(
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1)
3
156
58.
Chapter 3 Derivatives
y
dy
dt
(esin(t /2) )3
59. y
t2
t 3 4t
60. y
3t 4
5t 2
3
dy
dt
5
3(esin(t /2) )2 esin(t /2) cos 2t
3
t2
t 3 4t
2
(t 3 4t )(2t ) t 2 (3t 2 4)
( t 3 4t ) 2
dy
d cos (2t 5)
sin (cos (2t 5))
cos(cos (2t 5)) dt
dt
2 cos(cos(2t 5))(sin(2t 5))
62. y
cos 5sin 3t
5 sin
3
63. y
dy
dt
5sin 3t
d
dt
sin 5sin 3t
5sin 3t
1 cos 2 (7t )
64. y
1
6
65. y
(1 cos (t 2 ))1/2
2 cos
1
dy
dt
sin 5sin 3t
(3t 4)
(t 2 4)4
130(5t 2)4
26
(5t 2) 2
6
(3t 4)6
d t
dt 3
5cos 3t
1 (1
2
2
dy
dt
t
t
cos 1
t2 t
4cos
2
t
3 1 tan 4 12
4 tan 3 12t
2
t sec2 t
1 tan 4 12t
tan 3 12
12
7 1 cos 2 (7t )
cos2 (7t )]2 2 cos(7t )( sin(7t ))(7)
3 [1
6
dy
dt
1
d (1 cos (t 2 ))
cos (t 2 )) 1/2 dt
1
1 (1
2
t sin (t 2 )
2
d
dt
t
tan 12
(cos(7t ) sin(7t ))
d (t 2 )
sin(t 2 ) dt
cos (t 2 )) 1/2
1 cos (t 2 )
d
dt
t
1
t
4 cos
1
1
2 1 t
t
d
dt
1
t
t
t t t
dy
dt
67. y
tan 2 (sin 3 t )
68.
cos 4 (sec2 3t )
y
3
cos (t 2 )) 1/2 (sin (t )) 2t
1
(5t 2)6
3t 2 (t 2 4)
d (2t 5)
cos(cos (2t 5)) ( sin (2t 5)) dt
2
t
d 1 tan 4 t
3 1 tan 4 12
12
dt
dy
dt
2
t
t sec2 t
1
12 1 tan 4 12
tan 3 12
12 12
66. y 4sin
5
e3sin(t /2)
cos 3t
3
1 tan 4 12t
1 (1
2
t 4 (t 2 4t )4
6
5 53tt 42 15t 6 15t2 20
(5t 2)
61. y
3 cos t
2
2
3t 4 ( t 4 4t 2 )
3t 4
2t 4 8t 2 3t 4 4t 2
( t 3 4t ) 2
(t 3 4t ) 2
6 (5t 2) 3 (3t 4) 5
5 53tt 42
(5t 2)2
dy
dt
esin(t /2) e2sin(t /2)
3 cos t
2
2
1
2
2 tan(sin 3 t ) sec 2 (sin 3 t ) (3sin 2 t (cos t ))
dy
dt
4 cos3 sec 2 (3t )
6 tan(sin 3 t ) sec2 (sin 3 t )sin 2 t cos t
sin(sec2 (3t ) 2 sec(3t ) sec(3t ) tan(3t ) 3
24cos3 sec2 (3t ) sin sec2 (3t ) sec2 (3t ) tan(3t )
69. y
70. y
dy
dt
3t (2t 2 5)4
3t
2
1
2 3t
2
1 t
dy
dt
1 t
3
3t 4(2t 2 5)3 (4t ) 3 (2t 2 5)4
1
2 2
1 t
1
2
3t
1
2 1 t
2
1/2
1 t
2 3t
Copyright
3(2t 2 5)3 [16t 2
3 12 2
1
12 1 t 2
2
4 1 t 2
1 t
1 t
1/2
1 t 1
1 t
2014 Pearson Education, Inc.
1 (1
2
2t 2 5] 3(2t 2 5)3 (18t 2 5)
t ) 1/2 ( 1)
Section 3.6 The Chain Rule
71. y
72. y
3
1 1x
y
3
x2
21
1
x
1
x
1 x 3/2
2
1
2x
1
2
x
1
1
2 x
1)
2
1
x
1
2
1
2
2
1 1x
3
x2
6
x4
1
x 1/2
1
2
x 3/2
x 1/2 ( 2) 1
x 1 1
x
1
2
1
2x
1
1
1 csc 2 (3x
9
y
1
9 sec2 3x
1
2
d
dx
3
x2
2
6
x3
2
1 1x
1
x
1
2 d 3
1 1x
dx 2
x
1
x
6
x3
1
x
1
1 1x 1 2x
x 1/2
x 1/2
3
1
2
x 1/2 1
x
1
1
2 x
1 csc 2 (3 x
3
3sec2 3x
1
3
2
x
d csc(3 x 1))
(csc(3 x 1) dx
2 csc2 (3 x 1) cot(3x 1)
d (3 x 1))
1)( csc(3 x 1) cot(3 x 1) dx
y
1
3
1
2
1)(3)
1
x
x
x
1x 1
2
3 3
x
6
x3
1
2
x
3
y
1
x
1
2
x
3
x
9 tan 3x
74. y
75. y
1
2
1 cot (3 x
9
2 csc(3 x
3
73. y
y
1
1
x2
6
x3
1
x2
1
1
2
y
2
3 1 1x
157
1)
2
3
y
3 2 sec 3x sec 3x tan 3x
y
1
3
2sec2 3x tan 3x
x(2 x 1) 4
y x 4(2 x 1)3 (2) 1 (2 x 1)4 (2 x 1)3 (8 x (2 x 1)) (2 x 1)3 (10 x 1)
3
y (2 x 1) (10) 3(2 x 1)2 (2)(10 x 1) 2(2 x 1) 2 (5(2 x 1) 3(10 x 1)) 2(2 x 1) 2 (40 x 8)
16(2 x 1)2 (5 x 1)
76. y
x 2 ( x3 1)5
y x 2 5( x3 1) 4 (3x 2 ) 2 x( x3 1)5 x ( x3 1) 4 [15 x3 2 ( x3 1)] ( x3 1)4 (17 x 4 2 x)
3
4
y ( x 1) (68 x 3 2) 4 ( x3 1)3 (3 x 2 ) (17 x 4 2 x) 2 ( x3 1)3[( x3 1) (34 x3 1) 6 x 2 (17 x 4 2 x )]
2 x3 1
2
3
136 x 6
77.
y
ex
78.
y
sin( x 2 e x )
5x
47 x3 1
2 xe x
y
2
5
2
2 x e x (2 x ) 2e x
y
cos( x 2 e x ) ( x 2 e x
y
(Use triple product rule: D (uvw)
y
(x
2
x
x
1
2 x
g ( x)
(1 x ) 1
f ( g ( 1))
81. g ( x)
2 xe )) ( x 2
5 x
f ( g (1))
f ( g (1)) g (1)
f
1
2
g ( x)
f (5)
5
(1 x) 2 ( 1)
1
2
10
g(1)
csc 2 2
2)e x
2
2 x)e x cos( x 2 e x )
2 x)e x cos( x 2 e x ) (2 x 2)e x cos( x 2 e x )
4 x) sin( x 2 e x )
1;
2
f (u )
u5 1
f (u )
5u 4
1
(1 x )2
g ( 1)
1
2
and g ( 1)
f ( g ( 1)) g ( 1)
5;
2
4 14
1;
4
f (1)
5;
f (u )
2014 Pearson Education, Inc.
f (u ) 1 u1
f (u )
1
u2
1
cot 10u
f (u )
; therefore, ( f g ) (1) f ( g (1)) g (1)
10
5 and g (1)
Copyright
f ( g (1))
5
2
4; therefore, ( f g ) ( 1)
5
2 x
(4 x 2
uv w u vw from Exercise 75 part a in Section 3.3)
g (1) 1 and g (1)
g ( x)
( x2
x
4 x 2)e x cos( x 2 e x ) xe2 x ( x3 4 x 2
therefore, ( f g ) (1)
80. g ( x)
uvw
2 x
2 x)e ( sin( x e ) ( x e
( x2
79. g ( x)
2 x
2 xe x )
2
csc2 10u
5
10 2
10
4
10
csc2 10u
158
Chapter 3 Derivatives
82. g ( x)
x
g 14
g ( x)
1 2sec 2 u tan u
83. g ( x) 10 x 2
2u 2 2
(u 2 1)2
84. g ( x)
2
g 14
f
x 1
4
f
g ( x)
f ( g (0))
and g 14
1 2sec2 4 tan 4
4
20 x 1
f (1)
0; therefore, ( f g ) (0)
85. y
f ( g ( x)), f (3)
( 1) 5
5
86. r
sin( f (t )), f (0)
87. (a) y
( 4)(2)
f ( g ( 1)) g ( 1)
1, g (2)
5, g (2)
, f (0)
4
3
dy
dx
2 f ( x)
(b) y
f ( x) g ( x)
(c) y
f ( x) g ( x)
dy
dx
3
f ( g (0)) g (0)
g ( 1)
f (0)
g 14
u 2 1 (2) (2u )(2u )
f (u )
u2 1
2
d u 1
2 uu 11 du
u 1
f (u )
4; therefore,
dr
dt
f ( g ( x)) g ( x)
dr
dt t 0
cos( f (t )) f (t )
2 f (2)
2 13
dy
dx x 3
y
cos( f (0)) f (0)
f (3) 5
cos 3
1
2
4
2
3
f (3) g (3)
dy
dx x 3
f ( x) g ( x) g ( x ) f ( x)
f ( g (2)) g (2)
x 2
2
5
f (3) g (3) g (3) f (3)
3 5 ( 4)(2 ) 15 8
(d) y
f ( x)
g ( x)
(e) y
f ( g ( x))
(f ) y
( f ( x))1/2
(g) y
( g ( x )) 2
(h) y
(( f ( x)) 2
1 (( f (2))2
2
88. (a) y
(b)
y
g ( x) f ( x) f ( x ) g ( x )
dy
dx
dy
dx x 2
dy
dx x 2
2
[ g ( x )]
dy
dx
f ( g ( x)) g ( x)
dy
dx
dy
dx
1 ( f ( x)) 1/2
2
f ( x)
2( g ( x)) 3 g ( x)
( g ( x)) 2 )1/2
dy
dx
1 (( f
2
dy
dx
f ( x )( g ( x))3
dy
dx
(c) y
f ( x)
g ( x) 1
(d) y
f ( g ( x))
(e) y
g ( f ( x))
dy
dx
f ( g (2)) g (2)
( g ( x ) 1) 2
dy
dx
dy
dx
f ( g ( x)) g ( x)
g ( f ( x)) f ( x)
Copyright
37
6
1(
3
f (2)( 3)
1
3
f (2)
2 f (2)
dy
dx x 2
2( g (3)) 3 g (3)
3)
1
1
6 8
3
2 8
2( 4)
2
24
1
12 2
5
32
5
( g ( x))2 ) 1/2 (2 f ( x) f ( x) 2 g ( x) g ( x ))
dy
dx x 1
3(1)(1) 2 13
( g ( x ) 1) f ( x ) f ( x) g ( x)
(8)( 3)
22
[ g (2)]
1 (82
2
dy
dx x 1
dy
dx x 0
dy
dx x 0
22 ) 1/2 (2 8 13 2 2 ( 3))
5 f (1) g (1)
(1)3 (5)
5
1
3
8
3
1
( 4 1)
1
3
dy
dx x 2
5
3 17
dy
dx x 0
f ( x)(3( g ( x))2 g ( x)) ( g ( x ))3 f ( x)
3 f (0)( g (0)) 2 g (0) ( g (0))3 f (0)
1
3
2
f ( x)
2 f ( x)
dy
dx x 3
2
( x))
5 f ( x) g ( x)
(2)
g (2) f (2) f (2) g (2)
( g (2))2 ) 1/2 (2 f (2) f (2) 2 g (2) g (2))
5 f ( x) g ( x )
5
01 0
2
u 1
u 1
2; f (u )
f ( g ( 1))
y
f ( x) g ( x)
dy
dx
2u
u2 1
g 14
f
8
dy
dx x 2
2 f ( x)
f (u ) 1 2 sec u sec u tan u
5; therefore, ( f g ) 14
g (0) 1 and g (0) 1; f (u )
1 1
2
g ( x)
g ( 1) 0 and
x2
x3
(
u
1)(1)
(
u
1)(1)
2(
u
1)(2)
4(u 1)
u 1
u 1
(u 1)3
(u 1)3
(u 1)2
( f g ) ( 1)
u sec 2 u
; f (u )
6
( g (1) 1) f (1) f (1) g (1)
( g (1) 1)2
f ( g (0)) g (0)
f (1) 13
1
3
g ( f (0)) f (0)
g (1)(5)
8
3
2014 Pearson Education, Inc.
(3)
( 4 1)2
1
1
3
9
(5)
40
3
8
3
1
4
2
Section 3.6 The Chain Rule
( x11
(f ) y
dy
dx
f ( x )) 2
2( x11
2(1 3) 3 11 13
(g) y
89. ds
dt
dy
1
3
4
3
ds d
d
dt
:s
dy dx
:
dx dt
90. dt
91. With y
(a) y
x2
y
1
u
1
( x 1)
92. With y
(a) y
1 2
ds
d
sin
dy
dx
7x 5
dy
du
dy
du
7
(b) y 1
dy
dx x 0
f ( x g ( x))(1 g ( x))
ds
d
cos
dy
dx
dy
dx x 1
f ( x)
sin 32
3
2
dy
dx x 1
2x 7
1 so that ds
dt
ds d
d
dt
dy
dt
9
9 so that
dy dx
dx dt
1
3
dy
1
5
x 1
x 1
y
and x
4(0 1)
x 0
(0 1)3
x2
94. y
y
2
du
dx
y
0 1
0 1
4
4
y 1
2
y
x 7 and x
2(2) 1
x 2
95. y
(a)
3
6
2 (2)2 (2) 7
dy
dx
2
x
4
2 tan
dy
dx x 1
2
2
0
13
1
2
2sec2
sec ( 4 )
3x 2 ; therefore,
dy
dx
3u 2
1
2 x
1
2 u
3x 2
dy du
du dx
( 1)2
4( x 0)
2
y
( x 1) 1 ( x 1) 1
x 1
x 1
( x 1) 2
9
3. y
1 (x
2
y
1
2
4
2
2)
sec2
96. (a) y
y
y
2
2
dy du
du dx
3( x )2
1
2 x
3
3
2
1
2 x
3x 2
( x 1) 2
1) ( x 1)2
1
1
u 2 ( x 1) 2
x,
3 x1/2 ,
2
4( x 1)
2 (x
( x 1)3
1
2
x2
x 7
1/2
2x 1
(2 x 1)
2 x2 x 7
x 2
x
4
slope of tangent is
and that occurs when
dy
dx
4x 1
(2)2 (2) 7
y 3
x
4
1. y
; thus, y (1)
2 tan
4
2 and y (1)
y
x 2
given by y 2
( x 1)
(b) y 2 sec2 4x and the smallest value the secant function can have in 2
value of y is
3
5 1, as expected
; therefore
3 x1/2 for both (a) and (b):
2
dy du
du
1 ; therefore, dy
dx
dx
du dx
2 x
x3
;u
du
dx
1
1)2
again as expected.
93. y
15 5
1 for both (a) and (b):
x3/2 , we should get dx
dy
u3
3u 2 ; u
x
du
1
2 u
f (1))
f (1) 1 13
f (0 g (0))(1 g (0))
dy
du 5; therefore, dy
1 ; u 5 x 35
5
dx
dx
du
du
1 ; u ( x 1) 1
( x 1) 2 (1)
dx
u2
(x
2
1
1
(
x
1)
1,
again
as
expected
( x 1)2
( x 1) 2
as expected.
dy
(b) y
u
du
f (1)) 3 (11
2(1
1
3
4
9
x, we should get
u
5
32
3
2
43
dy
dx
f ( x g ( x))
f ( x)) 3 11x10
159
2
sec
2
x
4
1 sec
2
x
4
x
tangent line is
2 is 1
1 sec
x
4
the minimum
x
0.
sin 2 x
y 2 cos 2 x
y (0) 2cos(0) 2 tangent to y sin 2 x at the origin is y 2 x;
1 cos x
1 cos 0
1
sin 2x
y
y (0)
tangent to y
sin 2x at the origin is
2
2
2
2
1
2
x. The tangents are perpendicular to each other at the origin since the product of their slopes is 1.
Copyright
2014 Pearson Education, Inc.
160
Chapter 3 Derivatives
(b) y
sin(mx)
m cos(mx)
y
1 cos(0)
m
y (0)
y (0)
1 . Since
m
m cos 0
1
m
m
m; y
sin
x
m
1 cos x
m
m
y
1, the tangent lines are perpendicular at the origin.
y m cos( mx). The largest value cos(mx) can attain is 1 at x 0
(c) y sin(mx)
m cos (mx) m cos mx m 1 m . Also, y
sin
attain is | m | because y
1 cos x
m
m
y
1
m
x
m
cos
1
m
the largest value y can attain is
1
m
x
m
the largest value y can
1 cos x
y
m
m
.
(d) y sin(mx )
y m cos(mx)
y (0) m slope of curve at the origin is m. Also, sin(mx) completes m
periods on [0, 2 ]. Therefore the slope of the curve y sin(mx) at the origin is the same as the number of
periods it completes on [0, 2 ]. In particular, for large m, we can think of compressing the graph of
y sin x horizontally which gives more periods completed on [0, 2 ], but also increases the slope of the
graph at the origin.
97. s A cos(2 bt ) v ds
A sin(2 bt )(2 b)
2 bA sin(2 bt ). If we replace b with 2b to double the
dt
frequency, the velocity formula gives v
4 bA sin(4 bt ) doubling the frequency causes the velocity to
4 2b 2 A cos(2 bt ). If we replace b with 2b in the acceleration
double. Also v
2 bA sin(2 bt ) a dv
dt
formula, we get a
16 2b 2 A cos(4 bt )
2 2
4 b A cos(2 bt )
j da
Finally, a
dt
64 3b3 A sin(4 bt )
we get j
98. (a) y
37 sin
2 (x
365
doubling the frequency causes the acceleration to quadruple.
8 3b3 A sin(2 bt ). If we replace b with 2b in the jerk formula,
doubling the frequency multiplies the jerk by a factor of 8.
101)
25
y
37 cos
2 (x
365
2
365
101)
74
365
2 ( x 101)
365
2 (x
of cos 365
cos
is increasing the fastest when y is as large as possible. The largest value
2 (x
365
. The temperature
101) is l and
occurs when
101) 0 x 101 on day 101 of the year ( April 11), the temperature is
increasing the fastest.
2 (101 101)
74 cos(0) 74
cos 365
0.64 F/day
(b) y (101) 74
365
365
365
(1 4t )1/2
99. s
dv
dt
a
ds
dt
v
1
2
2(1 4t )
dv
dt
100. We need to show a
k
2 s
2
k
2
k s
4t )
(4)
1/2
(4)
2(1 4t )
4(1 4t )
3/2
dv
dt
is constant: a
1/2
v(6)
a(6)
dv dv
ds dt
and
4(1 4 6)
dv
ds
d
ds
1/2
2(1 4 6)
k s
3/2
4
125
k
2 s
2
5
2(1 4t ) 1/2
m/sec; v
m/sec 2
dv ds
ds dt
a
dv
ds
v
which is a constant.
101. v proportional to
k2 1
2 s2
1 (1
2
3/2
1
s
k
s
v
for some constant k
acceleration is a constant times
102. Let
dx
dt
f ( x). Then, a
103. T
2
L
g
dT
dL
2
dv
dt
1
2
L
g
dv dx
dx dt
1
g
g
dv
dx
L
g
1
s2
f ( x)
gL
dv
ds
k
2 s3/ 2
Thus, a
dv
dt
dv ds
ds dt
2
dv
ds
v
k
2 s3/ 2
so a is inversely proportional to s .
d dx
dx dt
Therefore,
f ( x)
dT
du
d ( f ( x))
dx
dT dL
dL du
gL
f ( x)
kL
f ( x) f ( x), as required.
k L
g
1
2
2 k
L
g
required.
104. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g (0), then f o g is
differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0
so there is no contradiction.
Copyright
k
s
2014 Pearson Education, Inc.
kT
2
, as
Section 3.6 The Chain Rule
sin 2( x h ) sin 2 x
105. As h 0, the graph of y
h
approaches the graph of y 2 cos 2 x because
sin 2( x h ) sin 2 x
d (sin 2 x) 2cos 2 x.
lim
dx
h
h
0
106. As h
0, the graph of y
approaches the graph of y
cos[( x h )2 ] cos( x 2 )
lim
h
h 0
cos[( x h)2 ] cos( x 2 )
h
2
2 x sin ( x ) because
d [cos
dx
1
2
x
d
dx
x
1
2
x
108. From the power rule, with y
dy
dx
1
2 x x
d
dx
x x
dy
dx
1
2 x x
x
1
2 x
x
2 x sin ( x 2 ).
dy
dx
3/4
1
4
dy
dx
3
4
x1/4 , we get
107. From the power rule, with y
dy
dx
( x 2 )]
1
2 x
1
4
x
x3/4 , we get
1
2 x x
3
2
x
3/4
. From the chain rule, y
, in agreement.
x
x
1/4
3 x
4 x x
. From the chain rule, y
3 x
4 x
x
3
4
x
1/4
109. (a)
(b)
df
dt
1.27324sin 2t 0.42444sin 6t
0.2546sin10t 0.18186sin14t
Copyright
x
2014 Pearson Education, Inc.
x x
, in agreement.
161
162
Chapter 3 Derivatives
df
dg
(c) The curve of y dt approximates y dt
the best when t is not , 2 , 0, 2 , nor .
110. (a)
(b)
(c)
dh
dt
2.5464cos(2t ) 2.5464cos (6t ) 2.5465cos (10t ) 2.54646cos(14t ) 2.54646cos (18t )
dh/dt
10
2
0
t
2
10
3.7
IMPLICIT DIFFERENTIATION
1. x 2 y xy 2
Step 1:
Step 2:
Step 3:
Step 4:
2. x3
3. 2 xy
y3
y2
Step 1:
Step 2:
Step 3:
Step 4:
6:
dy
x 2 dx y 2 x
dy
dy
x 2 dx 2 xy dx
dy 2
( x 2 xy)
dx
dy
2 xy y 2
dx
x 2 2 xy
2xy
2xy
3x 2 3 y 2
18 xy
x
dy
x 2 y dx
dy
dx
y2 1
0
y2
y2
dy
18 y 18 x dx
dy
(3 y 2 18 x) dx
18 y 3x 2
y:
dy
dy
dy
2 x dx 2 y 2 y dx 1 dx
dy
dy dy
2 x dx 2 y dx dx 1 2 y
dy
(2 x 2 y 1) 1 2 y
dx
dy
1 2y
dx
2x 2 y 1
Copyright
2014 Pearson Education, Inc.
dy
dx
6 y x2
y2 6x
Section 3.7 Implicit Differentiation
4. x3
xy
y3
5. x 2 ( x y )2
dy
3x 2
1
x2
Step 3:
dy
dx
Step 4:
dy
dx
6. (3xy 7) 2
2 x 2 x2 ( x
y ) 2 x( x
2 y dx
2 x [1 x( x
2 x 1 x( x y ) ( x y )2
dy
dy
dy
8. x3
2x y
x 4 3 x3 y
x 3y
2 4 x3 9 x 2 y
2
x y x
2(3 xy 7)(3x)
y (3 xy 7)
x (3 xy 7) 1
dy
dx
2
( x 1) 2
( x 1)2
6 dx
dy
dx
6 y(3xy 7)
( x 1) ( x 1)
2 y dx
x 1 x2 xy x 2 2 xy y 2
y x ( x y)
2(3 xy 7) 3x dx 3 y
7) 6]
y)2 ]
y) ( x
2
2x (x y) 2 y
6y
y )2
x 1 x( x y ) ( x y )2
2
x 1
x 1
x 2 x3 3 x 2 y xy 2
x 2 y x3 y
y
dy
6 dx
6 y (3xy 7)
1 3x2 y 7 x
1
y ( x 1)2
4 x3 9 x 2 y 3 x 3 y
2x y
3 xy
2
dy
dx
7y
3
2 y
(3x3 1) y
2 4 x3 9 x 2 y
3 x3 1
dy
dy
cot( xy )
dy
dx
x dx
0
1
x sec2 ( xy )
y
x
12. x 4 sin y
dy
dx
dy
dy
1
y
( 1)
cos
1
y
sin
y
x
2
x 1 csc ( xy )
dy
x sec2 ( xy) dx
0
1 y sec2 ( xy )
3x 2 y 2
1 dy
y 2 dx
dy
x3 2 y dx
sin
1
y
dy
dx
dy
(cos y 2 x3 y ) dx
1
y
x
dy
dx
1 y sec2 ( xy )
x sec 2 ( xy )
dy
x dx
y
dy
dx
3x 2 y 2
4 x3
1 cos 1
y
y
sin
y2
y
1
y
y csc2 ( xy ) 1
y csc2 ( xy ) y
cos2 ( xy ) y
x
y
x
4 x3 (cos y ) dx
y cos
x csc2 ( xy ) dx
x dx
dy
dx
y x dx
dy
dy
y
dy
sec 2 ( xy )
cos2 ( xy )
x
1 xy
cos2 y
y csc2 ( xy ) 1
1
x3 y 2
1
y
1
sec 2 y
csc2 ( xy ) x dx
y
x x csc2 ( xy)
11. x tan( xy )
13. y sin
dy
dx
1 (sec2 y ) dx
tan y
10. xy
3 y2 x
dy
2 x 2 y dx
y) 2 y
7. y 2
9. x
y 3 x2
dy
dx
y 3x 2
x ) dx
( x y ) 2 (2 x)
dy
y ) dx
2 x2 ( x
dy
[6 x(3xy
dx
y
dy
dx
dy
2 x2 ( x
Step 2:
dy
(3 y 2
0
y2 :
x 2 2( x y ) 1
Step 1:
dy
dx
y x dx 3 y 2
163
y sin
1
y
cos
1
y
xy
14. x cos(2 x 3 y ) y sin x
x sin(2 x 3 y )(2 3 y ) cos(2 x 3 y ) y cos x
2 x sin(2 x 3 y ) 3 xy sin(2 x 3 y ) cos(2 x 3 y ) y cos x y sin x
cos(2 x 3 y ) 2 x sin(2 x 3 y ) y cos x (sin x 3x sin(2 x 3 y )) y
cos(2 x 3 y ) 2 x sin(2 x 3 y ) y cos x
y
sin x 3 x sin(2 x 3 y )
Copyright
2014 Pearson Education, Inc.
y sin x
3 x 2 y 2 4 x3
dy
dx
1
y
cos y 2 x3 y
x
y
164
Chapter 3 Derivatives
15. e2 x
sin( x 3 y )
2
y
2
e x y ( x2 y
2x 2 y
2 2 xye x
y
2 x2 y
x e
17.
1/2
r1/2
er
1
y
22. x 2/3
dr
d
re r
csc2
2 x 2 yy
0
2 yy
x
y
y
x 1/3
2
3
1
1x
3
dx 2
y2
ex
2
2/3
2x
2
3
y 2x e
y
xe
x2
r
re
dr
d
1
2x 2
dx
y1/3 ( 13 x
2
y
2 xe x
1
2
2y y
1
y 1/ 2 1
1/ 2
(y
1/4
r cos( r )
cos( r )
( sin r ) ddr
csc2
r
re r
, cos(r )
0
e r ddr
csc2
x;
y
y 2 x2
y
y
y 1/3
d (y
dx
y 2 (1 y 2 )
1
y3
3
dx 2
y3
2
3
x1/3
)
d2y
,
now to find
x 1/3
dy
dx
y
1 y 2/3
3
x
y1/3
x1/3
2/3
y1/3
d
dx
)
1/3
x
y
1/3
x
y
y 1/3
;
x
1 x 2/3
3
1
3 y1/3 x 2/3
3 x 4/3
2
2
1/3
xe x 1
y
dy
dx
2
y 2 x 2e x
d2y
dx 2
2
ex
2
xe x
2
1 y
y2
y2 2x ex
2 x2 y2
2
x 2e 2 x
2
1
y3
2
d2y
dx 2
1 y
2y
y
y (2 y 2)
y
1)
2
1
y 1
y
( y 1) 1 ; then y
( y 1) 2 y
1
( y 1)3
y y 1/2 1
1 again to find y : y
2
dx 2
2 2 xye x
2
y 1/2 y
y
y
2y
r
dr
d
y
2/3
2 r
2
r cos(r )
sin r
2
dy 2
dx 3
0
y1/3
xe x 1
1
y
equation y y 1/2 1
d y
2
x2 e x y y
2 2y
1/2
dr
d
er
d2y
x
y
1 y 2/3 ) y
3
2/3
dr
d
r
dy
dx
2x
1 y1/3 x 4/3
3
( y 1) 2 ( y 1) 1
2
y
x
1/3
2x 1 2 y
x
2
dr
d
1
2
1/4
y2
25. 2 y
2 xye x
[ cos(r )]
er
dy
y 1/3 dx
x1/3 (
2 yy
2 x2
dr
d
since y
2
1
2 r
1/3
0
e r ddr
Differentiating again, y
d2y
1/2
dr
d
y x
dr
d
0
( sin r ) csc2
2
y 2/3
2
x2 e x y y
2 2y
1 r 1/2 dr
2
d
4 3/4
3
y ( 1) xy
y
2 xy )
cos(r ) r
sin r ddr
y2
1/2
1
2
1
20. cos r cot
24. y 2
2 e2 x
cos( x 3 y )
3y
2
1
2
19. sin(r )
23.
2e2 x
cos( x 3 y )
1 3y
2y
3 2/3
2
18. r 2
21. x 2
(1 3 y ) cos( x 3 y )
2e2 x cos( x 3 y )
3cos( x 3 y )
y
16. e x
2e2 x
1
1
2
y
dy
dx
3/2
y
y
y
1
y 1/ 2 1
1/2
1 y
3/ 2
1
2 y 3/ 2 ( y 1/ 2 1)3
Copyright
1
21
y
3
2014 Pearson Education, Inc.
y
y 1
0
; we can differentiate the
y 1/2 1 y
1 [ y ]2 y 3/2
2
Section 3.7 Implicit Differentiation
y2
26. xy
d2y
1
xy
y 2 yy
0
y
( x 2 y ) y y (1 2 y )
2 xy
2 y( x y)
( x 2 y )3
( x 2 y )3
dx
2
2y
2
27. x3
y3
2
y y
(x 2 y)
y [2 y y ]
2
d y
28. xy
y2
1
xy
1
2
y (0, 1)
29. y 2
x2
30. ( x 2
2
2x
dy
[2 y ( x 2
dx
dy
and dx
(1, 1)
( x 2 y)
3x2
x2
y2
y
y 2 yy
2 x 2 y[ y ]
0
y
y ( x 2 y)
1
2
( 2)
y
dy
dx
1
25
dy
2x
dy
dx ( 2, 1)
x 2 to find y :
2 x4
y3
2x
y
2 xy 3 2 x 4
2
y5
( x 2 y )( y ) ( y )(1 2 y )
y
dy
4 y3 dx 2
dy
1 and dx
( x 2 y )2
; since
y 2 ) 2 x 2 y dx
y2 ) (x
y)
dy
dy
2 y dx 4 y 3 dx
1
2 2x
( 2, 1)
dy
2( x 2
2 x ( x2
y )]
2x
y
xy
2 yy
0
dy
2( x
y ) 1 dx
2 x( x2 y2 ) ( x y)
dy
dx
2 y ( x2 y2 ) ( x y)
( x 2 y) y
2x
y
2 x 2 yy
0
y 4
dy
dx (1,0)
1
2 x 2 yy
0
3 (x
4
y
3)
2)
y
7
4
x 12
3)
1 (x
3
y
3
4
y
xy 2
( 1, 3)
1)
Copyright
y
3
4
x
y (3, 4)
(3, 4)
y
x 2 yy
(a) the slope of the tangent line m
(b) the normal line is y 3
2x y
;
2y x
7
3 4 (x
x;
y
y
4 (x
3
y 4
2 xy 2
y
y (2, 3) 74
the tangent line is y
4 ( x 2)
4 x 29
y
7
7
7
m
9
( x 2 y )3
1
4
2 y dx
x 1
2 y3 y
(a)
33. x 2 y 2
2
2
y
( x 2 y)
y
( 1)(0)
y ) 2 at (1, 0) and (1, 1)
y2 ) (x
y
4
(0, 1)
(b) the normal line is y 3
y2
x2
y2
2x 2 y
2y (x 2y ) 2y 2
2
; we differentiate y 2 y
(a) the slope of the tangent line m
32. x 2
2 y) y( x 2 y) 2 y2 ]
1
[ y( x
( x 2 y)
1
y2
xy
y
;
( x 2 y)
y
2
2 2x
(x
y
(x 2 y)
y
2
2
y y
we obtain y
4 y3 )
y 2 )2
31. x 2
3y 2 y
0
y ( x 2 y)
y1 2
(x 2 y)
y 4 2 x at ( 2, 1) and ( 2, 1)
dy
(2 y
dx
y
y
(x 2 y)
2
33 32
32
dx 2 (2,2)
2 yy
(x 2 y)
3x 2 3 y 2 y
16
xy
165
x 25
4
4
3
x
y
y
x ( 1, 3)
1x 8
3
3
y
;
x
3
the tangent line is y 3 3( x 1)
2014 Pearson Education, Inc.
y
3x 6
166
Chapter 3 Derivatives
34. y 2
2x 4 y 1 0
2 yy
2 4y
0
2( y 2) y
(a) the slope of the tangent line m
y ( 2, 1)
1
(b) the normal line is y 1 1( x 2)
y x 3
35. 6 x 2 3xy 2 y 2 17 y 6
0
12 x 3 y 3 xy
2
1 ;
y 2
y
the tangent line is y 1
4 yy
17 y
0
1( x 2)
y (3 x 4 y 17)
y
x 1
12 x 3 y
12 x 3 y
;
3 x 4 y 17
y
(a) the slope of the tangent line m
y
6
7
6
7
x
7 (x
6
(b) the normal line is y 0
36. x 2
3xy 2 y 2
5
y
2x
1)
3xy
37. 2 xy
sin y
2
7
6
y
3 y 4 yy
(a) the slope of the tangent line m
(b) the normal line is x
12 x 3 y
3 x 4 y 17 ( 1, 0)
( 1, 0)
y
6
7
6 (x
7
1)
x 76
0
y 4y
3x
3y 2x
4 y 3 x ( 3, 2)
( 3, 2)
the tangent line is y 0
0
3y
2x
y
3 y 2x
;
4 y 3x
the tangent line is y
2
3
2 xy
2y
(cos y ) y
(a) the slope of the tangent line m
y
1,
x
2
(b) the normal line is y
0
2x
2
y (2 x
cos y )
2y
cos y
2
1,
2y
y
2x
2y
;
cos y
the tangent line is y
2
2
( x 1)
2
y
38. x sin 2 y
y cos 2 x
2
2
( x 1)
x(cos 2 y )2 y
sin 2 y 2 y sin 2 x
2
y
sin 2 y
2 y sin 2 x
y
4 2
2
y
4
(b) the normal line is y
39. y
2sin( x
y)
y
2
1
2
2[cos( x
y 2 x 2
(b) the normal line is y 0
2 x cos 2 y
0
y
x
2
4
y )] (
1
2
y
2 x cos y sin y cos y
(a) the slope of the tangent line m
y)
,
the tangent line is
2
2
x 58
y [1 2 cos( x
2 cos( x y )
1 2 cos( x y ) (1, 0)
(1, 0)
( x 1)
2 x cos 2 y
2
1
2
y
4
x 2 (2 cos y )( sin y ) y
(b) the normal line is x
y (2 x cos 2 y cos 2 x)
2x
(a) the slope of the tangent line m
40. x 2 cos 2 y sin y
y cos 2 x
sin 2 y 2 y sin 2 x
cos 2 x 2 x cos 2 y
y
,
2 x
2
sin 2 y 2 y sin 2 x
;
cos 2 x 2 x cos 2 y
(a) the slope of the tangent line m
y
2
x
x
2
y
y )]
2
2 cos ( x
y)
y
the tangent line is y 0
2 x cos 2 y
y cos y
y [ 2 x 2 cos y sin y cos y ]
0
2 x cos 2 y
(0, )
2
2 x cos y sin y cos y (0, )
0
0
Copyright
2 ( x 1)
1
2
;
y
2 cos( x y )
;
1 2 cos( x y )
2014 Pearson Education, Inc.
the tangent line is y
Section 3.7 Implicit Differentiation
41. Solving x 2
y2
xy
7 and y
crosses the x-axis. Now x 2
2x y
x 2y
m
0
x2
7
x
y2
7
2x
y
xy
the slope at
7
xy
2 7
7
7,0 is m
7,0 and
2 yy
0
7,0 are the points where the curve
( x 2 y) y
2 and the slope at
2x
y
dy
dy
2x y
x 2y
y
2 7
7
7, 0 is m
slope is 2 in each case, the corresponding tangents must be parallel.
dy
167
2. Since the
y 2
42. xy 2 x y 0 x dx y 2 dx 0
; the slope of the line 2 x y 0 is 2. In order to be
dx
1 x
parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope
y 2
2y 4 1 x
x
3 2 y. Substituting in the original equation,
of the tangent is 12 . Therefore, 1 x 12
y ( 3 2 y ) 2( 3 2 y ) y 0
y2
y 3
y
2 x 3. If y
2( x 3)
43. y 4
y2
x2
3
, 23
4
x
y 2 y3
x
y 2 y3
is
1
2
3 1
,
4 2
44. y 2 (2 x)
4
2
45. y 4
2
2(2 y 3
2x
3
4
3 6 3
2
8
2 3
4 2
1
4
y (2 y 2 4)
3x 2
5
4
( 3)(18 9)
2(8 4)
and y (2, 4)
3 y x2
y
2
0
3x
y 2 3x 2
; the
2 y (2 x )
y
y
27 ; (
8
y (4 y 3 8 y )
0
x3/2 x
y2 3x
3 y x2
3/2
0
6 3
; the slope of the tangent line at
9 xy
9y
0
0
y
x2
3
y (3 y 2 9 x)
1 (x
2
4 x3 18 x
9 y 3x 2
y
y
4 x3 18 x
4 y3 8 y
y
27 ;(3,
8
1)
9 y 3 x2
3y
(1, 1)
3
2
x
2 x3 9 x
2 y3 4 y
27
8
2): m
2
1
2
3 y x2
y2 3x
9x
4;
5
3 y x2
y 2 3x
0
y2 3x2
2 y (2 x )
slope of the tangent line is m
27 ;(3, 2): m
8
3, 2): m
0
or x3/2
y
x2
3
x3
x3 ( x3 54) 0
x 0 or x 3 54 33 2
corresponding y -value, we will use part (c).
(c) dx
dy
x
y 2 y3
2 x 1; the normal line is y 1
4 x3 18 x
8 yy
3x 2 3 y 2 y
0
(a) y (4, 2)
0
4 y3 y
m; ( 3, 2): m
y 3 9 xy
y
3, then x 3 and
y
2 x 3.
3
2 yy (2 x ) y 2 ( 1)
x4 9 x2
2x
1. If y
2( x 1)
1; the slope of the tangent line at 43 , 12 is
1
2 3
3
4
1
2
y
3 or y
1 and y 1
y) y
the tangent line is y 1 2( x 1)
4 y2
(b) y
2 yy
3
, 23
4
3
4
2
8
x3
x (2 x 2 9)
46. x3
4 y3 y
4y 3 0
1, then x
2
9 x x3
x 6 54 x3
0
there is a horizontal tangent at x
3x ; y
0 or x3/2
3
6 3
3x
x
x3
0 or x
3x
3
3
108
9 x 3x
0
33 2. To find the
0
x3 6 3x3/2
0
33 4. Since the equation
x3 y 3 9 xy 0 is symmetric in x and y , the graph is symmetric about the line y x. That is, if ( a, b) is
1 . Thus, if the folium has
a point on the folium, then so is (b, a ). Moreover, if y ( a , b ) m, then y ( a , b) m
a horizontal tangent at (a, b), it has a vertical tangent at (b, a ) so one might expect that with a horizontal
tangent at x
3
54 and a vertical tangent at x
33 4, the points of tangency are
3 54,
33 4, 3 54 , respectively. One can check that these points do satisfy the equation x3
Copyright
2014 Pearson Education, Inc.
33 4 and
y3 9 xy
0.
168
Chapter 3 Derivatives
47. x 2
2 xy 3 y 2
0
2 x 2 xy
the tangent line m
y
(1, 1)
2 y 6 yy
x y
3 y x (1, 1)
0
1
y (2 x 6 y )
2x 2 y
x y
3y x
y
the slope of
the equation of the normal line at (1, 1) is y 1
1( x 1)
y
x 2. To find where the normal line intersects the curve we substitute into its equation:
2 x (2 x) 3(2 x)2 0
x 2 4 x 2 x 2 3(4 4 x x 2 ) 0
4 x 2 16 x 12 0 x 2 4 x 3 0
( x 3)( x 1) 0
x 3 and y
x 2
1. Therefore, the normal to the curve at (1, 1) intersects the
curve at the point (3, 1). Note that it also intersects the curve at (1, 1).
x2
48. Let p and q be integers with q
q
0 and suppose that y
d ( yq )
and assuming y is a differentiable function of x, dx
p
xp 1
q ( x p /q ) q
49. y 2
p xp 1
q x p p/q
1
dy
dx
p
. x p 1 ( p p /q )
q
p
q
y1
qy q
d (x p )
dx
x p Since p and q are integers
1 dy
dx
px p
px p
dy
dx
1
qy
1
p xp
q yq
q 1
1
1
1
y 0
1
2y
x
x1
2
x1
5
4 x 6 yy
50. 2 x 2 3 y 2
x3
(a x1 )2
1
2 yy
3x 2
x1
0
x1
2x
3y
y
3x 2
2y
y
1
2
x1
x1
a x1
x1 . For the normal to be perpendicular, x 1a
1
points on the parabola are x1 , x1 and x1 ,
y2
x p /q . Then y q
If a normal is drawn from (a, 0) to ( x1 , y1 ) on the curve its slope satisfies x1 a
2 y1
1
2 y1 ( x1 a) or a x1 12 Since x1 0 on the curve, we must have that a 12 . By symmetry, the two
x
( a x1 )
x ( p /q )
xp
y (1, 1)
2
1
4
x1
1
2
and y1
y (1, 1)
2x
3 y (1, 1)
3 x2
2 y (1, 1)
3
2
2
3
Therefore, 14 ,
3 x2
2 y (1, 1)
3.
4
and a
2x
3 y (1, 1)
and y (1, 1)
and y (1, 1)
1
2
1
2;
3
3 . Therefore
2
also,
the
tangents to the curves are perpendicular at (1, 1) and (1, 1) (i.e., the curves are orthogonal at these two points of
intersection).
51. (a) x 2 y 2 4, x 2 3 y 2
If y
1 x 2 ( 1)2
x2
y2
At
3, 1 : m1
At
At
At
4
dy
2 x 2 y dx
dy
dx
3, 1 : m1
dy
dx
3, 1 : m1
1 y2 ,
3
0
m1
3
1
dy
dx
3, 1 : m1
(b) x 1 y 2 , x
(3 y 2 ) y 2 4
y2 1
4
x2 3 x
3.
dy
dx
dy
dx
3
1
and x 2
3y2
dy
2x
6 y dx
m1 m2
3 and m2
3 and m2
dy
dx
3
3
3
1 y2
y2
2
3
3
1 . x 1 y2
x 1
2
2
4
dy
dy
1 23 y dx
m2 dx 23y
dy
1
1 and m
At 14 , 23 : m1 dx
2
2( 3/2)
3
dy
1
1 and m
At 14 , 23 : m1 dx
2
2( 3 /2)
3
If y
Copyright
3
3
3
3(1)
3
4
3
3
3( 1)
y
3
. If
2
dy
3
2
m1
dy
dx
1
2 y dx
dy
dx
dy
dx
3
3
2( 3 /2)
3
3
2( 3 /2)
3
3
3
3
1
2y
2014 Pearson Education, Inc.
m1 m2
3.
1
3
3
2
1
1.
4
1 y2
3
and x
m1 m2
3
3
3
2
x
1
3
3
3
x 1
3
x
3y
1
m1 m2
y
dy
dx
3
3
3
m1 m2
x2
4
m2
m1 m2
3
dy
dx
3 and m2
( 1)
1 y2
3
x 2 (1) 2
1. If y 1
3
3
3(1)
3
dy
3
dx
3( 1)
3 and m2
3
( 1)
x
y
dy
dx
y
3
3
1
3
1
3
1
3
3
1
Section 3.7 Implicit Differentiation
b, y 2
1x
3
52. y
x4
4
x4
x3
x3
dy
dx
4 x3
0
1
3
x3 ( x 4)
indeterminate at (0, 0). If x
53. xy 3
x2 y
also, xy 3
4
x2 y
y3 x 2 dx
x (3 y 2 )
6
(4)
2
y
2
dy
dx
x
0 or x
1
3
4. If x
x2
dy
dx
0
dx )
y (2 x dy
3 xy 2
b
x2
dx ( y 3
dy
0
3x2
2y
1
y
(0)2
2
0
1x
3
8. At (4, 8), y
2 xy
y3 dx
dy
3 x2
2y
3x2
0
dy
dy
x 3 y 2 dx
6
dy
and 2 y dx
8
x2
2
b
2 xy )
3xy 2
x3
1 is
28 .
3
b
y 3 2 xy
y 3 2 xy
dy
dx
y 3 2 xy
2
3x2
2y
1
3
0 and
1 (4)
3
x2
2
y
3 xy
x2
169
2
x
2
dx
dy
3 xy 2 x 2
3 xy 2 x 2
y 3 2 xy
;
;
dx appears to equal 1 The two different treatments view the graphs as functions symmetric across the
thus dy
dy
dx
line y
54. x3
y2
x, so their slopes are reciprocals of one another at the corresponding points (a, b) and (b, a ).
sin 2 y
dy
3x 2
2 y dx
dy
dy
dx
(2sin y )(cos y ) dx
dx 2 y
sin 2 y 3 x 2 dy
3 x2
; also, x3 y 2
2sin y cos y 2 y
1 The two different treatments
equal dy
dx
(2 y 2sin y cos y )
dx
dy
2 sin y cos y
dy
dx
2 sin y cos y 2 y
3x2
3x 2
3 x2
2 y 2sin y cos y
; thus dx
appears to
dy
view the graphs as functions symmetric across the line y
slopes are reciprocals of one another at the corresponding points (a, b) and (b, a ).
55.
y
sin 1 x
cos y
56. (a)
(b)
57-64.
x
1 sin 2 y
d
dx (sin
d
dx sin
1
1
x)2
1
x
d
dx ( x )
sin y
d
dx sin
x 2 . Therefore,
1
1
1
x
2
d 1
dx x
1
1
dy
dy
dx
x
1
1 x2
1 cos y dx
dy
dx
d
dx sin
1
1
cos y .
Since sin 2 y cos2 y 1,
.
2sin 1 x
1 x2
d sin 1 x
2sin 1 x dx
1
y
1
x2
1
x2
1
x2 1
1
x2
or
1
x x2 1
Example CAS commands:
Maple:
q1: x^3-x*y y^3 7;
pt : [x 2, y 1];
p1: implicitplot( q1, x -3..3, y -3..3 ):
p1;
eval( q1, pt );
q2 : implicitdiff( q1, y, x );
m : eval( q2, pt );
tan_line : y 1 m*(x-2);
p2 : implicitplot( tan_line, x -5..5, y -5..5, color green ):
p3 : pointplot( eval([x, y], pt), color blue):
display( [p1,p2,p3], "Section 3.7 #57(c)" );
Copyright
2014 Pearson Education, Inc.
x so their
170
Chapter 3 Derivatives
Mathematica: (functions and x0 may vary):
Note use of double equal sign (logic statement) in definition of eqn and tanline.
<<Graphics`ImplicitPlot`
Clear[x, y]
{x0, y0} {1, /4};
eqn x Tan[y/x] 2;
ImplicitPlot[eqn,{x, x0 3, x0 3},{y, y0 3, y0 3}]
eqn/.{x x0, y y0}
eqn/.{ y y[x]}
D[%, x]
Solve[%, y'[ x]]
slope y '[x]/.First[%]
m slope/.{x
x0, y[x] y0}
tanline y y0 m (x x0)
ImplicitPlot[{eqn, tanline}, {x, x0 3, x0 3},{y, y0 3, y0 3}]
3.8 DERIVATIVES OF INVERSE FUNCTIONS AND LOGARITHMS
1. (a) y = 2x + 3
x
(c)
2. (a)
df
dx x
1
df
dx x
3. (a) y = 5
x
(c)
3
1
( x)
f
2, dx
1
5
x 7
x = 5y
(c)
2x = y
3
2
df
1
5
y
y
2
1
df
dx x 1/2
x
y 7
(b)
f 1 ( x)
1
1 , df
dx
5
5
1
f
df
4, dx
5 x 35
x 34/5
4x = 5
y
4
3
2
1
2
x 1
35
4x
5
4
1
(b)
x
2
y
( x)
1
x 3
(b)
x
4
5
4
1
4
Copyright
2014 Pearson Education, Inc.
Section 3.8 Derivatives of Inverse Functions and Logarithms
4. (a)
2 x2
y
df
dx x 5
df 1
dx
5. (a)
(c)
1
2
y
f 1 ( x)
1
2
x
(c)
x2
4x x 5
20,
1
2 2
x 1/2
3
3
x 50
f ( g ( x))
x
3x 2
f ( x)
x
2
1
20
x 50
3 3
x, g ( f ( x))
f (1)
2/3
1x
3
g ( x)
(b)
y
x
3, f ( 1)
1,
3
g (1)
3;
1
3
g ( 1)
x3 at
(d) The line y = 0 is tangent to f ( x)
(0, 0); the line x = 0 is tangent to g ( x)
(0, 0).
1 ((4 x)1/3 )3
4
6. (a) h(k ( x))
3
4 x4
k (h( x))
(c)
3 x2
h (2)
4
4 (4 x ) 2/3
3
h ( x)
k ( x)
9.
11.
df 1
dx
y
3x 2 6 x
x 4
df 1
dx
ln 3x x
df 1
dx
1/3
x
3, h ( 2)
1,
3
k (2)
3;
k ( 2)
x3
4
x f (3)
df
dx
1
y
1
9
1
df
dx
x f (2)
1
3x
x at
(b)
(0, 0); the line x = 0 is tangent to k ( x )
at (0, 0).
df
dx
3
x,
(d) The line y = 0 is tangent to h( x)
7.
(b)
x
x 3
1
1
3
x 2
(3) 1
1
x
3
1
Copyright
1
3
at
(4 x)1/3
8.
10.
12.
df
dx
dg 1
dx
y
df 1
dx
2x 4
x 0
1
ln3x
dg 1
dx
1
x f (5)
df
dx
1
x f (0)
y
2014 Pearson Education, Inc.
dg
dx
x 0
x 5
1
6
1
2
1
1 3
(ln3 x )2 3x
1
x (ln3 x )2
171
172
Chapter 3 Derivatives
dy
dt
13.
y
ln(t 2 )
15.
y
ln 3x
ln 3 x 1
dy
dx
1
3x 1
17.
y
ln(
1) e
dy
d
1
18.
y
(cos )ln(2
2)
dy
d
19.
y
ln x3
dy
dx
1
x3
(3x 2 )
21.
y
t (ln t )2
22.
y
t ln t
23.
y
x4
4
24.
y
( x 2 ln x)4
25.
y
ln t
t
26.
y
t
ln t
27.
y
ln x
1 ln x
y
28.
y
x ln x
1 ln x
y
dy
dt
dy
dt
4
30. y = ln(ln(ln x))
y
1
x
1
(1) e
1
x4 1
4 x
x3 ln x
t
1
2 ln t
2
1
x
1
t
1
x
1
2 ln t
1
x
1
x
ln x
x
1
x
( x ln x )
1
x
2
(ln x)3
y
2t ln t
t
t 1
t
dy
dx
1
t 3/2
(ln t ) 2
1
2 t
ln t
dy
dx
Copyright
cos x
sin x
cot x
3(ln x )2
x
2 ln t
1
2
4 x 7 (ln x)3 8 x 7 (ln x )4
2ln t 1
2(ln t ) 3/2
1
x (1 ln x ) 2
(1 ln x )2 ln x
1
ln x
(1 ln x )2
1
1 d (ln x )
ln(ln x ) ln x dx
[sin(ln ) cos(ln )]
sin(ln ) cos(ln ) cos(ln ) sin(ln )
3
2t
1
2 t
2) (cos ) 1 1
d (ln x )
3(ln x)2 dx
1
x ln x
dy
d
3 t1/2
2
1 d (sin x )
sin x dx
( sin ) ln(2
4 x 6 (ln x)3 ( x 2 x ln x)
(1 ln x )2
d (ln(ln x ))
1
ln(ln x ) dx
[sin(ln ) cos(ln )]
ln x
x
2
(1 ln x )
(1 ln x )
y
ln(sin x )
dy
dt
x3 ln x
2 x ln x
ln t
2
(1 ln x ) ln x x
1
ln x
ln t
ln t
(ln x )
(1 ln x )
y
y
t
1 ln t
t2
ln t
(1 ln x )
16.
(ln t ) 2
4 x3
16
4( x 2 ln x)3 x 2 1x
(1) ln t
ln(t 3/2 )
20.
d t
t 1 dt
t
(ln t )(1)
y
2) (cos ) 2 1 2 (2)
3
x
(1) ln t
1
t
14.
e
1
( sin ) ln(2
d (ln t )
2t (ln t ) dt
t2
dy
dt
( 3x 2 )
(ln t )2
dy
dx
t
dy
dt
2
t
(2t )
dy
dx
x
ln x 16
29. y = ln(ln x)
31.
1
t2
1
x (ln x ) ln(ln x )
cos(ln ) 1 sin(ln ) 1
2 cos(ln )
2014 Pearson Education, Inc.
1
2 t
Section 3.8 Derivatives of Inverse Functions and Logarithms
32.
y
ln(sec
33.
y
ln
34.
y
1 ln 1 x
2 1 x
35.
y
1 ln t
1 ln t
36.
y
1
x x 1
ln x 12 ln( x 1)
1 [ln(1
2
y
ln(sec(ln ))
38.
y
cos
ln sin
1 2 ln
y
ln
40.
y
ln
41.
y
43.
t
t 1
y
dy
dt
1
2
3x 2
2 x ( x 1)
1 1 x 1 x
2 (1 x )(1 x )
( 1)
1
1 x2
2
t (1 ln t )2
1
2
1 2 ln
t 1/2
t 1
t 1
t 1 t
1
t 1
ln y
cos
x
x2 1
1 1
2 1 x
1 5
2 x 1
y
1))
( 1)
20
x 2
2ln y
tan(ln )
(ln )
y
y
(x
t
1
t 1 t (t 1)
Copyright
1
2(1 x )
5 ( x 2) 4( x 1)
2 ( x 1)( x 2)
2y
y
2x 1
2 x ( x 1)
1) 2 ln( x 1)]
ln(t 1)]
10 x
x2 1
ln( x) ln( x 1)
1 2x
2 x2 1
2
2
( x 2 1)( x 1)2 x 2 x x 1
1
x 1
1 [ln t
2
5 2x
x2 1
y
1 ln( x ( x
2
1 [ln( x 2
2
ln y
1
2
1
4t ln t
4
(1 2 ln )
tan
x ( x 1) (2 x 1)
2 x ( x 1)
1
x 1
ln y
1 (ln t1/2 ) 1/2 1 1 t 1/2
2
t1/ 2 2
sec(ln ) tan(ln ) d
sec(ln )
d
(sec(ln ))
1) 20 ln( x 2)]
( x( x 1))1/2
x( x 1) 1x
(t1/2 )
ln cos ) ln(1 2 ln )
2
( x 2 1)( x 1)2
y
1 ln t
t
t
2
1 (ln t1/2 ) 1/2 1 d
2
t1/ 2 dt
d
1
sec(ln ) d
( x 2 1)( x 1) 2
y
(ln t1/2 )
1 [5ln( x
2
( x 2) 20
1
2
1
1 x
(1 ln t )
5ln( x 2 1) 12 ln(1 x)
( x 1)5
x ( x 1)
dy
d
sin
cos
( x 2 1)5
1 x
y
1 ln t
t
t
2
1 (ln sin
2
1 cos
2 sin
39.
1
t
2( x 1) x
2 x ( x 1)
1 1
2 x 1
1 1
2 1 x
y
sec
(ln t1/2 )1/2
1 (ln t1/2 ) 1/2 d
2
dt
dy
d
(1 ln t )
(1 ln t )
37.
42.
1
t
sec (tan sec )
tan sec
1
x
y
x) ln(1 x )]
(1 ln t )
dy
dt
ln t
dy
dt
sec tan sec 2
sec tan
dy
d
tan )
1 dy
y dt
1 1
2 t
1)( x 1)
2
x 1
(2 x 2 x 1) x 1
x 2 1( x 1)
1
t 1
1
2 t (t 1)3/ 2
2014 Pearson Education, Inc.
5
3x 2
2 ( x 1)( x 2)
1
x
1
x 1
173
174
44.
Chapter 3 Derivatives
y
1
t (t 1)
[t (t 1)] 1/2
1
2
2t 1
1
t (t 1) t (t 1)
dy
dt
45.
y
3(sin )
dy
d
46.
y
dy
d
1
y
ln y
1)1/2
(tan )(2
2
1 sec
tan
(tan ) 2
47. y = t(t + 1)(t + 2)
48.
2
1
1 ln(
2
dy
dt
ln y
1
t (t 1)(t 2)
1
t
1
t 1
1
t 2
50.
y
sin
sec
ln y
ln
ln(sin ) 12 ln(sec )
1
cot
53.
y
( x 1)10
(2 x 1)
y
5
3 x ( x 2)
x2 1
y
54.
ln y
3
y
ln y
ln y
1 3 x ( x 2) 1
3
x2 1 x
x ( x 2)( x 2)
( x 2 1)(2 x 3)
1 dy
y dt
1 dy
yd
5) ln
x x2 1
( x 1)2/3
1
t
1
t 1
cos
sin
3)
ln(cos )
sec2
tan
1
2
2
2
1
1
t 2
1
t
1
t 1
1
1 dy
y d
5
1
6t 2
3t 2 6t 2
(t 3 3t 2 2t )2
dy
d
sin
cos
1
3t 2
1
t 2
(t 1)(t 2) t (t 2) t (t 1)
1
t (t 1)(t 2)
t (t 1)(t 2)
ln(
sin
sec
1 dy
y dt
ln1 ln t ln(t 1) ln(t 2)
ln y
y
1
1 dy
y d
1)
(t 1)(t 2) t (t 2) t (t 1)
t (t 1)(t 2)
t (t 1)(t 2)
5
cos
52.
2(
tan
2 1
1
ln y = ln t + ln(t + 1) + ln(t + 2)
y
y
1 dy
y d
ln(tan ) 12 ln(2
ln y
49.
51.
1
t 1
3) ln(sin )
(sec2 ) 2
1
t (t 1)(t 2) 1t t 11 t 12
1
t (t 1)(t 2)
dy
d
1 1
2 t
2t 1
2(t 2 t )3/ 2
3)1/2 sin
(
1 dy
y dt
ln(t 1)]
3(sin ) 2( 1 3) cot
(tan ) 2
dy
dt
1 [ln t
2
ln y
5
cos
1
5
1
tan
(sec )(tan )
2sec
cos
sin
1 tan
2
ln x 12 ln( x 2 1) 23 ln( x 1)
1 [10 ln( x
2
1 [ln x
3
1
x 2
ln y
1 3 x ( x 1)( x 2) 1
3 ( x 2 1)(2 x 3) x
1) 5ln(2 x 1)]
ln( x 2) ln( x 2 1)]
y
y
y
y
x
1
x2 x 1
5
x 1
y
y
1 1
3 x
2
3( x 1)
5
2x 1
1
x 2
y
y
( x 1)10
(2 x 1)5
2x
x2 1
2x
x2 1
1 [ln x
3
1
x 1
ln( x 1) ln( x 2) ln( x 2 1) ln(2 x 3)]
1
x 2
2x
x2 1
Copyright
2
2x 3
2014 Pearson Education, Inc.
x x2 1 1
( x 1)2/3 x
5
x 1
x
x2 1
5
2x 1
2
3( x 1)
Section 3.8 Derivatives of Inverse Functions and Logarithms
dy
d
55.
y
ln(cos 2 )
56.
y
ln(3 e
57.
y
ln(3te t )
58.
y
ln(2e t sin t )
dy
dt
)
ln 3 ln
59.
y
ln e
ln e
60.
y
ln
ln
1
1
y
e(cos t ln t )
62.
y
esin t (ln t 2 1)
e y sin x
64. ln xy
ex y
y
1 ye x
y
65.
xy
1
y
1 ye y sin x
y
y
yx
y
66. tan y
67.
y
2x
69.
y
5 s
ln x
ex
y
x
x
y
y
x
1
ln y x
xy ln y y 2
xy ln x x 2
ln x
ecos t
dy
ds
1
1/ 2
1
1 e
1
)
d (cos t )
tecos t dt
(1 t sin t )ecos t
esin t (ln t 2 1)(cos t )
e y sin x
y 1y
e
1 e
2
t
e y cos x
1 ye y sin x
ex y
1
x
1
y
y ( xe x
y
x (1 ye
x y
y ln x
(1 y )e x y
y
y 1y e x y
ex y
1
x
1)
)
y 1x
x ln y
y ln x
x 1y y
(1) ln y
ln x y
x
y
y
y x ln y y
x y ln x x
(sec2 y ) y
ex
1
x
y
( xe x 1) cos2 y
x
2 x ln 2
y
d
d
1
2 (1
(1 e ) 1
ye y cos x
y
y
d
d
1
( y e y )(sin x) e y cos x
y
xe x
d
d
1
dy
dt
1
1 e
1
esin t (cos t )(ln t 2 1) 2t esin t
ln x ln y
y
1 1t t
1
t
1
2 1
tecos t
e y cos x
ln x y
ln y
dy
dt
dy
d
ln(1 e )
2 1
e cos t eln t
1
cos t sin t
sin t
1
2
dy
dt
1
ln 2 t ln sin t
dy
d
1
1
61.
63. ln y
ln sin t
ln(1 e )
1
dy
d
ln 3 ln t t
t
1 cos
sin t
(sin t )
2 tan
ln 3 ln
ln 1
2
( sin )
ln e
ln 2 ln e t
d
1
sin t dt
1 e
2 cos
ln 3 ln t ln e t
1
1
1
cos2
175
68.
5 s (ln 5) 12 s 1/2
ln 5
2 s
Copyright
y
3 x
y
3 x (ln 3)( 1)
5 s
2014 Pearson Education, Inc.
3 x ln 3
ln y
y
x
176
Chapter 3 Derivatives
2
2
dy
ds
2s
71.
y
x
73.
y
log 2 5
ln 5
2
dy
d
74.
y
log3 (1
ln 3)
ln(1
75.
y
log 4 x log 4 x 2
76.
y
log 25 e x
77.
y
log 2 r log 4 r
ln r
ln 2
ln r
ln 4
ln 2 r
(ln 2)(ln 4)
dy
dr
1
(ln 2)(ln 4)
78.
y
log3 r log9 r
ln r
ln 3
ln r
ln 9
ln 2 r
(ln 3)(ln 9)
dy
dr
1
(ln 3)(ln 9)
y
dy
dx
80.
y
1
x 1
1 ln 7 x
2
y
sin(log7 )
82.
y
log 7 sin cos
e 2
dy
d
cos
(sin )(ln 7)
83.
y
log5 e x
84.
y
log 2
y
ln
ln e x
ln 5
x2e2
2 x 1
2
x ln 2
1
5
ln 3)
ln 3
ln x 2
ln 4
1
ln 3
ln x
ln 4
ln x
2 ln 5
x 1 ln 3
x 1
1
ln 3
1
x
2 ln
ln 4
(ln 3)
x
3 ln
ln 4
ln x
2 ln 5
x
2ln 5
(ln 3) ln
dy
dt
t1 e
y
(1 e)t e
1
ln 2
(5)
dy
d
ln 3
x 1
x 1
ln xx 11
ln 3
3
x ln 4
y
1
2 ln 5
1
ln 3
1
( x ln x)
1
2 ln 5
y
(2 ln r ) 1r
1 1x
x 1
2 x ln 5
2ln r
r (ln 2)(ln 4)
(2 ln r ) 1r
2 ln r
r (ln 3)(ln 9)
ln( x 1) ln( x 1)
2
( x 1)( x 1)
1 ln(3x
2
81.
1
ln 2
x ln e
ln 25
log5 x
7 x ln 5
3x 2
log5
72.
ln x
ln 4
x 1 ln 3
x 1
1
x 1
(ln 4) s 2 s
x ( 1)
y
log3
(ln 22 ) s 2 s
2
y
79.
2 s (ln 2)2s
2
70.
log 5 3 x7 x 2
2)
dy
dx
sin ln
ln 7
x
ln 5
y
7
2 7x
dy
d
3
2 (3 x 2)
1
ln 7
(3 x 2) 3 x
2 x (3 x 2)
sin ln
ln 7
ln 2
ln 2
ln 7
1
ln 7
ln
1 ln 7 x
2
3x 2
ln 5
1
x(3 x 2)
cos ln
ln 7
1
ln 7
ln(sin ) ln(cos )
ln 7
(cot
7x
3x 2
tan
sin(log 7 ) ln17 cos(log 7 )
ln 2
1 ln 2)
1
ln 5
2 ln x 2 ln 2
ln 2
ln x 2 ln e 2 ln 2 ln x 1
ln 2
1
2(ln 2)( x 1)
ln 5
2
ln 5
ln(sin ) ln(cos ) ln e
ln 7
sin
(cos )(ln 7)
7 x (ln 5)/ 2
3x 2
ln
(ln 5)/2
4( x 1) x
2 x ( x 1)(ln 2)
1 ln( x
2
1)
3x 4
2 x ( x 1) ln 2
Copyright
2014 Pearson Education, Inc.
Section 3.8 Derivatives of Inverse Functions and Logarithms
85.
y
3log 2 t
86.
y
3log8 (log 2 t )
87.
y
log 2 (8t ln 2 )
88.
y
t log 3 e(sin t )(ln 3)
89.
y
( x 1) x
90.
y
x ( x 1)
91.
y
t
92.
y
t t
93.
y
(sin x) x
94.
y
xsin x
y
1/ 2
t (t )
ln y
xsin x
x ln x , x
96.
y
(ln x)ln x
ln y
t
2
1 dy
y dt
y
y
x ln(sin x)
y
y
(sin x)(ln x)
dy
dt
t sin t
1
2
1 t 1/2
2
x
x cos
sin x
(sin x) 1x
sin t t cos t
( x 1) x xx 1 ln( x 1)
y
ln x 1 1x
ln x
1 dy
y dt
ln t
(t1/2 )(ln t )
1
t
x ( x1 1) ln( x 1)
( x 1) 1x
1
t (ln t )(ln 2)
t
2
(ln t )
1
t
(ln t ) t1/2 1t
ln(sin x)
y
y
x ( x 1) 1 1x ln x
t ln t
2
1
2
dy
dt
t
ln t 2
2 t
dy
dt
t ln t 2
2 t
ln t
2
sin x x (ln x )(cos x )
x
(cos x)(ln x)
y
y
(ln x )2
(ln x) ln(ln x)
2(ln x ) 1x
y
y
2
( x ln x ) lnxx
y
d (ln x )
(ln x ) ln1x dx
1
x
ln(ln x)
g ( f ( x))
lim
n
x
1 (n1/ x )
ln(ln x )
x
1
x
99. The derivative of x n at x
g ( f ( x)) f ( x) 1
(n/ x) x
e x for any x > 0.
(0 h )n
0 h
0 is given by lim
h
Copyright
lim h n 1. For n
h
0
2, n 1 1, so lim h n 1
2014 Pearson Education, Inc.
t
(sin x) x [ln(sin x) x cot x]
ln(ln x ) 1
x
x
n
1/ 2
ln t (t )
ln xsin x
ln y
(ln x)ln x
lim 1 nx
n
ln t t /2
3
t (ln t )(ln 8)
1
t ln 2
sin x x (ln x )(cos x )
x
0
97. ( g f )( x)
y
y
( x 1) ln x
ln y
dy
dt
t (sin t )(ln 3)
ln 3
y
y
x ln( x 1)
ln(sin x ) x
ln y
ln y
y
98.
ln x( x 1)
1
(ln t )/(ln 2)
3 ln t
t ln(3sin t )
ln 3
ln 3
1 (log 3)3log 2 t
2
t
1
t ln 2
3
ln 8
3ln 2 (ln 2)(ln t )
ln 2
t ln (eln 3 )sin t
t t /1
dy
dt
ln 8
ln( x 1) x
(t1/2 )t
ln t
ln 2
3ln
ln 8 ln(t ln 2 )
ln 2
ln y
t
3(ln t )/(ln 2) (ln 3)
3ln(log 2 t )
ln 8
ln y
95.
y
dy
dt
3(ln t )/(ln 2)
177
h
0
0.
1
2
178
Chapter 3 Derivatives
d ln x
100. Suppose n = 1. Then dx
1
x
( 1)0 0!1 and so the base case is established. Now if the statement holds for
x
n = k we have that, for n = k + 1, the following holds:
dn
dx n
(ln x )
dk
dx k
1
1
(ln x )
d dk
dx dx k
(ln x )
d
dx
( k 1)!
( 1)k 1 k
x
( 1)k k !
( 1) k1 1 (k 1)! ( k ) x k 1
Thus by mathematical induction the result is established for all n
x
( 1)n 1 ( n 1)!
k
xn
1.
101 108. Example CAS commands:
Maple:
with( plots );#101
f := x -> sqrt(3*x-2);
domain := 2/3 .. 4;
x0 := 3;
Df := D(f);
# (a)
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend= [" y=f(x)","y=f'(x)"],
title="#101(a) (Section 3.8)" );
q1 := solve( y=f(x), x );
# (b)
g := unapply( q1, y );
m1 := Df(x0);
# (c)
t1 := f(x0)+m1*(x-x0);
y=t1;
m2 := 1/Df(x0);
# (d)
t2 := g(f(x0)) + m2*(x-f(x0));
y=t2;
domaing := map(f,domain);
# (e)
p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ):
p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ):
p3 := plot( t1, x=x0-1..fx0+1, color=red, linestyle=4, thickness=0 ):
p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ):
p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ):
display( [p1,p2,p3,p4,p5], scaling=constrained, title="#101(e) (Section 3.8)" );
Mathematica: (assigned function and values for a, b, and x0 may vary)
If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows
Mathematica to do this.
<<Miscellaneous `RealOnly`
Clear[x, y]
{a,b} = { 2, 1}; x0 = 1/2 ;
f[x_] = (3x + 2) / (2x 11)
Plot[{f[x], f'[x]}, {x, a, b}]
solx = Solve[y == f[x], x]
g[y_] = x/. solx[[1]]
y0=f[x0]
ftan[x_] = y0+f'[x0] (x-x0)
gtan[y_] = x0 + 1/f'[x0] (y-y0)
Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Line[{{x0, y0},{y0, x0}}], PlotRange {{a, b},{a,b}}, AspectRatio
Copyright
2014 Pearson Education, Inc.
Automatic]
Section 3.8 Derivatives of Inverse Functions and Logarithms
179
109 110. Example CAS commands:
Maple:
with( plots );
eq := cos(y) = x^(1/5);
domain := 0 .. 1;
x0 := 1/2;
f := unapply( solve( eq, y ), x );
# (a)
Df := D(f);
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)",y=f'(x)"],
title="#110(a) (Section 3.8)" );
q1 := solve( eq, x );
# (b)
g := unapply( q1, y );
m1 := Df(x0);
# (c)
t1 := f(x0)+m1*(x-x0);
y=t1;
m2 := 1/Df(x0);
# (d)
t2 := g(f(x0)) + m2*(x-f(x0));
y=t2;
domaing := map(f,domain);
# (e)
p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ):
p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ):
p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ):
p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ):
p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ):
display( [p1,p2,p3,p4,p5], scaling=constrained, title="#110(e) (Section 3.8)" );
Mathematica: (Assigned function and values for a, b, and x0 may vary)
For problems 109 and 110, the code is just slightly altered. At times, different parts of solutions need to be
used as in the definitions of f[x] and g[y]
Clear[x, y]
{a,b} = {0, 1}; x0 = 1/2 ;
eqn = Cos[y] == x1/5
soly = Solve[eqn, y]
f[x_] = y /. soly[[2]]
Plot[{f[x], f'[x]}, {x, a, b}]
solx = Solve[eqn, x]
g[y_] = x /. solx[[1]]
y0 = f[x0]
ftan[x_] = y0+f'[x0] (x-x0)
gtan[y_] = x0 + 1/f'[x0] (y-y0)
Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]}, {x, a, b},
Epilog Line[{x0, y0}, {y0, x0}}], PlotRange {{a, b}, {a, b}}, AspectRatio
Copyright
2014 Pearson Education, Inc.
Automatic]
180
Chapter 3 Derivatives
3.9 INVERSE TRIGONOMETRIC FUNCTIONS
1. (a)
3. (a)
5. (a)
7. (a)
(b)
4
(b)
6
3
4
9. sin cos 1 22
11. tan sin 1
13.
15.
17.
19.
3
4
(b)
x 1
lim tan 1 x
x
lim sec 1 x
x
lim csc 1 x
x
21.
y
cos 1 ( x 2 )
23.
y
sin 1 2t
6
2
3
(c)
2.
(a)
4.
(a) 6
(b)
6.
(a) 4
(b)
8.
(a) 34
(b)
tan
12. cot sin 1
1
3
6
14.
2
16.
2
18.
2
lim sin 1 1x
0
x
dy
dx
dy
dt
25.
y
sec 1 (2 s 1)
26.
y
sec 1 5s
27.
y
csc 1 ( x 2 1)
dy
ds
20.
2x
2x
1 ( x 2 )2
1 x4
2
1
dy
ds
2
2t
2
1 2t
2
2
x
x
x
2 s 1 (2 s 1)
1
5
1
s 25s 2 1
2x
x 2 1 ( x 2 1) 2 1
Copyright
2s 1 4s
(c)
6
2
1
3
3
1
lim tan 1 x
lim sec 1 x
lim csc 1 x
22.
y
cos 1 1x
24.
y
sin 1 (1 t )
4s
(c)
3
cos
2
x
lim cos 1 1x
x
lim sin 1 1x
sec 1 x
1
2
(c)
4
sec 3
3
2
(c)
3
lim cos 1 x
x
2
2
5 s (5s )2 1
dy
dx
(b)
4
10. sec cos 1 12
1
2
sin 4
lim sin 1 x
3
(c)
6
1
2
6
(c)
4
(b)
3
(c)
3
2s 1 s2 s
2x
( x 2 1) x 4 2 x 2
2014 Pearson Education, Inc.
dy
dt
dy
dx
2
0
1
x x2 1
1
1
1 (1 t ) 2
2t t 2
6
3
6
2
3
Section 3.9 Inverse Trigonometric Functions
28.
y
29.
y
sec 1 1t
30.
y
sin 1 32
1
2
dy
dx
csc 1 2x
1
x 2
2
x
2
dy
dt
cos 1 t
1
2t
3
dy
dt
t2
t2
3
31.
y
cot 1 t
32.
y
cot 1 t 1
33.
y
ln(tan
x)
34.
y
tan 1 (ln x)
dy
dx
35.
y
csc 1 (et )
36.
y
cos 1 (e t )
37.
y
s 1 s2
38.
y
39.
y
y
e
1 [( x
cot 1 1x
2
(t 1)
1/ 2
1
2 t 1(1 t 1)
1/ 2 2
1 [(t 1)
1
]
1
2t t 1
1
x )(1 x2 )
s (1 s 2 )1/2
1 s2
1 s
2
]
tan 1 x
t
1 e
2t
dy
ds
cos 1 s
s2 1
1 s
tan 1 ( x 2 1)1/2
(1 s 2 )1/2
1 s2 s2 1
2
( s 2 1)1/2 sec 1 s
(2 x)
1/ 2 2
e2 t 1
e
1 (e )
1
1)
t
t 2
cos 1 s
1/ 2
1
2
1
( et ) 2 1
e
dy
dt
( x 2 1)
1
2 t (1 t )
)
et
t
2
t t4 9
1
x[1 (ln x )2 ]
1 (ln x )2
tan 1 x 2 1 csc 1 x
1
2
(tan
x
1
x
s 2 1 sec 1 s
dy
dx
40.
tan
dy
dt
1 s
1/ 2
1/ 2 2
dy
dt
t
1
6
t4 9
9
2
1
1 x2
1
s2
1 s2
2t
2
3
t
1 (t
cot 1 (t 1)1/2
dy
dx
1
1
2
dy
dt
cot 1 t1/2
x x2 4
1 t2
2
csc 1 t3
t
2
x2 4
4
x
1
181
1 s
dy
dx
1 s2
( s 2 1) 1/2 (2 s)
1
2
1
ss 1
s s2 1
s s2 1
s
1
s s2 1
s2 1
csc 1 x
1
1
1
x x2 1
x x2 1
tan 1 ( x 1 ) tan 1 x
Copyright
1
1 s2
2s2
2
x x2 1
2
s 12 (1 s 2 ) 1/2 ( 2 s)
0, for x 1
dy
dx
0
x 2
1 ( x 1 )2
1
1 x2
2014 Pearson Education, Inc.
1
x2 1
1
1 x2
0
182
41.
Chapter 3 Derivatives
x sin 1 x
y
dy
dx
42.
sin 1 x x
ln( x 2
y
1 x2
x sin 1 x (1 x 2 )1/2
1
1 x2
4) x tan 1 2x
43. The angle
(1 x 2 ) 1/2 ( 2 x )
1
2
dy
dx
tan 1 2x
2x
x
2
sin 1 x
4
x
x
1 x2
1
2
1
tan 1 2x
2x
x2 4
x 2
2
sin 1 x
2x
4 x2
tan 1 2x
is the large angle between the wall and the right end of the blackboard minus the small angle
x
cot 1 15
between the left end of the blackboard and the wall
44. 65
x
1 x2
(90
) (90
) 180
65
65
cot 1 3x .
21
tan 1 50
65
22.78
42.22
45. Take each square as a unit square. From the diagram we have the following: the smallest angle
1
of 1
tan 1; the middle angle
of 3
tan 1 3. The sum of these three angles is
has a tangent of 2
y
sec
x to the line y =
sec 1 x
x; i.e.,
(b) cos 1 ( x)
2; and the largest angle has a tangent
sec 1 x above the x-axis at
sec 1 ( x ).
x
1
cos 1
cos 1 1x , where x 1 or x
1
x
sec 1 x
47. (a) Defined; there is an angle whose tangent is 2.
(b) Not defined; there is no angle whose cosine is 2.
48. (a) Not defined; there is no angle whose cosecant is 12 .
(b) Defined; there is an angle whose cosecant is 2.
49. (a) Not defined; there is no angle whose secant is 0.
(b) Not defined; there is no angle whose sine is 2.
50. (a) Defined; there is an angle whose cotangent is
1.
2
(b) Not defined; there is no angle whose cosine is 5.
51. csc 1 u
52.
y
sec 1 u
2
tan 1 x
tan y
dy
(sec2 y ) dx
dy
dx
1
sec2 y
d (csc 1 u )
dx
x
d (tan
dx
y)
d
dx 2
sec 1 u
0
du
dx
2
u u
1
1
2
1
1 x2
d ( x)
dx
, as indicated by the
triangle.
Copyright
du
dx
2
u u
1
1 x2
.
sec 1 x is the vertical distance from the graph of
and this distance is the same as the height of y
cos 1 x, where 1
sec 1 ( x)
has a tangent
tan 1 1 tan 1 2 tan 1 3
46. (a) From the symmetry of the diagram, we see that
1
tan
1
2014 Pearson Education, Inc.
1
, u
1
1
Section 3.9 Inverse Trigonometric Functions
53. f(x) = sec x
f ( x)
df 1
dx
sec x tan x
1
1
sec(sec 1 b) tan(sec 1 b)
df
dx x f 1 ( b )
x b
1
b
b2 1
d sec 1 x
Since the slope of sec 1 x is always positive, we choose the right sign by writing dx
54. cot 1 u
tan 1 u
2
d (cot 1 u )
dx
d
dx 2
tan 1 u
55. The functions f and g have the same derivative (for x
0
du
dx
du
dx
1 u2
1 u2
0), namely
1
.
x ( x 1)
183
1
x x2 1
.
The functions therefore differ by a
constant. To identify the constant we can set x equal to 0 in the equation f(x) = g(x) + C, obtaining
sin 1 ( 1)
2 tan 1 (0) C
0 C
2
C
2
. For x
0, we have sin 1 xx 11
56. The functions f and g have the same derivative for x > 0, namely
1 .
1 x2
2 tan 1 x
2
.
The functions therefore differ by a
constant for x > 0. To identify the constant we can set x equal to 1 in the equation f(x) = g(x) + C, obtaining
sin 1 1
2
57. (a)
(c)
tan 1 1 C
sec 1 1.5
cot 1 2
58. (a)
sec 1 ( 3)
(c)
cot 1 ( 2)
4
1
cos 1 1.5
2
2
C
0.84107
tan 1 2
cos 1
C
4
0. For x > 0, we have sin 1
(b) csc 1 ( 1.5)
sin 1
tan 1 1x .
1
1.5
1
3
1
1.91063
tan ( 2)
(b) csc 1 1.7
1
sin 1 1.7
0.62887
2.67795
<y<
tan 1 (tan x) is periodic, the graph of y
k
where k is an integer. Range:
< x < ; Range:
The graph of y
x< .
0.72973
0.46365
59. (a) Domain: all real numbers except those having the form 2
(b) Domain:
1
x2 1
Copyright
tan(tan 1 x)
2014 Pearson Education, Inc.
x for
2
y
2
184
Chapter 3 Derivatives
60. (a) Domain:
(b) Domain: 1
< x < ; Range:
x
The graph of y
1 x 1.
61. (a) Domain:
(b) Domain: 1
1; Range: 1
The graph of y
1 x 1.
y
2
1
1
sin (sin x) is periodic; the graph of y
< x < ; Range: 0
x
y
2
1; Range: 1
sin(sin 1 x)
x for
y
y
1
1
cos (cos x) is periodic; the graph of y
Copyright
cos(cos 1 x)
2014 Pearson Education, Inc.
x for
Section 3.9 Inverse Trigonometric Functions
185
62. Since the domain of sec 1 x is
( , 1] [1, ), we have sec(sec 1 x) x for |x|
open line segment from ( 1, 1) to (1, 1) removed.
63. The graphs are identical for y
1. The graph of y
sec(sec 1 x ) is the line y = x with the
2sin(2 tan 1 x )
4[sin(tan 1 x )][cos(tan 1 x )]
4
4x
x2 1
64. The graphs are identical for y
x
x2 1
1
x2 1
from the triangle
cos(2sec 1 x )
cos2 (sec 1 x ) sin 2 (sec 1 x )
1
x2
x2 1
x2
2 x2
x2
from the triangle
Copyright
2014 Pearson Education, Inc.
186
Chapter 3 Derivatives
65. The values of f increase over the interval [ 1, 1] because f
0, and the graph of f steepens as the values of
f increase towards the ends of the interval. The graph of f is concave down to the left of the origin where
f
0, and concave up to the right of the origin where f
f
0 and f has a local minimum value.
66. The values of f increase throughout the interval (
0. There is an inflection point at x = 0 where
, ) because f
0, and they increase most rapidly near
the origin where the values of f are relatively large. The graph of f is concave up to the left of the origin
where f
0, and concave down to the right of the origin where f
where f
0 and f has a local maximum value.
3.10
0. There is an inflection point at x = 0
RELATED RATES
1. A
r2
2. S
4 r2
3. y
5 x, dx
dt
dA
dt
dS
dt
2
dy
4. 2 x 3 y 12, dt
2 r dr
dt
8 r dr
dt
dy
dt
5 dx
dt
2
dy
dt
5(2) 10
dy
2 dx
3 dt
dt
0
2 dx
3( 2)
dt
Copyright
0
dx
dt
3
2014 Pearson Education, Inc.
Section 3.10 Related Rates
5. y
x 2 , dx
dt
6. x
y3
7. x 2
dy
y, dt
y2
8. x 2 y 3
Thus
9. L
dy
dt
3
25, dx
dt
dy
y 2 , dx
dt
(5)( 1) (12)(3)
0; when x
3 and y
v3
12, dr
dt
S
3(2) 2 dv
dt
2
13. (a) V
r2h
(c) V
r2h
(c) dV
dt
15. (a) dV
dt
1
3
dS
dt
1
2 x2 y2
r 2 dh
dt
dV
dt
dV
dt
r 2 dh
dt
dV
dt
2 dh 2
r dt 3
dr
dt
1
6
2 x dx
dt
2 s ds
3v 2 dv
dt
dt
12 x dx
; when x
dt
3 x 2 dx
; when x
dt
12 x dx
dt
3
r 2h
1
3
dS
dt
dV
dt
54 in
sec
3(3) (2)
dv
dt
0
m
5 min
72 in
sec
2
3
m
5 min
x3 , dx
dt
6 x 2 , dS
dt
14. (a) V
4, ds
dt
6 x 2 , dx
dt
dL
dt
3
dl
dt
1
2
P
RI 2
dP
dt
(b) P
RI 2
0
R
1
I
12
2
2(3)( 2) 2( 4) dt
4
27
y
0
dy
dt
3
2
1.
3
dy
x dx
y dt
dt
dy
2 y dt
x2 y2
; when x
3
dS
dt
dV
dt
dx
dt
3 and s 1
12(3)( 5)
3(3)2 ( 5)
r2h
2 rh dr
dt
(b) dl
dt
dR 1 dV V dI
dR
1 dV R dI
I dt
dt
dt
I dt
I dt
dt
1 (3) 3 ohms/sec, R is increasing
2
2
I 2 dR
dt
5 and y 12
1
3
(3) (1)2
v3 12
I 2 dR
dt
2 RI dI
dt
dR
dt
x
dx
x 2 y 2 dt
x
dx
x 2 y 2 dt
17. (a) s
x2
y2
( x2
y 2 )1/2
ds
dt
(b) s
x2
y2
( x2
y 2 )1/2
ds
dt
(c) s
x2
y2
s2
y2
2 s ds
dt
Copyright
2 x dx
dt
2 RI dI
I 2 dt
2
I
y
dy
x 2 y 2 dt
dy
2 y dt
2s
P
I
2
2
3
m
135 min
3 x 2 dx
; when x
dt
dV
dt
dV
dt
2 rh dr
dt
dV
dt
2
3
rh dr
dt
amp/sec
0
dI
dt
2 x dx
dt
2014 Pearson Education, Inc.
v
m
180 min
2 RI dI
dt
dP
dt
x2
r 2h
1
3
(b) V
rh dr
dt
dR
dt
1
3
x3
in ; V
2 sec
(b) V
r 2 dh
dt
1
3
0; when r
3
72 12(3) dx
dt
1 volt/sec
dV
dt
dR
dt
dy
4
(2) 2 y 3
2
55
31
13
(5)2 (12)2
dV
dt
16. (a)
2 y dt
6
3(2)2 (5) (5)
dx
dt
2
1, dt
(b) V
(d)
dy
2 x dx
dt
2
when y
x2
4 2(1)( 3)
(c)
dy
;
dt
2( 1)(3)
dy
3 x 2 y 2 dt 2 xy 3 dx
0; when x
dt
3
9
dx
2(2) 13 dx
0
2
dt
dt
10. r s 2
12. S
dy
3 y 2 dt
dx
dt
5
dy
dt
1
4 , dy
1
27 dt
2
2 1 2 1
3(2) 3
2
dL
dt
11. (a)
2 x dx
; when x
dt
187
2 P dI
I 3 dt
dy
2 y dt
dx
dt
y dy
x dt
3
2
188
Chapter 3 Derivatives
x2
18. (a) s
y2
z2
s2
x
dx
x 2 y 2 z 2 dt
From part (a) with dx
dt
(c) From part (a) with ds
dt
dA
dt
dA
dt
1 ab sin
2
1 ab sin
2
19. (a) A
(c) A
r 2 , dr
dt
20. Given A
y2
z2
2s ds
dt
dy
2 x dx
dt
1 ab cos d
2
dt
1 ab cos d
1 b sin
2
2
dt
y dy
x dt
db
dt
1 a sin
2
50 cm. Since dA
dt
0.01 cm/sec, and r
dz
dt
(b) A
da
dt
2 z dz
dt
2 y dt
y
dy
dz
z
x 2 y 2 z 2 dt
x 2 y 2 z 2 dt
y
dy
ds
z
0
dt
x 2 y 2 z 2 dt
x2 y 2 z 2
dy
dx
0 0 2 x dx
2 y dt 2 z dz
dt
dt
dt
ds
dt
(b)
x2
z dz
x dt
0
dA
dt
1 ab sin
2
2 r dr
, then dA
dt
dt r 50
21. Given ddt
2 cm/sec, dw
2 cm/sec,
12 cm and w 5 cm.
dt
dA
dw
d
dA
(a) A w
w dt
12(2) 5( 2) 14 cm 2 /sec, increasing
dt
dt
dt
dP 2 d
2 dw
2( 2) 2(2) 0 cm/sec, constant
(b) P 2 2w
dt
dt
dt
w2
(c) D
14
13
22. (a) V
dV
dt
2 1/2
dD
dt
)
w2
1
2
1/2
2
2 w dw
2 ddt
dt
w dw
dt
dD
dt
d
dt
w
2
d
dt
2
1 b sin
2
(5)(2) (12)( 2)
25 144
dy
x
y
z2
d
dt (4, 3, 2)
( x2
4
29
2 m3 /sec
0 m 2 /sec
(10)(1) (12)( 2) (14)(1)
(1)
y2
z 2 )1/2
d
dt
( 2)
2
29
3
29
x
x
(1)
2
y
2
y
dx
z dt
2
x
2
y
dy
z dt
2
dz
z
x 2 y 2 z 2 dt
2
0 m/sec
23. Given: dx
5 ft/sec, the ladder is 13 ft long, and x 12, y 5 at the instant of time
dt
dy
x dx
12 (5)
12 ft/sec, the ladder is sliding down the wall
(a) Since x 2 y 2 169
5
dt
y dt
(b) The area of the triangle formed by the ladder and walls is A
changing at 12 [12( 12) 5(5)]
x
13
(c) cos
24. s 2
y2
x2
sin ddt
2 s ds
dt
2 x dx
dt
1 dx
13 dt
dy
2 y dt
119
2
d
dt
ds
dt
1
2
xy
dA
dt
1
2
dy
x dt
y dx
. The area is
dt
59.5 ft 2 /sec.
1
13sin
1
s
x dx
dt
dx
dt
1
5
dy
(5)
ds
dt
y dt
1 rad / sec
1 [5(
169
442) 12( 481)]
614 knots
25. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the
400(25)
ds
x dx
20 ft/sec.
girl and kite s 2 (300)2 x 2
s dt
dt
500
26. When the diameter is 3.8 in., the radius is 1.9 in. and dr
dt
dV
dt
1
12 (1.9) 3000
1
3000
6 r2
in/min. Also V
3
0.0076 . The volume is changing at about 0.0239 in /min.
Copyright
2014 Pearson Education, Inc.
dV
dt
da
dt
cm 2 /min.
1
2 (50) 100
dV
xz dt xy dz
(3)(2)(1) (4)(2)( 2) (4)(3)(1)
dt (4, 3, 2)
dt
dS (2 y 2 z ) dx (2 x 2 z ) dy (2 x 2 y ) dz
dt
dt
dt
dt
yz dx
dt
2 xy 2 xz 2 yz
dS
dt (4, 3, 2)
2
2
(c)
( w2
cos
cm/sec, decreasing
xyz
(b) S
2
1 ab
2
12 r dr
dt
Section 3.10 Related Rates
1 r 2 h, h 3 (2r ) 3r
r 43h V
4
3
8
dh
9
90
(10) 256
0.1119
dt h 4
16 42
dr
15
4 dh 4 90
r 43h
3 dt
3 256
dt
32
27. V
(a)
(b)
28. (a) V
(b) r
4h
3
1
3
3
y 2 (3R
dV
dt
y)
3
[2 y (3R
16 h3
27
h
16 h 2 dh
9 dt
dV
dt
m/sec 11.19 cm/sec
0.1492 m/sec 14.92 cm/sec
2
1 r 2 h and r 15h
V 13 152h h
3
2
8
0.0113 m/min
1.13 cm/min
225
15h
dr
15 dh
dr
15
8
2
dt
2 dt
dt h 5
2 225
29. (a) V
2
189
75 h3
4
4
15
dy
dy
dt
3
4( 50)
dh
dt h 5
0.0849 m/sec
y 2 ( 1)] dt
y)
225 h 2 dh
4
dt
dV
dt
225 (5) 2
8.49 cm/sec
(6 Ry 3 y 2 )
1
dV
dt
at R 13 and y
8
dy
we have dt 1441 ( 6) 241 m/min
(b) The hemisphere is one the circle r 2 (13 y )2
(c) r
(26 y
dr
dt
y 2 )1/2
dr
dt
5
288
dS
dt
4
3
r3, r
8 r dr
dt
5, and dV
dt
8 (5)(1)
26 y
dr
dt
y2m
13 y dy
26 y y 2 dt
m/min
30. If V 43 r 3 , S 4 r 2 , and dV
kS 4k r 2 , then dV
dt
dt
Therefore, the radius is increasing at a constant rate.
31. If V
r
dy
y 2 ) 1/2 (26 2 y ) dt
1 (26 y
2
13 8
1
26 8 64 24
y 8
169
4 r 2 dr
dt
4k r 2
4 r 2 dr
dt
dr
dt
k , a constant.
dr 1 ft/min. Then S 4 r 2
4 r 2 dr
dt
dt
40 ft 2 /min, the rate at which the surface area is increasing.
100 ft 3 /min, then dV
dt
32. Let s represent the length of the rope and x the horizontal distance of the boat from the dock.
s ds
s
ds . Therefore, the boat is approaching the dock at
dx
(a) We have s 2 x 2 36
x dt
dt
dt
2
s
dx
dt
10
102 36
s 10
(b) cos
d
dt
6
r
6
8
102 10
( 2)
sin ddt
( 2)
36
2.5 ft/sec.
6 dr
r 2 dt
3
20
d
dt
rad/sec
6
r 2 sin
dr . Thus,
dt
r 10, x
8, and sin
8
10
33. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal
distance between the balloon and the bicycle. The relationship between the variables is s 2 h 2 x 2
ds 1 h dh x dx
ds
1 [68(1) 51(17)] 11 ft/sec.
dt
s
dt
dt
dt
85
34. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is
dV
10 in/min.
1 dV
9 dh
the rate the coffee is rising is dh
V 9 h
dt
dt
dt
9 dt
9
(b) Let h the height of the coffee in the pot. From the figure, the radius of the filter r
the volume of the filter. The rate the coffee is falling is
Copyright
dh
dt
4 dV
h2 dt
2014 Pearson Education, Inc.
4
25
( 10)
h
2
8
5
V
1
3
in/min.
r 2h
h3
12
,
190
Chapter 3 Derivatives
dy
dt
QD 1
35. y
dQ
D 1 dt QD 2 dD
dt
origin. Consequently, tan
cos
2
x
x 3
2
2
3
y 2 x2
92 32
tan
we have
x2
37. The distance from the origin is s
1 ( x2
2
ds
dt (5,12)
38. Let s
2
y 2 ) 1/2 2 x dx
dt
x
x
x
d
dt x 3
466
1681
( 2)
x 2 and
36. Let P ( x, y ) represent a point on the curve y
y
x
1 ,
10
233
(41)2
1 (0)
41
L/min
increasing about 0.2772 L/min
the angle of inclination of a line containing P and the
sec 2
d
dt
dx
dt
cos2
d
dt
dx . Since dx
dt
dt
1 rad/sec.
y 2 and we wish to find
dy
2 y dt
(5, 12)
(5)( 1) (12)( 5)
25 144
5m/sec
distance of the car from the foot of perpendicular in the textbook diagram
sec 2
d
dt
1 ds
132 dt
cos2 ds ; ds
dt
132 dt
d
dt
traveled 132 ft right of the perpendicular
( 12 )
d
dt
132
10 m/sec and
264 and
| |
4
d
dt
0
, cos
2
s
132
tan
2 rad/sec. A half second later the car has
1 , and ds
2
dt
264 (since s increases)
(264) 1 rad/sec.
39. Let s 16t 2 represent the distance the ball has
fallen, h the distance between the ball and the
ground, and I the distance between the shadow and
the point directly beneath the ball. Accordingly,
s h 50 and since the triangle LOQ and triangle
30 h
PRQ are similar we have I 50
h 50 16t 2
h
30(50 16t 2 )
and I
50 (50 16t 2 )
dI
dt t
1
2
1500
16t 2
dI
dt
30
1500
8t 3
1500 ft/sec.
40. When x represents the length of the shadow, then tan
We are given that ddt
3
16
ft/min
3
2000
0.27
0.589 ft/min
rad/ min . At x
80
x
sec 2
d
dt
3
5
dx
dt
60, cos
80 dx
x 2 dt
x 2 sec 2
80
x 2 sec 2
80
dx
dt
d
dt
d
dt
48
5
72
10
3
.
and sec
5
3
7.1 in./min.
dr
5 in./min when dV
4 r 2 dr
dt
dt r 6 72
dt
ds 8 r dr
the thickness of the ice is decreasing at 725 in/min. The surface area is S 4 r 2
dt
dt
41. The volume of the ice is V
3
2000
d
dt
4
3
r3
4
3
43
dV
dt
in 2 /min, the outer surface area of the ice is decreasing at 10
in 2 /min.
3
10 in 3 /min,
dS
dt r 6
42. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between
dr
ds
ds
5 ( 160)
r
the car and plane 9 s 2 r 2
200 mph
speed of plane speed
dt
dt
dt
2
r
of car
200 mph
9
r 5
16
the speed of the car is 80 mph.
Copyright
2014 Pearson Education, Inc.
Section 3.10 Related Rates
43. Let x represent distance of the player from second base and s the distance to third base. Then dx
dt
(a) s 2
x 2 8100
and
ds
dt
60 (
30 13
d 1
dt
16)
90
s 2 cos
1
90
x 2 8100
0
90
dx
x 2 8100 dt
x dx . When the
s dt
d 1
dt
90 ds
s 2 dt
32
13
ds
dt
ds
dt
8
65
90
s 2 cos
30 13
90
s 2 xs
dx
dt
x
s
d
1
6
lim 2
x 0 dt
d 2
dt
90
s2
1 rad/sec; d 2
6
dt
1
90
s
rad/sec; cos 2
60 and s
( 15)
player is 30 ft from first base, x
16 ft/sec
60
s
30 13
8.875 ft/sec
cos 1 dt1
Therefore, x
lim
x
32
13
90
30 13 (60)
90 ds .
s k dt
d 1
dt
2 x dx
dt
d
90
s
(b) sin 1
(c)
2 s ds
dt
191
90 ds .
s k dt
2
Therefore, x
d
32
13
8
65
90
s 2 xs
x
s
90
s 2 sin
30 13
2
ds
dt
rad / sec.
d
90
dx
x 2 8100 dt
ds
dt
60 and s
d 2
dt
90 ds
s 2 dt
sin 2 dt2
90
30 13 (60)
dx
dt
90
s 2 sin
ds
dt
lim dt1
0
x
dx
dt
90
s2
dx
dt
rad/sec
44. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the
dD
1 2a da 2b db
distance between the ships. By the Law of Cosines, D 2 a 2 b 2 2ab cos120
dt
dt
2D
dt
a db
b da
. When a
dt
dt
ships are moving dD
29.5 knots apart.
dt
1
2D
2a da
dt
2b db
dt
5, da
dt
14, b
3, and db
dt
21, then dD
dt
413
2D
where D
7. The
0.5 /min. The minute hand, starting at 12, chases
45. The hour hand moves clockwise from 4 at 30 /hr
6 /min. Thus, the angle between them is decreasing and is changing at 0.5 /min
the hour hand at 360 /hr
6 /min
5.5 /min.
46. The volume of the slick in cubic feet is V
length of the minor axis.
substitute:
dV
dt
3
16
dV
dt
3
4
2(5280)(10)
a d
2 dt
3
4
b
2
3
4
b d
2 dt
(5280)(30)
Copyright
a
2
a
2
b
2
3
16
3
16
, where a is the length of the major axis and b is the
a db
b da
dt
dt . Convert all measurements to feet and
(224,400) 132,183 ft 3 /hr
2014 Pearson Education, Inc.
192
Chapter 3 Derivatives
3.11
LINEARIZATION AND DIFFERENTIALS
x3 2 x 3
1. f ( x)
x2
2. f ( x)
4 (x
5
4. f ( x)
x1/3
5. f ( x)
tan x
6. (a) f ( x)
(b) f ( x)
(c) f ( x)
f (2)( x 2)
L( x)
( x 2 9)
1
2
at x
4
L( x)
f (1)
L( x)
sec2 x
f ( x)
1/2
x
(2 x)
f( )
2 0( x 1)
1 (x
12
f ( 8)
f ( )( x
L( x)
x2 9
f (1)( x 1)
f ( 8)( x ( 8))
L( x)
f (2) 10( x 2) 7
)
L( x) 10 x 13 at x
f ( 4)( x 4)
ex
f ( x)
f ( x ) ln(1 x )
7. f ( x)
x2
8. f ( x)
x 1
f ( x)
9. f ( x)
2 x2
4x 3
2x
f ( x)
f ( x)
10.
f ( x) 1 x
f ( 4)
2
8) 2
0 1( x
11.
f ( x)
3
12.
f ( x)
x
x 1
f ( x)
f ( x)
x1/3
L( x )
f (0)
f (0)( x 0) 1 x
1
1 x
L( x )
f (0)
2x 2
L( x)
x 2
L ( x)
f ( x)
4x 4
f ( x) 1
)
L( x )
1
12
x 43
x
L( x )
f (0)( x 0)
f (1)( x 1)
L( x)
(1)( x 1) (1)( x )
f (1)
L( x )
( 1)( x 1) 1
f ( 1)
f (8)( x 8)
L( x )
x
x
2( x 0) 0
f (8) 1( x 8) 9
L( x)
1
( x 1)2
( x 1)2
f (0)
f ( 1)( x 1)
f (8)( x 8)
x 2/3
1
3
f ( x)
f (0)( x 0)
L( x ) 1 x
f (8)
f (1)( x 1)
L( x)
L ( x)
2 x at x
x 2 at x 1
0( x 1) ( 5)
L( x)
L( x)
8
1 (x
12
f (1)
x 1 at x
8) 2
1 (x
4
0
L( x)
1) 12
1
12
L ( x)
5 at x
1
4
3
at x
8
x
1
4
x
1
4
at x 1
13.
f ( x)
e x
14.
f ( x)
sin 1 x
15. f ( x)
16. (a)
e x
(b) f ( x)
(c) f ( x)
1
f ( x)
(1 x)6
f (0)
L( x)
1 x2
2
1 x
[1 ( x)]6
2 1 ( x)
1 x
1/2
(d) f ( x)
2 x2
(e) f ( x)
(4 3 x)1 3
(f)
L ( x)
f (0)( x 0)
f (0)
f ( x)
1
x
2 x
1
k . L( x )
1
2 1
x2
2
x 1
1/2
1
f (0)
f (0)( x 0) 1 k ( x 0) 1 kx
2 2x
x
2
2
2 1 12 x2
13
41 3 1 34x
2/3
x at x = 0.
1 6( x) 1 6 x
2[1 ( 1)( x)]
1
2
x 1 at x = 0.
f (0)( x 0)
k (1 x) k 1. We have f (0) 1 and f (0)
f ( x)
2
f ( x ) cos x
L( x) f (0) f (0)( x 0) x
L( x) x
f ( x)
sin x
L( x ) f (0) f (0)( x 0) 1 L ( x) 1
f ( x) sec 2 x
L ( x) f (0) f (0)( x 0) x
L( x) x
ex
x
9
5
x
1
3 x 2/3
f ( x)
sin x
cos x
tan x
(d) f ( x )
4
5
L ( x)
2
f ( x)
f ( x) 1 x 2
1
x
x
(e)
( x 2 9)1/2
9
4) 5
3. f ( x)
3x 2
f ( x)
41 3 1 13 34x
x
2 x
Copyright
2/3
1 23
2 1
x2
4
41 3 1 4x
x
1 62 3x x
2 x
2014 Pearson Education, Inc.
Section 3.11 Linearization and Differentials
193
17. (a) (1.0002)50 (1 0.0002)50 1 50(0.0002) 1 .01 1.01
(b) 3 1.009 (1 0.009)1/3 1 13 (0.009) 1 0.003 1.003
18. f ( x)
( x 1)1/2 sin x
x 1 sin x
3 (x
2
0) 1
3
2
L f ( x)
1
2
f ( x)
( x 1) 1/2
cos x
x 1, the linearization of f ( x); g ( x )
L f ( x)
f (0)( x 0)
1/2
x 1
( x 1)
g ( x)
f (0)
1
2
( x 1) 1/2
Lg ( x) g (0)( x 0) g (0) 12 ( x 0) 1 Lg ( x) 12 x 1, the linearization of g ( x ); h( x ) sin x
h ( x) cos x
Lh ( x) h (0)( x 0) h(0) (1)( x 0) 0 Lh ( x) x, the linearization of h( x).
L f ( x) Lg ( x) Lh ( x) implies that the linearization of a sum is equal to the sum of the linearizations.
19. y
x3 3 x
x3 3 x1/2
20. y
x 1 x2
x (1 x 2 )1/2
1/2
1 x2
21. y
2x
1 x2
22. y
2 x
31 x
23. 2 y 3/2
24. xy 2
xy x
0
x
dy
26. y
cos ( x 2 )
27. y
3
4 tan x3
2 2 x2
(1 x 2 )2
3 1 x1/ 2
dx
2 x1/ 2
y dx x dy dx
3 x 1/ 2
2
dx
0
28. y
sec x 2 1
(3 y1/2
0
dy [ sin ( x 2 )](2 x)dx
3x 1 2 3 3
9(1 x1/ 2 )2
x) dy
(2 xy 1) dy
(cos (5 x1/2 )) 52 x 1/2 dx
3
4 sec2 x3
dx
( x) 12 (1 x 2 ) 1/2 ( 2 x) dx
1/ 2 2
dy
3
2 x
dx
dx
1/ 2
3x2
dy
y 2 dx 2 xy dy 6 x1/2 dx dy
sin (5 x ) sin (5 x1/2 )
dy
x 1/2 dx
91 x
3 y1/2 dy
0
y
1 x2
(1 x 2 )2
25. y
29. y
1 2 x2
(2)(1 x 2 ) (2 x )(2 x )
2 x1/ 2
3 1 x1/ 2
3
2
(1) (1 x 2 )1/2
dy
(1 x 2 ) x 2 dx
dy
4 x3/2
3x 2
dy
dy
dx
dy
1
3 x (1 x ) 2
1 y
dx
3 y x
(1 y ) dx
dy
(6 x1/2
y 2 )dx
5 cos 5 x
2 x
dy
dx
2 x sin ( x 2 ) dx
( x 2 ) dx
dy
3
4 x 2 sec2 x3 dx
dy [sec ( x 2 1) tan ( x 2 1)](2 x) dx
3csc (1 2 x ) 3csc (1 2 x1/2 ) dy
dy 3 csc (1 2 x ) cot (1 2 x ) dx
2 x [sec ( x 2 1) tan ( x 2 1)]dx
3( csc (1 2 x1/2 )) cot (1 2 x1/2 ) ( x 1/2 ) dx
x
30. y
2 cot 1
31.
e x
y
x
dy
2cot x 1/2
e x
2 x
dy
2 csc2 ( x 1/2 )
1
2
( x 3/2 ) dx
dy
dx
Copyright
2014 Pearson Education, Inc.
1
x
3
csc 2
dx
1
x
dx
6 x y2
2 xy 1
dx
194
Chapter 3 Derivatives
32.
y
xe x
33.
y
ln(1 x 2 )
34.
y
ln x 1
35.
y
2
tan 1 e x
dy
x 1
( xe x
e x )dx
2x
1 x2
dy
y
cot 1 12
x
Note: dd cos 1
37.
38.
y
y
sec 1 (e x )
e tan
1
x2 1
dx
ln( x 1) 12 ln( x 1)
1
dy
ex
1
36.
(1 x)e x dx
ex
2 2
1
x 1
dy
2
2 xe x
2 x dx
1
dy
dy
2
x
e tan
(e
1
x 2
)
1
2
d cot 1 1
, so that dx
2
1
x2 1
2
12
2
1 4 x2
1 4 x2
1
1
ex
1
1
dx
2
1 ( x2
2
1
ex
1 e2 x
1) 1/2 2 x dx
2
(x
2
40. f ( x) 2 x 2 4 x 3, x0
1, dx 0.1
f ( x) 4 x 4
(a) f f ( x0 dx ) f ( x0 ) f ( .9) f ( 1) .02
(b) df f ( x0 ) dx [4( 1) 4](.1) 0
(c) | f df | |.02 0| .02
41. f ( x) x3 x, x0 1, dx 0.1
f ( x) 3x 2 1
(a) f f ( x0 dx ) f ( x0 ) f (1.1) f (1) .231
(b) df f ( x0 )dx [3(1) 2 1](.1) .2
(c) | f df | |.231 .2| .031
42. f ( x) x 4 , x0 1, dx 0.1
f ( x ) 4 x3
(a) f f ( x0 dx ) f ( x0 ) f (1.1) f (1) .4641
(b) df f ( x0 )dx 4(1)3 (.1) .4
(c) | f df | |.4641 .4| .0641
2014 Pearson Education, Inc.
1 x2 1
2) x2 1
dx
2
x3
x4 1
x4
2x
x4 1
dx
xe tan
1
x3
1
x4
. Thus dy
39. f ( x) x 2 2 x, x0 1, dx 0.1
f ( x) 2 x 2
(a) f f ( x0 dx ) f ( x0 ) f (1.1) f (1) 3.41 3 0.41
(b) df f ( x0 ) dx [2(1) 2](0.1) 0.4
(c) | f df | |0.41 0.4| 0.01
Copyright
1
x
( e x ) dx
x2 1
x 3 dx
2( x2 1)
dx
1d
1
dx
2
d (cos 1 (2 x ))
d , so that dx
1
e
2
1 e2 x
cos 1 (2 x) Note: dd cot 1
1
1 1
2 x 1
2
1 4 x2
2x
x4 1
dx
Section 3.11 Linearization and Differentials
43. f ( x) x 1 , x0 0.5, dx 0.1
(a) f f ( x0 dx ) f ( x0 )
1
( 4) 10
1
df | | 13 52 | 15
(b) df
44. f ( x) x3
(a) f
(b) df
(c) | f
2 x 3, x0 2, dx 0.1
f ( x) 3 x 2 2
f ( x0 dx) f ( x0 ) f (2.1) f (2) 1.061
f ( x0 )dx (10)(0.10) 1
df | |1.061 1| .061
r3
45. V
4
3
47. S
6 x2
4 r02 dr
dV
dS
r r2
dS
h2
r (r 2
2 r02
h
r02
h2
h 2 )1/2 , h constant
(r 2
h 2 )1/2
50. S
2 rh
dS
dr
r r (r 2
h2 ) 1/2
dS
dr
r 2 h2
r2
r 2 h2
2
2 r and dC
dA
3x02 dx
dV
12 x0 dx
dr , h constant
dV
51. Given r 2 m, dr .02 m
(a) A
r2
dA 2 r dr
.08
(b) 4 (100%) 2%
52. C
x3
46. V
r 2 h, height constant
49. V
1
3
2
5
f ( x0 )dx
(c) | f
48. S
f ( x)
x 2
f (.6) f (.5)
195
2 r dr
2 in.
dS
2 r dh
2 (2)(.02) .08 m 2
dC
2 dr
1
dr
the diameter grew about
2
in.; A
r2
10 in.2
1
2 (5)
2 r0 h dr
53. The volume of a cylinder is V
r 2 h. When h is held fixed, we have dV
2 rh, and so dV 2 rh dr .
dr
For h 30 in., r 6 in., and dr 0.5 in., the volume of the material in the shell is approximately
dV 2 rh dr 2 (6)(30)(0.5) 180
565.5 in 3 .
54. Let
angle of elevation and h
height of building. Then h
| dh | 0.04h, which gives: |30sec2 d | 0.04 |30 tan |
| d | 0.04sin
5
12
cos
5
12
30 tan , so dh
1
cos 2
|d |
0.04 sin
cos
30 sec 2
d . We want
| d | 0.04sin cos
0.01 radian. The angle should be measured with an error of less than
0.01 radian (or approximately 0.57 degrees), which is a percentage error of approximately 0.76%.
dr
55. The percentage error in the radius is dtr
(a) Since C
dr
dt
2 r
2 r
2
100
dC
dt
dr
dt
r
2
100
dr . The
dt
100
2%.
percentage error in calculating the circle s circumference is
2%.
Copyright
2014 Pearson Education, Inc.
dC
dt
C
100
196
Chapter 3 Derivatives
r2
(b) Since A
r dr
dt
2
2
dA
dt
100
r
2
dr
dt
100
r
2(2%)
6 x2
dS
dt
A
100
4%.
dx
dt
56. The percentage error in the edge of the cube is
(a) Since S
dA
dt
2 r dr
. The percentage error in calculating the circle s area is given by
dt
x
100
0.5%.
dS
dt
12 x dx
. The percentage error in the cube s surface area is
dt
12 x dx
dt
100
S
6 x2
100
dx
2 dtx
100
x3
(b) Since V
dx
dt
3 x
h3
57. V
100
1 h
300
| dh |
1%
3
5 Di dDi . Recall that V
Di2
40
5 Di dDi
(1)( h3 )
100
dDi
Di
(1)( h3 )
100
| dV |
Di 2
2
r2h
Di2 h
4
h
dV . We want | V | (1%)(V )
and h 10
| dV |
1
100
V
5 Di2
2
dS
s
hdDe
5 Di2
2
Di2
40
200. The inside diameter must be measured to within 0.5%.
(b) Let De represent the exterior diameter, h the height and S the area of the painted surface. S
dS
100
x3
h. Therefore the greatest tolerated error in the measurement
58. (a) Let Di represent the interior diameter. Then V
dV
100
3(0.5%) 1.5%
(1)( h3 )
100
1 %.
3
3 x 2 dx
dt
dV
3 x 2 dx
. The percentage error in the cube s volume is Vdt
dt
dV
dt
3 h 2 dh ; recall that V dV. Then | V | (1%)(V )
dV
|3 h 2 dh |
of h is
2(0.5%) 1%
dDe
De
De h
. Thus for small changes in exterior diameter, the approximate percentage
change in the exterior diameter is equal to the approximate percentage change in the area painted, and to
estimate
the amount of paint required to within 5%, the tank s exterior diameter must be measured to within 5%.
104
2
106
106
2
106
102 %
6
60. V
4
3
%
3%
6
r3
dV
4
3
D 3
2
D3
200
D2
2
D 3
2
4
3
59. Given D 100 cm, dD 1 cm, V
D3
6
dV
2
D 2 dD
2
(100)2 (1)
104
2
. Then dV
(100%)
V
104
2
106
6
D3
6
dD
D2
2
dV
D3
200
dD; recall that V
dD
D
100
(1%) D
dV . Then
V
(3%)V
3
100
D3
6
D3
200
the allowable percentage error in measuring the
diameter is 1%.
b dg
61. W
a
b
g
a bg
1
dW
2
bg dg
b dg
g
2
dWmoon
dWearth
(5.2)2
b dg
(32)2
32
5.2
2
37.87, so a change of gravity
on the moon has about 38 times the effect that a change of the same magnitude has on Earth.
Copyright
2014 Pearson Education, Inc.
Section 3.11 Linearization and Differentials
62. C (t )
4 8t 3
0.06 t
0.06e
(1 t 3 )2
hours changes from
1
3
C
1
3
1
3
1
2
1
3
1
6
C
, where t is measured in hours. When the time changes from 20 min to 30 min, t in
to 12 , so the differential estimate for the change in C is
0.584 mg/mL.
63. The relative change in V is estimated by
1.1r and
64. (a) T
r
2
dV /dr
V
r
4 kr3
kr4
r
4 r.
r
If the radius increases by 10%, r changes to
0.1r. The approximate relative increase in V is thus
L
g
1/2
197
dT
2
L
1
2
g 3/2 dg
4(0.1r )
r
0.4 or 40%.
Lg 3/2 dg
(b) If g increases, then dg 0 dT 0. The period T decreases and the clock ticks more frequently. Both the
pendulum speed and clock speed increase.
(c) 0.001
100(980 3/2 ) dg dg
0.977 cm/sec2 the new g 979 cm/sec 2
65. (a) i.
ii.
Q ( a ) f ( a) implies that b0 f (a).
Since Q ( x) b1 2b2 ( x a ), Q ( a )
iii. Since Q ( x)
2b2 , Q (a )
In summary, b0
f (a), b1
f ( a ) implies that b1
f (a ) implies that b2
f (a )
.
2
f (a), and b2
f (a).
f (a )
.
2
(b) f ( x) (1 x) 1 ; f ( x)
1(1 x ) 2 ( 1) (1 x ) 2 ; f ( x)
2(1 x ) 3 ( 1) 2(1 x) 3 Since
f (0) 1, f (0) 1, and f (0) 2, the coefficients are b0 1, b1 1, b2 22 1. The quadratic
approximation is Q ( x) 1 x x 2 .
(c)
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
1x 2 ; g ( x) 2 x 3
(d) g ( x) x 1; g ( x )
Since g (1) 1, g (1)
1, and g (1) 2, the coefficients are b0
2
1, b1
1, b2
2
2
1. The quadratic
approximation is Q ( x) 1 ( x 1) ( x 1) .
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(e) h( x)
(1 x)1/2 ; h ( x )
1 (1
2
x ) 1/2 ; h ( x)
1 , and h (0)
2
2
is Q ( x) 1 2x x8 .
Since h (0) 1, h (0)
approximation
Copyright
1 , the
4
1 (1
4
x) 3/2
coefficients are b0
1, b1
2014 Pearson Education, Inc.
1,b
2 2
1
4
2
1 . The
8
quadratic
198
Chapter 3 Derivatives
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f ) The linearization of any differentiable function u ( x) at x a is L( x) u (a ) u (a )( x a) b0 b1 ( x a ),
where b0 and b1 are the coefficients of the constant and linear terms of the quadratic approximation. Thus,
the linearization for f ( x ) at x 0 is 1 x; the linearization for g ( x) at x 1 is 1 ( x 1) or 2 x; and the
linearization for h( x) at x 0 is 1 2x .
66. E ( x)
f ( x) g ( x )
E ( x)
Next we calculate m:
f (a) m
as claimed.
67. (a)
0
f ( x)
2x
f ( x)
log3 x
f ( x) m( x a ) c. Then E ( a)
E ( x)
lim
x a x a
m
0
f ( x ) m( x a ) c
lim
x a
x a
f (a ). Therefore, g ( x)
f ( x)
2 x ln 2; L( x)
0
m( x a ) c
(20 ln 2) x 20
f ( a ) m( a a ) c
0
lim
x
a
f ( x) f ( a)
x a
f (a)( x a)
m
c
0
0 (since c
f ( x)
1 ,
x ln 3
and f (3)
ln 3
ln 3
x ln 2 1 0.69 x 1
L( x)
1 (x
3ln 3
ln 3
3) ln
3
(b)
69 74.
Example CAS commands:
Maple:
with(plots):
a : 1: f : x -> x^3 x^2 2*x;
plot(f(x), x 1..2);
diff (f(x), x);
fp : unapply ( , x);
L: x ->f(a) fp(a)*(x a);
plot({f(x), L(x)}, x 1..2);
err: x -> abs(f(x) L(x));
Copyright
2014 Pearson Education, Inc.
f (a))
f ( a) is the linear approximation,
(b)
68. (a)
f ( a).
x
3ln 3
1
ln 3
1
Chapter 3 Practice Exercises
plot(err(x), x
err( 1);
1..2, title
199
#absolute error function#);
Mathematica: (function, x1, x2, and a may vary):
Clear[f , x]
{x1, x2} { 1, 2}; a 1;
f[x_ ]: x 3 x 2 2x
Plot [f[x], {x, x1, x2}]
lin[x_ ] f[a] f [a](x a)
Plot[{f[x], lin[x]},{x, x1, x2}]
err[x_ ] Abs [f[x] lin[x]]
Plot[err[x], {x, x1, x2}]
err//N
After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and
delta (del)
eps 0.5; del 0.4
Plot[{err[x], eps}, {x, a del, a del}]
CHAPTER 3
PRACTICE EXERCISES
1. y
x5 0.125 x 2 0.25 x
2. y
3 0.7 x3 0.3 x7
3. y
x3 3( x 2
4. y
x7
5. y
( x 1)2 ( x2
2
1
7x
2( x 1)(2 x
)
7. y
( 2 sec
8. y
1 csc2
9. s
t
10. s
11. y
t
1
t 1
ds
dt
2.1x 2
2.1x 6
3x 2 3(2 x 0)
dy
dx
7 x6
dy
dx
3x2 6 x
3x( x 2)
7
( x 1)2 (2 x 2) ( x 2
2 x)(2( x 1))
2( x 1)[( x 1) 2
x( x 2)]
4 x 1)
(2 x 5)(4 x ) 1
1
5 x 4 0.25 x 0.25
dy
dx
2 x)
6. y
ds
dt
dy
dx
1
2
dy
dx
dy
dx
dy
d
1)3
2
(2 x 5)( 1)(4 x) 2 ( 1) (4 x) 1 (2)
2
dy
d
4
1
3( 2 sec
t
1
t
2
t
( t 1) (0) 1
2 tan 2 x sec2 x
t 1
dy
dx
2
2
1 csc2
2
1
2 t
1)2 (2
1
2 t
1
t
t
2 t 1
1
2 t
1
2 t
t 1
t
sec tan )
csc cot
2
4
2
2
1
2 t 1
t
1 csc2
2
4
(csc cot
2
2
(4 tan x)(sec2 x) (2sec x )(sec x tan x)
Copyright
(4 x) 2 [(2 x 5) 2(4 x)] 3(4 x ) 2
2sec2 x tan x
2014 Pearson Education, Inc.
)
200
Chapter 3 Derivatives
dy
dx
csc2 x 2 csc x
12. y
1
sin 2 x
2
sin x
13. s
cos 4 (1 2t )
14. s
cot 3 2t
15. s
(sec t tan t )5
16. s
csc5 (1 t 3t 2 )
(2 csc x cot x )(1 csc x)
4 cos3 (1 2t )( sin(1 2t ))( 2) 8cos3 (1 2t )sin(1 2t )
ds
dt
3cot 2 2t
ds
dt
(2 csc x)( csc x cot x ) 2( csc x cot x)
csc 2 2t
2
t2
6
t2
cot 2 2t csc2 2t
5(sec t tan t ) 4 sec t tan t sec2 t
ds
dt
5(sec t )(sec t tan t )5
5 csc4 (1 t 3t 2 ) ( csc (1 t 3t 2 ) cot (1 t 3t 2 )) ( 1 6t )
ds
dt
5(6t 1) csc5 (1 t 3t 2 ) cot (1 t 3t 2 )
17. r
2 sin
(2 sin )1/2
dr
d
1 (2
2
cos
2 (cos )1/2
dr
d
2
18. r
2
19. r
sin 2
20. r
sin
21. y
1 x 2 csc 2
2
x
22. y
2 x sin x
23. y
x 1/2 sec (2 x) 2
sin(2 )1/2
dy
dx
cos
1 x2
2
dy
dx
1
2 x3/ 2
dy
dx
or
2 x
3
sin x
1
2
2x
2
2 x
cos
csc 2x cot 2x
cos x
x 3/2 sec (2 x) 2
sin
cos
1 x1/2 sec (2 x ) 2
2
x csc 2x
sin x
x
x 1/2 sec (2 x )2 tan(2 x )2 (2(2 x) 2) sec (2 x)2
1
2
2 cos
1
16 tan (2 x )
1
2
2
x 3/2
x 2
dy
x1/2 ( csc ( x 1)3 cot ( x 1)3 ) (3( x 1) 2 ) csc ( x
dx
csc( x 1)3
1 x csc ( x 1)3 1 6( x 1) 2 cot ( x 1)3
1)3 cot ( x 1)3
x
2
2 x
2
3
csc ( x 1) 1 6 x ( x 1) cot ( x 1)
dy
dx
25. y
5cot x 2
26. y
x 2 cot 5 x
27.
x 2 sin 2 (2 x 2 )
y
1
2 x
csc 2x
1 1
1
x1/2 csc ( x 1)3
3 x ( x 1)2 csc ( x
1
2
x2
1
2 cos
cos 2
2
2
2
1
2
sin
cos
sec (2 x ) 2 16 x 2 tan(2 x )2 1
x csc( x 1)3
24. y
csc 2x cot 2x
cos sin
2 sin
2sin )
(cos ) 1/2 ( sin ) 2(cos )1/2
1 1
2 x cos x
8 x1/2 sec (2 x) 2 tan (2 x)2
or
1
2
cos(2 )1/2 12 (2 ) 1/2 (2)
dr
d
dr
d
1
sin ) 1/2 (2 cos
dy
dx
5( csc2 x 2 )(2 x )
10 x csc2 ( x 2 )
x 2 ( csc2 5 x)(5) (cot 5 x )(2 x)
dy
dx
2
5 x 2 csc2 5 x 2 x cot 5 x
x 2 (2sin (2 x 2 )) (cos (2 x 2 ))(4 x) sin 2 (2 x 2 )(2 x)
8 x3 sin(2 x 2 ) cos(2 x ) 2 x sin 2 (2 x 2 )
Copyright
2014 Pearson Education, Inc.
1)3 12 x 1 2
or 1 csc ( x 1)3
2 x
Chapter 3 Practice Exercises
2
4t
t 1
29. s
30. s
2
x
x 1
32. y
2 x
2 x 1
33. y
x2 x
x2
35. r
36. r
2
dy
dx
x (1)
( x 1)3
(2 x 1)
dy
dx
1
2 x
1
x
1
x
2 x
1
2
1/2
1 1x
dy
dx
4( x
8t 3
1 x
( x 1)3
1
x
1
2 x2 1
4
(2 x 1)3
1
x
4 x 12 ( x x1/2 ) 1/2 1 12 x 1/2
x)
(cos
sin
2 cos
1
(t 1)
(2 x 1)3
1
x2
6sin ( x3 ) cos ( x3 ) 2 x 3 sin 2 ( x3 )
3
(15t 1)4
4 x
(2 x 1) 2
2 x 1
2
sin
dr
cos 1
d
(2sin )(1 cos )
( x 1) 2 x
( x 1)2
4 x( x x1/2 )1/2
x
3
4
2 t4t1
(t 1) 2
3)(15t 1) 4 (15)
1 (
15
1
2 x
2 2 x
1/2
1 1x
x ) 1 2 (2 x
(x
1)(cos ) (sin )( sin )
(cos
1)
2
x
( x x1/2 )1/2 (4)
6x 5 x
4x 4 x )
x
x
cos2
cos sin 2
(cos 1)2
(cos
cos 2
sin
2 cos
1
2 sin
(cos 1)2
1)3
2
sin 1
1 cos
2(sin 1)(cos
ds
dt
( x 1)
2 x x1
x ) 1 2 2x 1
(cos
1) 3
1 (15t
15
dy
dx
4x x
(x
x 2 (2sin ( x 3 )) (cos ( x3 ))(3x 2 ) sin 2 ( x3 )( 2 x 3 )
3 (t 1)(4) (4t )(1)
2 t4t1
(t 1)2
ds
dt
1
15(15t 1)3
31. y
34. y
dy
dx
x 2 sin 2 ( x3 )
28. y
(1 cos )(cos ) (sin
dr 2 sin 1
1 cos
d
sin 1)
1)(sin )
(1 cos )2
2(sin
1)
(1 cos )3
sin 2
(1 cos )3
(2 x 1) 2 x 1
38. y
20(3x 4)1/4 (3x 4) 1/5
39. y
3(5 x 2 sin 2 x) 3/2
40. y
(3 cos3 3 x) 1/3
dy
dx
(2 x 1)3/2
37. y
dy
dx
dy
dx
41.
y 10e x /5
dy
dx
(10)
42.
y
2e 2 x
dy
dx
2
3
3
2
e x /5
2 e 2x
(5 x 2
1)1/2 (2)
dy
dx
20(3x 4)1/20
1 (3
3
1
5
3 (2 x
2
3 2x 1
1 (3 x 4) 19/20 (3)
20 20
sin 2 x ) 5/2 [10 x (cos 2 x )(2)]
(3 x
9(5 x
5x
cos3 3 x ) 4/3 (3cos 2 3x )( sin 3 x)(3)
2
3
4)19/ 20
cos 2 x )
sin 2 x
3cos2 3 x sin 3 x
(3 cos3 3 x )4/3
2e x /5
2e 2 x
Copyright
201
2014 Pearson Education, Inc.
5/ 2
sin )
202
Chapter 3 Derivatives
xe 4 x
dy
dx
1 e4 x
16
1 [ x (4e 4 x )
4
1 (4e 4 x )
e 4 x (1)] 16
43.
y
1
4
44.
y
x 2 e 2/ x
45.
y
ln(sin 2 )
dy
d
2(sin )(cos )
46.
y
ln(sec2 )
dy
d
2(sec )(sec tan )
47.
y
2
log 2 x2
48.
y
log5 (3 x 7)
49.
y
8 t
51.
y
5 x3.6
52.
y
2x
53.
y
y
55.
y
dy
dx
dy
dx
2
3
(ln 5)(3 x 7)
50.
dy
dt
9 2t
y
92t (ln 9)(2)
2 x
ln( x 2) x 2
2 1
92t (2 ln 9)
2 1
2x
y
y
( x 2) ln( x 2)
( x 2) x 1 2
(1) ln( x 2)
( x 2) x 2 [ln( x 2) 1]
ln y
1
2 ln x
1
2
1 (1
2
u2 )
sin 1 1
y
1
v
y
y
ln(ln x ) 2(ln x) x /2
(ln x ) x /2 ln(ln x)
1
ln x
x
2
x
2
0
ln(cos
x)
1
x
ln x
(ln(ln x)) 12
sin 1 (1 u 2 )1/2
1/ 2
( 2u )
2
1/ 2 2
1 (1 u )
y
ln(ln x )
ln[2(ln x) x /2 ] ln(2)
u
u
1 u 2 1 (1 u 2 )
u 1 u2
dy
dv
sin 1 v 1/2
1 v 3/ 2
2
1/ 2 2
1 (v
)
u
1
u 1 u2
1 u2
1
2v3/ 2 1 v
, 0<u<1
1
1
2v3/ 2
v 1
v
1
57.
23 2/ x (1 x)
18 x 2.6
2
ln y
3
3x 7
1
ln 5
8 t (ln 8)
5(3.6) x 2.6
1
(2 2 x)e 2 x
xe4 x
2
(ln 2) x
x2
2
dy
dx
1 e4 x
4
2 tan
x
1
ln 2
1 e4 x
4
2 cot
sec2
8 t (ln 8)( 1)
dy
dx
2 cos
sin
sin 2
ln(3 x 7)
ln 5
sin 1 1 u 2
dy
du
56.
dy
dt
1
1
x 2 [(2 x 2 )e 2 x ] e 2 x (2 x)
dy
dx
ln 2
2(ln x) x /2
y
1
x2
2
ln
( x 2) x 2
dy
dx
54.
x2e 2 x
xe4 x
y
1 x2
1
cos
1
x
1 x 2 cos
Copyright
1
x
2014 Pearson Education, Inc.
v
2v3/ 2 v v v1
1
2
1
Chapter 3 Practice Exercises
58.
y
z cos 1 z
dy
dz
1 z2
cos 1 z
t tan 1 t
60.
y
(1 t 2 ) cot 1 2t
61.
y
z sec 1 z
62.
y
1
2
dy
dt
ln t
dy
dt
z2 1
z z2 1
y
csc 1 (sec )
64.
y
1
(1 x 2 )e tan x
( xy
66. x 2
2
67. x3
y2 5x
69. ( xy )1/2
71. y 2
sec tan
sec
sec2
tan 1 t
1
t
t
1 t2
cos 1 z
1
2t
2
1 4t 2
1
2
1
2 xe tan x
2x
dy
x dx
x
z
sec 1 z
z
z z2 1
z2 1
xy
sec 1 z , z 1
1 z
z2 1
1
2 xe tan x
3y
2 y
x
x 1x
2 y
y
1
1
e tan x
y ( x 3)
dy
2 y dx 5 0
1
sec
2
2
dy
y
x
1
2 x 1 x 2
1x
1 x
1
2 sec
1, 0
tan
(1 x 2 ) e
0
1/ 2
x x 1
tan
tan
1
y) 2 3 y
1
1
x
x 1
2 3x2
1 ( xy ) 1/2
2
dy
x 2 2 y dx
dy
4 x dx
dy
x dx
dy
dx
4y
dy
x dx
y
y 2 (2 x )
( x 1)(1) ( x )(1)
( x 1)2
0
dy
4 y1/3 dx
4y
2 3x
2 y dx
5 2x
y
y
x
dy
(x
dx
2
3
2 y ) 5 2x
2
dy
12 y1/5 dx
0
dy
x1/2 y 1/2 dx
dy
dy
dx
dy
4 y1/3 dx
dy
4 x 1/5
dy
dx
4 x dx
2 3x2
4y
4y
2 x 2 y dx
0
2
4 x 4 y1/3
dy
4 x 1/5 12 y1/5 dx
15
2 y dx
dy
3x2
2x
(4 x 4 y1/3 )
68. 5 x 4/5 10 y 6/5
70. x 2 y 2
z
1 z2
5 2x y
x 2y
4 xy 3 y 4/3
dy
dx
2t cot 1 2t (1 t 2 )
y
65. xy 2 x 3 y 1
dy
dx
1
2
1
1 t2
z
1 z2
2( x 1)1/2 sec 1 ( x1/2 )
dy
d
63.
xy
tan 1 t t
( x 1) 1/2 sec 1 ( x1/2 ) ( x 1)1/2
1
2
2
cos 1 z
z sec 1 z ( z 2 1)1/2
2 x 1 sec 1 x
dy
dx
(1 z 2 ) 1/2 ( 2 z )
(sec 1 z )(1) 12 ( z 2 1) 1/2 (2 z )
1
z
1
2
z
y
dy
dz
z cos 1 z (1 z 2 )1/2
1 z2
59.
203
2 xy 2
dy
dx
x 1/2 y1/2
dy
dx
y
x
1
2 y ( x 1)2
Copyright
2014 Pearson Education, Inc.
1 x 1/5 y 1/5
3
x 1y
dy
dx
1
3( xy )1/5
y
x
y
204
Chapter 3 Derivatives
72. y 2
1 x
1 x
73. e x 2 y
74.
y4
dy
2e 1/ x
y2
77.
1
ye tan x
78.
xy
2
4 pq 3q 2
dp
dq
dp
2 p ) 3/2
d ( y ln x)
dx
ln 2
2
dp
4 p q dq
6q
1
2e tan x
0
y 1x
dp
4q dp
3 p 2 dq
0
dp
2 p) 5/2 10 p dq
3 (5 p 2
2
1
dp
1x
2e tan
1 x2
1
1 x2
dy
ln x dx
0
6q 4 p
dy
dx
y
x ln x
dp
(3 p 2
dq
4q )
dp
(10 p
dq
2)
6q 4 p
dp
2 (5 p 2
3
2 dq
2 p)5/2
5/ 2
5 p2 2 p
3(5 p 1)
81. r cos 2 s sin 2 s
r ( sin 2 s )(2) (cos 2 s ) dr
ds
2 r sin 2 s sin 2 s
cos 2 s
82. 2rs r s s 2
y3
d2y
dx
(b) y 2
2
cos x x 1
6q 4 p
dp
dq
83. (a) x3
x2 1
x
x)
3 p 2 4q
80. q (5 p 2
dr
ds
y
x
1
d ( tan 1 x )
2e tan x dx
y ln x
3 p 2 dq
2
dy
dx
(1 x 2 ) cos( x 1
dy
dx
ln(21/2 )
e 1/ x
yx 2
dy
dx
x)
x)
1
2e tan x
ln( x y )
2
79. p 3
d (x 1
x) dx
y
2e 1/ x
x2
0
y2
sin( x 1
1
2 y 3 (1 x )2
1
2
y (1) x dx
0
y
cos( x 1
dy
dx
0
dy
dx
(1 x ) 2
dy
1 d x
x / y dx y
x sin 1 y 1 x 2
(1 x )(1) (1 x )( 1)
d ( x 1)
2e 1/ x dx
2 y dx
1
dy
dx
dy
4 y 3 dx
1 x
1 x
dy
e x 2 y 1 2 dx
1
75. ln xy
76.
1/2
d2y
dx
2 r s dr
ds
3
dy
2
x2
2 xy 2 (2 yx2 )
2 xy
y
y
dy
dx
x2
2 4
y x
2
x2
1
yx 2
dy
dx
0
x2
y2
2 x4
y
2 xy 2
y2
4
2y
1 2s
dy
dx
0
0
dr
ds
(cos 2 s)
2r sin 2 s
(2r 1)(tan 2 s)
dr
ds
3 x 2 3 y 2 dx
1
2
1 2x
(2 r 1)(sin 2 s )
cos 2 s
2sin s cos s
1 2 s 2r
2s 1
dy
d2y
y 2 ( 2 x ) ( x 2 ) 2 y dx
dx 2
y4
dy
dx
y5
2
( yx ) 1
d2y
dx 2
dy
( yx 2 ) 2 y (2 x) x 2 dx
2 xy 2 1
y3 x4
Copyright
dr
ds
(2 s 1) 1 2 s 2r
2 xy 3 2 x 4
4
1
yx 2
dr
ds
2014 Pearson Education, Inc.
2sin s cos s
Chapter 3 Practice Exercises
84. (a) x 2
dy
(b) dx
dy
y2
1
x
y
d2y
dx
85. (a) Let h( x)
dy
2 x 2 y dx
0
dy
y
y (1)
2
y
x dx
2 y dx
x
2
y
6 f ( x) g ( x)
2
(b) Let h( x) f ( x) g ( x)
2(1)(1) 12 (1) 2 ( 3)
x
y
y 2 x2
2
h ( x)
dy
dx
2x
y3
1
y3
x
y
(since y 2
6 f ( x ) g ( x)
x2
1)
6 12
h (1) 6 f (1) g (1)
2
h ( x)
2
f ( x) (2 g ( x)) g ( x) g ( x ) f ( x )
2 f (0) g (0) g (0) g 2 (0) f (0)
h (0)
(5 1)
( g (1) 1) f (1) f (1) g (1)
1
2
3( 4)
(d) Let h( x)
1
f (1) 12
2
g ( f ( x)) h ( x) g ( f ( x)) f ( x) h (0) g ( f (0)) f (0) g (1) f (0) ( 4)( 3) 12
( x f ( x))3/2
h ( x) 32 ( x f ( x))1/2 (1 f ( x)) h (1) 32 (1 f (1))1/2 (1 f (1))
f ( g ( x))
86. (a) Let h( x)
x f ( x)
(b) Let h( x)
( f ( x))1/2
(c) Let h( x)
f
( g ( x ) 1)
h ( x)
9
3)1/2 1 12
2
(g) Let h( x) f ( x g ( x ))
1 3
f (1) 1 12
2 2
3 (1
2
h ( x)
(e) Let h( x)
2
(5 1)
1
1
2
4
f ( g (0)) g (0)
f ( x g ( x))(1 g ( x))
x f ( x)
h ( x)
h ( x)
f ( x)
2 cos x
h (0)
( g (1) 1)
2
h (0)
f ( g (0))(1 g (0))
f
f ( x)
1
2 x
h (1)
1 ( f ( x)) 1/2 ( f ( x ))
2
1
x
h (1)
2 x
f
1
f (1 5 tan x)( 5sec 2 x )
1
2 1
1 ( 3) 1
5
2
1 (9) 1/2 ( 2)
2
f (1 5 tan 0)( 5sec2 0)
h (0)
h (0)
(2 cos x )2
f (1)
1 ( f (0)) 1/2 f (0)
2
1
1 1
1
2 1 5 2 10
h (0)
(2 cos x ) f ( x ) f ( x )( sin x )
h ( x)
1 f (1)
(2 1) f (0) f (0)(0)
(2 1)2
(f ) Let h( x) 10sin 2x f 2 ( x) h ( x) 10sin 2x (2 f ( x) f ( x)) f 2 ( x) 10 cos 2x
h (1) 10sin 2 (2 f (1) f (1)) f 2 (1) 10cos 2
20( 3) 15 0
12
2
t2
87. x
dx
dt
dy dx
dx dt
dy
thus, dt
88. t
(u 2
2u )1/3
2
1/3
2(u
dw dr
dr ds
dw
ds
3sin 2 x
6 cos (2t 2 ) 2t
dy
3(cos 2 x)(2)
dx
dy
6 cos(0) 0
dt t 0
6 cos 2 x
6 cos(2t 2
2 )
5;
[2(2
2
cos e 3/2 e 3/2
e r 1
2 r
1
2 3/2
3cos s
3cos 6
Copyright
6
; at x
3 3e 3/2
4 3/2
2
3
2
6 cos (2t 2 );
ds
1 (u 2 2u ) 2/3 (2u 2) 2 (u 2 2u ) 2/3 (u 1); s t 2 5t
3
3
dt
ds
ds dt [2(u 2 2u )1/3 5] 2 (u 2 2u ) 2/3 (u 1)
thus du
3
dt du
2(2))1/3 5] 23 (22 2(2)) 2/3 (2 1) 2(2 81/3 5)(8 2/3 ) 2(2 2
cos e r
3( 2)
9
0
dt
du
2u )
ds
du u 2
dw
ds
2t ; y
5
12
3
4
h ( x)
x
h (1)
2
f ( g ( x )) g ( x )
(d) Let h( x) f (1 5 tan x) h ( x)
f (1)( 5) 15 ( 5)
1
89.
7
f ( x)
g ( x) 1
(e) Let h( x)
(f ) Let h( x)
( g ( x ) 1) f ( x) f ( x ) g ( x )
( 4)
(c) Let h( x)
h ( x)
205
0, r
cos e 3/2
3
2
3sin 6
3 2e
4
3/2
cos e 3/2
2014 Pearson Education, Inc.
2t 5
5) 14
9
2
13
10
1
3
206
90.
Chapter 3 Derivatives
2
t
91. y3
y
7) 2/3
2 (1
3
1
dy
dy
dx
dy
(3 y
92. x1/3
y1/3
4
x 2/3
d2y
2
1)
1
3
y 2/3
x 2/3
1 and f (t
2t 1
2
(2t 2h 1)(2t 1)
dy
y 2/3 dx
2h2
h
d2y
2
dx 2
h)
f (t )
4 x 2h
2sin x
dy
dx
x
82/3
2
3
(8, 8)
1; dx
1/3
8
( 1)
2( x h)2 1 2 x 2
g ( x)
g ( x h) g ( x )
h
0
lim
h
2 sin(0)
3 1
0;
x 2/3
82/3
2
3
8
1/3
1
3
1
3
2/3
8
1
g ( x h) g ( x )
h
lim (4 x 2h)
h
1
y 2/3
84/3
4 xh 2h 2 1
1
1
1
1
2
dy
dy
dx (8, 8)
2/3
t 0,
dy
dx (0, 1)
2sin x
3 y2 1
f (t h ) f (t )
2t 1 (2t 2 h 1)
2( t h ) 1 2 t 1
1
2(t h) 1
h
h
(2t 2 h 1)(2t 1) h
f (t h ) f ( t )
2
2
lim
lim
2
h
h 0 (2t 2h 1)(2t 1) (2t 1)
h 0
2 x 2 1 and g ( x h)
4 xh
y 2/3
1
93. f (t )
94. g ( x)
2 x 1/3
3
1
6
7)1/3
1 so that ddt
1
( 1)
( 2
;r
(3 1)2
dy
dx
0
2 t 1
(3 1)( 2 cos 0)( 2sin 0)(6 0)
dx 2 (0,1)
1/3 dy
dx
dx 2
1
6
(3 y 2 1)
d2y
2
1 x 2/3
3
2y
3
dy
dx
2sin x
(3 y 2 1)( 2cos x ) ( 2sin x) 6 y dx
2
dr
dr
d t 0 d t 0
dr
dt t 0
1
6
3 y 2 dx
2cos x
d2y
dx
2
d
d (2 t 1)
d
t 2 ddt
0
dt
dt
dt
7) 2/3 (2 ) 23 ( 2 7) 2/3 ; now t 0 and 2 t
1( 2
3
dr
d
and ddr
2
2
1
2
3
4
1
6
2h
(2t 2 h 1)(2t 1) h
(2 x 2 4 xh
2 h2 1) (2 x 2 1)
h
4x
0
95. (a)
(b)
(c)
lim f ( x)
lim x 2
x
0
x
0
x
0
x
0
0 and lim f ( x)
0
x
follows that f is continuous at x 0.
lim f ( x ) lim (2 x) 0 and lim f ( x)
x
follows that f is differentiable at x
0
0.
x2
lim
x
0
0
lim ( 2 x)
x
0
lim f ( x)
x
0
0
lim f ( x)
x
0
96. (a)
Copyright
0. Since lim f ( x)
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x
0
0
f (0) it
0. Since this limit exists, it
Chapter 3 Practice Exercises
(b)
lim f ( x)
lim x
x
0
x
0
x
0
x
0
0 and lim f ( x)
lim tan x
x
0
x
x
0
x
follows that f is continuous at x 0.
lim f ( x) lim 1 1 and lim f ( x )
(c)
that f is differentiable at x
0.
0
0
lim f ( x)
x
lim sec2 x 1
0. Since lim f ( x)
0
x
0
0
207
f (0), it
lim f ( x ) 1. Since this limit exists it follows
x
0
0
97. (a)
(b)
lim f ( x)
lim x 1 and lim f ( x)
x 1
x 1
x 1
x 1
x 1
x 1
follows that f is continuous at x 1.
lim f ( x) lim 1 1 and lim f ( x)
(c)
98. (a)
lim f ( x)
x
0
f (0)
(b)
lim sin 2 x
x
0
x
0
lim (sin 2 x)
0
x
0
2
1
x
lim f ( x)
0
0
1
2x 4
1
2
0
x
x (2 x 4) 1
2(2 x 4) 2
(2 x 5)(2 x 3)
lim 2 cos 2 x
x
1
0
x
dy
dx
1
(2 x 4)2
5 or x
2
dy
x e x ; dx 1 e x 2 e x
slope of 2 at the point (0, 1).
100. y
101. y
2 x 3 3 x 2 12 x 20
102. y
x3
1
2
x
0, independent of m; since
0
0
0
lim ( mx)
x
0
m
2.
lim m
x
3
2
5, 5
2 9
x
4 x 2 16 x 16 1
and 32 ,
0
y
1
4
m
0
2(2 x 4) 2 ; the slope of the tangent is 32
(2 x 4)2 1
1
x 1
0 for all values of m.
lim f ( x)
0
lim f ( x ), so lim f ( x) does not
x 1
lim f ( x)
x
2 and lim f ( x)
x
f (1), it
x 1
x 1
lim mx
0
0 provided that lim f ( x)
differentiable at x
x
2
0 and lim f ( x)
x
lim f ( x) 1. Since lim f ( x) 1
x 1
lim f ( x) it follows that f is continuous at x
0
lim f ( x)
x
lim 1
x 1
f is not differentiable at x 1.
exist
99. y
lim (2 x) 1
x 1
4 x 2 16 x 15
3
2
f is
2(2 x 4) 2
1
2
0
are points on the curve where the slope is
0 e0
3.
2
1. Therefore, the curve has a tangent with a
dy
dx
dy
6 x 2 6 x 12; the tangent is parallel to the x -axis when dx 0
6 x 2 6 x 12 0
x 2 x 2 0 ( x 2)( x 1) 0 x 2 or x
1 (2, 0) and ( 1, 27) are points
on the curve where the tangent is parallel to the x-axis.
dy
dx
3x 2
dy
dx ( 2, 8)
12; an equation of the tangent line at ( 2, 8) is y 8 12( x 2)
y 12 x 16; x-intercept : 0 12 x 16
103. y
2 x 3 3 x 2 12 x 20
dy
dx
4
3
x
4,0
3
; y -intercept: y 12(0) 16 16
6 x 2 6 x 12
dy
x when
(a) The tangent is perpendicular to the line y 1 24
dx
x2 x 2 4
x 2 x 6 0 ( x 3)( x 2) 0
x .
where the tangent is perpendicular to y 1 24
0
x
0 or x 1
x
24; 6 x 2
1
1
24
2 or x
dy
3
6 x 12
24
( 2, 16) and (3, 11) are points
2 12 x when dx
12 6 x 2 6 x 12
12 x 2 x 0
(0, 20) and (1, 7) are points where the tangent is parallel to y
2 12 x.
(b) The tangent is parallel to the line y
x ( x 1)
(0, 16)
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2014 Pearson Education, Inc.
208
Chapter 3 Derivatives
104. y
sin x
x
x ( cos x ) ( sin x )(1)
dy
dx
2
dy
dx x
m1
x2
2
dy
dx x
1 and m2
2
1
m2
1. Since m1
2
the tangents intersect at right angles.
dy
sec2 x; now the slope
105. y tan x, 2 x 2
dx
x
2
of y
y
is 12
the normal line is parallel to
x when dy
2
dx
2
1
cos x 2
for
2. Thus, sec 2 x
1
2
cos x
1
cos 2 x
2
x
4
and x
2
4
x
, 1 and 4 , 1 are points
2
2
4
x.
where the normal is parallel to y
2
dy
dx
106. y 1 cos x
dy
dx
sin x
2
1
,1
the tangent at 2 , 1 is the line y 1
y
x 2 1; the normal at 2 , 1 is
y 1 (1) x
107. y
1
2
108. y
dy
dx
x2 C
1 2
2
x3
y
2
C
C
dy
dx
3x 2
x
2 x and y
2
( x a ) ( x 2a )
2
1
2
dy
dx
x
1; the parabola is tangent to y
dy
dx x a
3
3
0
3a 2
3a 2 ( x a )
a
x
the tangent line at (a, a 3 ) is y a 3
a or x
( x a) ( x 2
2a. Now
is 4 times as large as the slope at (a, a 3 ) where x
( x 1) 2
x
c, x
c
( x 1)2
b
1. Thus c c ( x 1)
0
x 1 (since x
1)
a 2 b2
b
( x b)
y
1
2
1 ; thus,
2
y
1
2
12a
2
( x a )( x 2
xa 2a 2 )
2
4 (3a ), so the slope at x
0
2a
the line through (0, 3) and (5, 2) is
c
x 1
x 3
x
1
x 3, x
c
( x 1)2
1
( x 1)[ x 1 ( x 3)] 0,
y2
a 2 b2 has slope
a 2 b2
b
dy
dx
x 3
a 2 . Then x 2
a 2 b2
y
3a 2 ( x a). The tangent line
3a 2 ( x a )
( x 1)( x 3)
c 4.
y2
normal line through b,
a2 b2
a 2 b2
x
a.
c intersects
x 1
2
a 2 b2 be a point on the circle x 2
dy
dx x a
3( 2a )
2a
, so the curve is tangent to y
1. Moreover, y
c ( x 1)( x 3), x
1 ( x 1)(2 x 2)
110. Let b,
y
c
x 1
x 3; y
dy
dx
dy
dx x
xa a 2 )
3 ( 2)
0 5
109. The line through (0, 3) and (5, 2) has slope m
y
x when 2 x 1
1
4
x3 when x
intersects y
x
a2
dy
2 x 2 y dx
a 2 b2
b
a2 b2
0
dy
dx
x
y
normal line is
y
a 2 b2
b
x which passes
through the origin.
111. x 2
2 y2
9
dy
2 x 4 y dx
and the normal line is y
0
dy
dx
2 4( x 1)
dy
dx (1, 2)
x
2y
1
4
the tangent line is y
4 x 2.
Copyright
2014 Pearson Education, Inc.
2 14 ( x 1)
1
4
x
9
4
Chapter 3 Practice Exercises
112. e x
y2
d (e x
dx
2
1
mtan
m
114. ( y x)2
line is y
115. x
xy
dy
2
x dx
2 2( x 3)
is y 1 54 ( x 4)
116. x3/2
y
2 y 3/2
y2
x
3
4
1
2 xy
5
4
1
4
y
dy
2
( y x ) dx
dy
0
x dx
y
dy
dx
1 ( x 3)
2
2
1 ( y x)
dy
3 y1/2. dx
dy
dx
0
x1/ 2
2 y1/ 2
1 3x 2 y3
y 3 (3x 2 )
dy
dx
dy
2 y dx
1
2
dy
dx
4
3
4
5
1
4
4 4( x 1)
4 x.
dy
1 3 x2 y3
3 x3 y 2 2 y 1
2
dy
dx (6, 2)
the tangent
3
4
the tangent
x 10.
dy
dx (4, 1)
5
4
the tangent line
the tangent line is
dy
dy
3 x3 y 2 dx
dy
dx (1, 1)
y = 2x + 1
x 11
.
5
dy
dx (1, 4)
1 dx
1;
2
7.
2
x
2 xy y
x
dy
dx
0)
dy
dx (3, 2)
1 y x
y x
2 43 ( x 6)
2 xy
1 = 2(x
y 2
x 5
y 2
x 6 and the normal line is y 1 45 ( x 4)
dy
2 y 1)
sin( x sin x)
dy
x3 3 y 2 dx
1
2
5)
x 17
and the normal line is y
4
y
curve has slope
dy
(x
dx
0
e0
2(1)
dy
dx (0, 1)
mtan
y 1 2x ; normal line: y
0)
x 52 and the normal line is y
x dx
3 x1/2
2
17
dy
(3x3 y 2
dx
118. y
2( y x) dx 1
4 14 ( x 1)
117. x3 y 3
dy
2 5 dx
ex
2y
dy
dx
0
2 x 4 and the normal line is y
2 34 ( x 6)
1
2 y dx
1 (x
2
dy
2x 4
6
y
dy
ex
d (2)
dx
2; tangent line: y 1
113. xy 2 x 5 y
line is y
y2 )
209
2 y dx
2 , but dy
4
dx (1, 1)
dy
dx
1 3x 2 y 3
is undefined. Therefore, the
at (1, 1) but the slope is undefined at (1, 1).
dy
dx
[cos( x sin x)](1 cos x); y 0
sin( x sin x)
0
x sin x
k ,k
2, 1, 0, 1, 2
dy
dx
0 and y 0 when 1 cos x 0 and x k .
(for our interval) cos( x sin x ) cos(k )
1. Therefore,
For 2
x 2 , these equations hold when k
2, 0, and 2(since cos( ) cos
1.) Thus the curve has
the
horizontal tangents at the x-axis for the x-values 2 , 0, and 2 (which are even integer multiples of )
curve has an infinite number of horizontal tangents.
119. B graph of f , A graph of f . Curve B cannot be the derivative of A because A has only negative slopes
while some of B s values are positive.
120. A graph of f , B graph of f . Curve A cannot be the derivative of B because B has only negative slopes
while A has positive values for x 0.
121.
122.
123. (a) 0, 0
(b) largest 1700, smallest about 1400
Copyright
2014 Pearson Education, Inc.
210
Chapter 3 Derivatives
124. rabbits/day and foxes/day
125. lim sin2 x
x
0 2x
x
0
7x
126. lim 3 x 2tan
x
x 0
sin r
127. lim tan
2r
r 0
0
lim
0
lim 32 xx
0
sin 7 x
2 x cos 7 x
lim sinr r tan2 r2 r 12
0
tan
tan 2 5
1
x
131. lim 2 x2sin
cos x
x 0
r
1
tan
0
lim
2
1
2sin 2
2
2
0
2
7
cot
(0 2)
(5 0 0)
8
cot 2
x sin x
0 2 2sin 2 2x
lim
x
lim
sin
0
2
sin
2
1
2
2
2
2
7
Let x
tan(tan x )
sin . Then x
0
4
0
x x
2 2
sin 2 2x
sin x
x
(1)(1) 12
2( x 2 1)
cos 2 x
2x
x2 1
10 3 x 4
2x 4
y
x
2
lim
x
0 sin
x
2
x
2
sin
x
2
sin x
x
(1)(1)(1) 1
1
2
lim g ( x )
x
0
tan(tan x )
0 tan x
lim
x
lim tan
0
tan(tan x )
x
define f (0) 1.
136. y
0
x
sin x
y
0 as
1.
sin x (using the result of # 103); let
sin x
1
lim sin(sin x ) lim
1 lim sin(sin
x)
tan x
sin(sin x ) cos x
x 0
x 0
0
sin
x
0 as x 0
lim sin(sin x ) lim sin
1. Therefore, to make f continuous at the origin,
134. lim f ( x)
135. y
2
1
2
133. lim tanx x lim cos1 x sinx x 1; let
tan x
0 as x 0
x 0
x 0
Therefore, to make g continuous at the origin, define g (0) 1.
x
1 1 72
2
5
lim
x
3
2
1
(1) 11
(4 0 0)
(1 0)
5
tan 2
cot 2
5
2r
tan 2
1
8
1
2
sin (sin )
.
0 sin
1
lim
0
lim
4
lim
lim x sin x
x 0 2(1 cos x )
1 cos
0
2r
(1) lim cos
sin 2 r
1
2
2
1 2 cot 2
lim
2
5cot
7 cot
0
132. lim
lim cos17 x sin7 x7 x
0
sin(sin ) sin
sin
0
sin
lim x 1
x 0 x
sin(sin )
sin
1
x
lim
2
130.
3
2
r
2
lim 4 tan
129.
(1) 11
1
(2 x 1)
x
sin(sin )
128. lim
sin x
x
lim
x
ln y
2( x 2 1)
cos 2 x
ln
tan 2 x y
ln y
3
1
10 3 x 4
y
ln(2) ln( x 2 1) 12 ln(cos 2 x)
2( x 2 1) 2 x
cos 2 x x 2 1
ln 10 23xx 44
1
x 2
0
0
0
2x
x2 1
tan 2 x
1 [ln(3 x
10
10 3 x 4 1
2 x 4 10
y
y
4) ln(2 x 4)]
3
3x 4
Copyright
y
y
3
1
10 3 x 4
1
x 2
2014 Pearson Education, Inc.
2
2x 4
1 ( 2sin 2 x )
2
cos 2 x
Chapter 3 Practice Exercises
(t 1)(t 1) 5
(t 2)(t 3)
137. y
dy
dt
1
y
138. y
2u 2u
1
u2 1 u
(sin )
140. y
1
ln x
ln y
ln(ln x)
y
y
dS
dt
dS
dt
2 r2
2 rh and h constant
(b) S
2 r2
2 rh and r constant
(c) S
2
142. S
r r2
dS
dt
2 rh
dS
dt
(d) S constant
h2
dS
dt
0
(b) r constant
dr
dt
0
144. V
s3
dA
dt
dV
dt
dR
dt
r
dS
dt
2
ln(sin )
1
ln x
1
x
r2
ds
dt
h
r
r2
h2
h 2 dr
dt
dr
dt
(ln x)
x 1 ln(ln x )
x (ln x )2
(4 r 2 h) dr
dt
2 r dh
dt
dr
r dh
dt
dt
r dh
2r h dt
r2
r2
h2
r
2
h2
dr
dt
rh dh
r 2 h 2 dt
10 and dr
dt
2
1 dV
3s 2 dt
20 and dV
dt
; so s
y
h 2 dr
;
dt
2
r2
2
1/ln
1
x
1
(ln x )2
ln(ln x)
rh dh
r 2 h2 dt
h2
2 r dr
; so r
dt
3s 2 ds
dt
1/2
1
2
4 r dr
2 h dr
dt
dt
2 r dh
dt
r 2 h2
r 2 dr
dS
dt
r2
(c) In general, dS
dt
cos
sin
1
ln x
r dr
h dh
dt
dt
r
dh
dt
r2
1
t 3
ln 2 u 12 22
u 1
1
u
4 r dr
2 r dh
h dr
(4 r 2 h) dr
dt
dt
dt
dt
dh
dr
0 (4 r 2 h) dr
2
r
(2
r
h
)
dt
dt
dt
0
(a) h constant
143. A
1
t 2
ln(sin )
2
cot
141. (a) S
2 r
1
t 1
dy
du
1
y
dy
d
1
y
ln(sin )
(sin )
1
t 1
5 (t 2)(t 3)
u
u2 1
ln 2
ln y
(ln x)1/ln x
(t 1)(t 1) 5
dy
dt
1
t 3
ln 2 ln u u ln 2 12 ln(u 2 1)
ln y
u2 1
dy
d
5[ln(t 1) ln(t 1) ln(t 2) ln(t 3)]
5 t 11 t 11 t 12
2u 2u
dy
du
139. y
ln y
211
m /sec
dA
dt
40 m 2 /sec
2
(2 )(10)
1200 cm3 /min
dR
ds
dt
dR
1
30(20)2
(1200) 1 cm/min
dR
1
1
1 dR
1 1
1
2
1 ohm/sec, dt2 0.5 ohm/sec; and R1
Also, R1 75 ohms
145. dt1
R1 R2
R12 dt
R22 dt
R 2 dt
1
1
1
R 30 ohms. Therefore, from the derivative equation,
and R2 50 ohms
R
75 50
1 dR
(30) 2 dt
146. dR
dt
X
1
(75)2
( 1)
1
(50)2
(0.5)
3 ohms/sec and dX
dt
20 ohms
dZ
dt
1
5625
1
5000
2 ohms/sec; Z
(10)(3)
10
(20)( 2)
2
202
1
5
Copyright
R2
dR
dt
X2
5000 5625
( 900) 5625 5000
dZ
dt
R dR
dt
R
X dX
dt
2
X2
0.45 ohm/sec.
2014 Pearson Education, Inc.
9(625)
50(5625)
1
50
0.02 ohm/sec.
so that R 10 ohms and
212
Chapter 3 Derivatives
dy
10 m/sec and dt 5 m/sec, let D be the distance from the origin
D2 x2 y2
147. Given dx
dt
dy
dy
2 D dD
2 x dx
2 y dt
D dD
x dx
y dt . When ( x, y ) (3, 4), D
32 ( 4)2 5 and
dt
dt
dt
dt
dD
10 2. Therefore, the particle is moving away from the origin at 2 m/sec
5 dD
(3)(10) ( 4)(5)
dt
dt
5
(because the distance D is increasing).
11 units/sec. Then D 2
148. Let D be the distance from the origin. We are given that dD
dt
x2
x3
2 x dx
3x 2 dx
dt
dt
2 D dD
dt
equation gives (2)(6)(11)
x(2 3 x) dx
;x
dt
9) dx
dt
(3)(2
149. (a) From the diagram we have 10
h
(b) V
r2h
1
3
1
3
2
2h
5
4
r
6 ft/sec and r
ds
dt
1.2 ft
151. (a) From the sketch in the text, ddt
x 0
0
reaches point A.
152. From the figure, ar
d
dt
r ddt
b
BC
a
r
( x3/2 )2
6 and substitution in the derivative
dr .
dt
5 and h
6
Also r 1.2 is constant
5 rad/sec
0.6 rad/sec and x
(sec 0)( 0.6)
18
32 33
x2
h.
4 h 2 dh , so dV
25 dt
dt
2
dx
dt
(3/5) rad 1 rev 60 sec
sec
2 rad min
(b)
2
5
dV
dt
r
D
y2
4 units/sec.
r
4 h3
75
h
150. From the sketch in the text, s
Therefore, ds
dt
dx
dt
3
x2
tan . Also x
tan
dh
dt
dr
dt
125
144
ft/min.
0
ds
dt
dx
dt
0.6. Therefore the speed of the light is 0.6
sec2
3
5
r ddt
d
dt
(1.2) ddt .
; at point A,
km/sec when it
revs/min
b
b2 r 2
. We are given
that r is constant. Differentiation gives,
b2 r 2
1 da
r dt
db
dt
b
(b )
db
dt
b2 r 2
b2 r 2
. Then, b
2 r ( 0.3 r )
(2 r )2 r 2 ( 0.3r ) (2 r )
db
dt
da
dt
0.3r
3r 2 ( 0.3r )
r
(2r )
4 r 2 (0.3 r )
3r 2
3 3r 2
increasing when OB
f
4
r
(2 r )2 r 2
2
(3r 2 )( 0.3r ) (4 r 2 )(0.3r )
3r
153. (a) If f ( x)
2
2r and
r
10 3
m/sec. Since da
is positive, the distance OA is
dt
2r , and B is moving toward O at the rate of 0.3r m/sec.
tan x and x
4
1 and f
f ( x) is L( x)
0.3r
3 3
2 x
4
4
, then f ( x) sec2 x,
2. The linearization of
( 1)
2x
Copyright
2
2.
2014 Pearson Education, Inc.
Chapter 3 Practice Exercises
(b) if f ( x )
sec x and x
f
2 and f
4
f ( x) is L( x )
1
1 tan x
154. f ( x)
155. f ( x)
2 x
f (0)( x 0)
2
1
(1 x)
2 (4
4
2x
)
.
linearization at x
( x 1)1/2 sin x 0.5
f (0) 1.5( x 0) 0.5
L( x)
2 1 x
r r2
h0 to h0
2
4
f (0)( x 0)
h2 , r constant
dh
( x 1) 1/2
cos x
h 2 ) 1/2 2h dh
rh
r 2 h2
dh. Height changes from
r 2 h02
r2
|12r dr | 12
100
r . The measurement of the
| dr | 100
(100%)
(b) When V r 3 , then dV 3r 2 dr. The accuracy of the volume is dV
V
159. C
2 r
C
2
dV
(a) dr
(b) dS
(c) dV
2
2
3
r
(dr )(100%)
r
C
2
4 r2
,S
2
10
2 2
dr
r
cm
(0.4)
8
(100%)
C2
20
2
120a 2 da
cm
4
3
15
6
120 da
a2
(100%)
2
10
8
(100%)
r3
C 3 . It
6 2
(100%) (.04)(100%)
Copyright
1
12
dC , dS
2C
dC and
4%
(100%) 8%
100
20 6 2
2
1000
(100%) 12%
h 14 ft. The same triangles imply that 20h a
120
a2
1
2
also follows that dr
0.4 cm.
0.2
(100%)
dV
V
3r 2 dr
r3
3%
, and V
(100%)
dS
S
cm
160. Similar triangles yield 35
h
dh
r
100
dC. Recall that C 10 cm and dC
0.4
0.2
2
20 (0.4)
( x 1) 1/2
1
2
2.5 x 0.1 , the linearization of f ( x) .
158. (a) S 6r 2
dS 12r dr. We want | dS | (2%) S
edge r must have an error less than 1%.
3
r
f (0) 1 x.
2(1 x) 2 ( 1)
f ( x)
r h0 ( dh )
dS
1
2
f (0)( x 0)
L( x) 1.5 x 0.5 , the linearization of f ( x) .
f (0)
r 12 (r 2
dS
0 is L( x )
f ( x)
1 x 3.1 2(1 x ) 1 ( x 1)1/2 3.1
1 x
2
sec x tan x,
2. The linearization of
sec2 x . The
(1 tan x )2
f ( x)
2
156. f ( x)
, then f ( x)
4
x 1 sin x 0.5
L( x)
157. S
4
213
120
152
1
12
2
45
.0444 ft
2014 Pearson Education, Inc.
a
6
h 120a 1 6
0.53 inches.
214
Chapter 3 Derivatives
CHAPTER 3
ADDITIONAL AND ADVANCED EXERCISES
1. (a) sin 2
d
d
2
2sin cos
2
(sin 2 )
d
d
(2sin cos )
2cos 2
2[(sin )( sin ) (cos )(cos )]
cos 2
cos
sin
d (cos 2 )
d (cos 2
(b) cos 2 cos 2
sin 2
sin 2 )
d
d
2sin 2 (2 cos )( sin ) (2sin )(cos )
sin 2
cos sin
sin cos
sin 2
2sin cos
2. The derivative of sin ( x a) sin x cos a cos x sin a with respect to x is cos( x a) cos x cos a sin x sin a,
which is also an identity. This principle does not apply to the equation x 2 2 x 8 0, since x 2 2 x 8 0 is
not an identity: it holds for 2 values of x ( 2 and 4), but not for all x.
3. (a) f ( x) cos x
f ( x)
sin x
f ( x)
cos x, and g ( x) a bx cx 2
g ( x) b 2cx
also, f (0) g (0) cos(0) a a 1; f (0) g (0)
sin(0) b b 0; f (0) g (0)
1 . Therefore, g ( x) 1 1 x 2 .
cos(0) 2c c
2
2
g ( x)
2c;
(b) f ( x) sin( x a )
f ( x) cos( x a), and g ( x) b sin x c cos x
g ( x ) b cos x c sin x; also
f (0) g (0) sin(a ) b sin(0) c cos(0) c sin a; f (0) g (0) cos(a ) b cos(0) c sin(0)
b cos a. Therefore, g ( x) sin x cos a cos x sin a.
(c) When f ( x) cos x, f ( x) sin x and f (4) ( x) cos x; when g ( x) 1 12 x 2 , g ( x ) 0 and g (4) ( x) 0.
Thus f (0) 0 g (0) so the third derivatives agree at x 0 . However, the fourth derivatives do not
agree since f (4) (0) 1 but g (4) (0) 0. In case (b), when f ( x) sin( x a) and
g ( x)
sin x cos a cos x sin a , notice that f ( x)
have f
(n)
( x)
g
(n)
g ( x) for all x, not just x
0. Since this is an identity, we
( x) for any x and any positive integer n.
4. (a) y
sin x
y cos x
y
sin x
y y
sin x sin x 0; y cos x
y
sin x
y
cos x
y y
cos x cos x 0; y a cos x b sin x
y
a sin x b cos x
y
a cos x b sin x
y y ( a cos x b sin x) (a cos x b sin x) 0
(b) y sin(2 x)
y 2 cos(2 x)
y
4sin(2 x)
y 4y
4sin(2 x) 4sin(2 x) 0. Similarly,
y cos(2 x) and y a cos(2 x ) b sin(2 x) satisfy the differential equation y 4 y 0. In general,
y cos(mx), y sin(mx ) and y a cos ( mx) b sin ( mx) satisfy the differential equation y m 2 y
0.
5. If the circle ( x h) 2 ( y k )2 a 2 and y x 2 1 are tangent at (1, 2), then the slope of this tangent is
m 2 x (1, 2) 2 and the tangent line is y 2 x. The line containing (h, k) and (1, 2) is perpendicular to
y
2x
k 2
h 1
1
2
x h ( y k) y
0
tangent line and that y
2
1 (2)2
k 2
5 2k
the location of the center is (5 2k , k ). Also, ( x h)2
1 ( y )2
2
h
we have that a
0
1 (y )
k y
y
. At the point (1, 2) we know y
5 2k
4
the circle is ( x 4)2
2
y 92
6. The total revenue is the number of people times the price of the fare: r ( x)
0
x
2 from the
2
a 2 . Since (1, 2) lies on the circle
xp
x
x 3 40
2
, where 0
x
60.
dr
x
x
x
x
2x
x 1 x .
1
3 40
2 x 3 40
3 40
3 40
3 3 40
40
40
40
dx
40 (since x 120 does not belong to the domain). When 40 people are on the bus the
The marginal revenue is dr
dx
dr
Then dx
a2
2 from the parabola. Since the second derivatives are equal at (1, 2) we obtain
9 . Then h
2
5 5
.
2
k
( y k) y
( y k )2
marginal revenue is zero and the fare is p (40)
Copyright
x
3 40
2
$4.00.
( x 40)
2014 Pearson Education, Inc.
Chapter 3 Additional and Advanced Exercises
215
dy
du v u dv (0.04u )v u (0.05v ) 0.09uv 0.09 y
7. (a) y uv
the rate of growth of the total
dt
dt
dt
production is 9% per year.
dy
0.02u and dv
0.03v, then dt ( 0.02u )v (0.03v )u 0.01uv 0.01y, increasing at 1% per year.
(b) If du
dt
dt
8. When x 2
y2
225, then y
to the balloon at (12, 9) is y 9
4
3
y
x . The
y
4 (x
3
tangent line
12)
x 25. The top of the gondola is
15 8 23 ft below the center of the balloon. The
intersection of y
23 and y 43 x 25 is at the far
23 43 x 25
right edge of the gondola
Thus the gondola is 2 x 3 ft wide.
3.
2
x
9. Answers will vary. Here is one possibility.
10. s (t ) 10 cos t
ds
dt
v (t )
4
10sin t
a (t )
4
d 2s
dt 2
dv
dt
10
2
(a) s (0) 10 cos 4
(b) Left : 10, Right:10
(c) Solving 10 cos t 4
10 cos t
10
4
10
cos t
the right. Thus, v 34
(d) Solving 10 cos t
0, v
0
4
cos t
1
4
7
4
t
t
3
4
, but t
0
1
4
t
0, a 34
4
4
when the particle is farthest to the left. Solving
t
2
10, and a 74
v 4
4
10 cos t
10, v 4
7
4
4
when the particle is farthest to
10.
10 and a 4
0.
s (t ) 64t 16t 2
v(t ) ds
64 32t 32(2 t ). The maximum height is reached when v(t ) 0
dt
t 2 sec. The velocity when it leaves the hand is v (0) 64 ft/sec.
(b) s (t ) 64t 2.6t 2
v(t ) ds
64 5.2t. The maximum height is reached when v (t ) 0 t 12.31sec.
dt
The maximum height is about s (12.31) 393.85 ft.
11. (a)
12. s1
3t 3 12t 2 18t 5 and s2
t 3 9t 2 12t
9t 2 24t 18
3t 2 18t 12 2t 2 7t 5
13. m v 2
dx
dt
v02
v
14. (a) x
k x02
m dv
dt
At 2
x2
m 2v dv
dt
kx, as claimed.
Bt C on [t1 , t2 ]
v
k
dx
dt
v1
0
2 x dx
dt
2 At
9t 2 24t 18 and v2
3t 2 18t 12; v1 v2
(t 1)(2t 5) 0
t 1 sec and t 2.5 sec.
m dv
dt
B
v
k
t1 t2
2
2 x dx
2v dt
2A
m dv
dt
t1 t2
2
B
kx 1v dx
. Then substituting
dt
A(t1 t2 ) B is
the instantaneous velocity at the midpoint. The average velocity over the time interval is
vav
x
t
At22 Bt2 C
At12 Bt1 C
t2 t1
Copyright
t2 t1 [ A t2 t1
t2 t1
B]
A(t2 t1 ) B.
2014 Pearson Education, Inc.
216
Chapter 3 Derivatives
(b) On the graph of the parabola x At 2 Bt C , the slope of the curve at the midpoint of the interval [t1 , t2 ]
is the same as the average slope of the curve over the interval.
requires that lim sin x
15. (a) To be continuous at x
cos x, x
(b) If y
x
is differentiable at x
m, x
x
16. f ( x) is continuous at 0 because lim 1 cos
x
x
0
2
1
lim sinx x
1 cos x
x 0
x 1 cos x
lim 1 cos
1 cos x
x2
x 0
x
lim (mx b)
, then lim cos x
x
0
f (0). f (0)
1 . Therefore
2
m
0
m
b
lim
x
.
1 cos x
x
0
b;
m
1 and b
f ( x) f (0)
x 0
0
lim
x
m
0
x
f (0) exists with value 12 .
f continuous at
17. (a) For all a, b and for all x 2, f is differentiable at x. Next, f differentiable at x 2
x 2
lim f ( x) f (2) 2a 4a 2b 3 2a 2b 3 0. Also, f differentiable at x 2
x
2
a, x
2
. In order that f (2) exist we must have a
2ax b, x 2
Then 2a 2b 3 0 and 3a b a 34 and b 94 .
f ( x)
(b) For x
2a (2) b
a
4a b
3a
2, the graph of f is a straight line having a slope of 34 and passing through the origin; for x
b.
2, the
3
2
graph of f is a parabola. At x 2, the value of the y -coordinate on the parabola is which matches the
y -coordinate of the point on the straight line at x 2. In addition, the slope of the parabola at the match up
point is 34 which is equal to the slope of the straight line. Therefore, since the graph is differentiable at the
match up point, the graph is smooth there.
1 g continuous
18. (a) For any a, b and for any x
1, g is differentiable at x. Next, g differentiable at x
1
at x
1
lim g ( x) g ( 1)
a 1 2b
a b b 1. Also, g differentiable at x
x
1
a, x
3ax 2 1, x
g ( x)
(b) For x
1
. In
1
order that g ( 1) exist we must have a
3a ( 1)2 1
a
3a 1
1, the graph of g is a straight line having a slope of 12 and a y -intercept of 1. For x
a
1.
2
1, the
graph of g is a cubic. At x
1, the value of the y -coordinate on the cubic is 32 which matches the
y -coordinate of the point on the straight line at x
1. In addition, the slope of the cubic at the match up
point is 12 which is equal to the slope of the straight line. Therefore, since the graph is differentiable at
the match up point, the graph is smooth there.
19. f odd
20. f even
f ( x)
f ( x)
f ( x)
d (f(
dx
x ))
d (
dx
f ( x)
d (f(
dx
x))
d ( f ( x))
dx
f ( x ))
f ( x )( 1)
f ( x)( 1)
f ( x)
f ( x)
f ( x)
f ( x)
f ( x)
f ( x)
f is even.
f is odd.
h ( x ) h ( x0 )
f ( x ) g ( x ) f ( x0 ) g ( x0 )
lim
lim
x x0
x x0
x x0
x x0
f ( x ) g ( x ) f ( x ) g ( x0 ) f ( x ) g ( x0 ) f ( x0 ) g ( x0 )
g ( x ) g ( x0 )
f ( x ) f ( x0 )
lim
lim f ( x)
lim g ( x0 )
x x0
x x0
x x0
x x0
x x0
x x0
g ( x ) g ( x0 )
g ( x ) g ( x0 )
f ( x0 ) lim
g ( x0 ) f ( x0 ) 0 lim
g ( x0 ) f ( x0 ) g ( x0 ) f ( x0 ), if g is
x x0
x x0
x x0
x x0
21. Let h( x)
( fg )( x)
f ( x) g ( x)
h ( x)
continuous at x0 . Therefore ( fg ) ( x) is differentiable at x0 if f ( x0 )
Copyright
0, and ( fg ) ( x0 )
2014 Pearson Education, Inc.
g ( x0 ) f ( x0 ).
Chapter 3 Additional and Advanced Exercises
217
22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f (0) 0 and g is continuous at 0.
(a) If f ( x ) sin x and g ( x ) | x |, then | x | sin x is differentiable because f (0) cos(0) 1, f (0) sin (0) 0
and g ( x) | x | is continuous at x 0.
(b) If f ( x) sin x and g ( x) x 2/3 , then x 2/3 sin x is differentiable because
f (0) cos (0) 1, f (0) sin (0) 0 and g ( x) x 2/3 is continuous at x 0.
(c) If f ( x) 1 cos x and g ( x) 3 x, then 3 x (1 cos x ) is differentiable because f (0) sin (0) 0,
f (0) 1 cos (0) 0 and g ( x) x1/3 is continuous at x 0.
(d) If f ( x) x and g ( x) x sin 12 , then x 2 sin 1x is differentiable because f (0) 1, f (0) 0 and
lim x sin 1x
x
0
23. If f ( x)
h (0)
lim
x
0
0
lim sint t
1
x
0
g (0) f (0)
lim h ( x)
x
x
1
x
sin
lim
0
0 (so g is continuous at x
x
0 (so g is continuous at x
t
25. Step 1:
Step 2:
2 x sin 1x . But
does not exist because cos 1x has no limit as x
f ( x h) f ( x )
h
0
lim
f ( x)
1
x2
cos 1x
2 x sin 1x
f ( x) f (h), f (h) 1 hg (h) and lim g (h) 1. Therefore,
h
f ( x ) f ( h) f ( x )
h
0
f (h) 1
h
lim f ( x )
h
0
2 (a single product) since y
Assume the formula holds for n k :
du1
du
dy
y u1u2 uk
u u u k u1 dx2 u3
dx
dx 2 3
If y
u1u2
du1
u u
dx 2 3
du1
u u
dx 2 3
u k uk 1
u1u2
du
u1 dx2 u3
uk
uk
h
u1u2
... u1u2
d (u u
dy
0
f ( x) lim g (h)
f ( x) and f ( x) exists at every value of x.
The formula holds for n
0. Therefore,
0 because it has no limit there.
lim
h
0 and
0 ). In fact, from Exercise 21,
0, h ( x)
24. From the given conditions we have f ( x h)
h
x 2 cos 1x
0 because f (0) 1, f (0)
0. However, for x
the derivative is not continuous at x
f ( x)
0 ).
x sin 1x , then x 2 sin 1x is differentiable at x
x and g ( x)
lim x sin 1x
x
lim sint t
1
x
0
x
1
x
sin
lim
f ( x) 1
0
du1
u
dx 2
dy
dx
f ( x)
du
u1 dx2
du
u k 1 dxk .
u )
du
1 2
k
uk uk 1, then dx
uk 1 u1u2 uk dxk 1
dx
du
du
uk
u1u2 uk 1 dxk uk 1 u1u2 uk dxk 1
du
uk 1 u1 dx2 u3 uk 1
Thus the original formula holds for n
du
du
uk 1 dxk uk 1 u1u2 uk dxk 1 .
(k 1) whenever it holds for n k .
u1u2
m !( k 1) m !( m k )
m
m!
m!
m ! . Then m
m!
m and m
k
1
k 1
k !( m k )! ( k 1)!( m k 1)!
( k 1)!( m k )!
k !( m k )!
1!( m 1)!
m !( m 1)
( m 1)!
m 1
k 1 . Now, we prove Leibniz s rule by mathematical induction.
( k 1)!( m k )! ( k 1)!(( m 1) ( k 1))!
d (uv )
u dv
v du
. Assume that the statement is true for n k , that is:
Step 1: If n 1, then dx
dx
dx
k
1
k
k
d (uv )
k d k 2u d 2 v
d u v k d u dv
d k 1v u d k v .
... kk 1 du
2
2
k
k
k 1 dx
k 2
dv
dx
dx
dx
dx
dx
dx k 1
dx k
26. Recall m
k
Step 2:
If n
k 1, then
k
2
k 1
d k 1 (uv )
dx
2
k 2
ud
k 1
dx k
d k ( uv )
dx k
3
k d u d v
2
dx k 2 dx3
d u d v
dx k 1 dx 2
du d k v
dx dx k
d
dx
k 1
u
1
d k 1u v
dx k 1
...
( k 1)
Copyright
d k 1u v
dx k 1
k
2
d k u dv
dx k dx
k 1
d u d v
k 1 dx 2 dx k 1
k
k
d k u dv
1
2
dx k dx
k
k d ku dv
dx dx
k
k
du d u
k 1 dx dx k v
d k 1u d 2v
dx k 1 dx 2
2014 Pearson Education, Inc.
k 1
2
k d k u1 d 2v
dx
dx
218
Chapter 3 Derivatives
k
k 1
k 1 du
k
dx
du d k v
dx dx k
k 1
u d k v1 .
dx
k
k
k
d v
dx k
ud
k 1
d k 1u v
dk k 1
v
dx
k 1
Therefore the formula (c) holds for n
27. (a) T 2
4
(b) T 2
4
2
g
2
g
L
L
L
T
T 2g
2
4
2
g
L
k
dv
(k 1) d ku dx
dx
(k 1) whenever it holds for n
(1sec2 )(32.2 ft/sec2 )
4
2
L; dT
L
2
1
g 2 L
k 1 d k 1u d 2v
2
dx k 1 dx 2
dL
Lg
...
k.
0.8156 ft
dL; dT
(0.8156ft)(32.2ft/sec2 )
(0.01 ft)
(c) Since there are 86, 400 sec in a day, we have we have (0.00613 sec)(86, 400 sec/day)
8.83 min/day; the clock will lose about 8.83 min/day.
s3
28. v
2k
1
1
dv
dt
s0
3 1/3
4
3s 2 ds
dt
s1. Let t
k (6 s 2 )
ds
dt
2k . If s0
s0
2k
11 hr.
Copyright
529.6 sec/day, or
the initial length of the cube s side, then s1
the time it will take the ice cube to melt. Now, t
2014 Pearson Education, Inc.
s0
s0 s1
0.00613 sec.
1/3
s0
(v0 )
1/3
(v0 )
1/3
3
v
4 0
2k
CHAPTER 4
4.1
APPLICATIONS OF DERIVATIVES
EXTREME VALUES OF FUNCTIONS
1. An absolute minimum at x c2 , an absolute maximum at x
extreme values because h is continuous on [a, b].
2. An absolute minimum at x b, an absolute maximum at x
extreme values because f is continuous on [a, b].
b. Theorem 1 guarantees the existence of such
c. Theorem 1 guarantees the existence of such
3. No absolute minimum. An absolute maximum at x c. Since the function s domain is an open interval, the
function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values.
4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill
the conclusions of Theorem 1.
5. An absolute minimum at x a and an absolute maximum at x c. Note that y g ( x) is not continuous but still
has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the
hypothesis is not satisfied, absolute extrema may or may not occur.
6. Absolute minimum at x c and an absolute maximum at x a. Note that y g ( x) is not continuous but still has
absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the
hypothesis is not satisfied, absolute extrema may or may not occur.
7. Local minimum at ( 1, 0), local maximum at (1, 0).
8. Minima at ( 2, 0) and (2, 0), maximum at (0, 2).
9. Maximum at (0, 5). Note that there is no minimum since the endpoint (2, 0) is excluded from the graph.
10. Local maximum at ( 3, 0), local minimum at (2, 0), maximum at (1, 2), minimum at (0, 1).
11. Graph (c), since this is the only graph that has positive slope at c.
12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative
slope at c.
13. Graph (d), since this is the only graph representing a function that is differentiable at b but not at a.
14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b.
15. f has an absolute min at x 0 but does not have
an absolute max. Since the interval on which f is
defined, 1 x 2, is an open interval, we do not
meet the conditions of Theorem 1.
Copyright
2014 Pearson Education, Inc.
219
220
Chapter 4 Applications of Derivatives
16. f has an absolute max at x 0 but does not have an
absolute min. Since the interval on which f is defined,
1 x 1, is an open interval, we do not meet the
conditions of Theorem 1.
17. f has an absolute max at x 2 but does not have an
absolute min. Since the function is not continuous at
x 1, we do not meet the conditions of Theorem 1.
18. f has an absolute max at x 4 but does not have an
absolute min. Since the function is not continuous at
x 0, we do not meet the conditions of Theorem 1.
19. f has an absolute max at x 2 and an absolute min at
x 32 . Since the interval on which f is defined,
0 x 2 , is an open interval we do not meet the
conditions of Theorem 1.
20. f has an absolute max at x 0 and an absolute min
1 but does not have an absolute
at x 2 and x
y
( 0, 1)
maximum. Since f is defined on a union of halfopen intervals, we do not meet the conditions of
Theorem 1.
y
1
f ( x)
x
0
2
21. f ( x)
f ( 2)
2
3
x 5
19 ,
3
f ( x)
f (3)
3
2
3
no critical points;
the absolute maximum
is 3 at x 3 and the absolute minimum is 19
3
at x
2
Copyright
2014 Pearson Education, Inc.
Section 4.1 Extreme Values of Functions
x 4
f (x)
22. f ( x )
1 no critical points;
f ( 4) 0, f (1)
5 the absolute maximum is 0
at x
4 and the absolute minimum is 5 at x 1
23. f ( x) x 2 1
f ( x) 2 x
a critical point at
x 0; f ( 1) 0, f (0)
1, f (2) 3 the absolute
maximum is 3 at x 2 and the absolute minimum is
1 at x 0
24.
f ( x) 4 x3
f ( x)
3x2
a critical point at
the absolute
x 0; f ( 2) 12, f (0) 4, f (1) 3
maximum is 12 at x
2 and the absolute
minimum is 3 at x 1
y
( 2, 12)
10
5
f ( x ) 4 x3
(1, 3)
2
25. F ( x)
1
x2
x
2
F ( x)
2x
3
2 ,
x3
1
0
however
x 0 is not a critical point since 0 is not in the domain;
F (0.5)
4, F (2)
0.25 the absolute maximum
is 0.25 at x 2 and the absolute minimum is 4 at
x 0.5
26. F ( x)
1
x
x
1
F ( x)
x
2
1
x2
, however
x 0 is not a critical point since 0 is not in the
domain; F ( 2) 12 , F ( 1) 1 the absolute
maximum is 1 at x
1 and the absolute minimum
1
2
is 2 at x
Copyright
2014 Pearson Education, Inc.
1
x
221
222
Chapter 4 Applications of Derivatives
27. h( x) 3 x x1/3
h ( x) 13 x 2/3
a critical point
at x 0; h( 1)
1, h(0) 0, h(8) 2 the
absolute maximum is 2 at x 8 and the absolute
1
minimum is 1 at x
28. h( x)
3 x 2/3
h ( x)
2 x 1/3
a critical point at
x 0; h( 1)
3, h(0) 0, h(1)
3 the absolute
maximum is 0 at x 0 and the absolute minimum is
1
3 at x 1 and x
4 x2
29. g ( x)
(4 x 2 )1/2
1 (4
2
g ( x)
x 2 ) 1/2 ( 2 x)
x
critical
4 x2
points at x
2 and x 0, but not at x 2 because 2
is not in the domain;
g ( 2) 0, g (0) 2, g (1)
3 the absolute
maximum is 2 at x 0 and the absolute minimum is
0 at x
2
30.
5 x2
g ( x)
1
2
g ( x)
x
f
(5 x 2 )1/2
(5 x 2 )
1/2
( 2 x)
x
5 x2
critical points at x
5 and x 0, but not at
5 because 5 is not in the domain;
5 0, f (0)
5
the absolute maximum is 0 at x
absolute minimum is 5 at x 0
31. f ( )
sin
point, but
f ( ) cos
is a critical
2
is not a critical point because 2 is
2
not interior to the domain; f
f
5
6
1
2
5 and the
1, f
2
1,
2
the absolute maximum is 1 at
and the absolute minimum is 1 at
2
2
32. f ( ) tan
f ( ) sec2
f has no critical
points in 3 , 4 . The extreme values therefore
occur at the endpoints: f
3 and f
3
the absolute maximum is 1 at
the absolute minimum is 3 at
4
4
1
and
3
Copyright
2014 Pearson Education, Inc.
Section 4.1 Extreme Values of Functions
g ( x)
33. g ( x) csc x
point at x 2 ; g 3
(csc x)(cot x)
g 2 1, g
2 ,
3
the absolute maximum is
2
3
at x
and the absolute minimum is 1 at x
34. g ( x ) sec x g ( x )
point at x 0; g 3
35.
f (t )
2 |t |
2
t2
3
2
3
,
2
a critical
2
3
6
the
and the absolute
2 (t 2 )1/2
1 (t 2 ) 1/2 (2t )
2
f (t )
2
3
and x
3
(sec x )(tan x )
2, g (0) 1, g
absolute maximum is 2 at x
minimum is 1 at x 0
a critical
2
3
t
t
2
t
|t |
a critical
point at t 0; f ( 1) 1, f (0) 2, f (3)
1 the
absolute maximum is 2 at t 0 and the absolute
minimum is 1 at t 3
36. f (t ) | t 5|
f (t )
t 5
| t 5|
1 ((t
2
(t 5)2
((t 5) 2 )1/2
5)2 ) 1/2 (2(t 5))
a critical point at t
t 5
(t 5)2
5; f (4) 1, f (5)
f (7) 2 the absolute maximum is 2 at t
the absolute minimum is 0 at t 5
37.
g ( x)
xe x
g ( x)
e x
0,
7 and
xe x
a critical point at x = 1; g( 1) = e, and g (1)
1,
e
the absolute maximum is 1e at x = 1 and the
absolute minimum is e at x = 1
38. The first derivative h ( x )
1
x 1
has no zeros, so we
need only consider the endpoints, h(0) = ln 1;
h(3) = ln 4
Maximum value is ln 4 at x = 3;
Minimum value is ln 1 at x = 0.
Copyright
2014 Pearson Education, Inc.
223
224
Chapter 4 Applications of Derivatives
1
x2
39. The first derivative f ( x)
1
x
has a zero at
x = 1.
Critical point value: f(1) = 1 + ln 1 = 1
Endpoint values: f(0.5) = 2 + ln 0.5 1.307;
f (4) 14 ln 4 1.636;
Absolute maximum value is 14 ln 4 at x = 4;
Absolute Minimum value is 1 at x = 1;
Local maximum at 12 , 2 ln 2
40.
g ( x)
2
e x
g ( x)
2
2 xe x
a critical point at
4
x = 0; g ( 2) e , g(0) = 1, and g (1) e 1
the absolute maximum is 1 at x = 0 and the
absolute minimum is e 4 at x = 2
41. f ( x) x 4/3
f ( x) 43 x1/3
a critical point at x 0; f ( 1) 1, f (0)
maximum is 16 at x 8 and the absolute minimum is 0 at x 0
0, f (8) 16
42. f ( x) x5/3
f ( x) 53 x 2/3
a critical point at x 0; f ( 1)
1, f (0)
1
maximum is 32 at x 8 and the absolute minimum is 1 at x
32
the absolute
8, g (0)
0, g (1) 1
the absolute
2 1/3
a critical point at
0; h( 27) 27, h(0)
0
27 and the absolute minimum is 0 at
0, h (8) 12
3/5
43. g ( )
g ( ) 35 2/5
a critical point at
0; g ( 32)
32
maximum is 1 at
1 and the absolute minimum is 8 at
44. h( ) 3 2/3
h( )
maximum is 27 at
45. y
x2 6 x 7
y
2x 6
2x 6
0
x
12 x 3 x 2
0, f (8)
the absolute
3. The critical point is x
the absolute
3.
46. f ( x) 6 x 2 x3
x 0 and x 4.
f ( x) 12 x 3 x 2
47. f ( x) x(4 x)3
4(4 x) 2 (1 x )
f ( x) x[3(4 x)2 ( 1)] (4 x)3 (4 x)2 [ 3x (4 x)] (4 x) 2 (4 4 x)
4(4 x) 2 (1 x) 0 x 1 or x 4. The critical points are x 1 and x 4.
48. g ( x) ( x 1) 2 ( x 3) 2
4( x 3)( x 1) ( x 2)
and x 3.
49. y
x2
2
x
y
2x
0
3 x(4 x )
0
x
0 or x
4. The critical points are
g ( x ) ( x 1) 2 2( x 3)(1) 2( x 1)(1) ( x 3) 2 2( x 3) ( x 1)[( x 1) ( x 3)]
4( x 3)( x 1)( x 2) 0
x 3 or x 1 or x 2. The critical points are x 1, x 2,
2
x2
2 x3 2
x2
The domain of the function is (
, 0)
2 x3 2
x2
0
2 x3 2
(0, ), thus x
Copyright
0
3
x 1; 2 x 2 2
x
undefined
x2
0
x
0.
0 is not the domain, so the only critical point is x 1.
2014 Pearson Education, Inc.
Section 4.1 Extreme Values of Functions
x2
x 2
2
50. f ( x)
( x 2 )2 x x 2 (1)
f ( x)
( x 2)
2
x2 4 x
( x 2) 2
x2 4 x
( x 2)2
( x 2)
0 x 2. The domain of the function is (
critical points are x 0 and x 4
51. y
x 2 32 x
x
0
y
x
2 x x2
0
2 x3/2 16
x
16
x
x
g ( x)
0 or x
53. Minimum value is 1 at x
1 x
1 x
2 x x2
which is zero when x
0
, 2)
0
x
(2, ), thus x
0
0
x 1;
0, x 1, and x
1 x
2 x x2
2.
2.
2.
3
3x2
2,
Local maximum at
2, 4
3
4 6
9
( 0.816, 5.089); local minimum
2, 4
3
4 6
9
(0.816, 2.911)
55. To find the exact values, note that y 3x 2 2 x 8
(3 x 4)( x 2), which is zero when x
2 or x 43 .
Local maximum at ( 2, 17); local minimum at
4 , 41
3
27
56. Note that y 5 x 2 ( x 5)( x 3), which is zero at
x 0, x 3, and x 5. Local maximum at (3, 108);
local minimum at (5, 0); (0, 0) is neither a maximum
nor a minimum.
Copyright
x
2
4; x 4 x2
0 or x
( x 2)
undefined
2 is not the domain, so the only
3/2
4; 2 x 16 undefined
x
0.
1 x
2. The critical points are x
4x
2 x3/2 16
0
4 and x
2 x x2
54. To find the exact values, note that y
at
2 x3/2 16
x
0. The critical points are x
2x x2
52. g ( x)
2x
x2
0
225
2014 Pearson Education, Inc.
undefined
2 x x2
0
226
Chapter 4 Applications of Derivatives
57. Minimum value is 0 when x
58. Note that y
x 2
,
x
1 or x 1.
which is zero at x
4 and is
undefined when x 0. Local maximum at (0, 0);
absolute minimum at (4, 4)
59. The actual graph of the function has asymptotes
at x
1, so there are no extrema near these values.
(This is an example of grapher failure.) There is
a local minimum at (0, 1).
60. Maximum value is 2 at x 1;
minimum value is 0 at x
1 and x
61. Maximum value is 12 at x 1;
minimum value is
1
2
at x
3.
1.
Copyright
2014 Pearson Education, Inc.
Section 4.1 Extreme Values of Functions
62. Maximum value is 12 at x
minimum value is
63.
y
ex
e x
y
1
2
0;
as x
ex
2.
e x
ex
e2 x 1 ,
ex
1
ex
which is 0 at x = 0; an absolute minimum value is 2 at
x = 0.
64.
y
ex
e x
y
ex
( e x)
ex
1
ex
e2 x 1 ,
ex
which is never zero; there are no extreme values.
65. y = x ln x
e 1; an absolute minimum value is
zero at x
66.
x 1x (1) ln x 1 ln x, which is
y
x
1
e
y
x 2 ln x
y
which is zero at x
value is
1
2e
at x
x 2 1x
2 x ln x
1
e
at
x(1 2 ln x),
e 1/2 ; an absolute minimum
1
e
Copyright
2014 Pearson Education, Inc.
227
228
67.
Chapter 4 Applications of Derivatives
cos 1 ( x 2 )
y
y
1
2x
(2 x)
1 ( x 2 )2
1 x4
, which
is zero at x = 0; an absolute maximum value is 2 at
x = 0; an absolute minimum value is 0 at x = 1 and
x = 1.
68.
sin 1 (e x )
y
y
ex
(e x )
1
x 2
1 (e )
1 e2 x
, which
is never zero; an absolute maximum value is 2 at
x = 0.
x 2/3 (1)
69. y
crit.pt.
0
crit.pt.
x
0
local max
12 101/3
25
undefined
local min
0
0
x
x 1/3 ( x 2
4)
0
minimum
3
undefined
local max
0
0
minimum
3
1
2 4 x
2
value
x 2 (4 x 2 )
( 2 x) (1) 4 x 2
4 x
2
crit.pt.
derivative
extremum
value
x
undefined
local max
0
0
minimum
0
maximum
2
2
undefined
local min
0
2
x
2
x
x
2
2
1.034
8x2 8
33 x
extremum
x 1
71. y
2
3
value
derivative
1
x
5x 4
33 x
extremum
x 2/3 (2 x)
70. y
x 1/3 ( x 2)
derivative
4
5
x
x
2
3
Copyright
4 2 x2
4 x2
2014 Pearson Education, Inc.
Section 4.1 Extreme Values of Functions
x2
1
2 3 x
5 x 2 12 x
2 3 x
72. y
crit.pt.
x 2 (4 x )(3 x )
2 3 x
( 1) 2 x 3 x
derivative
extremum
value
0
minimum
0
x
0
x
12
5
0
local max
144 151/2
125
x
3
undefined
minimum
0
73. y
2,
x 1
1,
x 1
crit.pt.
derivative
extremum
value
x 1
undefined
minimum
2
derivative
extremum
value
undefined
local min
3
0
local mix
4
derivative
extremum
value
0
maximum
5
undefined
local min
1
0
maximum
5
74. y
1,
x
0
2 2 x,
x
0
crit.pt.
x
0
x 1
75. y
2 x 2,
x 1
2 x 6,
x 1
crit.pt.
x
1
x 1
x
3
4.462
76. We begin by determining whether f ( x) is defined at x 1, where f ( x)
Clearly, f ( x)
lim f (1 h)
h
0
Thus, f ( x)
1
2
x 12 if x 1, and lim f (1 h)
h
0
1. Also, f ( x)
1. Since f is continuous at x 1, we have that f (1)
1
2
x 12 ,
1 x2
4
x 15
, x 1
4
x3 6 x 2 8 x ,
x 1
2
3 x 12 x 8 if x 1, and
1.
x 1
3 x 2 12 x 8, x 1
Copyright
1
2
2014 Pearson Education, Inc.
229
230
Chapter 4 Applications of Derivatives
Note that
But 2 2 3 3
crit.pt.
1
2
x 12
1, and 3 x 2 12 x 8
0 when x
0.845 1, so the critical points occur at x
derivative
extremum
value
x
1
0
local max
4
x
3.155
0
local min
3.079
122 4(3)(8)
2(3)
12
0 when x
2 2 33
1 and x
12
48
6
2 2 33 .
3.155.
77. (a) No, since f ( x) 23 ( x 2) 1/3 , which is undefined at x 2.
(b) The derivative is defined and nonzero for all x 2. Also, f (2) 0 and f ( x) 0 for all x 2.
(c) No, f ( x ) need not have a global maximum because its domain is all real numbers. Any restriction of f to a
closed interval of the form [a, b] would have both a maximum value and minimum value on the interval.
(d) The answers are the same as (a) and (b) with 2 replaced by a.
(a)
(b)
(c)
(d)
x 3 9 x, x
3 or 0
x
3
79. Yes, since f ( x) | x |
3
x2
( x 2 )1/2
f ( x)
. Therefore, f ( x)
3 x3 9,
x
3 or 0
x
3
.
x 9 x, 3 x 0 or x 3
3x 9,
3 x 0 or x 3
No, since the left- and right-hand derivatives at x 0, are 9 and 9, respectively.
No, since the left- and right-hand derivatives at x 3, are 18 and 18, respectively.
No, since the left- and right-hand derivatives at x
3, are 18 and 18, respectively.
3) and when f ( x) is undefined (at x 0 and x
3).
The critical points occur when f ( x) 0 (at x
The minimum value is 0 at x
3, at x 0, and at x 3; local maxima occur at
3, 6 3 and 3, 6 3 .
78. Note that f ( x)
1 ( x 2 ) 1/2 (2 x )
2
3
x
( x 2 )1/ 2
x
| x|
is not defined at x
0. Thus it is
not required that f be zero at a local extreme point since f may be undefined there.
80. If f (c) is a local maximum value of f, then f ( x) f (c) for all x in some open interval (a, b) containing c. Since
f is even, f ( x) f ( x) f (c) f ( c) for all x in the open interval ( b, a ) containing c. That is, f assumes
a local maximum at the point c. This is also clear from the graph of f because the graph of an even function is
symmetric about the y -axis.
81. If g (c) is a local minimum value of g, then g ( x) g (c) for all x in some open interval (a, b) containing c.
g ( x)
g (c) g ( c) for all x in the open interval ( b, a) containing c. That is,
Since g is odd, g ( x)
g assumes a local maximum at the point c. This is also clear from the graph of g because the graph of an odd
function is symmetric about the origin.
82. If there are no boundary points or critical points the function will have no extreme values in its domain. Such
x
. (Any other linear function f ( x) mx b with
functions do indeed exist, for example f ( x) x for
m 0 will do as well.)
Copyright
2014 Pearson Education, Inc.
Section 4.1 Extreme Values of Functions
83. (a) V ( x) 160 x 52 x 2 4 x3
V ( x) 160 104 x 12 x 2 4( x 2)(3x 20)
The only critical point in the interval (0, 5) is at x 2. The maximum value of V ( x) is 144 at x
(b) The largest possible volume of the box is 144 cubic units, and it occurs when x 2 units.
231
2.
84. (a) f ( x) 3ax 2 2bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f.
1 and x 1. The function f ( x) x3 1 has one
The function f ( x) x3 3 x has two critical points at x
critical point at x 0. The function f ( x) x3 x has no critical points.
(b) The function can have either two local extreme values or no extreme values. (If there is only one critical
point, the cubic function has no extreme values.)
85. s
1
2
gt 2
v
Thus s g0
v0 t
1
2
ds
dt
s0
v
g g0
2
gt v0
v
v0 g0
s0
0
v02
2g
v0
. Now
g
t
s0
s (t )
s0
t
gt
2
0
t
2v0
.
g
0 or t
s0 is the maximum height over the interval 0
86. dI
2sin t 2 cos t , solving dI
0 tan t 1 t 4
dt
dt
the peak current is 2 2 amps.
t is never negative)
t
2v0
.
g
n where n is a nonnegative integer (in this exercise
87. Maximum value is 11 at x 5; minimum value is 5
on the interval [ 3, 2]; local maximum at ( 5, 9)
88. Maximum value is 4 on the interval [5, 7];
minimum value is 4 on the interval [ 2, 1].
Copyright
v0
2014 Pearson Education, Inc.
232
Chapter 4 Applications of Derivatives
89. Maximum value is 5 on the interval [3, );
minimum value is 5 on the interval ( , 2].
90. Minimum value is 4 on the interval [ 1, 3]
91-98.
Example CAS commands:
Maple:
with(student):
f : x - x^4-8*x^2 4*x 2;
domain : x -20/25..64/25;
plot( f(x), domain, color black, title "Section 4.1 #91(a)" );
Df : D(f );
plot( Df(x), domain, color black, title "Section 4.1 #91(b)" )
StatPt : fsolve( Df(x) 0, domain )
SingPt : NULL;
EndPt : op(rhs(domain));
Pts : evalf ([EndPt,StatPt,SingPt]);
Values : [seq( f(x), x Pts )];
Maximum value is 2.7608 and occurs at x 2.56 (right endpoint).
Minimum value is -6.2680 and occurs at x 1.86081 (singular point).
Mathematica: (functions may vary) (see Section 2.5 re. RealsOnly ):
<<Miscellaneous `RealOnly`
Clear[f,x]
a
1; b 10/3;
f[x_ ] 2 2x 3 x 2/3
f '[ x]
Plot[{f[x], f '[x]}, {x, a, b}]
NSolve[f '[x]
0, x]
{f[a], f[0], f [x]/.%, f[b]}//N
In more complicated expressions, NSolve may not yield results. In this case, an approximate solution
(say 1.1 here) is observed from the graph and the following command is used:
FindRoot[f '[x] 0, {x, 1.1}]
Copyright
2014 Pearson Education, Inc.
Section 4.2 The Mean Value Theorem
4.2
1.
233
THE MEAN VALUE THEOREM
x2
When f ( x)
2 x 1 for 0
2. When f ( x)
x 2/3 for 0
3. When f ( x)
x
x 1, then
f (1) f (0)
1 0
f (1) f (0)
1 0
x 1, then
f (c )
x
2, then
f (2) f (1/2)
2 1/2
4. When f ( x)
x 1 for 1 x
3, then
f (3) f (1)
3 1
5. When f ( x)
sin 1 ( x) for 1
c2
1
1
x
4
c
2
6. When f(x) = ln(x
x3
7. When f ( x)
1
7
3
x
3
c
x
x 2 for 1
2
x
f (c )
0 x 2
1. Only c
2c 2
c 1/3
2
2
3.
2
c
1
1 c
8 .
27
c 1.
1
2 c 1
f (1) f ( 1)
1 ( 1)
1.
2
c
c
1
c2
0 1
f (c )
1, then f (c)
f (4) f (2)
4 2
4, then f (c)
x
0
2
3
1
2
1 c2
2
2
2
2, then
f (2) f ( 1)
2 ( 1)
, then
g (2) g ( 2)
2 ( 2)
f (c)
g (c )
1 is in the interval. If 0
9. Does not; f ( x) is not differentiable at x
ln 3 ln1
2
1
c 1
0.549 are both in the interval 1
2
3
1 c2
2
4
2
0.771
2
1) for 2
x3
x
4
1
1.22 and 1 3 7
8. When g ( x)
3c 2
for 12
f (c )
2
x
3c 2
2
ln 3
c 1
2c
c
1
7
3
c 1 ln23
2.820
.
2.
3
g (c ). If 2
x
2, then g ( x)
x
0, then g ( x)
2x
3
3x 2
g (c )
2c
3
3
g (c)
c
3.
2
0 in ( 1, 8).
10. Does; f ( x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1).
11. Does; f ( x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1).
12. Does not; f ( x) is not continuous at x
0 because lim f ( x) 1 0
13. Does not; f is not differentiable at x
1 in ( 2, 0).
x
0
f (0).
14. Does; f ( x) is continuous for every point of [0, 3] and differentiable for every point in (0, 3).
15. Since f ( x) is not continuous on 0
16. Since f ( x) must be continuous at x
lim f ( x)
x 1
lim f ( x )
x 1
we have lim f ( x)
x 1
1 3 a
lim f ( x)
x 1
x 1, Rolle s Theorem does not apply: lim f ( x)
x 1
0 and x 1 we have lim f ( x)
m b
5
2 x 3| x 1
Copyright
x
0
a
f (0)
a
lim x 1 0
x 1
3 and
m b. Since f ( x) must also be differentiable at x 1
m |x 1
1 m. Therefore, a
2014 Pearson Education, Inc.
3, m 1 and b
4.
f (1).
234
Chapter 4 Applications of Derivatives
17. (a)
a1 x a0 , then P (r1 ) P (r2 ) 0.
(b) Let r1 and r2 be zeros of the polynomial P ( x) x n an 1 x n 1
Since polynomials are everywhere continuous and differentiable, by Rolle s Theorem P (r ) 0 for some r
a1.
between r1 and r2 , where P ( x) nx n 1 (n 1)an 1 x n 2
18. With f both differentiable and continuous on [a, b] and f (r1 ) f (r2 ) f (r3 ) 0 where r1 , r2 and r3 are in [a, b],
then by Rolle s Theorem there exists a c1 between r1 and r2 such that f (c1 ) 0 and a c2 between r2 and r3 such
that f (c2 ) 0. Since f is both differentiable and continuous on [a, b], Rolle s Theorem again applies and we
have a c3 between c1 and c2 such that f (c3 ) 0. To generalize, if f has n 1 zeros in [a, b] and f ( n ) is continuous
on [a, b], then f ( n ) has at least one zero between a and b.
19. Since f exists throughout [a, b] the derivative function f is continuous there. If f has more than one zero
in [a, b], say f (r1 ) f (r2 ) 0 for r1 r2 , then by Rolle s Theorem there is a c between r1 and r2 such that
f (c) 0, contrary to f
0 throughout [a, b]. Therefore f has at most one zero in [a, b]. The same argument
holds if f
0 throughout [a, b].
20. If f ( x) is a cubic polynomial with four or more zeros, then by Rolle s Theorem f ( x) has three or more zeros,
f ( x) has 2 or more zeros and f ( x) has at least one zero. This is a contradiction since f ( x ) is a non-zero
constant when f ( x) is a cubic polynomial.
1 0 we conclude from the Intermediate Value Theorem that
21. With f ( 2) 11 0 and f ( 1)
4
f ( x) x 3 x 1 has at least one zero between 2 and 1. Then 2 x
1
8 x3
1
32 4 x3
4
3
29 4 x 3
1
f ( x) 0 for 2 x
1
f ( x) is decreasing on [ 2, 1]
f ( x) 0 has exactly one
solution in the interval ( 2, 1).
22. f ( x)
x3
and f ( x)
23. g (t )
g (0.99)
25. r ( )
(
f ( x)
x
0
t 1 4
15
1
1 t
7
0 if 2
t
g (15)
24. g (t )
4
x2
0
1 t
98.3
3x 2
8
x3
, 0)
f ( x) is increasing on (
f ( x) has exactly one zero in (
g (t )
1
2 t
1
2 t 1
0
, 0). Also, f ( x)
0 if x
2
, 0).
g (t ) is increasing for t in (0, ); g (3)
3 2
0 and
g (t ) has exactly one zero in (0, ).
3.1
g (t )
1
(1 t )2
1
2 1 t
0
g (t ) is increasing for t in ( 1, 1); g ( 0.99)
2.5 and
g (t ) has exactly one zero in ( 1, 1).
sin 2 3 8 r ( ) 1
, ); r (0)
8 and r (8) sin 2
26. r ( ) 2 cos2
( , ); r ( 2 )
zero in ( , ).
0 on (
2
4
2 sin
3
3
8
0
3
cos 3 1 13 sin 23
0 on (
r ( ) has exactly one zero in (
, )
, ).
r ( ) 2 2sin cos
2 sin 2
0 on ( , )
cos( 2 )
2
4 1
2 0 and r (2 ) 4 1
Copyright
2014 Pearson Education, Inc.
r ( ) is increasing on
r ( ) is increasing on
2 0 r ( ) has exactly one
Section 4.2 The Mean Value Theorem
27. r ( )
1
sec
5
3
r( )
0 on 0, 2
r ( ) has exactly one zero in 0, 2 .
and r (1.57) 1260.5
28. r ( )
tan
cot
0, 2 ; r 4
0 and r (1.57) 1254.2
4
r ( ) is increasing on 0, 2 ; r (0.1)
4
csc 2
sec 2
r( )
29. By Corollary 1, f ( x) 0 for all x
C 3
f ( x ) 3 for all x.
30. g ( x)
f (0)
3
(sec )(tan )
f ( x)
1 sec 2
cot 2
0 on 0, 2
r ( ) has exactly one zero in 0, 2 .
235
994
r ( ) is increasing on
C , where C is a constant. Since f ( 1)
3 we have
2 x 5 g ( x) 2 f ( x) for all x. By Corollary 2, f ( x ) g ( x ) C for some constant C. Then
g (0) C 5 5 C C 0
f ( x ) g ( x ) 2 x 5 for all x.
31. g ( x) x 2
g ( x) 2 x f ( x) for all x. By Corollary 2, f ( x) g ( x) C.
(a) f (0) 0 0 g (0) C 0 C C 0
f ( x) x 2
f (2) 4
(b) f (1) 0 0 g (1) C 1 C
C
1
f ( x) x 2 1
f (2) 3
(c) f ( 2) 3 3 g ( 2) C 3 4 C C
1
f ( x) x2 1
f (2)
3
32. g ( x) mx
g ( x) m, a constant. If f ( x) m, then by Corollary 2, f ( x) g ( x) b mx b where b is a
constant. Therefore all functions whose derivatives are constant can be graphed as straight lines y mx b.
33. (a) y
x2
2
C
(b) y
x3
3
C
(c) y
x4
4
C
34. (a) y
x2 C
(b) y
x2
x C
(c) y
x3
x2
35. (a) y
x 2
(b) y
x
C
(c) y
5 x 1x C
36. (a) y
(c) y
1
2
x2
40. g ( x)
1
x
f ( x)
42. r (t )
43. v
e2 x
2
x C
C
(b) y
2 x C
(b) y
2sin 2t C
2 sin 2t C
C
x C; 0
x2
C; 1
C ; f (0)
9.8t 5
1/2
(b) y
sec t t C; 0
ds
dt
y
1
x
2 x C
tan
39. f ( x)
41.
2
C
x1/2 C
y
1 cos 2t
2
1 cos 2t
2
(c) y
38. (a) y
x 1/2
2x
37. (a) y
1
x
y
x C
s
f (0)
02 0 C
1
1
g ( 1)
3
2
r (0)
e 2(0)
2
C
C
( 1)2
3
2
sec(0) 0 C
0
C
C
1
C
4.9t 2 5t C; at s 10 and t
Copyright
x2
f ( x)
C
2 3/2
3
y
1
f ( x) 1
1
r (t )
C
(c)
y
2 3/2
3
x
g ( x)
1
x
x2 1
e2 x
2
sec t t 1
0 we have C
10
2014 Pearson Education, Inc.
s
4.9t 2
5t 10
tan
C
236
Chapter 4 Applications of Derivatives
44. v
ds
dt
32t 2
s 16t 2
45. v
ds
dt
sin( t )
s
46. v
ds
dt
2
47. a
dv
dt
ds
dt
et
v
cos 2t
1
s
2t C ; at s
cos( t ) C ; at s
sin 2t
1
2
4 and t
0 and t
C; at s 1 and t
we have C
0 we have C
1
2
1
we have C
v
et
C; at v = 20 and t = 0 we have C = 19
et 19
s
et 19t C; at s = 5 and t = 0 we have C = 4
48. a 9.8 v 9.8t C1; at v
3 and t
t 0 we have C2 0 s 4.9t 2 3t
0 we have C1
3
v
4sin(2t ) v 2 cos(2t ) C1 ; at v 2 and t 0 we have C1
3 s sin(2t ) 3
3 and t 0 we have C2
50. a
9
s
cos 3t
1 and t
v
3 sin 3t
0 we have C2
C1; at v
0
s
0 and t
2t 1
1 cos( t )
s
sin 2t
s
1
et 19
v
49. a
s
2
s 16t 2
1
s
9.8t 3
0
0 we have C1
v
0
et 19t 4
s
4.9t 2
2cos(2t )
v
3t C2 ; at s
s
3 sin 3t
0 and
sin(2t ) C2 ; at
cos 3t
s
C2 ; at
cos 3t
51. If T (t ) is the temperature of the thermometer at time t, then T (0)
19 C and T (14)
100 C. From the Mean
T (14) T (0)
14 0
8.5 C / sec T (t0 ), the rate at which the
Value Theorem there exists a 0 t0 14 such that
temperature was changing at t t0 as measured by the rising mercury on the thermometer.
52. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been
going that speed at least once during the trip.
53. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been
going that speed at least once during the trip.
54. The runner s average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value
Theorem, the runner must have been going that speed at least once during the marathon. Since the initial speed
and final speed are both 0 mph and the runner s speed is continuous, by the Intermediate Value Theorem, the
runner s speed must have been 11 mph at least twice.
55. Let d (t ) represent the distance the automobile traveled in time t. The average speed over 0
The Mean Value Theorem says that for some 0 t0 2, d (t0 )
automobile at time t0 (which is read on the speedometer).
56. a(t )
v (t ) 1.6
v(t ) 1.6t C ; at (0, 0) we have C
57. The conclusion of the Mean Value Theorem yields
1
b
0
1
a
b a
2
58. The conclusion of the Mean Value Theorem yields bb
a2
a
d (2) d (0)
. The
2 0
(t ) 1.6t. When t
1
c2
c2
a b
ab
2c
c
a b.
2
t
2014 Pearson Education, Inc.
d (2) d (0)
.
2 0
value d ( t0 ) is the speed of the
30, then v(30)
a b
c
48 m/sec.
ab .
59. f ( x) [cos x sin( x 2) sin x cos( x 2)] 2 sin( x 1) cos( x 1) sin( x x 2) sin 2( x 1)
sin(2 x 2) sin (2 x 2) 0. Therefore, the function has the constant value f (0)
sin 2 1
Copyright
2 is
0.7081
Section 4.2 The Mean Value Theorem
237
which explains why the graph is a horizontal line.
60. (a) f ( x) ( x 2)( x 1) x ( x 1)( x 2) x5 5 x3 4x is one possibility.
(b) Graphing f ( x) x5 5 x3 4 x and f ( x) 5 x 4 15 x 2 4 on [ 3, 3] by [ 7, 7] we see that each
x-intercept of f ( x) lies between a pair of x -intercepts of f ( x ), as expected by Rolle s Theorem.
(c) Yes, since sin is continuous and differentiable on (
, ).
61. f ( x) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f ( x ) is
zero twice between a and b. Then by the Mean Value Theorem, f ( x ) would have to be zero at least once
between the two zeros of f ( x), but this can t be true since we are given that f ( x) 0 on this interval.
Therefore, f ( x) is zero once and only once between a and b.
62. Consider the function k ( x) f ( x ) g ( x). k ( x) is
continuous and differentiable on [a, b], and since
k (a) f (a) g (a) and k (b) f (b) g (b), by the
Mean Value Theorem, there must be a point c in
(a, b) where k (c) 0. But since k (c) f (c ) g (c),
this means that f (c ) g (c ), and c is a point where
the graphs of f and g have tangent lines with the
same slope, so these lines are either parallel or are
the same line.
63. f ( x) 1 for 1
x
4
f ( x ) is differentiable on 1 x
conditions of the Mean Value Theorem
f (4) f (1) 3
64. 0
1
2
f ( x)
for all x
f (4) f (1)
4 1
4
f (c) for some c in 1 x
f ( x) exists for all x, thus f is differentiable on ( 1, 1)
f satisfies the conditions of the Mean Value Theorem
0
since 0
f (1)
f ( 1)
2
f (1)
f is continuous on 1 x
f (1) f ( 1)
1 ( 1)
4
4
f (c ) 1
f (4) f (1)
3
f is continuous on [ 1, 1]
f (c) for some c in [ 1, 1]
1
2
0 f (1) f ( 1) 1. Since f (1) f ( 1) 1
f (1) 1 f ( 1) 2
f ( 1) we have f ( 1) f (1). Together we have f ( 1) f (1) 2 f ( 1).
Copyright
f satisfies the
2014 Pearson Education, Inc.
f ( 1), and
1
238
Chapter 4 Applications of Derivatives
65. Let f (t ) cos t and consider the interval [0, x] where x is a real number. f is continuous on [0, x] and f is
differentiable on (0, x) since f (t )
sin t
f satisfies the conditions of the Mean Value Theorem
f ( x) f (0)
x (0)
1 cosxx 1
cos x 1
x
f (c) for some c in [0, x]
sin c. Since 1 sin c 1
1
sin c 1
1. If x 0, 1 cosxx 1 1
x cos x 1 x |cos x 1| x | x | . If x 0, 1 cosxx 1 1
x cos x 1 x x cos x 1
x
( x) cos x 1
x |cos x 1|
x | x | . Thus, in both cases,
we have |cos x 1| | x | . If x 0, then |cos 0 1| |1 1| |0| |0|, thus |cos x 1| | x | is true for all x.
66. Let f ( x)
sin b sin a
b a
sin x for a x b. From the Mean Value Theorem there exists a c between a and b such that
sin b sin a
sin b sin a
cos c
1
1
1 |sin b sin a | | b a | .
b a
b a
67. Yes. By Corollary 2 we have f ( x) g ( x) c since f ( x)
then f (a ) g (a) c 0
f ( x) g ( x).
g ( x). If the graphs start at the same point x
a,
68. Assume f is differentiable and | f (w) f ( x)| | w x | for all values of w and x. Since f is differentiable,
f ( w) f ( x )
f ( x) exists and f ( x) lim
using the alternative formula for the derivative. Let g ( x) x ,
w x
w
x
which is continuous for all x. By Theorem 10 from Chapter 2, | f ( x)|
lim
w
x
| f ( w) f ( x)|
.
|w x|
Since f ( w)
from Chapter 2, f ( x )
lim
w
x
f ( x)
w x for all w and x
| f ( w) f ( x )|
|w x|
lim 1 1
w
f (b ) f ( a )
b a
69. By the Mean Value Theorem we have
f (b) f ( a), we have f (b) f ( a) 0
f ( x)
x
f (c)
lim
w x
| f ( w) f ( x )|
| w x|
1
1
f ( w) f ( x )
w x
lim
w
x
1 as long as w
f ( w) f ( x )
w x
x. By Theorem 5
f ( x ) 1.
f (c) for some point c between a and b. Since b a
0 and
0.
70. The condition is that f should be continuous over [a, b]. The Mean Value Theorem then guarantees the
f (b ) f ( a )
existence of a point c in (a, b) such that b a
f (c). If f is continuous, then it has a minimum and
f (c) max f , as required.
maximum value on [a, b], and min f
71.
f ( x)
(1 x 4 cos x)
1
(1 x 4 cos x) 2 (4 x3 cos x x 4 sin x)
f ( x)
x3 (1 x 4 cos x) 2 (4 cos x x sin x) 0 for 0
min f
1.09999
72. f ( x)
0
0.9999 and max f
x
0.1
1. Now we have 0.9999
1
0.09999
x
0.1
f (0.1) 1 0.1
f (0.1) 1.1.
(1 x 4 )
1
f ( x)
4 x3
(1 x 4 )3
(1 x 4 ) 2 ( 4 x3 )
x 0.1 min f 1 and max f
0.1 f (0.1) 2 0.10001 2.1
0 for 0 x
f ( x ) f (1)
Mean Value Theorem x 1
0
f ( x ) f (1)
Yes. From part (a), lim
x 1
x 1
f ( x)
Copyright
f (1)
f ( x ) f (1)
x 1
0
f ( x) is increasing when
1.0001
f ( x)
f (1). Suppose x 1, then by the
f (1). Therefore f ( x) 1 for all x since f (1) 1.
0 and lim
are equal and have the value f (1)
0.1
f (0.1) 2
0.1
1.0001. Now we have 1
f (0.1) 2.10001.
73. (a) Suppose x 1, then by the Mean Value Theorem
(b)
f ( x) is decreasing when 0
f (0.1) 1
0.1
x 1
f ( x ) f (1)
x 1
0 and f (1)
0. Since f (1) exists, these two one-sided limits
0
f (1)
2014 Pearson Education, Inc.
0.
Section 4.3 Monotonic Functions and the First Derivative Test
74. From the Mean Value Theorem we have
q
.
2p
has only one solution c
75. Proof that ln bx
f (b ) f ( a )
b a
(Note: p
d ln b
ln b ln x: Note that dx
x
76. (a)
ln b + C
d (tan 1 x
dx
cot 1 x)
tan 1 x cot 1 x
1
77. e x e
78.
y
b
1
b / x x2
ln bx
1
1 x2
0
4
C
2
1
tan
x
C for some constant C; if x
sec
1
2 csc
1
1
(e x1 ) x2
2
e0
ln y
e x2 x1
x)
1
1
1
x x2 1
x x2 1
4
4
x
e
x2 ln e x1
2
C
0
sec
1
ex
for all x;
x2 x1
x1 x2
1
x cot
1
x)
1
4
1
e( x
1;
x
ln b ln x C; if x = b, then
x
2
.
by Corollary 2 of the Mean Value Theorem that
e x1
e x2
2, then
x csc
e x1
eln y
1
e x2
e x1x2
1
x
2
.
e x1 e x2
y
e x1 x2
e x1x2
(e x1 ) x2
e x1x2 . Likewise,
e x1x2 .
MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST
1. (a) f ( x)
(b) f
x ( x 1)
critical points at 0 and 1
|
|
0
1
(c) Local maximum at x
2. (a) f ( x)
(b) f
( x 1)( x 2)
|
|
2
1
(c) Local maximum at x
3. (a) f ( x)
(b) f
( x 1)2 ( x 2)
|
|
2
increasing on (
critical points at 2 and 1
increasing on ( , 2) and (1, ), decreasing on ( 2, 1)
2 and a local minimum at x 1
critical points at 2 and 1
increasing on ( 2, 1) and (1, ), decreasing on (
1
( x 1)2 ( x 2) 2
|
|
2
(c) No local extrema
, 0) and (1, ), decreasing on (0, 1)
0 and a local minimum at x 1
(c) No local maximum and a local minimum at x
4. (a) f ( x)
(b) f
0
by Corollary 2 of the
C for some constant C; if x = 1, then
sec
x
d (ln b ln x )
and dx
by Corollary 2 of the Mean Value Theorem that
csc
x csc
1
x
ln b ln x.
d (sec 1 x
dx
(e x2 ) x1
4.3
1
1 x2
1
tan 1 cot 1
(b)
ln bx
C=0
2 pc q
0 since f is a quadratic function.)
Mean Value Theorem there is a constant C so that
ln 1 = ln b
f (c) where c is between a and b. But f (c)
239
2
critical points at 2 and 1
increasing on ( , 2) ( 2, 1)
(1, ), never decreasing
1
Copyright
, 2)
2014 Pearson Education, Inc.
240
Chapter 4 Applications of Derivatives
5. (a)
(b)
f ( x)
( x 1)e
x
critical point at x = 1
|
f
decreasing on (
, 1), increasing on (1, )
1
(c) Local (and absolute) minimum at x = 1
6. (a) f ( x)
(b) f
( x 7)( x 1)( x 5)
|
|
|
5
1
7
(c) Local maximum at x
1, local minima at x
7. (a) f ( x)
x 2 ( x 1)
( x 2)
(b) f
)(
|
|
2
0
1
critical points at x
(c) Local minimum at x 1
8. (a) f ( x)
( x 2)( x 4)
( x 1)( x 3)
(b) f
|
(b) f
|
)(
1
2
3
x2 4
x2
)(
|
2
0
2
6
x
10. (a) f ( x) 3
(
|
0
4
x
1/3
|
1, and x
3
, 4), ( 1, 2) and (3, ), decreasing on
2
2, x
increasing on (
critical points at x
2 and x
0.
, 2) and (2, ), decreasing on ( 2,0) and (0, 2)
2
4 and x
0
increasing on (4, ), decreasing on (0, 4)
(c) Local minimum at x
11. (a) f ( x)
(b) f
4, x
2, local minimum at x
3 x 6
x
2
, 2) and (1, ), decreasing on ( 2, 0) and (0, 1)
2, x
critical points at x
(c) Local maximum at x
, 5) and ( 1, 7)
7
increasing on (
4 and x
4
x2
|
(b) f
increasing on (
)(
( 4, 1) and (2, 3)
(c) Local maximum at x
5 and x
0, x 1 and x
critical points at x
4
9. (a) f ( x) 1
critical points at 5, 1 and 7
increasing on ( 5, 1) and (7, ), decreasing on (
( x 2)
)(
2
0
(c) Local maximum at x
4
critical points at x
2 and x 0
increasing on ( , 2) and (0, ), decreasing on ( 2, 0)
2, local minimum at x
0
12. (a) f ( x) x 1/2 ( x 3) critical points at x 0 and x 3
(b) f
increasing on (3, ), decreasing on (0, 3)
(
|
0
3
(c) No local maximum and a local minimum at x
13. (a) f ( x)
(b) f
[
0
,
2
2
3
(sin x 1)(2cos x 1), 0
|
|
2
and
|
2
3
4
3
4
3
2
x
]
2
3
critical points at x
increasing on
2
3
,
2
,x
2
3
4
3
, decreasing on 0,
2
3
and x
,2
(c) Local maximum at x
4
3
and x
0, local minimum at x
Copyright
2014 Pearson Education, Inc.
2
4
3
, and x
2
,
Section 4.3 Monotonic Functions and the First Derivative Test
14. (a) f ( x)
(b) f
(sin x cos x)(sin x cos x), 0
[
|
0
3 ,5
4
4
|
4
7
4
and
|
3
4
|
5
4
2
x
critical points at x
increasing on
]
7
4
2
4
4
3 ,
4
5 ,7
4
4
,x
, 34 and
5
4
x
, and x
241
7
4
, decreasing on 0,
4
,
,2
(c) Local maximum at x
3
4
0, x
7
4
and x
, local minimum at x
4
5
4
,x
and x
2
15. (a) Increasing on ( 2, 0) and (2, 4), decreasing on ( 4, 2) and (0, 2)
(b) Absolute maximum at ( 4, 2), local maximum at (0, 1) and (4, 1); Absolute minimum at (2, 3), local
minimum at ( 2, 0)
16. (a) Increasing on ( 4, 3.25), ( 1.5, 1), and (2, 4), decreasing on ( 3.25, 1.5) and (1, 2)
(b) Absolute maximum at (4, 2), local maximum at ( 3.25, 1) and (1, 1); Absolute minimum at ( 1.5, 1), local
minimum at ( 4, 0) and (2, 0)
17. (a) Increasing on ( 4, 1), (0.5, 2), and (2, 4), decreasing on ( 1, 0.5)
(b) Absolute maximum at (4, 3), local maximum at ( 1, 2) and (2, 1); No absolute minimum, local minimum
at ( 4, 1) and (0.5, 1)
18. (a) Increasing on ( 4, 2.5), ( 1, 1), and (3, 4), decreasing on ( 2.5, 1) and (1, 3)
(b) No absolute maximum, local maximum at ( 2.5, 1), (1, 2) and (4, 2); No absolute minimum, local
minimum at ( 1, 0) and (3, 1)
t 2 3t 3
19. (a) g (t )
,
3
2
g (t )
3,
2
3
2
, decreasing on
(b) local maximum value of g
20. (a) g (t )
3t 2
9t 5
g (t )
3,
2
decreasing on
x3
2 x2
increasing on 0,
3x2
2 x3 18 x
|
3
f
3
2
4
4;
3
, increasing on
, increasing on
|
3/2
21
4
3
2
at t
|
g
, increasing on
47
4
at t
critical points at x
local minimum value of h (0)
3
,
3 and
0,
4
3
h
0 at x
critical points at x
3,
3
2
,
3
2
4,
3
3 x
,
3/2
|
|
0
4/3
0, no absolute
3
, decreasing on
3, 3
3
(b) a local maximum is h
extrema
23. (a) f ( )
, 0) and
at x
3;
2
absolute maximum is
x(4 3 x)
6 x 2 18 6 x
h ( x)
|
h
4x
32
27
4
3
3,
2
at t
3;g
2
absolute maximum is
a critical point at t
, decreasing on (
(b) local maximum value of h
extrema
22. (a) h( x)
47
4
3,
2
at t
6t 9
h ( x)
4
3
21
4
3
2
(b) local maximum value of g
21. (a) h( x)
a critical point at t
2t 3
3
3
f ( )
|
|
0
1/2
(b) a local maximum is f
12 3 at x
6
2
12
3; local minimum is h
6 (1 2 )
critical points at
, increasing on 0, 12 , decreasing on (
1
2
1
4
at
1,
2
Copyright
3
a local minimum is f (0)
12 3 at x
0, 12
, 0) and
0 at
2014 Pearson Education, Inc.
3, no absolute
1,
2
0, no absolute extrema
,
242
Chapter 4 Applications of Derivatives
24. (a) f ( )
6
3
f ( )
|
f
6 3
|
2
2
3
2
2
, increasing on
critical points at
2, 2 , decreasing on
2
4 2 at
2, a local minimum is f
25. (a) f (r ) 3r 3 16r
f (r ) 9r 2 16 no critical points
decreasing
(b) no local extrema, no absolute extrema
(r 7)3
3(r 7)2
h (r )
2
a critical point at r
x 4 8 x 2 16
|
|
x4
7
2
0
4 x3
|
4 x2
|
0
1
3 t4
2
t6
|
H (t )
|
1
2
0, local minima are f ( 2)
2
0
|
h
0
x 6 x 1
, increasing on
0 at x
2, no absolute maximum;
0 at x
0 and g (2)
0 at x
2, no absolute
1
2
at t
1 and H (1) 12 at t 1, the local minimum is H (0)
1; no absolute minimum
45t 2 5t 4 5t 2 (3 t )(3 t )
|
, increasing on ( 3, 0)
0 at t
0,
critical points at t 0, 3
(0, 3), decreasing on ( , 3) and (3, )
3
(b) a local maximum is K (3) 162 at t
31. (a) f ( x)
, ), never
6t 3 6t 5 6t 3 (1 t )(1 t ) critical points at t 0, 1
|
, increasing on ( , 1) and (0, 1), decreasing on ( 1, 0) and (1, )
1
K (t )
|
3
2, no
g ( x) 4 x3 12 x 2 8 x 4 x ( x 2)( x 1) critical points at x 0, 1, 2
|
, increasing on (0, 1) and (2, ), decreasing on ( , 0) and (1, 2)
(b) the local maxima are H ( 1)
absolute maximum is 12 at t
30. (a) K (t ) 15t 3 t 5
K
|
4 2 at
7
(b) a local maximum is g (1) 1 at x 1, local minima are g (0)
maximum; absolute minimum is 0 at x 0, 2
29. (a) H (t )
H
2,
4 x3 16 x 4 x( x 2)( x 2)
critical points at x 0 and x
2
, increasing on ( 2, 0) and (2, ), decreasing on ( , 2) and (0, 2)
f ( x)
|
(b) a local maximum is f (0) 16 at x
absolute minimum is 0 at x
2
28. (a) g ( x)
g
2 and
, increasing on (
f
( , 7) ( 7, ), never decreasing
(b) no local extrema, no absolute extrema
27. (a) f ( x)
f
,
2
(b) a local maximum is f
absolute extrema
26. (a) h(r )
2
3, a local minimum is K ( 3)
3
x 1
f ( x) 1
x 1 3
x 1
162 at t
3, no absolute extrema
critical points at x 1 and x 10
f
(
|
1
10
,
increasing on (10, ), decreasing on (1, 10)
(b) a local minimum is f (10)
8, a local and absolute maximum is f (1) 1, absolute minimum of 8 at x 10
32. (a) g ( x)
4 x
x2
3
g ( x)
2
x
2x
2 2 x3/ 2
x
critical points at x 1 and x
increasing on (0, 1), decreasing on (1, )
(b) a local minimum is f (0) 3, a local maximum is f (1)
Copyright
0
g
6, absolute maximum of 6 at x 1
2014 Pearson Education, Inc.
(
|
0
1
,
Section 4.3 Monotonic Functions and the First Derivative Test
x 8 x2
33. (a) g ( x)
x (8 x 2 )1/2
critical points at x
(8 x 2 )1/2
g ( x)
2, 2 2
(8 x 2 )
1
2
x
(
|
|
2 2
2
2
2 2
0 at x
g
1/2
2(2
( 2 x)
x )(2
(b) local maxima are g (2)
x
4 at x
2 and g 2 2
0 at x
x2 5 x
x 2 (5 x)1/2
34. (a) g ( x)
at x
0, 4 and 5
2 and g
x
) , increasing on ( 2, 2), decreasing
2 2
x2 3
x 2
35. (a) f ( x)
g
f ( x)
f
|
), increasing on (0, 4), decreasing on (
4
5
)(
|
2
3
f ( x)
4, a local minimum is 0 at x
( x 2)2
|
5 x(4 x)
2 5 x
x 2 12 (5 x) 1/2 ( 1)
|
( x 3)( x 1)
, increasing on (
0 and x
x1/3 ( x 8)
, 0) and (4, 5)
5, no absolute maximum;
critical points at x 1, 3
, 1) and (3, ), decreasing on (1, 2) and (2, 3),
2 at x 1, a local minimum is f (3)
3 x 2 (3 x 2 1) x3 (6 x )
3 x 2 ( x 2 1)
(3 x 2 1)2
(3 x 2 1)2
f
x 4/3 8 x1/3
|
)(
2
0
4 x1/3
3
f ( x)
6 at x
3, no absolute extrema
a critical point at x
0
|
f
x 2/3 ( x 5)
38. (a) g ( x)
2 and x
x
0
0,
x1/3 ( x 2
x5/3 5 x 2/3
g
2
7
4)
, 0 and 0,
2
7
(b) local maximum is h
absolute extrema
2
7
5 x 2/3
3
g ( x)
|
)(
2
0
33 4
7.56 at x
0, 2
, 2)
2, no absolute maximum; absolute
)(
|
2/ 7
0
2/ 7
3.12 at x
Copyright
x 1/3
5( x 2)
5( x 2)
33 x
4
3
x 2/3
, 2) and (0, ), decreasing on ( 2, 0)
7x 2
3 x
,
2
2
7
, the local minimum is h
2014 Pearson Education, Inc.
0 at x
7x 2
3
, increasing on
2
7
critical points at
33 x
2, a local minimum is g (0)
7 x 4/3
3
h ( x)
|
24 3 2
77 / 6
10
3
, increasing on (
4.762 at x
x 7/3 4 x1/3
h
critical points at x
3 x 2/3
2
(b) local maximum is g ( 2)
extrema
39. (a) h( x)
4( x 2)
(0, ), decreasing on (
63 2
(b) no local maximum, a local minimum is f ( 2)
,
0
8 x 2/3
3
, increasing on ( 2, 0)
minimum is 6 3 2 at x
2
critical points
increasing on ( , 0) (0, ), and never decreasing
(b) no local extrema, no absolute extrema
37. (a) f ( x)
4 at
2; absolute minimum is 4 at x
0
( x 2)2
1
x3
3 x2 1
36. (a) f ( x)
2 x (5 x)1/2
g ( x)
2 x ( x 2) ( x2 3)(1)
discontinuous at x 2
(b) a local maximum is f (1)
2 2, local minima are g ( 2)
2 2, absolute maximum is 4 at x
(b) a local maximum is g (4) 16 at x
absolute minimum is 0 at x 0, 5
2
7
x)
2 2 x 2 2
2 2, 2 and 2, 2 2
on
x
243
and
2
7
0, no absolute
critical points at
2
7
,
24 3 2
77/6
, decreasing on
3.12, no
244
Chapter 4 Applications of Derivatives
x8/3 4 x 2/3
40. (a) k ( x)
x 2/3 ( x 2
k
|
)(
|
1
0
1
4)
k ( x)
8 x 5/3
3
8( x 1)( x 1)
33 x
8 x 1/3
3
, increasing on ( 1, 0) and (1, ), decreasing on (
(b) local maximum is k (0) 0 at x 0, local minima are k ( 1)
absolute minimum is 3 at x
1
41. (a)
f ( x)
e2 x
f
e x
2e 2 x
f ( x)
e x
0
e3 x
, increasing on 13 ln 12 ,
|
1 ln 1
3
2
3 at x
(b) a local minimum is 2/3
2
critical points at x
1 ln 1
3
2
1
2
3 at x
0, 1
, 1) and (0, 1)
1, no absolute maximum;
1 ln 1
3
2
a critical point at x
, 13 ln 12
, decreasing on
3 at x
; no local maximum; an absolute minimum 2/3
1 ln 1
3
2
2
;
no absolute maximum
42. (a)
f ( x)
e x
e x
2 x
f ( x)
no critical points
f
|
, increasing on (0, )
0
(b) A local minimum is 1 at x = 0, no local maximum; an absolute minimum is 1 at x = 0, no absolute
maximum
43. (a) f(x) = x ln x
f ( x) 1 ln x
e 1
a critical point at x
f
[
|
0
(e 1 ,
f ( x)
, increasing on
1
), decreasing on (0, e 1 )
(b) A local minimum is
absolute maximum
44. (a)
e
x 2 ln x
e 1 at x
f ( x)
increasing on (e 1/2 ,
(b) A local minimum is
e 1 , no local maximum, an absolute minimum is
x 2 x ln x
x (1 2 ln x)
a critical point at x
e 1 , no
e 1 at x
e 1/2
f
[
|
0
e
e 1
2
at x
,
1/ 2
), decreasing on (0, e 1/2 )
e 1
2
at x
e 1/2 , no local maximum; an absolute minimum is
e 1/2 ,
no absolute maximum
45. (a) f ( x)
2 x x2
f ( x)
2 2x
a critical point at x 1
f
a local maximum is 1 at x 1, a local minimum is 0 at x 2.
(b) There is an absolute maximum of 1 at x 1; no absolute minimum.
(c)
46. (a) f ( x)
( x 1) 2
f ( x)
2( x 1)
a critical point at x
1
f
|
] and f (1) 1 and f (2)
1
2
|
] and
1
0
f ( 1) 0, f (0) 1 a local maximum is 1 at x 0, a local minimum is 0 at x
(b) no absolute maximum; absolute minimum is 0 at x
1
(c)
Copyright
2014 Pearson Education, Inc.
1
0
Section 4.3 Monotonic Functions and the First Derivative Test
47. (a) g ( x)
x2
4x 4
g ( x)
2x 4
2( x 2)
a critical point at x
2
g
g (1) 1, g (2) 0 a local maximum is 1 at x 1, a local minimum is g (2)
(b) no absolute maximum; absolute minimum is 0 at x 2
(c)
48. (a) g ( x)
x2 6 x 9
g ( x)
2x 6
2( x 3)
a critical point at x
[
|
1
2
0 at x
3
f (t ) 12 3t 2
3(2 t )(2 t )
critical points at t
and f ( 3)
9, f ( 2)
16, f (2) 16 local maxima are 9 at t
is 16 at t
2
(b) absolute maximum is 16 at t 2; no absolute minimum
(c)
Copyright
2014 Pearson Education, Inc.
2
g
f
3 and 16 at t
and
2
1, g ( 3) 0 a local maximum is 0 at x
g ( 4)
3, a local minimum is 1 at x
(b) absolute maximum is 0 at x
3; no absolute minimum
(c)
49. (a) f (t ) 12t t 3
245
[
|
4
3
4
and
[
|
|
3
2
2
2, a local minimum
246
Chapter 4 Applications of Derivatives
50. (a) f (t )
t 3 3t 2
f (t )
3t 2 6t
3t (t 2)
critical points at t
and f (0) 0, f (2)
4, f (3) 0 a local maximum is 0 at t
(b) absolute maximum is 0 at t 0, 3; no absolute minimum
(c)
51. (a) h( x)
x3
3
2x2
4x
h ( x)
x2
4x 4
( x 2)2
x3 3 x 2
3x 1
3x 2
k ( x)
0 and t
2
6 x 3 3( x 1)2
f
|
|
]
0
2
3
3, a local minimum is 4 at t
a critical point at x
h(0) 0 no local maximum, a local minimum is 0 at x
(b) no absolute maximum; absolute minimum is 0 at x 0
(c)
52. (a) k ( x)
0 and t
2
h
0
a critical point at x
1
[
|
0
2
and
k
and k ( 1) 0, k (0) 1 a local maximum is 1 at x 0, no local minimum
(b) absolute maximum is 1 at x 0; no absolute minimum
(c)
53. (a) f ( x)
25 x 2
f ( x)
x
25 x 2
critical points at x
0, x
5, and x
5
f
f ( 5) 0, f (0) 5, f (5) 0 local maximum is 5 at x 0; local minimum of 0 at x
(b) absolute maximum is 5 at x 0; absolute minimum of 0 at x
5 and x 5
(c)
Copyright
2014 Pearson Education, Inc.
2
|
]
1
0
(
|
),
5
0
5
5 and x
5
Section 4.3 Monotonic Functions and the First Derivative Test
x2
54. (a) f ( x)
f
2 x 3,3
[
, f (3)
x
0
2x 2
f ( x)
only critical point in 3
x2 2 x 3
local minimum of 0 at x
x
is at x
247
3
3, no local maximum
3
3, no absolute maximum
(b) absolute minimum of 0 at x
(c)
55. (a) g ( x)
g
x 2 ,0
x2 1
x 1
[
|
0
0.268
maximum at x
), g 2
(c)
x2
4 x2
g
f
, 2
3
4 3 6
x 1
at x
|
], g (0)
2
0
1
sin 2 x, 0
0
and 0 at x
3
4 3 6
2
3, no absolute maximum
|
4
x
8x
(4 x 2 )2
g ( x)
(
[
3
1
(b) absolute minimum of 0 at x
(c)
57. (a) f ( x)
only critical point in 0
1.866
x 1 is x
local minimum of
2
3
4 3 6
3
at x
0.268
2
3, local
0.
(b) absolute minimum of
56. (a) g ( x)
x2 4 x 1
( x 2 1)2
g ( x)
0
|
x 1 is x
0
0, local maximum of 13 at x 1.
local minimum of 0 at x
0, no absolute maximum
f ( x)
3
4
only critical point in 2
2 cos 2 x, f ( x )
] , f (0)
cos 2 x
1, f 34
0, f 4
, and local minima are 1 at x
0
3
4
and 0 at x
0
1, f ( )
0.
(b) The graph of f rises when f
0, falls when f 0, and has local
extreme values where f
0. The function f has a local minimum
value at x 0 and x 34 , where the values
of f change from negative to positive. The function f has a local
maximum value at x
and x 4 , where the values of f change
from positive to negative.
Copyright
2014 Pearson Education, Inc.
critical points are x
0
4
and x
local maxima are 1 at x
3
4
4
248
Chapter 4 Applications of Derivatives
58. (a) f ( x)
sin x cos x, 0
7
4
and x
f
2
x
[
f ( x)
|
0
7
4
0
tan x
1, f 34
] , f (0)
|
3
4
cos x sin x, f ( x )
2
3 cos x
are x
6
sin x, 0
x
2
f ( x)
7
6
f
[
|
and x
0
3 sin x cos x, f ( x)
|
] , f (0)
7
6
6
2 at x
0
x
2 x tan x,
4
and x
x
2
f
4
(
|
4
2
is 2 1 at x
4
2 sec2 x, f ( x)
f ( x)
2
|
), f
4
2
, and local minimum is 1 2 at x
4
4
1
3
2, f 76
local maxima are 2at x 6 and 3 at x 2 , and local minima are 2 at x
(b) The graph of f rises when f
0, falls when
f 0, and has local extreme values where
f
0. The function f has a local minimum
value at x 0 and x 76 , where the values
of f change from negative to positive. The
function f has a local maximum value at
x 2 and x 6 , where the values of f
change from positive to negative.
60. (a) f ( x)
and 1 at x
tan x
3, f 6
2
2, f (2 )
7
4
0
sec2 x
2
1, f 4
7
6
2
1
f
0 at x
x
2
2sin 2x
[
|
0
2 /3
0 and
at x
f ( x)
1
2
cos 2x , f ( x)
] and f (0)
2
0, f 23
2 , a local minimum is 3
Copyright
0
critical points
2, f (2 )
and 3 at x
0.
critical points are
1 2
local maximum
.
3
cos 2x
1
2
3, f (2 )
3 at x
2
3
2014 Pearson Education, Inc.
local
0.
(b) The graph of f rises when f
0, falls when
f 0, and has local extreme values where
f
0. The function f has a local minimum
value at x 4 , where the values of f change
from negative to positive. The function f has a
, where the
local maximum value at x
4
values of f change from positive to negative.
61. (a) f ( x)
3
4
critical points are x
2, f 74
maxima are 2 at x 34 and 1 at x 2 , and local minima are
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f
0. The function f has a local minimum
value at x 0 and x 74 , where the values
of f change from negative to positive. The
function f has a local maximum value at x 2
and x 34 , where the values of f change from
positive to negative.
59. (a) f ( x)
1
a critical point at x
local maxima are
2
3
3
Section 4.3 Monotonic Functions and the First Derivative Test
249
(b) The graph of f rises when f
0, falls when
f 0, and has a local minimum value at the
point where f changes from negative to
positive.
62. (a) f ( x)
2 cos x cos 2 x
x
, 0,
f
[
f ( x) 2 sin x 2 cos x sin x 2(sin x)(1 cos x ) critical points at
3, f ( ) 1 a local maximum is
|
] and f ( ) 1, f (0)
0
1 at x
, a local minimum is 3 at x 0
(b) The graph of f rises when f
0, falls when
f 0, and has local extreme values where
f
0. The function f has a local minimum
value at x 0, where the values of f change
from negative to positive.
63. (a) f ( x) csc 2 x 2 cot x
f
point at x 4
(
f ( x)
|
/4
0
2(csc x)( csc x )(cot x ) 2( csc2 x)
2(csc2 x) (cot x 1)
a critical
0 no local maximum, a local minimum is 0 at x 4
) and f 4
(b) The graph of f rises when f
0, falls when
f 0, and has a local minimum value at the
0 and the values of f change
point where f
from negative to positive. The graph of f
steepens as f ( x)
.
64. (a) f ( x) sec2 x 2 tan x
f
at x 4
(
/2
f ( x)
|
/4
2(sec x)(sec x)(tan x) 2sec2 x (2sec2 x) (tan x 1)
a critical point
0 no local maximum, a local minimum is 0 at x 4
) and f 4
/2
(b) The graph of f rises when f
0, falls when
f 0, and has a local minimum value where
f
0 and the values of f change from
negative to positive.
Copyright
2014 Pearson Education, Inc.
250
Chapter 4 Applications of Derivatives
65. h( )
3cos 2
3 sin
2
2
h( )
a local minimum is 3 at
66. h( )
5sin 2
minimum is 0 at
67. (a)
h( )
h
[
0
2
5
2
cos 2
h
0
(b)
[
] , (0, 3) and (2 , 3)
], (0, 0) and ( , 5)
a local maximum is 5 at
0
(c)
68. (a)
(b)
(c)
(d)
69. (a)
(b)
70. (a)
(b)
Copyright
a local maximum is 3 at
0,
2
2014 Pearson Education, Inc.
(d)
, a local
Section 4.3 Monotonic Functions and the First Derivative Test
251
x sin 1x has an infinite number of local maxima and minima on its domain, which is
71. The function f ( x)
( , 0) (0, ). The function sin x has the following properties: a) it is continuous on ( , ); b) it is
periodic; and c) its range is [ 1, 1]. Also, for a 0, the function 1x has a range of ( , a ] [a, )
0, 1a . In particular, if a 1, then 1x
1,0
a
on
1 or 1x
(0, 1], namely at 1x
takes on the values of 1 and 1 infinitely many times on [ 1, 0)
2,
x
2
3
2
5
,
,
1
x
. Thus sin
1
x
interval (0, 1], 1 sin
(0, 1]. This means sin 1x
1 when x is in [ 1, 0)
2
3
2
,
has infinitely many local maxima and minima in [ 1, 0)
1 and since x
0 we have x
x sin
1
x
5
2
,
,
.
(0, 1]. On the
x. On the interval
[ 1, 0), 1 sin 1x 1 and since x 0 we have x x sin 1x
x. Thus f ( x) is bounded by the lines
y x and y
x. Since sin 1x oscillates between 1 and 1 infinitely many times on [ 1, 0) (0, 1] then f will
oscillate between y x and y
x infinitely many times. Thus f has infinitely many local maxima and minima.
We can see from the graph (and verify later in Chapter 7) that lim x sin 1x 1 and lim x sin 1x 1. The
x
x
graph of f does not have any absolute maxima, but it does have two absolute minima.
a x2
ax 2 bx c
72. f ( x)
b
a
x
a x2
c
b . Thus when a 0,
2a
, 2 ab and decreasing
vertex is at x
increasing on
a 0, f
; for a
|
b
a
ax 2 bx
f ( x)
2x
2
b2
4a 2
b2
4a
b,
2a
f is increasing on
b,
2a
on
0, f
b /2 a
73. f ( x)
x
f
2
b 2 4ac ,
4a
, 2 ab ; when a
and decreasing on
2ax b
a parabola whose
b
2a
2a x
0, f is
for
2a x b, f (1)
2
a b
2, f (1)
0
2a b
0
a
2, b
4
4x
f ( x)
sin x
cos x
tan x
0
0 d 0, f (1)
3, c 0, d 0
x = 0; f ( x)
0 for
1 a b c d
f ( x ) 2 x3 3 x 2
4
x
0 and f ( x)
1,
0 for
there is a relative maximum at x = 0 with f(0) = ln(cos 0) = ln 1 = 0;
3
ln cos
4
b
2a
b /2 a
f ( x)
75. (a) f(x) = ln(cos x)
x
a x
. Also note that f ( x)
|
.
74. f ( x) ax3 bx 2 cx d
f ( x) 3ax 2 2bx c, f (0)
f (0) 0 c 0, f (1) 0 3a 2b c 0 a 2, b
0
c
ln 1
4
1 ln 2
2
2
absolute minimum occurs at x
3
and f 3
with f 3
ln cos 3
ln 12
ln 2. Therefore, the
ln 2 and the absolute maximum occurs at x = 0 with
f(0) = 0.
sin(ln x )
x
0 for 12 x 1 and f ( x)
there is a relative maximum at x = 1 with f(1) = cos(ln 1) = cos 0 = 1;
(b) f(x) = cos(ln x)
f 12
at x
cos ln 12
1
2
76. (a) f(x) = x
(b) f(1) = 1
f ( x)
cos( ln 2)
and x = 2 with f 12
0
x = 1; f ( x)
0 for 1 < x
2
cos(ln 2) and f(2) = cos(ln 2). Therefore, the absolute minimum occurs
f (2)
cos(ln 2), and the absolute maximum occurs at x = 1 with f(1) = 1.
f ( x) 1 1x ; if x > 1, then f ( x) 0 which means that f(x) is increasing
ln 1 = 1
f(x) = x ln x > 0, if x > 1 by part (a) x > ln x if x > 1
ln x
Copyright
2014 Pearson Education, Inc.
252
77.
Chapter 4 Applications of Derivatives
ex
f ( x)
2x
ex
f ( x)
2; f ( x)
ex
0
2
x
ln 2; f(0) = 1, the absolute maximum;
f(ln 2) = 2 2 ln 2 0.613706, the absolute minimum; f(1) = e
since f is increasing on the interval (ln 2, 1).
2
0.71828, a relative or local maximum
2esin( x /2) has a maximum whenever sin 2x 1 and a minimum whenever sin 2x
1.
Therefore the maximums occur at x = + 2k(2 ) and the minimums occur at x = 3 + 2k(2 ), where k is any
integer. The maximum is 2e 5.43656 and the minimum is 2e 0.73576.
78. The function f ( x)
79.
x 2 ln 1x
f ( x)
ln x
1.
2
f ( x)
0 for x
x 2 11 ( x 2 )
2 x ln 1x
f ( x)
2 x ln 1x
x
e 1/2
Since x = 0 is not in the domain of f, x
1 .
e
1
e
Therefore, f
1 ln
e
x
x(2 ln x 1); f ( x)
1 .
e
1 ln e1/2
e
e
Also, f ( x)
1 ln e
2e
1
2e
0
0 for 0
x = 0 or
1
e
x
and
is the absolute maximum value of f
1 .
e
assumed at x
ex
80. (a) Let f ( x)
x 1
ex 1
f ( x)
a critical point at x = 0
f
|
, so f is increasing
0
on (0, ); since f(0) = 0 it follows that f ( x)
ex
(b) Let f ( x)
1 x2
2
x 1
ex
f(0) = 0 it follows that f ( x)
81. Let x1
ex
f ( x)
1 x2
2
ex
x 1 0 for x
x 1 0 for x
x 1 0 for x
ex
0
x 1 for x
0 by part (a), so f is increasing on (0, ); since
ex
0
1 x2
2
x 1 for x
x2 be two numbers in the domain of an increasing function f. Then, either x1
implies f ( x1 ) f ( x2 ) or f ( x1 ) f ( x2 ), since f(x) is increasing. In either case, f ( x1 )
to-one. Similar arguments hold if f is decreasing.
5
6
82. f(x) is increasing since x2
x1
1x
3 2
83. f(x) is increasing since x2
x1
27 x23
df
dx
81x 2
df 1
dx
1
81x 2
1 1/3
x
3
84. f(x) is decreasing since x2
df
dx
24 x 2
df 1
dx
df
dx
3(1 x) 2
df 1
dx
1
(1
2
x1
27 x13 ; y
1
9
1/3
x)
5 ; df
6
dx
df 1
dx
1
3
27 x3
x
1
1
3
1 y1/3
3
0.
x2 or x1
x2 which
f ( x2 ) and f is one-
3
f 1 ( x)
1 x1/3 ;
3
x 2/3
1 8 x13 ; y 1 8 x3
1 8 x23
x1
1
24 x 2
85. f(x) is decreasing since x2
1
9 x 2/3
1x
3 1
0.
1
6(1 x ) 2/3
(1 x2 )3
1
3(1 x ) 2 1 x1/3
Copyright
1 (1
6
1 (1
2
y )1/3
f 1 ( x)
1 (1
2
x)1/3 ;
x) 2/3
(1 x1 )3 ; y
1
3 x 2/3
x
(1 x)3
x 1 y1/3
1 x 2/3
3
2014 Pearson Education, Inc.
f 1 ( x) 1 x1/3 ;
Section 4.4 Concavity and Curve Sketching
86. f(x) is increasing since x2
df
dx
4.4
df 1
dx
5 x 2/3
3
x1
1
5 2/3
x
3
x
3/5
x25/3
x15/3 ; y
x5/3
y3/5
x
f 1 ( x)
253
x3/5 ;
3 x 2/5
5
3
5 x 2/5
CONCAVITY AND CURVE SKETCHING
x3
3
1. y
x2
2
2 x 13
x2
y
x 2
( x 2)( x 1)
, 1) and (2, ), falling on ( 1, 2), concave up on 12 ,
(
a local maximum is 32 at x
x4
4
2. y
2x2
4
x3 4 x
y
x( x 2
4)
x( x 2)( x 2)
3 ( x2
4
3. y
and
2 , 2 . Consequently, a local
3
3
2 , 16 are points of inflection.
3 9
1)2/3
3
4
y
( x 2 1) 1/3 (2 x)
2
3
y
|
3
down on
|
1
3
7)
y
at x 1; y
3
14
x
(x
2
1) 3x
x sin 2 x
y
1 2 cos 2 x, y
2 /3
falling on
are
3
2
x
, 0)
5/3
3
2
|
)(
1
0
1
3 at
4
x
5/3
x
(2 x
2
3
3 3 ( x 2 1) 4
2, and
x
0, local
,
,
3 and
1), y
3,
, concave
|
|
/3
/3
]
|
)(
|
1
0
1
the graph is
0
the graph is rising on
2 /3
2 /3
2
3
2
3
and 3
|
/2
3
2
2014 Pearson Education, Inc.
at x
3
|
|
0
/2
, 2 and 0, 2
, 2 , (0, 0), and 2 , 2
the
1, a local minimum is
)(
a point of inflection at (0, 0).
, 0 and 2 , 23 , concave down on
Copyright
2
a local maximum is 27
at x
7
3
, 3 and 3 , 23
local maxima are 23
at x
2
3
3
at x
and 23
at x 23 ; y
4sin 2 x, y
[
2
3
2
2
and
the
) (
x 2/3 ( x 2 1), y
2
3
the graph is concave up on
2 ,
3
and
3
2x
[
,
3 x 2 . The graph
3, 3 4 4
1/3
1/3
2
3
a local maximum is
( x 2 1) 4/3 (2 x)
9 x1/3 (2 x )
x 2/3 ( x 2 7) 14
concave up on (0, ), concave down on (
5. y
1
3
, 1) and (1, ), falling on ( 1, 1)
5/3
3x 2
0, local minima are 0 at x
the graph is concave up on
points of inflection at
graph is rising on (
27
7
)(
1
3, 3
9 x1/3 ( x 2
14
4. y
)(
maximum is 4 at x
, 1) and (0, 1)
( x 2 1) 1/3 ( x)
1; y
4
x ( x 2 1) 1/3 , y
graph is rising on ( 1, 0) and (1, ), falling on (
minima are 0 at x
3x2
y
, 2) and (0, 2), concave up on
concave down on
2 , 16
3 9
2, and 12 , 34 is a point of inflection.
1, a local minimum is 3 at x
is rising on ( 2, 0) and (2, ), falling on (
at
2 x 1 2 x 12 . The graph is rising on
and concave down on
, 12 . Consequently,
y
3
,3 ,
, local minima
]
2 /3
points of inflection
254
Chapter 4 Applications of Derivatives
6. y
tan x 4 x
y
sec2 x 4, y
(
|
/3
/2
,
3 2
y
, falling on
,
3 3
2(sec x)(sec x)(tan x)
a local maximum is
2
2(sec x)(tan x), y
)
/3
/2
4
3
3
at x
(
/2
concave down on
2
,0
the graph is rising on
|
3
, a local minimum is 3
|
)
0
/2
2
4
3
, 3 and
at x
3
the graph is concave up on 0, 2 ,
a point of inflection at (0, 0)
7. If x
0, sin x sin x and if x 0, sin x sin( x)
sin x. From the sketch the graph is rising on
3 ,
, 0, 2 and 32 , 2 , falling on
2
2
2 , 32 ,
are 1 at x
2
3
2
, 0 and 2 , 32 ; local minima
and 0 at x 0; local maxima are
1 at x
and 0 at x
2 ; concave up on
2
( 2 , ) and ( , 2 ), and concave down on
points of inflection are
( , 0) and (0, )
( , 0) and ( , 0)
8. y
2 cos x
3
4
2
at x
2x
y
2 sin x
, 4 and 54 , 32 , falling on
2
at x
and 3 2 2 at x
4
4
5
4
;y
2 cos x, y
concave down on
[
2, y
[
, 34 and
3
2
|
/4
|
|
/2
points of inflection at
]
3 /2
, 22
2
|
]
5 /4
3 /2
local maxima are 2
, and local minima are
/2
,
2 2
|
3 /4
,5
4 4
2
3
2
4
at x
concave up on
and 2 ,
9. When y x 2 4 x 3, then y 2 x 4 2( x 2)
and y 2. The curve rises on (2, ) and falls on
( , 2). At x 2 there is a minimum. Since y 0,
the curve is concave up for all x.
10. When y 6 2 x x 2 , then y
2 2x
2(1 x)
and y
2. The curve rises on ( , 1) and falls on
1 there is a maximum. Since y 0,
( 1, ). At x
the curve is concave down for all x.
Copyright
;
2014 Pearson Education, Inc.
2
2
rising on
2 at x
3
4
and
,
2
5
, 2 and 2 , 32 ,
2
4
Section 4.4 Concavity and Curve Sketching
11. When y x3 3 x 3, then y 3x 2 3
3( x 1)( x 1) and y 6 x. The curve rises on
1 there is
( , 1) (1, ) and falls on ( 1, 1). At x
a local maximum and at x 1 a local minimum. The
curve is concave down on ( , 0) and concave up on
(0, ). There is a point on inflection at x 0.
12. When y
y
x(6 2 x ) 2 , then
4 x(6 2 x) (6 2 x) 2
12(3 x)(1 x) and
y
12(3 x) 12(1 x) 24( x 2). The curve
rises on ( , 1) (3, ) and falls on (1, 3). The curve
is concave down on ( , 2) and concave up on
(2, ). At x 2 there is a point of inflection.
13. When y
2 x3 6 x 2 3, then y
6 x 2 12 x
6 x( x 2) and y
12 x 12
12( x 1). The
curve rises on (0, 2) and falls on ( , 0) and (2, ).
At x 0 there is a local minimum and at x 2 a local
maximum. The curve is concave up on ( , 1) and
concave down on (1, ). At x 1 there is a point of
inflection.
14. When y 1 9 x 6 x 2 x3 , then y
9 12 x 3 x 2
3( x 3)( x 1) and y
12 6 x
6( x 2). The
curve rises on ( 3, 1) and falls on ( , 3) and
1 there is a local maximum and at
( 1, ). At x
x
3 a local minimum. The curve is concave up on
2
( , 2) and concave down on ( 2, ). At x
there is a point of inflection.
15. When y ( x 2)3 1, then y 3( x 2) 2 and
y 6( x 2). The curve never falls and there are no
local extrema. The curve is concave down on ( , 2)
and concave up on (2, ). At x 2 there is a point of
inflection.
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Chapter 4 Applications of Derivatives
16. When y 1 ( x 1)3 , then y
3( x 1)2 and
y
6( x 1). The curve never rises and there are no
local extrema. The curve is concave up on ( , 1)
and concave down on ( 1, ). At x
1 there is a
point of inflection.
17. When y x 4 2 x 2 , then y 4 x3 4 x
4 x( x 1)( x 1) and y 12 x 2 4
12 x
1
3
x
1
3
. The curve rises on ( 1, 0)
and (1, ) and falls on ( , 1) and (0, 1). At x
1
there are local minima and at x 0 a local maximum.
The curve is concave up on
and concave down on
points of inflection.
18. When y
4x x
,
1 , 1
3
3
x 4 6 x 2 4, then y
3 x
3 and y
1
3
1
3
. At x
there are
4 x3 12 x
12 x 2 12
12( x 1)( x 1). The curve rises on
and 0, 3 , and falls on
1 ,
3
and
3, 0 and
,
3,
3
. At
x
3 there are local maxima and at x 0 a local
minimum. The curve is concave up on ( 1,1) and
1 there
concave down on ( , 1) and (1, ). At x
are points of inflection.
19. When y 4 x 3 x 4 , then
y 12 x 2 4 x3 4 x 2 (3 x) and y 24 x 12 x 2
12 x(2 x ). The curve rises on ( , 3) and falls on
(3, ). At x 3 there is a local maximum, but there is
no local minimum. The graph is concave up on (0, 2)
and concave down on ( , 0) and (2, ). There are
inflection points at x 0 and x 2.
20. When y x 4 2 x3 , then y 4 x3 6 x 2 2 x 2 (2 x 3)
and y 12 x 2 12 x 12 x( x 1). The curve rises on
3,
and falls on
, 32 . There is a local
2
3 , but no local maximum. The
minimum at x
2
curve is concave up on ( , 1) and (0, ), and
1 and x 0 there
concave down on ( 1, 0). At x
are points of inflection.
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
21. When y
x5 5 x 4 , then
5 x4
y
20 x 3
5 x3 ( x 4) and y
20 x3 60 x 2
20 x 2 ( x 3). The curve rises on ( , 0) and (4, ),
and falls on (0, 4). There is a local maximum at x 0,
and a local minimum at x 4. The curve is concave
down on ( , 3) and concave up on (3, ). At x 3
there is a point of inflection.
22. When y
x
2
y
and y
4
x 2x 5 , then
5
4
3
x(4) 2x 5 12
2
3 2x 5 12
5x
2
x
2
x
2
5
5
5
3 5x
2
5 ,
3 5
2
2
5 2x 5 ( x 4). The curve is rising on (
, 2) and
(10, ), and falling on (2, 10). There is a local
maximum at x 2 and a local minimum at x 10.
The curve is concave down on ( , 4) and concave
up on (4, ). At x 4 there is a point of inflection.
23. When y x sin x, then y 1 cos x and y
sin x.
The curve rises on (0, 2 ). At x 0 there is a local
and absolute minimum and at x 2 there is a local
and absolute maximum. The curve is concave down
there is
on (0, ) and concave up on ( , 2 ). At x
a point of inflection.
24. When y x sin x, then y 1 cos x and y sin x.
The curve rises on (0, 2 ). At x 0 there is a local
and absolute minimum and at x 2 there is a local
and absolute maximum. The curve is concave up on
there is
(0, ) and concave down on ( , 2 ). At x
a point of inflection.
25. When y
3 x 2 cos x, then y
3 2sin x and
y 2 cos x. The curve is increasing on 0, 43 and
5
3
, and decreasing on 43 , 53 . At x 0 there
is a local and absolute minimum, at x 43 there is a
,2
local maximum, at x 53 there is a local minimum,
and at x 2 there is a local and absolute maximum.
The curve is concave up on 0, 2 and 32 , 2 , and
is concave down on 2 , 32 . At x
there are points of inflection.
2
and x
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Chapter 4 Applications of Derivatives
26. When y
y
4x
3
2
4
3
tan x, then y
sec2 x and
2sec x tan x. The curve is increasing on
, , and decreasing on 2 , 6 and 6 , 2 .
6 6
At x
there is a local minimum, at x 6 there is
6
a local maximum, there are no absolute maxima or
absolute minima. The curve is concave up on
, 0 , and is concave down on 0, 2 . At x 0
2
there is a point of inflection.
27. When y sin x cos x, then y
sin 2 x cos 2 x
cos 2x and y
2sin 2 x. The curve is increasing
3
on 0, 4 and 4 , , and decreasing on 4 , 34 . At
x
0 there is a local minimum, at x
4
3
4
there is
there is a
a local and absolute maximum, at x
there is
local and absolute minimum, and at x
a local maximum. The curve is concave down on
0, 2 , and is concave up on 2 , . At x 2 there is
a point of inflection.
28. When y cos x
3 sin x, then y
sin x
3 cos x
and y
cos x
3 sin x. The curve is increasing on
0, 3 and 43 , 2 , and decreasing on 3 , 43 . At
x
0 there is a local minimum, at x
3
there is
a local and absolute maximum, at x 43 there is a
local and absolute minimum, and at x 2 there is
a local maximum. The curve is concave down on
0, 56 and 116 , 2 , and is concave up on
5
6
, 116 . At x
of inflection.
5
6
and x
11
6
there are points
4 x 9/5 .
29. When y x1/5 , then y 15 x 4/5 and y
25
The curve rises on ( , ) and there are no extrema.
The curve is concave up on ( , 0) and concave
down on (0, ). At x 0 there is a point of inflection.
6 x 8/5 .
30. When y x 2/5 , then y 52 x 3/5 and y
25
The curve is rising on (0, ) and falling on ( , 0).
At x 0 there is a local and absolute minimum.
There is no local or absolute maximum. The curve is
concave down on ( , 0) and (0, ). There are no
points of inflection, but a cusp exists at x 0.
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
31. When y
x
x2 1
3 x . The
( x 2 1)5/ 2
y
1
( x 2 1)3/ 2
, then y
and
curve is increasing on (
, ).
There are no local or absolute extrema. The curve is
concave up on ( , 0) and concave down on (0, ).
At x 0 there is a point of inflection.
32. When y
y
1,
1 x2
2x 1
( x 2)
, then y
(2 x 1)2 1 x2
and
4 x3 12 x 2 7 . The curve is decreasing on
(2 x 1)3 (1 x 2 )3/ 2
1 and
1 , 1 . There are no absolute extrema,
2
2
there is a local maximum at x
1 and a local
minimum at x 1. The curve is concave up on
( 1, 0.92) and 12 , 0.69 , and concave down on
0.92, 12 and (0.69, 1). At x
0.92 and x
0.69
there are points of inflection.
33. When y 2 x 3 x 2/3 , then y 2 2 x 1/3 and
2 x 4/3 . The curve is rising on (
y
, 0) and (1, ),
3
and falling on (0, 1). There is a local maximum at
x 0 and a local minimum at x 1. The curve is
concave up on ( , 0) and (0, ). There are no points
of inflection, but a cusp exists at x 0.
34. When y
2 x
3/5
5 x 2/5 2 x, then y
6
5
1 and y
x
2 x 3/5 2
8/5
. The curve is rising
on (0, 1) and falling on ( , 0) and (1, ). There is
a local minimum at x 0 and a local maximum at
x 1. The curve is concave down on ( , 0) and
(0, ). There are no points of inflection, but a cusp
exists at x 0.
y
x 2/3 52
5 x 2/3 x5/3 , then
2
5 x 1/3 5 x 2/3 5 x 1/3 (1 x ) and
3
3
3
5 x 4/3 10 x 1/3
5 x 4/3 (1 2 x ). The
9
9
9
35. When y
x
y
curve
is rising on (0, 1) and falling on ( , 0) and (1, ).
There is a local minimum at x 0 and a local
maximum at x 1. The curve is concave up on
, 12 and concave down on 12 , 0 and (0, ).
There is a point of inflection at x
at x 0.
1
2
and a cusp
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Chapter 4 Applications of Derivatives
36. When y x 2/3 ( x 5)
y 53 x 2/3 10
x 1/3
3
x5/3 5 x 2/3 , then
5 x 1/3 ( x 2) and
3
37. When y x 8 x 2
y (8 x 2 )1/2 ( x)
x(8 x 2 )1/2 , then
(8 x 2 ) 1/2 ( 2 x)
y 10
x 1/3 10
x 4/3 10
x 4/3 ( x 1). The curve
9
9
9
is rising on ( , 0) and (2, ), and falling on (0, 2).
There is a local minimum at x 2 and a local
maximum at x 0. The curve is concave up on
( 1, 0) and (0, ), and concave down on ( , 1).
There is a point of inflection at x
1 and a cusp
at x 0.
(8 x 2 )
y
1/2
on
2(2 x )(2 x )
(8 2 x )
x2 )
1 (8
2
2 x ( x 2 12)
(8 x 2 )3
1
2
2
3
2
2 2 x 2 2 x
and
( 2 x )(8 2 x 2 ) (8 x 2 )
1
2
( 4 x)
. The curve is rising on ( 2, 2), and falling
2 2, 2 and 2, 2 2 . There are local minima
x
2 and x 2 2, and local maxima at x
2 2
and x 2. The curve is concave up on 2 2, 0 and
concave down on 0, 2 2 . There is
a point of inflection at x
38. When y
(2 x 2 )3/2 , then y
3x 2 x 2
y
3x
2 1/2
( 3)(2 x )
6(1 x )(1 x )
2 x
0.
2 x
2
( 3x)
3
2
x
1
2
(2 x 2 )1/2 ( 2 x)
2
x and
(2 x 2 )
1/2
. The curve is rising on
( 2 x)
2, 0 and
falling on 0, 2 . There is a local maximum at x
0,
and local minima at x
2. The curve is concave
down on ( 1, 1) and concave up on
2, 1 and
1, 2 . There are points of inflection at x
39. When y
y
16 x 2 , then y
16
(16 x 2 )3/ 2
x
16 x 2
1.
and
. The curve is rising on ( 4, 0) and
falling on (0, 4). There is a local and absolute
maximum at x 0 and local and absolute minima at
x
4 and x 4. The curve is concave down on
( 4, 4). There are no points of inflection.
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
x2
40. When y
y
, then y
3
2 x 4 . The
x3
4
x3
2
2
x
2 x3 2
x2
2
x2
2x
and
curve is falling on (
, 0) and
(0, 1), and rising on (1, ). There is a local minimum at
x 1. There are no absolute maxima or absolute minima.
, 3 2 and (0, ), and
The curve is concave up on
3
concave down on
3
at x
41. When y
and y
2, 0 . There is a point of inflection
2.
x2 3 ,
x 2
2 x ( x 2) ( x 2 3)(1)
then y
(2 x 4)( x 2)
2
(x
2
( x 3)( x 1)
( x 2) 2
( x 2) 2
4 x 3)2( x 2)
2 . The
( x 2)3
( x 2) 4
curve
is rising on ( , 1) and (3, ), and falling on (1, 2) and
(2, 3). There is a local maximum at x 1 and a local
minimum at x 3. The curve is concave down on
( , 2) and concave up on (2, ). There are no points
of inflection because x 2 is not in the domain.
42. When y
3 3
x
x2
( x 1)2/3
1, then y
3
2x
.
( x3 1)5/3
and y
The curve is rising on ( , 1), ( 1, 0), and (0, ). There
are no local or absolute extrema. The curve is concave up
on ( , 1) and (0, ), and concave down on ( 1, 0).
There are points of inflection at x
1 and x 0.
43. When y
8x
x
2
4
, then y
8( x 2 4)
(x
2
4)
2
16 x ( x 2 12)
and y
( x 2 4)3
.
The curve is falling on ( , 2) and (2, ), and is rising
on ( 2, 2). There is a local and absolute minimum at
x
2, and a local and absolute maximum at x 2. The
curve is concave down on
, 2 3 and 0, 2 3 , and
concave up on
2 3, 0 and 2 3,
of inflection at x
2 3, x
horizontal asymptote.
44. When y
5
x
4
5
, then y
0, and x
20 x3
( x 5) 2
4
. There are points
2 3. y
and y
0 is a
100 x 2 ( x 4 3)
( x 4 5)3
.
The curve is rising on ( , 0), and is falling on (0, ).
There is a local and absolute maximum at x 0, and there
is no local or absolute minimum. The curve is concave up
, 4 3 and 4 3, , and concave down on 4 3, 0
on
and 0, 4 3 . There are points of inflection at x
x
4
3. There is a horizontal asymptote of y
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0.
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Chapter 4 Applications of Derivatives
x 2 1, | x | 1
, then y
1 x2 , | x | 1
| x 2 1|
45. When y
2 x, | x | 1
2 x, | x | 1
2, | x | 1
. The curve rises on ( 1, 0) and (1, )
2, | x | 1
and falls on ( , 1) and (0, 1). There is a local maximum
1. The curve is concave
at x 0 and local minima at x
up on ( , 1) and (1, ), and concave down on ( 1, 1).
There are no points of inflection because y is not
differentiable at x
1 (so there is no tangent line at
those points).
and y
x2
46. When y
|x
then y
2
2 x, x
0
2
2x |
2 x 2, x
0
2 2 x, 0
x
2x x , 0
x
x2
2
2 x, x
2, and y
2,
2, x
0
2, 0
x
2.
2, x 2
2 x 2, x 2
The curve is rising on (0, 1) and (2, ), and falling on
( , 0) and (1, 2). There is a local maximum at x 1 and
local minima at x 0 and x 2. The curve is concave up
on ( , 0) and (2, ), and concave down on (0, 2). There
are no points of inflection because y is not differentiable
at x 0 and x 2 (so there is no tangent at those points).
47. When y
x,
| x|
x
and y
3/ 2
4
( x)
4
,
3/ 2
Since lim y
x
x
0
x, x
0
x
0
, x
0
, then y
x
x
0
x
0
.
and lim y
0
1 ,
2 x
1 ,
2 x
there is a cusp at
0
x 0. There is a local minimum at x 0, but no local
maximum. The curve is concave down on ( , 0) and
(0, ). There are no points of inflection.
48. When y
y
| x 4|
1
2 x 4
1
2 4 x
,x
,x
Since lim y
x
4
x 4, x
4
4 x, x
4
4
4
and y
and lim y
x
4
, then
( x 4)
4
3/ 2
(4 x )
4
3/ 2
,x
4
,x
4
.
there is a cusp at
x 4. There is a local minimum at x 4, but no local
maximum. The curve is concave down on ( , 4) and
(4, ). There are no points of inflection.
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
xe1/ x
x2
xe1/ x , then y
49. When y
e1/ x 12
and y
e1/ x
x2
1 1x
x
e1/ x
e1/ x 1 1x
e1/ x 1
x2 x
e1/ x
x3
.
The curve is rising on (1, ) and ( , 0) and falling on
(0, 1). The curve is concave down on ( , 0) and
concave up on (0, ). There is a local minimum of e at
x = 1, but there are no inflection points.
50.
ex
x
y
( x 1) e x
xe x e x
x2
y
y
|
|
0
1
the graph is rising on
(1, ), falling on (
at x = 1; y
x2
, 0) and (0, 1); a local minimum is e
x 2 ( xe x e x e x ) ( xe x e x )(2 x )
( x 2 2 x 2)e x
4
x3
x
y
|
the graph is concave up on
0
(0, ), concave down on (
points.
51.
ln(3 x 2 )
y
y
2x
3 x2
y
(
3
, 0), but has no inflection
|
)
0
3
2x
x2 3
the graph is rising on
3, 0 , falling on 0, 3 ; a local minimum is ln 3 at
( x 2 3)(2) (2 x )(2 x )
x = 0; y
(x
y
(
2
3)
2
)
3
2( x 2 3)
( x 2 3)2
the graph is concave down on
3
3, 3 .
52.
y
x(ln x) 2
y
y
(1) (ln x)2
x 2 ln x 1x
(
0
|
e
|
2
ln x(2 ln x )
the graph is rising on
1
(0, e 2 ) and (1, ), falling on (e 2 , 1); a local
maximum is 4e 2 at x
at x = 1; y
y
(
0
(e 1 ,
ln x 1x
|
e
e 2 and a local minimum is 0
1
x
(2 ln x)
the graph is concave up on
1
), concave down on (0, e 1 )
1
2(1 ln x )
x
point of
1
inflection at (e , e ).
Copyright
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264
53.
Chapter 4 Applications of Derivatives
ex
y
2e x
ex
y
3x
2e x 3
y
(e x )2 3e x 2
(e x 2)(e x 1)
x
ex
e
|
|
0
ln 2
the graph is increasing
on ( , 0) and (ln 2, ), decreasing on (0, ln 2); a local
maximum is 1 at x = 0 and a local minimum is
1
ex
y
the graph is concave up on
|
1 ln 2
2
1 ln 2,
2
xe x
y
2e x
y
y
e
3 ln 2
2
x
x
point of
.
(1 x)e x
xe
|
ex
, 12 ln 2
, concave down on
inflection at 12 ln 2,
54.
(e x ) 2 2
3 ln 2 at x = ln 2; y
the graph is increasing on (
, 1)
1
and decreasing on (1, ); a local maximum is e 1 at
( x 1)e x
x = 1; y
y
( 1)e x
|
( x 2)e x
the graph is concave up on
2
(2, ), concave down on (
, 2)
point of inflection at
(2, 2e 2 ).
55. y = ln(cos x)
y
sin x
cos x
y
tan x
.... ) none (
7
2
5
2
|
) none (
|
) none (
|
2
3
2
0
3
2
2
5
2
the graph is increasing on ...,
3
2
,2
, ..., decreasing on
2 ,
2
,
2
,
3
2
, 0, 2 ,
2
5
2
the graph is concave down on
y
3
2
2 , 52 ;
1
cos2 x
,
, 2 , 32 , 52 , ...
2
56.
,
,0 ,
sec2 x
local maxima are 0 at x = 0, 2 , 4 , ...; y
2
ln x
x
1
x
x
y
ln x
x
1
2 x
2 ln x
2 x3/ 2
2
y
(
0
|
e2
the graph is increasing on (0, e2 ), decreasing on (e2 ,
local maximum is
2 x3/ 2
y
1
x
2
e
at x
y
(
0
e ;
(2 ln x ) 2 32 x1/ 2
(2 x3/ 2 )2
|
8/3
); a
2
3ln x 8
4 x5/ 2
the graph is concave up on (e8/3 ,
),
point of inflection is e8/3 ,
.
e
concave down on (0, e8/3 )
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8
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) none ( ...
5
2
7
2
Section 4.4 Concavity and Curve Sketching
57.
y
1
1 e
ex
ex 1
x
y
y
ex
(e x 1)2
( e x 1)2
the graph is increasing on (
(e
y
( e x 1)e x e x e x
x
2
1) e
x
x
e 2(e
x
1)e
x
(e x 1)4
y
x
, );
x
e (1 e )
(e x 1)3
|
the graph is concave up on
0
(
58.
y
point of inflection is 0, 12 .
, 0), concave down on (0, )
ex
1 ex
(e x 1)e x e x e x
y
y
the graph is increasing on (
(e
y
ex
(1 e x )2
(1 e x )2
x
2
1) e
x
x
e 2(e
x
1)e
x
(1 e x )4
y
|
x
, );
x
e (1 e )
(1 e x )3
the graph is concave up on
0
(
59. y
, 0), concave down on (0, )
2 x x2
point of inflection is 0, 12 .
(1 x)(2 x), y
rising on ( 1, 2), falling on (
|
|
1
2
, 1) and (2, )
there is a local maximum at x 2 and a local
minimum at x
1; y 1 2 x, y
|
1/2
, 12 , concave down on 12 ,
concave up on
a point of inflection at x
60. y
x2 x 6
1
2
( x 3)( x 2), y
|
|
2
3
rising on ( , 2) and (3, ), falling on ( 2, 3)
there is a local maximum at x
2 and a local minimum at
x 3; y 2 x 1, y
|
1/2
concave up on 12 ,
a point of inflection at x
61. y
x( x 3) 2 , y
, 12
, concave down on
1
2
rising on (0, ), falling
|
|
0
3
on ( , 0) no local maximum, but there is a local minimum at
x 0; y ( x 3) 2 x (2) ( x 3) 3( x 3)( x 1), y
concave up on ( , 1) and (3, ), concave
|
|
1
3
down on (1, 3)
62. y
points of inflection at x 1 and x
x 2 (2 x), y
on (2, )
minimum; y
|
|
0
4/3
and 43 ,
|
|
0
2
rising on (
there is a local maximum at x
2 x(2 x) x 2 ( 1)
3
, 2), falling
2, but no local
x(4 3x ), y
concave up on 0, 43 , concave down on
points of inflection at x
0 and x
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Chapter 4 Applications of Derivatives
63. y
x( x 2 12)
y
2 3,
x x 2 3 x 2 3 ,
|
|
|
2 3
0
2 3
, falling on
rising on
2 3, 0 and
, 2 3 and 0, 2 3
a local
maximum at x 0, local minima at
x
2 3; y 1 ( x 2 12) x(2 x) 3( x 2)( x 2),
concave up on ( , 2) and (2, ),
y
|
|
2
2
concave down on ( 2, 2)
at x
2
64. y
points of inflection
( x 1) 2 (2 x 3), y
3,
2
, 32
, falling on
|
|
3/2
1
no local maximum,
3;
2
a local minimum at x
2( x 1)(2 x 3) ( x 1)2 (2)
y
y
|
|
2/3
1
concave down on
rising on
2( x 1)(3 x 2),
, 23 and (1, ),
concave up on
2,1
3
2
3
points of inflection at x
and
x 1
65. y
y
(8 x 5 x 2 )(4 x) 2
|
|
0
, 0) and 85 ,
(
x (8 5 x)(4 x)2 ,
rising on 0, 85 , falling on
|
8/5
4
a local minimum at x
y
0;
(8 10 x)(4 x) 2
(8 x 5 x 2 )(2)(4 x )( 1)
4(4 x)(5 x 2 16 x 8), y
concave
up on
66. y
y
|
|
|
8 2 6
5
8 2 6
5
4
, 8 25 6 and 8 25 6 , 4 , concave down on
8 2 6 8 2 6
, 5
5
x
and (4, )
( x2
2 x)( x 5) 2 x( x 2)( x 5)2 ,
rising on (
|
|
|
2
5
falling on (0, 2) a local maximum at x
a local minimum at x 2;
(2 x 2)( x 5)2
2( x 2
and
|
6
2
concave up on
, 4 2 6 and
5
6
2
4
2
,
4
6
6
,5
2
0,
|
4
4
, 0) and (2, ),
2 x)( x 5)
2( x 5)(2 x 2 8 x 5), y
x
8 2 6
5
points of inflection at x
4
0
y
8,
5
a local maximum at x
4
|
6
2
5
and (5, ), concave down on
points of inflection at x
Copyright
4
6
2
and
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
sec2 x, y
67. y
(
(
/2
on
)
/2
tan x, y
(
/2
2
x
concave up on 0, 2 , concave down
, 0 , 0 is a point of inflection.
2
68. y
|
0
,0
|
)
0
/2
rising on 0, 2 , falling on
no local maximum, a local minimum at
sec2 x, y
0; y
(
)
/2
concave up on
69. y
cot 2 , y
/2
,
2 2
(
no points of inflection
|
)
0
rising on (0, ), falling on
2
( , 2 ) a local maximum at
1 csc 2 , y
local minimum; y
2
2
up, concave down on (0, 2 )
70. y
csc2 2 , y
(
)
0
2
no local extrema;
y
, 2 , never falling
2 (sec2 x ) (tan x),
2(sec x)(sec x)(tan x)
y
2
/2
no local extrema;
y
rising on
)
/2
2 csc 2
never concave
)
0
2
no points of inflection
rising on (0, 2 ) , never falling
csc 2 cot 2
csc 2 2 cot 2 , y
, no
(
(
1
2
|
)
0
2
concave up on ( , 2 ), concave down on (0, )
a point of inflection at
71. y
y
tan 2
(
1 (tan
|
, 2 , falling on
local minimum at
4
y
4
on
2
1),
)
/4
/2
/4
/2
(
1)(tan
|
,4
;y
4
rising on
a local maximum at
2
, 4 and
4
,a
2 tan sec2 ,
concave up on 0, 2 , concave down
|
)
/2
0
/2
,0
a point of inflection at
0
Copyright
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Chapter 4 Applications of Derivatives
72. y
y
1 cot 2
(
|
(1 cot )(1 cot ),
rising on 4 , 34 , falling on
|
)
/4
0
3 /4
0, 4 and 34 ,
a local maximum
3
at
, a local minimum at
;
4
4
2(cot )( csc 2 ), y
y
(
|
0
/2
)
concave up on 0, 2 , concave down on 2 ,
a point of infection at
2
73. y
3
2
cos t , y
[
|
0
/2
|
]
3 /2
2
rising on 0, 2 and
, falling on 2 , 32
local maxima at t 2 and t
3
local minima at t 0 and t 2 ; y
sin t , y [
|
,2
0
concave up on ( , 2 ), concave down on (0, )
inflection at t
74. y
sin t , y
[
|
]
2
a point of
rising on (0, ), falling on
]
0
2 ,
2
( , 2 ) a local maximum at t
,
local minima at t 0 and t 2 ; y cos t ,
concave up on 0, 2 and
y [
|
|
]
3
2
75. y
0
/2
3 /2
2
, concave down on 2 , 32
points of inflection at t 2 and t
,2
( x 1) 2/3 , y
)(
1
falling
no local extrema; y
3
2
rising on (
2 (x
3
, ), never
1) 5/3 , y
concave up on ( , 1), concave down on ( 1, )
inflection and vertical tangent at x
1
76. y
( x 2) 1/3 , y
(
x
, 2)
2; y
(
x
, 2) and (2, )
2
)(
)(
1
a point of
rising on (2, ), falling on
2
no local maximum, but a local minimum at
1 ( x 2) 4/3 , y
)(
concave down on
3
2
no points of inflection, but there is a cusp at
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
77. y
x 2/3 ( x 1), y
)(
|
0
1
rising on (1, ),
falling on ( , 1) no local maximum, but
a local minimum at x 1; y 13 x 2/3 23 x 5/3
1 x 5/3 ( x
3
2), y
|
)(
2
0
concave up on
( , 2) and (0, ), concave down on ( 2, 0)
points of
inflection at x
2 and x 0, and a vertical tangent at x 0
78. y
x 4/5 ( x 1), y
|
)(
1
0
rising on ( 1, 0) and
(0, ), falling on ( , 1) no
local maximum, but a local minimum at x
y 15 x 4/5 54 x 9/5 15 x 9/5 ( x 4), y
1;
)(
|
0
4
concave up on ( , 0) and (4, ), concave down on (0, 4)
points of inflection at x 0 and x 4, and a vertical tangent
at x 0
79. y
2 x, x
0
2 x, x
0
,y
local extrema; y
rising on (
|
2, x
0
,y
2, x 0
on (0, ), concave down on ( , 0)
x 0
80. y
(
x2 , x
0
,y
|
0
x ,x 0
, 0) no local maximum,
2
y
|
no
concave up
)(
0
a point of inflection at
rising on (0, ), falling on
2 x, x
0
2 x, x
concave up on ( , )
0
but a local minimum at x
, )
0
0; y
,
0
no point of inflection
81. The graph of y f ( x ) the graph of y f ( x) is concave up on
(0, ), concave down on ( , 0) a point of inflection at x 0;
the graph of y f ( x )
y
|
|
the graph
y f ( x) has both a local maximum and a local minimum
Copyright
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Chapter 4 Applications of Derivatives
82. The graph of y f ( x)
|
y
the graph of
y f ( x ) has a point of inflection, the graph of
y f ( x)
y
|
the graph of y f ( x) has
|
both a local maximum and a local minimum
83. The graph of y f ( x)
y
|
|
the graph of y f ( x) has two points of inflection, the graph of
y f ( x)
y
|
the graph of y f ( x) has a local
minimum
84. The graph of y f ( x)
y
|
the graph of
y f ( x ) has a point of inflection; the graph of
y f ( x)
y
the graph of y f ( x) has
|
|
both a local maximum and a local minimum
85. y
2 x2 x 1
x2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
1
2
y
; y
(x
1)
2
( x 1)
( x 1)3
There are no critical points. The function is decreasing on its
domain. There are no inflection points. The function is concave
down on ( , 1) ( 1, 1) and concave up on (1, ). The
numerator and denominator share a factor of x 1. Dividing out
this common factor gives y
2x 1
x 1
( x 1), which shows that
x 1 is a vertical asymptote. Now dividing numerator and
denominator by x gives y
2 (1/ x )
,
1 (1/ x )
which shows that y
horizontal asymptote. The graph will have a hole at x
y
2( 1) 1
1( 1) 1
3 . The
2
x-intercept is
2 is a
1,
1.
2
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
86.
y
x 2 49
x 2 5 x 14
Since 7 and 2 are roots of the denominator, the domain is
( , 7) ( 7, 2) (2, ).
5
10
y
; y
(x
7)
( x 2)2
( x 1)3
There are no critical points. The function is increasing on its
domain. There are no inflection points. The function is concave
up on ( , 7) ( 7, 2) and concave down on (2, ). The
numerator and denominator share a factor of x 7. Dividing out
this common factor gives y
x 7
x 2
(x
7), which shows that
x 1 is a vertical asymptote. Now dividing numerator and
1 (7/ x )
,
1 (2/ x )
denominator by x gives y
which shows that y 1 is a
horizontal asymptote. The graph will have a hole at x
87.
y
( 1) 7
( 7) 2
y
x4 1
x2
14 . The
9
x-intercept is
7,
7.
2
Since 0 is a root of the denominator, the domain is
( , 0) (0, ).
y
2 x4 2
3
; y
2
6
x
x4
There are critical points at x
1. The function is increasing on
( 1, 0) (1, ) and decreasing on ( , 1) (0, 1). There are no
inflection points. The function is concave up on its domain. The
y-axis is a vertical asymptote. Dividing numerator and
denominator by x 2 gives y
x 2 1/ x 2
1
, which shows that there
are no horizontal asymptotes. For large x , the graph is close to
the graph of y
88.
y
x2.
x2 4
2x
Since 0 is a root of the denominator, the domain is
( , 0) (0, ).
y
x2 4 ;
2 x2
y
4
x3
There are no critical points at x
2. The function is increasing
on ( , 2) (2, ) and decreasing on ( 2, 0) (0, 2). There
are no inflection points. The function is concave down on ( , 0)
and concave up on (0, ). The y-axis is a vertical asymptote.
Dividing numerator and denominator by x gives y
shows that the line y
x
2
x 4/ x ,
2
which
is an asymptote.
Copyright
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Chapter 4 Applications of Derivatives
89. y
1
x2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2x ;
( x 1)2
2
y
6 x2 2
( x 2 1)3
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). The function is concave up on
( , 1) (1, ) and concave down on ( 1, 1). The lines x 1
and x
1 are vertical asymptotes. The x-axis is a horizontal
asymptote.
90.
y
x2
x2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2x ;
( x 2 1)2
y
6 x2 2
( x 2 1)3
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). There are no inflection points. The
function is concave up on ( , 1) (1, ) and concave down on
1 are vertical asymptotes.
( 1, 1). The lines x 1 and x
Dividing numerator and denominator by x 2 gives y
1
1 (1/ x 2 )
which shows that the line y 1 is a horizontal asymptote. The xintercept is 0 and the y-intercept is 0.
91.
y
x2 2
x2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2x ;
( x 2 1)2
y
6 x2 2
( x 2 1)3
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). There are no inflection points. The
function is concave up on ( , 1) (1, ) and concave down on
1 are vertical asymptotes.
( 1, 1). The lines x 1 and x
Dividing numerator and denominator by x 2 gives y
which shows that the line y
x-intercepts are
1 (2/ x 2 )
1 (1/ x 2 )
1 is a horizontal asymptote. The
2 and the y-intercept is 2 .
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
92.
y
x2 4
x2 2
Since
,
y
2 and
2 are roots of the denominator, the domain is
2
2, 2
4x ;
( x 2)2
.
4(3 x 2 2)
( x 2 2)3
y
2
2,
There is a critical point at x
0, where the function has a local
minimum. The function is increasing on 0, 2
decreasing on
,
2
,
2
and
2, 0 . There are no inflection
points. The function is concave up on
down on
2,
2,
2, 2 and concave
. The lines x
2 and x
2
are vertical asymptotes. Dividing numerator and denominator by
1 (4/ x 2 )
1 (2/ x 2 )
x 2 gives y
which shows that the line y 1 is a
horizontal asymptote. The x-intercepts are
intercept is 2 .
93.
y
2 and the y-
x2
x 1
Since 1 is a root of the denominator, the domain is
( , 1) ( 1, ).
y
x2 2 x ;
( x 1)2
2
( x 1)3
y
There is a critical point at x 0, where the function has a local
minimum, and a critical point at x 2 where the functions has a
local maximum. The function is increasing on ( , 2) (0, )
and decreasing on ( 2, 1) ( 1, 0). There are no inflection
points. The function is concave up on ( 1, ) and concave down
1 is a vertical asymptote. Dividing
on ( , 1) . The line x
x 1 x1 1 , which shows
that the line y x 1 is an oblique asymptote. (See Section 2.6.)
The x-intercept is 0 and the y-intercept is 0.
numerator by denominator gives y
94.
y
x2 4
x 1
Since 1 is a root of the denominator, the domain is
( , 1) ( 1, ).
y
x2 2 x 4 ;
( x 1)2
y
6
( x 1)3
There are no critical points. The function is decreasing on its
domain. There are no inflection points. The function is concave up
1 is a
on ( 1, ) and concave down on ( , 1) . The line x
vertical asymptote. Dividing numerator by denominator gives
y 1 x x3 1 , which shows that the line y 1 x is an oblique
asymptote. (See Section 2.6.) The x-intercepts are
intercept is 4.
Copyright
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274
95.
Chapter 4 Applications of Derivatives
y
x2 x 1
x 1
Since 1 is a root of the denominator, the domain is
( , 1) (1, ).
y
x2 2 x ;
2
x 1
y
2
( x 1)3
There is a critical point at x 0, where the function has a local
maximum, and a critical point at x 2 where the function has a
local minimum. The function is increasing on ( , 0) (2, )
and decreasing on (0, 1) (1, 2). There are no inflection points.
The function is concave up on (1, ) and concave down on
( , 1). The line x 1 is a vertical asymptote. Dividing
numerator by denominator gives y
x
1
x 1
which shows that
the line y x is an oblique asymptote. (See Section 2.6.) The yintercept is 1.
x2 x 1
x 1
96. y
Since 1 is a root of the denominator, the domain is
( , 1) (1, ).
y
2 x x2 ;
2
x 1
y
2
( x 1)3
There is a critical point at x 0, where the function has a local
minimum, and a critical point at x 2 where the function has a
local maximum. The function is increasing on (0, 1) (1, 2) and
decreasing on ( , 0) (2, ). There are no inflection points.
The function is concave up on ( , 1) and concave down on
(1, ). The line x 1 is a vertical asymptote. Dividing numerator
by denominator gives y
x
1
x 1
which shows that the line
y
x is an oblique asymptote. (See Section 2.6.) The yintercept is 1.
97.
y
x 3 3x 2 3 x 1
x2 x 2
( x 1)3
( x 1)( x 2)
Since 1 and 2 are roots of the denominator, the domain is
( , 2) ( 2, 1) (1, ).
y
( x 1)( x 5)
,
( x 2)2
x 1; y
18 ,
( x 2)3
x 1
Since 1 is not in the domain, the only critical point is at x
5,
where the function has a local maximum. The function is
increasing on ( , 5) (1, ) and decreasing on
( 5, 2) ( 2, 1). There are no inflection points. The function is
concave up on ( 2, 1) (1, ) and concave down on ( , 2).
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
2 is a vertical asymptote. Dividing numerator by
The line x
the denominator gives y
y
9
x 2
x 4
which shows that the line
x 4 is an oblique asymptote. (See Section 2.6.) The y-
intercept is 12 . The graph has a hole at the point (1, 0).
98.
y
x3 x 2
x x2
( x 1)( x 2 x 2)
( x 1)( x )
Since 1 and 0 are roots of the denominator, the domain is
( , 0) (0, 1) (1, ).
y
x2 2 ,
x2
4 ,
x2
x 1; y
There is a critical point at x
x 1
2 where the function has a local
minimum, and a critical point at x
2 where the function has a
local maximum. The function is increasing on
and decreasing on
,
2
2,
2, 0
0, 2
. There are no inflection
points. The function is concave up on ( , 0) and concave down
on (0, 1) (1, ). The y-axis is a vertical asymptote. Dividing
x 1 2x which shows that
the line y
x 1 is an oblique asymptote. (See Section 2.6.)
The graph has a hole at the point (1, 4).
numerator by denominator gives y
99. y
x
x2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
x2 1 ;
( x 2 1) 2
2 x3 6 x
( x 2 1) 3
y
There are no critical points. The function is decreasing on its
domain. There is an inflection point at x 0. The function is
concave up on ( 1, 0) (1, ) and concave down on
1 are vertical
( , 1) (0, 1). The lines x 1 and x
asymptotes. Dividing numerator and denominator by x 2 gives
y
1/ x
1 (1/ x 2 )
which show that the x-axis is a horizontal asymptote.
The x-intercept is 0 and the y-intercept is 0.
100. y
x 1
x 2 ( x 2)
Since 0 and 2 are roots of the denominator, the domain is
( , 0) (0, 2) (2, ).
y
2 x2 5 x 4 ;
x 3 ( x 2)2
y
6 x 3 24 x 2 40 x 24
x 4 ( x 2) 3
There are no critical points. The function is increasing on ( , 0)
and decreasing on (0, 2) (2, ). There is an inflection point at
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276
Chapter 4 Applications of Derivatives
approximately x 1.223. The function is concave up on
( , 0) (0, 1.223) (2, ) and concave down on (1.223, 2).
The lines x 0 (the y-axis) and x 2 are vertical asymptotes.
Dividing numerator and denominator by x 3 gives
y
(1/ x 2 ) (1/ x 3 )
1 (2/ x )
which shows that the x-axis is a horizontal
asymptote. The x-intercept is 1.
101. y
8
x2 4
The domain is (
16 x ;
( x 2 4)2
y
, ).
16(3 x 2 4)
( x 2 4)3
y
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 0) and decreasing
on (0, ). There are inflection points at x
x
2 / 3 and at
2 / 3. The function is concave up on
, 2/ 3
2 / 3,
and concave down on
2 / 3, 2 / 3 . Dividing numerator and denominator by x 2
8/ x 2
1 (4/ x 2 )
gives y
which shows that the x-axis is a horizontal
asymptote. The y-intercept is 2.
102. y
4x
x2 4
The domain is (
2
4( x 4)
;
( x 2 4)2
y
, ).
8 x ( x 2 12)
( x 2 4)3
y
There is a critical point at x
2, where the function has a local
minimum, and at x 2, where the function has a local maximum.
The function is increasing on ( 2, 2) and decreasing on
( , 2) (2, ). There are inflection points at
x
2 3, x
2 3, 0
, 2 3
x 2 gives y
0, and x
2 3,
2 3. The function is concave up on
and concave down on
0, 2 3 . Dividing numerator and denominator by
4/ x
1 (4/ x 2 )
which shows that the x-axis is a horizontal
asymptote. The x-intercept is 0 and the y-intercept is 0.
103.
Point
P
Q
R
S
T
y
y
0
0
Copyright
2014 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
277
105.
104.
106.
107. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here.
(a) The body is moving away from the origin when |displacement| is increasing as t increases, 0 t 2 and
6 t 9.5; the body is moving toward the origin when |displacement| is decreasing as t increases, 2 t 6
and 9.5 t 15.
(b) The velocity will be zero when the slope of the tangent line for y s (t ) is horizontal. The velocity is zero
when t is approximately 2, 6, or 9.5 sec.
(c) The acceleration will be zero at those values of t where the curve y s (t ) has points of inflection. The
acceleration is zero when t is approximately 4, 7.5, or 12.5 sec.
(d) The acceleration is positive when the concavity is up, 4 t 7.5 and 12.5 t 15; the acceleration is
negative when the concavity is down, 0 t 4 and 7.5 t 12.5.
108. (a) The body is moving away from the origin when |displacement| is increasing as t increases, 1.5 t 4,
10 t 12 and 13.5 t 16; the body is moving toward the origin when |displacement| is decreasing as
t increases, 0 t 1.5, 4 t 10 and 12 t 13.5 .
(b) The velocity will be zero when the slope of the tangent line for y s (t ) is horizontal. The velocity is zero
when t is approximately 0, 4, 12 or 16 sec.
(c) The acceleration will be zero at those values of t where the curve y s (t ) has points of inflection. The
acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 sec.
(d) The acceleration is positive when the concavity is up, 0 t 1.5, 6 t 8 and 10 t 13.5, the
acceleration is negative when the concavity is down, 1.5 t 6, 8 t 10 and 13.5 t 16.
2
109. The marginal cost is dc
which changes from decreasing to increasing when its derivative d 2c is zero. This is a
dx
dx
point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units.
d2y
dy
110. The marginal revenue is dx and it is increasing when its derivative 2 is positive
the curve is concave up
dx
2
d y
0 t 2 and 5 t 9; marginal revenue is decreasing when 2 0 the curve is concave down
2 t
dx
5 and 9 t 12.
111. When y ( x 1) 2 ( x 2), then y 2( x 1)( x 2) ( x 1) 2 . The curve falls on ( , 2) and rises on (2, ).
At x 2 there is a local minimum. There is no local maximum. The curve is concave upward on ( , 1) and
5,
, and concave downward on 1, 53 . At x 1 or x 53 there are inflection points.
3
Copyright
2014 Pearson Education, Inc.
278
Chapter 4 Applications of Derivatives
( x 1) 2 ( x 2)( x 4), then y
112. When y
2( x 1)( x 2)( x 4) ( x 1)2 ( x 4) ( x 1)2 ( x 2)
( x 1)[2( x 2 6 x 8) ( x 2 5 x 4) ( x 2 3x 2)] 2( x 1)(2 x 2 10 x 11). The curve rises on ( , 2) and
(4, ) and falls on (2, 4). At x 2 there is a local maximum and at x 4 a local minimum. The curve is concave
downward on (
5
3
, 1) and 5 2 3 , 5 2 3 and concave upward on 1, 5 2 3 and 5 2 3 ,
. At x 1, 5 2 3 and
there are inflection points.
2
113. The graph must be concave down for x
1
f ( x)
0.
2
0 because
x
114. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always
be concave up or concave down so it will have no inflection points and no cusps or corners.
115. The curve will have a point of inflection at x 1 if 1 is a solution of y
y
3x
116. (a) f ( x)
2
2bx c
y
ax 2 bx c
vertex is at x
b
2a
6 x 2b and 6(1) 2b
a x2
b
a
x
c
a x2
b
a
0
b
x
b2
4a 2
x3 bx 2
0; y
cx d
3.
b2
4a
c
(b) The second derivative, f ( x) 2a, describes concavity
when a 0 the parabola is concave down.
b
2a
2
b2 4 ac
4a
a parabola whose
b 2 4ac
4a
b ,
2a
the coordinates of the vertex are
a x
when a
0 the parabola is concave up and
117. A quadratic curve never has an inflection point. If y ax 2 bx c where a
Since 2a is a constant, it is not possible for y to change signs.
0, then y
2ax b and y
2a.
118. A cubic curve always has exactly one inflection point. If y ax3 bx 2 cx d where a 0, then
y 3ax 2 2bx c and y 6ax 2b. Since 3ab is a solution of y 0, we have that y changes its sign at
b and y exists everywhere (so there is a tangent at x
b ). Thus the curve has
3a
3a
b . There are no other inflection points because y changes sign only at this zero.
3a
x
x
119. y
( x 1)( x 2), when y
and x
120. y
0
x
1 or x
2; y
2
x 2 ( x 2)3 ( x 3), when y
inflection at x
3 and x
0
x
2
Copyright
3, x
0 or x
|
|
1
2
2; y
an inflection point at
points of inflection at x
|
|
|
3
0
2
2014 Pearson Education, Inc.
1
points of
Section 4.4 Concavity and Curve Sketching
a x3 bx 2 cx
y 3a x 2 2bx c and y 6a x 2b; local maximum at x 3
3a(3)2 2b(3) c 0 27a 6b c 0; local minimum at x
1 3a ( 1) 2 2b( 1) c 0
3
2
3a 2b c 0; point of inflection at (1, 11) a (1) b(1) c (1) 11 a b c 11 and
6a (1) 2b 0 6a 2b 0. Solving 27 a 6b c 0, 3a 2b c 0, a b c 11, and 6a 2b 0
a
1, b 3, and c 9
y
x3 3 x 2 9 x
121. y
122. y
x2 a
bx c
y
bx 2 2cx ab ; local
(bx c ) 2
minimum at ( 1, 2)
Solving 9b 6c ab
maximum at x
b ( 1)2 2c ( 1) a b
(b ( 1) c )
2
0, b 2c a b
0
3
b(3)2 2c (3) ab
(b (3) c )2
b 2c a b
0, and a 2b 2c 1
0
( 1)2 a
0 and b( 1) c
a
124. If y x3 12 x 2 then y 3x( x 8) and y 6( x 4).
The zeros of y and y are extrema, and points of
inflection, respectively.
4 x5
5
3
16 x 2
25, then y
4 x( x 3 8) and
y 16( x 2). The zeros of y and y are extrema,
and points of inflection, respectively.
126. If y
x 4 x3
4
3
3
2
4 x 2 12 x 20, then
y x x 8 x 12 ( x 3)( x 2)2 . So y has a
local minimum at x
3 as its only extreme value.
Also y 3 x 2 2x 8 (3x 4)( x 2) and there
are inflection points at both zeros, 43 and 2, of y .
Copyright
2
3, b 1, and c
123. If y x5 5 x 4 240, then y 5 x3 ( x 4) and
y 20 x 2 ( x 3). The zeros of y' are extrema, and
there is a point of inflection at x 3.
125. If y
9b 6c ab
2014 Pearson Education, Inc.
0; local
a 2b 2c 1.
1
y
x2 3 .
x 1
279
280
Chapter 4 Applications of Derivatives
127. The graph of f falls where f 0, rises where f
0,
and has horizontal tangents where f
0. It has
local minima at points where f changes from
negative to positive and local maxima where f
changes from positive to negative. The graph of f is
0 and concave up where
concave down where f
f
0. It has an inflection point each time f
changes sign, provided a tangent line exists there.
128. The graph f is concave down where f
0, and
concave up where f
0. It has an inflection point
each time f changes sign, provided a tangent line
exists there.
4.5 INDETERMINATE FORMS AND L HÔPITAL S RULE
1. l Hôpital: lim x2 2
1
2x
x 2
1
4
2. l Hôpital: lim sinx5 x
5cos 5 x
1
x 0
2 x
x
x
4
0
2
3. l Hôpital: lim 5 x 2 3x
7x
x
1
x3 1
3
x 1 4x x 3
2
lim 3 x2
x
5. l Hôpital: lim 1 cos
2
sin 2 x
0 x (1 cos x )
lim
x
x
0
2
x
7.
9.
lim x2 2
x
t
2 x
4
3
lim t 2 4t 15
3 t
t 12
2
x
sin x
x
3x
3
11
1
7x
x3 1
3
x 1 4x x 3
0
1
1 cos x
lim 64x
x
1
4
2
lim 32t t 14
3
t
2
or lim 5 x 2 3 x
x
1
2
23
7
Copyright
3
x
1
x2
5
7
5
7
( x 1) x 2 x 1
x 1 ( x 1) 4 x
x
0
lim
x
0
2
2
lim x2 x 1
x 1 4x
4x 3
(1 cos x ) 1 cos x
1 cos x
x2
1
2
2
0 or lim 2 3x 3x
x
x
8.
3( 3) 2 4
2( 3) 1
x
x
or lim 1 cos
2
x
51 5
lim
1
lim
1
4
x
5 lim sin5 x5 x
5x 0
or lim
lim cos2 x
lim 4 x2 3
x 1
lim 21x
x
x
0
sin x
x
5
7
x
lim x 1 2
2
x
0
10
lim 14
x
lim sin
2x
0
2
6. l Hôpital: lim 2 3x 3x
x
x
1
lim
x
x 2
lim ( x 2)(
x 2)
2
4
5 or lim sinx5 x
x 1 12 x
x
2x
x
lim 1014x x 3
x
4. l Hôpital: lim
x
or lim x2 2
10.
x 1
lim
x
2
lim x x 525
x
5
t
3
lim 33t 3
1 4t
t 3
2
x
1
3
x2
1
x2
1
x3
0
1
lim 21x
x
t
5
9t 2
2
1 12t 1
lim
2014 Pearson Education, Inc.
0
10
9
11
4x 3
3
11
Section 4.5 Indeterminate Forms and L Hôpital s Rule
11.
12.
13.
15.
16.
3
lim 5 x 3 2 x
2
lim 15 x 2 2
x
7x
lim
x 8 x2
12 x 2 5 x
x
lim sint t
t
3
2
1 16 x
lim 24
x 5
t
8 x2
lim cos
x 1
x 0
2
lim cos(2
17.
3
3
sin
3
sin
lim 11 cos
2
19.
20.
21.
22.
23.
24.
2
x
lim ln(sec
x)
0
ln(csc x )
lim
x
2
x
2
x
2
t (1 cos t )
lim t sin t
0
lim
t
lim 1t sin
cos t
25.
0
lim
x
26.
0
2
2
1
cos( x )
2
1
( 4)( 1)
1
4
2x
lim tan
x
0
csc x cot x
csc x
2 x
lim
x
2
lim
0
sec x
t
lim
x
x tan x
2
0
x
2
0
2
2
x
x
12
2
1
2
2
sin t (sin t t cos t )
sin t
lim
x
cot x
2
12
2
lim csc2 x
cos t (cos t t sin t )
cos t
cos x
2
2
2
0 sec x
lim
x
cot x
2 x 2
lim
t
2
lim
x
1
x
t
lim sin tsint cos
t
t
1
(1 cos t ) t (sin t )
1 cos t
2
lim
x
x
t
1
x
2x
2
2
lim 4sin
cos 2
sec x tan x
sec x
0
lim
t
t
lim
x
5
2
2
x 1
lim
x
0
3
2
x 1
lim
x 1 ln x sin( x )
5t
lim 5 cos
2
t
3
cos
cos
lim 2sin
2
2
0
1
6
0
3
2
sin
3
lim
3
x
2
)
t
16
x
lim cos
6
0
2
lim
16
1
x
lim sin
6x
x
lim sin(22
)
2
18.
16
lim cos
x
0
0 3x
lim sin2t5t
14.
x
lim cos x2 1
x
2
3
0
x
lim 16
0 sin x
x
0
16
lim 24
1
x
lim sin x3 x
x
5
7
x
x
cos t 2 (2t )
0
lim 30
42
x
x
lim
0
x
lim 30
42 x
21x
x
t cos t t sin t
lim cos t coscos
t
0
t
1 (1 0)
1
1
sin x
2
1
1
1
2
lim
x
Copyright
2
1
csc2 x
lim sin 2 x 1
x
2
2014 Pearson Education, Inc.
1 1 1 0
1
3
281
282
Chapter 4 Applications of Derivatives
27.
sin
1
lim 3
28.
lim
29.
lim
30.
0
1
2
1
x 2x
02
1
x
ln( x 1)
lim log x
2
x
32.
lim log ( 2x 3)
3
x
34.
35.
36.
37.
38.
39.
x
0
lim
x
0
x
x
5 y 25 5
y
0
0
ay a 2 a
y
0
lim
y
lim (ln x ln sin x )
0
(ln x )2
lim
x
0
1/ 2
y
a
lim ln x2 x1
x
x
lim ln sinx x
0
2(ln x )
cos x
sin x
1
x
lim
x
0
1/ 2
(5)
ay a 2
Copyright
1
2
a
0 2 ay a 2
1 ,a
2
ln lim 12
x
lim sinx x
0
ln
x
2(ln x )(sin x )
x cos x
5
0 2 5 y 25
lim
y
ln lim x2 x1
x
ln
(a )
1
0
1
y
1/ 2
lim
x
0
1
0
lim
1
ln 2
ln 3 lim x 3
ln 2 x
x
lim 22
x
1 0
1
e
(5 y 25)
1
2
x 3
x
0
0
lim
y
1
2
1
x
1
ln 3 lim
ln 2 x
x
x
lim e xxe
x
(ln 2) lim 11
x
lim 24 xx 22
0
2x
1
lim
y
0
lim ln 2 x ln( x 1)
lim ln(sin x)
x 0
x
ay a 2
x
x
0 e
(5 y 25)1/ 2 5
y
0
y
x
0
lim
lim
y
x
x
lim xe
x
ex 1
1
x
lim
(ln 2) lim xx 1
x
2
lim 2 x2 2 x
ex
ln e x 1
lim
y
0
1
ln 2
ln 3 lim ln x
ln 2 x
ln( x 3)
x2 2 x
1
x
lim
x
1
x 1
1
x
x
2x 2
ln x
ln 2
ln 3
ln 2
ln x
ln 2
ln( x 3)
ln 3
lim
1 20 0
(ln 2) 20
(ln 2) lim
ln x
ln 2
x
ln x 2 2 x
ln x
ln( x 1)
lim
ln1 ln 2
x
30 ln 3
20 ln 2
0 2 ln 2
log x
lim
ln 12
(ln 2) 2
0
ln 3
1
( x ) (ln 2) 2 x
x
lim 3x ln 3
x
31.
33.
(1) 2 x
lim
x
02 1
1
2
1
0
x
lim 3 x 1
x
1
2
ln
lim
0
x
30 (ln 3)(1)
3sin (ln 3)(cos )
1
0
lim
ln 2
lim cos1 x
0
x
ln1 0
2(ln x ) sin x
cos x
x
2014 Pearson Education, Inc.
1
0
ln 3 lim 1
ln 2 x
1
ln 3
ln 2
Section 4.5 Indeterminate Forms and L Hôpital s Rule
40.
lim 3 xx 1 sin1 x
x 0
3 3 (1)(0)
1 1 0
41.
x
6
2
0
ln x ( x 1)
lim x1 1 ln1x
x 1
lim ( x 1)(ln x)
x 1
lim (ln x 11) 1
1
(0 1) 1
44.
lim
45.
46.
47.
h
t
lim e t 2t
e 1
t
2
lim xx
x
x
sin x
lim xx tan
x
0
x
x
ex 1
lim
x 0 x sin x
49.
lim
0
t
lim e t 2
t
lim et
e
t
2 ex 1 ex
lim
x 0 x cos x sin x
lim 2x
sec
cos x
(1 cos x ) (sin x )(cos x)
sin x
lim
x
sin x
2
2
0 2 x sec x tan x 2sec x
lim
x
e2 x 2e x
lim x2cos
x sin x
x 0
0
1
2
lim 2 sin2
0 tan
3x 3 2 x
lim 2sin x3cos
cos 2 x cos x sin 2 x
0
x
0
2
0
e 2 x 2e x
lim x 4sin
x 2cos x
x 0
lim 2 cos2
0
3x 3 2 x
lim sin3cos
x cos 2 x sin 3 x
0
x
2
2
1
2
3x 2
lim 2sin x sin 2 x9 sin
cos x cos 2 x 3cos 3 x
0
x
1
2
51. The limit leads to the indeterminate form 1 . Let f ( x)
lim ln f ( x )
x 1
x 1
0
e
x
2
2
lim 1 sin 2 cos
0
lim ( x ln1x )x x 1
1
x
1
e
t
e
x
1 cos x
2
0 x sec x tan x
3 x x2
lim sinsin3 xx sin
2x
0
2
4
0
1
2
0
lim
sin cos
tan
x
h
lim 2 xx
e
2
48.
50.
e
t
lim x 2 e x
x
3cos x 3cos x (3 x 1)( sin x )
cos x cos x x sin x
1
h
lim e2
h
t
2
lim e t t
lim
1
e
0
h
lim e2h1
0
h2
cos x
sin x
lim cos
1
0 e
eh (1 h)
0
0 1 0
1
lim sin
1
0e
1
(ln x ) ( x 1)
x 1
x
2
x sin 2 x
lim sin x cos
cos x
x 0
lim cos 1
0
1
x
lim
lim sin1 x
0
0
43.
x
3sin x (3 x 1)(cos x ) 1
sin x x cos x
1
2
lim (csc x cot x cos x)
x
lim
3
x 1
42.
(3 x 1)(sin x ) x
x sin x
lim
283
lim 1ln xx
x 1
lim
x 1
1
x
1
x1/(1 x )
1. Therefore lim x1/(1 x )
Copyright
x 1
ln f ( x )
lim f ( x)
x 1
2014 Pearson Education, Inc.
ln x1/(1 x )
lim eln f ( x )
x 1
ln x .
1 x
Now
e 1
1
e
284
Chapter 4 Applications of Derivatives
x1/( x 1)
52. The limit leads to the indeterminate form 1 . Let f ( x)
1
x
lim ln
x 1
x 1
lim ln f ( x)
x 1
1. Therefore lim x1/( x 1)
lim 1x
x 1
lim ln f ( x)
ln(ln x )
x
lim
x
x
1
x ln x
lim
0
. Let f ( x)
ln(ln x )
x e
lim
x
e
lim
x
55. The limit leads to the indeterminate form 00. Let f ( x)
lim eln f ( x)
e
56. The limit leads to the indeterminate form
0
lim x
x
1/ln x
lim f ( x)
0
x
lim x1/ln x
x
0
x
0
lim e1n f ( x )
lim f ( x)
x
x
e1
x
lim eln f ( x )
x
lim
x
1/2
ln(1 2 x )
2 ln x
0
x
1 2x
lim
x
x
0
. Let f ( x)
. Let f ( x)
lim
x
1
2
1.
2
x
0
e0
lim
x
ln e
x
x
0
ex 1
x
0e x
ln x
ln x
ln x
ln x
(1 2 x)1/(2 ln x )
1. Therefore
1. Therefore
ln(1 2 x )
2 ln x
ln f ( x )
Therefore lim (1 2 x )1/(2 ln x )
lim f ( x)
x
ex
x
lim
x
0
ln x
x
1
x
lim
1
x
x
lim ( x)
1
x2
0
1/ x
x
2. Therefore lim e x
lim
x
x
x
0
0
xx
ln e x x
ln f ( x)
1/ x
x
0
x ln x
0. Therefore lim x x
x
lim eln f ( x)
lim f ( x)
x
ln f ( x)
x
0
x
0
ln x
ln f ( x)
1
x
lim eln f ( x )
lim f ( x)
x
e2
0
x
0
1
0
60. The limit leads to the indeterminate form
lim
x
e
e
59. The limit leads to the indeterminate form 00. Let f ( x)
lim ln f ( x)
ln f ( x)
lim ln f ( x)
x
1
e
58. The limit leads to the indeterminate form 1 . Let f ( x)
lim ln f ( x)
x1/ln x
1
e1/ e
e
ln f ( x)
Now
e
57. The limit leads to the indeterminate form
lim ln f ( x)
1
x
1/ln x
x
ln(ln x )
x e
lim eln f ( x )
e
e
e0
x
ln f ( x)
lim f ( x )
x
e1
lim eln f ( x )
lim f ( x)
Now
ln(ln x )
.
x
ln(ln x)1/ x
x
(ln x)1/( x e )
Therefore (ln x)1/( x e)
1.
e
1
e
x 1
ln f ( x)
x
54. The limit leads to the indeterminate form 1 . Let f ( x)
1
x ln x
(ln x)1/ x
ln x .
x 1
eln f ( x )
lim
x 1
0. Therefore lim (ln x )1/ x
1
x
lim f ( x)
x 1
53. The limit leads to the indeterminate form
ln x1/( x 1)
ln f ( x )
0
x 2
1 x 1
2
x
1
0 1 x
lim
x
1
lim
x
0
x
x 1
. Let f ( x)
1
1
x
0. Therefore lim 1
Copyright
x
0
x
1
x
ln f ( x)
x
lim f ( x)
x
2014 Pearson Education, Inc.
0
1
ln 1 x
x
lim ln f ( x )
1
x
0
lim eln f ( x )
x
0
e0
1
Section 4.5 Indeterminate Forms and L Hôpital s Rule
x 2 x
x 1
61. The limit leads to the indeterminate form 1 . Let f ( x)
lim ln f ( x)
x
x
3x 2
( x 2)( x 1)
lim
x
x 2
x 1
lim x ln
x
6x
2x 1
lim
x
x 2
x 1
1
x
ln
lim
6
2
lim
x
lim ln f ( x)
lim
0
x
x
x2 4 x 1
x3 2 x 2 x 2
e0
1
lim
63.
64.
65.
66.
67.
68.
lim x 2 ln x
x
x
0
0
lim
9x 1
x 1
lim
x
sin x
x
0
lim
x
71.
cot x
csc x
lim
2 x 3x
3x 4 x
x
0
9x 1
x 1
lim
x
lim
1
1
1
cos x
cos x
sin x
1
sin x
0
x
x
csc 2
0
2
9
1
9
x
1
1
lim
csc x cot x
0
0
1
x
1
1
x
lim
x
0
lim
x
0
cos x
sin x
2 x
3
1
x
1
sin x
1
4 x
3
sin x tan x
x
1
x 2
1/ x
1 ln x 2 1
x
x 2
x
lim f ( x)
lim eln f ( x )
1
lim
x
lim
x
lim
x
1
2
0
0
Copyright
x2 1
x 2
e3
x
0
2 x2
x
lim 2 x
x
0
0
1
lim cos x 1
x
lim eln f ( x)
x
x 24x 1
( x 2 1)( x 2)
x2 1
1/ x
2
x
1
x2
3
lim
x
3
( x 2)( x 1)
1
x2
0
1
2
lim
x
2x
0
lim
x
2
x
lim
x
x
1
lim sinx x
lim
2
ln x
csc x
0
x
sec x
tan x
lim
x
x
x
x
cot
0
lim
0
x
x
2 ln x
lim
1
x2
0
x
ln
x
3x2
2
lim
1
ln f ( x)
x2 1
x 2
lim
x2
lim f ( x)
lim
x ln xx 12
1
x 1
x
x
x3
2x
0
x
1/ x
x
lim
0
x
x 2 x
x 1
0. Therefore, lim
2(ln x ) 1x
lim
1
x
lim
x
1
x
2
x3
1
x 2
x
ln x 2 1 ln( x 2)
x
2
6x 4
x
0
(ln x )2
0
x
2
x
lim
x
lim
lim
1
x2
x2 1
x 2
x
2x 4
3 x2 4 x 1
ln x
lim
lim sin x ln x
x
x
0
x
0
lim x tan
x
lim
lim
lim x (ln x)2
69.
70.
x
x2 1
x 2
. Let f ( x)
x 2 x
x 1
ln
lim
1
x
x
x
ln
lim
ln( x 2) ln( x 1)
lim
3. Therefore, lim
62. The limit leads to the indeterminate form
1 ln x 2 1
x
x 2
ln f ( x)
285
2014 Pearson Education, Inc.
0
sin x sec2 x cos x tan x
1
0
1
0
286
72.
73.
74.
Chapter 4 Applications of Derivatives
x
lim
2x 4x
5x 2 x
2
lim
ex
xe x
lim
x
x
lim
ex
x
x
lim
x
x
e1/ x
5 x
2
1)
1 0
0 1
1
e x( x
lim
1
1)
(2 x 1)
1
x
e1/ x
1
x2
lim e1/ x
1
x2
0
x
1 2x
lim
e x( x
x
lim
1
x
0
x
4 x
2
5 x 1
2
2 x
lim
1/ x
0 e
x
x
1
lim
0
x
0
0
75. Part (b) is correct because part (a) is neither in the
2x 2
0 2 x cos x
76. Part (b) is correct; the step lim
x
nor
2
0 2 sin x
lim
x
form and so l Hôpital s rule may not be used.
2x 2
0 2 x cos x
in part (a) is false because lim
x
is not an
indeterminate quotient form.
77. Part (d) is correct, the other parts are indeterminate forms and cannot be calculated by the incorrect arithmetic
78. (a) We seek c in
1
2c
1
2
f (c )
g ( c)
2,0 so that
c
f (0) f ( 2)
g (0) g ( 2)
1
b a
f (c )
f ( b) f ( a )
g (b) g ( a )
f (c )
f (3) f (0)
g (3) g (0)
(c) We seek c in (0, 3) so that g ( c)
1
3
1
c
37
80.
lim tan32 x
0
x
lim 81cos 3x
x 0 30
a
x2
x
c
3
79. If f ( x) is to be continuous at x
lim 27 sin 3 x
x 0 30 x
37
3
1 .
b a
Since f (c) 1 and g (c)
2c we have that
x
0
Since f (c )
c2
4 and g (c)
2c we have
f (0)
c
f (0)
3x
lim 9 x 3sin
3
x
0
5x
lim 9 9 cos2 3 x
x
0
15 x
27 .
10
0
b
2
2
bx 2 x sin bx will be in 0 form if
lim 2 sec 2 x a bx cos
2
0
x
2 2
16 6b
1.
3
.
0
3x
0
x
a 2
bx 4bx cos bx 2sin bx
lim 8sec 2 x tan 2 x b x sin
6x
x 0
0
b a
b2 a2
3 0
9 0
2
lim tan 2 x ax3 x sin bx
sin bx
x
2
1
x
0
16 6b
6
2c we have that
0, then lim f ( x)
lim (2sec 2 2 x a bx 2 cos bx 2 x sin bx)
x
Since f (c) 1 and g (c)
b a.
2
c
2
that c 2c 4
1.
2
1.
(b) We seek c in (a, b) so that g ( c)
1
2c
0 2
0 4
0
2
2
bx 2 x sin bx
2; lim 2sec 2 x 2 bx cos
2
a
x
2
2
lim 32 sec 2 x tan 2 x 16 sec 2 x b6 x cos bx 6b x sin bx 6b cos bx
x
0
8
3
81. (a)
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3x
0
2
2014 Pearson Education, Inc.
4
3 2
Section 4.5 Indeterminate Forms and L Hôpital s Rule
(b) The limit leads to the indeterminate form
x2
lim x
x
lim
1
x
82.
1
1
x
x
x2
lim x
x
1
1 0
1
1
x
x2 1
lim
x
287
:
x
x
x2 x
x
x2 x
lim
x2
x2 x
x
x2 x
x
x
lim
x
x2 x
x
1
2
x2 1
x
lim x
x
x
x
x2 1
x2
lim x
x
x
x2
lim x
1
x
1
x2
1
x
83. The graph indicates a limit near 1. The limit leads
2 x 2 (3 x 1) x 2
to the indeterminate form 00 : lim
x 1
x 1
2
3/ 2
1/ 2
lim 2 x 3 x x 1 x 2
x 1
4
9
2
1
1
2
4 5
1
9 1/ 2
x
2
4x
lim
1 x 1/ 2
2
1
x 1
1
84. (a) The limit leads to the indeterminate form 1 . Let f ( x)
lim
1
x
ln 1
x
lim
1
x
lim eln f ( x )
(b)
x
x
e1
x
1
ln 1 x
lim
1
x
x 2
1 x 1
2
x
lim
x
1
1
1
x
x
1 1x
ln f ( x)
1
1 0
lim 1 1x
x
1
x ln 1 1x
x
lim f ( x)
x
e
1 1x x
x
10
2.5937424601
100
2.70481382942
1000
2.71692393224
10,000
2.71814592683
100,000
2.71826823717
Both functions have limits as x approaches
infinity. The function f has a maximum but no
minimum while g has no extrema. The limit of
f ( x) leads to the indeterminate form 1 .
(c) Let f ( x)
1
lim ln f ( x)
x
Therefore lim 1
x
x
1
x2
lim
ln f ( x )
ln 1 x
x
x
1
x2
x
1
x ln 1 x 2
2
lim
x
lim f ( x)
x
Copyright
2x 3
1 x 2
2
x
2
lim 23x
x
lim eln f ( x )
x
x
e0
lim
x
x
4x
3x2 1
1
2014 Pearson Education, Inc.
lim 64x
x
lim ln f ( x)
x
0.
288
Chapter 4 Applications of Derivatives
85. Let f (k )
1
ln 1 rk
ln f (k )
k
x1/ x
y
ln y
lim
(c)
2
y
x1/ x
( x ) ln x
y
|
|
0
e
y
x1/ x
y
|
n
(d)
87. (a)
n
lim x1/ x
e1/e when x
y
x
3
2
88.
f (0)
0
1
x
1
x
x
1
x
tan
lim
r
1 rk
lim krkr
1
k
|
|
0
e
e
2
x1/ x . The sign pattern is
e1/(2e) when x
x2 n
1/( ne)
e
exp lim lnnx
e
x1/ x . The sign pattern is
n
when x
e
1
x
1
x
sec 2
lim
x
1
x
x
e0
1
nx n
exp lim
x
x
1
x
x
sec 2
lim
1
1
x2
lim sec2 1x
1; lim
x
1
x2
x
x tan 1x
1
x2
x
1
x2
lim sec2 1x
1
2x
lim 4e3 x
lim
the horizontal asymptote is y 1 as
.
3 x e2 x
2 x e3 x
f (0 h ) f (0)
h
0
lim
2
lim h2 e 1/ h
h
n
lim e(ln x ) x
2x
lim 3 2e3 x
2 3e
x
the horizontal asymptotes are y
h
k
er .
x n 1 (1 nn ln x )
y
x
x
3 x e2 x , lim
2 x e3 x x
y
1/ x n
, lim x tan
tan
1
lim
k
e
and as x
x
(b)
1
x
(ln x ) nx n
which indicates a maximum of y
x
x tan
lim
xn
rk 2
1 rk 1
2
x1/ x . The sign pattern is y
1 2 ln x
x3
y
4
x2 n
lim e ln x
x
1 ln x
x2
2 x ln x
x
1
x
ln x
xn
|
0
x
2
k
which indicates a maximum of y
ln y
n
1
x
lim
1
k
y
x2
1
lim eln f (k )
lim f (k )
1
x
y
y
ln x
x2
ln y
k
k
which indicates a maximum value of y
(b)
ln 1 rk
k
k
y
y
ln x
x
1
1
k
r. Therefore lim 1 kr
k
lim 1r
86. (a)
k
r
k
2
1/ h
lim e h 0
h 0
9e
x
0 as x
1/ h
lim e h
h 0
2
x
4
9e x
lim
h
as x
2
lim
h
0 e1/ h
0
Copyright
x
lim
.
1
1
h
0 e1/ h
x
3
2
and y
3 x e2 x
2 x e3 x
0; lim
2014 Pearson Education, Inc.
h2
2
2
h3
h
2
0 2e1/ h
lim
h
3 e2 x
2 3ex
Section 4.5 Indeterminate Forms and L Hôpital s Rule
89. (a) We should assign the value 1 to
(sin x ) x to make it continuous at x
f ( x)
(b) ln f ( x)
2x
2
0 sec x
lim
x
ln(sin x )
x ln(sin x )
0
lim ln f ( x)
1
x
x
e0
lim f ( x)
x
0.
0
0
lim
x
0
ln(sin x )
1
x
1
sin x
lim
x
(cos x )
1
x2
0
2
lim tanx x
0
x
1
(c) The maximum value of f ( x) is close to 1 near the point x 1.55 (see the graph in part (a)).
(d) The root in question is near 1.57.
90. (a) When sin x 0 there are gaps in the sketch.
The width of each gap is .
(sin x ) tan x
(b) Let f ( x)
ln f ( x)
(tan x ) ln(sin x)
lim ln f ( x)
x
lim
x
x
ln(sin x )
cot x
2
x
lim ( cos
csc x )
lim
x
lim
x
csc2 x
lim
x
f ( x)
2
(cos x )
2
0
2
Similarly,
1
sin x
f ( x)
e0
1.
2
e
0
1. Therefore,
2
lim f ( x) 1.
x
2
(c) From the graph in part (b) we have a minimum of about 0.665 at x
1.491 at x 2.66.
Copyright
0.47 and the maximum is about
2014 Pearson Education, Inc.
289
290
4.6
Chapter 4 Applications of Derivatives
APPLIED OPTIMIZATION
1. Let and w represent the length and width of the rectangle, respectively. With an area of 16 in.2 , we have that
( )( w) 16
P( )
0
1
w 16
2(
4)(
4)
2
the perimeter is P
0
2
2
32
1
and P ( )
2
32
2
2(
2
16)
2
. Solving
0 for the length of a rectangle, must be 4 and w
4, 4. Since
perimeter is 16 in., a minimum since P ( )
2w
16
3
4
the
0.
2. Let x represent the length of the rectangle in meters (0 x 4). Then the width is 4 x and the area is
A( x) x(4 x) 4 x x 2 . Since A ( x) 4 2 x, the critical point occurs at x 2. Since, A ( x) 0 for 0 x 2
and A ( x) 0 for 2 x 4, this critical point corresponds to the maximum area. The rectangle with the largest
area measures 2 m by 4 2 2 m, so it is a square.
Graphical Support:
3. (a) The line containing point P also contains the points (0, 1) and (1, 0) the line containing P is y 1 x
a general point on that line is ( x, 1 x).
(b) The area A( x) 2 x(1 x), where 0 x 1.
(c) When A( x) 2x 2 x 2 , then A ( x) 0 2 4 x 0 x 12 . Since A(0) 0 and A(1) 0, we conclude
that A
1
2
1
2
sq units is the largest area. The dimensions are 1 unit by 12 unit.
4. The area of the rectangle is A 2 xy 2 x (12 x 2 ),
where 0 x
12. Solving A ( x) 0 24 6 x2 0
x
2 or 2. Now 2 is not in the domain, and
since A(0) 0 and A 12 0, we conclude that
A(2) 32 square units is the maximum area. The
dimensions are 4 units by 8 units.
5. The volume of the box is V ( x) x (15 2 x)(8 2 x)
120 x 46 x 2 4 x3 , where 0 x 4. Solving
V ( x) 0 120 92 x 12 x 2 4(6 x )(5 3 x) 0
x 53 or 6, but 6 is not in the domain. Since
V (0) V (4)
0, V
5
3
2450
27
91 in3 must be the
maximum volume of the box with dimensions
35 5 inches.
3 3
14
3
Copyright
2014 Pearson Education, Inc.
Section 4.6 Applied Optimization
1 ba
2
6. The area of the triangle is A
where 0
200 b 2
400 b2
When b
20. Then dA
db
b
0
1
2
b
2
291
400 b 2 ,
400 b 2
b2
2 400 b2
the interior critical point is b 10 2.
0 or 20, the area is zero
2
A 10 2 is the
2
400 and b 10 2,
maximum area. When a b
the value of a is also 10 2
the maximum area
occurs when a b.
7. The area is A( x) x(800 2 x), where 0 x 400.
Solving A ( x ) 800 4 x 0
x 200. With
A(0) A(400) 0, the maximum area is
A(200) 80, 000 m 2 . The dimensions are 200 m by
400 m.
8. The area is 2 xy
216
y
108 . The
x
amount of
fence needed is P 4 x 3 y 4 x 324 x 1 , where
4 324
0
x 2 81 0 the critical
0 x; dP
2
dx
x
points are 0 and 9, but 0 and 9 are not in the
domain. Then P (9) 0 at x 9 there is a
the dimensions of the outer rectangle are
minimum
18 m by 12 m 72 meters of fence will be needed.
9. (a) We minimize the weight tS where S is the surface area, and t is the thickness of the steel walls of the
tank. The surface area is S x 2 4 xy where x is the length of a side of the square base of the tank, and y
y 500
. Therefore, the weight of the tank is
is its depth. The volume of the tank must be 500 ft 3
2
w( x)
t x2
x
2000
x
. Treating the thickness as a constant gives w ( x)
at x 10. Since w (10)
t 2
4000
103
t 2x
2000
x2
. The critical value is
0, there is a minimum at x 10. Therefore, the optimum dimensions
of the tank are 10 ft on the base edges and 5 ft deep.
(b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of
the steel walls would likely be determined by other considerations such as structural requirements.
10. (a) The volume of the tank being 1125 ft 3 , we have that yx 2
tank is c ( x)
5 x 2 30 x
1125
x2
, where 0
but 0 is not in the domain. Thus, c (15)
y 5 ft will minimize the cost.
1125
x. Then c ( x) 10 x
0
y
33750
x2
1125 . The
x2
0
cost of building the
the critical points are 0 and 15,
at x 15 we have a minimum. The values of x 15 ft and
(b) The cost function c 5( x 2 4 xy ) 10 xy, can be separated into two items: (1) the cost of the materials and
labor to fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of
the tanks is ( x 2 4 xy ), it can be deduced that the unit cost to fabricate the tanks is $5/ft 2 . Normally,
excavation costs are per unit volume of excavated material. Consequently, the total excavation cost can be
taken as 10 xy
10
x
2
( x 2 y ). This suggests that the unit cost of excavation is $10/ft
where x is the length of
x
a side of the square base of the tank in feet. For the least expensive tank, the unit cost for the excavation is
Copyright
2014 Pearson Education, Inc.
292
Chapter 4 Applications of Derivatives
$10/ft 2
15 ft
$0.67
ft 3
$18
.
yd 3
The total cost of the least expensive tank is $3375, which is the sum of $2625 for
fabrication and $750 for the excavation.
11. The area of the printing is ( y 4)( x 8) 50.
50
Consequently, y
4. The area of the paper is
x 8
A( x)
x x508 4 , where 8
50
x 8
A ( x)
4
x. Then
4( x 8)2 400
50
( x 8) 2
x
0
( x 8) 2
the critical points are 2 and 18, but 2 is not in the
domain. Thus A (18) 0 at x 18 we have a
minimum. Therefore the dimensions 18 by 9 inches
minimize the amount of paper.
2
V ( y)
3
(9 y )( y 3)
r 2 h, where r
1
3
12. The volume of the cone is V
(27 9 y 32 y
3
2
9 y 2 and h
x
3
y )
V ( y)
points are 3 and 1, but 3 is not in the domain. Thus V (1)
volume of V (1)
3
32
3
(8)(4)
cubic units.
13. The area of the triangle is A( )
. Solving A ( )
0
0
ab sin
2
Since A ( )
maximum at
2
A
3
(9 6 y 3 y 2 )
(1 y )(3 y ). The critical
( 6 6(1))
at y 1 we have a maximum
0
, where
0
2
.
0, there is a
2
.
r 2 h 100
14. A volume V
ab sin
2
ab cos
2
3
y 3 (from the figure in the text). Thus,
1000 . The
r2
h
amount
of material is the surface area given by the sides and
bottom of the can S 2 rh
r 2 2000
r2,
r
0
r. Then dS
dr
2000
r2
2 r
The critical points are 0 and
2
domain. Since d 2s
dr
4000
r3
10
3
2
minimum surface area when r
h
1000
r2
10
3
r 3 1000
r2
0
0.
, but 0 is not in the
0, we have a
10
3
cm and
cm. Comparing this result to the result
found in Example 2, if we include both ends of the
can, then we have a minimum surface area when
the can is shorter specifically, when the height of
the can is the same as its diameter.
15. With a volume of 1000 cm3 and V
A 8r
r
2
2 rh
8r
2
2000 . Then
r
r 2 h, then h
A (r ) 16r
0 results in no can. Since A ( r ) 16
1000
r3
Copyright
1000 . The amount
r2
2000 0
8r 3 1000
r2
r2
of aluminum used per can is
0
0 we have a minimum at r
2014 Pearson Education, Inc.
the critical points are 0 and 5, but
5
h
40
and h:r
8: .
Section 4.6 Applied Optimization
293
x (10 2 x )(15 2 x )
2
16. (a) The base measures 10 2x in. by 15 22 x in., so the volume formula is V ( x)
2 x3 25 x 2 75 x.
(b) We require x 0, 2 x 10, and 2 x 15. Combining these requirements, the domain is the interval (0, 5).
(c) The maximum volume is approximately 66.02 in.3 when x 1.96 in.
(d) V ( x)
6 x 2 50 x 75. The critical point occurs when V ( x)
0, at x
( 50)2 4(6)(75)
2(6)
50
50
700
12
25 5 7
6
, that is, x 1.96 or x 6.37. We discard the larger value because it is not in the domain. Since
V ( x) 12 x 50, which is negative when x 1.96, the critical point corresponds to the maximum volume.
The maximum volume occurs when x
25 5 7
6
1.96, which confirms the result in (c).
17. (a) The sides of the suitcase will measure 24 2x in. by 18 2x in. and will be 2x in. apart, so the volume
formula is V ( x) 2 x(24 2 x)(18 2 x ) 8 x3 168 x 2 862 x.
(b) We require x 0, 2 x 18, and 2 x 12. Combining these requirements, the domain is the interval (0, 9).
(c) The maximum volume is approximately 1309.95 in.3 when x
(d) V ( x)
24 x 2 336 x 864
3.39 in.
24( x 2 14 x 36). The critical point is at x
14
( 14)2 4(1)(36)
2(1)
14
52
2
7 13, that is, x 3.39 or x 10.61. We discard the larger value because it is not in the domain. Since
V ( x) 24(2 x 14) which is negative when x 3.39, the critical point corresponds to the maximum
volume. The maximum value occurs at x 7 13 3.39, which confirms the results in (c).
(e) 8 x3 168 x 2 862 x 1120 8( x3 21x 2 108 x 140) 0 8( x 2)( x 5)( x 14) 0. Since 14 is not in
the domain, the possible values of x are x 2 in. or x 5 in.
(f ) The dimensions of the resulting box are 2x in., (24 2 x) in., and (18 2 x). Each of these measurements
must be positive, so that gives the domain of (0, 9).
18. If the upper right vertex of the rectangle is located at ( x, 4cos 0.5 x ) for 0 x
, then the rectangle has width
, we find
2x and height 4 cos 0.5x, so the area is A( x ) 8x cos 0.5 x.. Solving A ( x) 0 graphically for 0 x
that x 2.214. Evaluating 2x and 4 cos 0.5x for x 2.214, the dimensions of the rectangle are approximately
4.43 (width) by 1.79 (height), and the maximum area is approximately 7.923.
19. Let the radius of the cylinder be r cm, 0
V (r )
2 r 2 100 r 2 cm3 . Then, V (r )
Copyright
r 10. Then the height is 2 100 r 2 and the volume is
2 r2
1
2 100 r
2
( 2r )
2
100 r 2 (2r )
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2 r 3 4 r (100 r 2 )
100 r 2
294
Chapter 4 Applications of Derivatives
2 r (200 3r 2 )
100 r 2
and V (r )
2
3
0 for 10
2
3
r 10
. The critical point for 0
r 10 occurs at r
200
3
10
2 . Since V
3
20
3
11.55 cm, and the volume is
4000
3 3
2418.40 cm3 .
20. (a) From the diagram we have 4 x
108 and
V x 2 . The volume of the box is
V ( x) x 2 (108 4 x), where 0 x 27. Then
V ( x) 216 x 12 x 2 12 x(18 x) 0
the
critical points are 0 and 18, but x 0 results in
no box. Since V ( x) 216 24 x 0 at x 18
we have a maximum. The dimensions of the
box are 18 18 36 in.
(b) In terms of length, V ( )
x2
2
108
4
. The
graph indicates that the maximum volume
occurs near
36, which is consistent with the
result of part (a).
21. (a) From the diagram we have 3h 2w 108 and
V h 2 w V (h) h 2 54 32 h 54h 2 32 h3 .
Then V (h) 108h 92 h 2 92 h(24 h) 0
h 0 or h 24, but h 0 results in no box.
Since V (h ) 108 9h 0 at h 24, we
have a maximum volume at h 24 and
w 54 32 h 18.
(b)
22. From the diagram the perimeter is P 2r 2h
r,
where r is the radius of the semicircle and h is the
height of the rectangle. The amount of light
transmitted proportional to A
dA
dr
2r
h
3
2
P 4r
2h
4
P
8
1
4
r)
4P
8 3
r
r2
0
2 P
8 3
2rh
rP 2r 2
r
2P
8 3
(4 ) P
.
8 3
1
4
3
4
r2
r 2 . Then
Therefore,
gives the proportions that admit the most
2
light since d 2A
dr
4
3
2
0 for 0
r 10
2
3
r 10, the critical point corresponds to the maximum volume. The dimensions are
8.16 cm and h
r ( P 2r
(r )
0.
Copyright
2014 Pearson Education, Inc.
Section 4.6 Applied Optimization
r2h
23. The fixed volume is V
r3
2
3
V
r2
h
2r
3
295
, where h is the height of the cylinder and r is the radius
of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the
2 rh 4 r 2
surface area of the hemisphere. Thus, we minimize C
Then dC
dr
2V
r2
4V 1/3
32/3
16
3
r
2 31/3 V 1/3
3 2 1/3
1/3
the cost.
0
r3
8
3
V
r
31/3 2 4 V 1/3 2 31/3 V 1/3
3 2 1/3
2 r V2
3V 1/3 . From the volume
8
3V 1/3 . Since d 2C
4V 16
3
dr 2
r3
4 r2
2r
3
r
equation, h
V
r2
2V
r
r2.
8
3
2r
3
0, these dimensions do minimize
24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram
the area of the cross section is A( ) cos
sin cos , 0
. Then A ( )
sin
cos 2
sin 2
2
(2sin 2
sin
1)
(2sin
sin
1 when 0
is a maximum.
2
1)(sin
. Also, A ( )
0 for 0
25. (a) From the diagram we have: AP
CH
L x2
x2
L2
x2
172 x 2
f ( x)
L2
4x
4 x 17
17 2
2
x
RH
2
x2
4 x3
4 x 17
2
2 r
r
2
PQ
2
x2
17 x (8.5)2
there
RQ
2
(8.5)2
172 x 2
4[17 x (8.5) 2 ]
4 x 2 (8 x 51)
and h
y
y
f ( x)
0 when x
51
8
and
51 .
8
P
2
x. If P
36, then y 18 x. When the
h 18 x. The volume of the cylinder is
3 x (12 x )
4
0
(4 x 17)2
0
x
0 or 12; but when x
0 there is no
there is a maximum at x 12. The values of x 12 cm
6 cm give the largest volume.
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6
(8.5)2
2x 2 y
2
3
V
r h V ( x) 18 x4 x . Solving V ( x)
cylinder. Then V ( x) 3 3 2x
V (12)
. Therefore, at
2
then L 11.0 in.
cylinder is formed, x
and y
HQ
is minimized when x
x
2
2
because
(8.5 x)2
2 x3 .
2 x 8.5
51 . Thus L2
8
6
6
(8.5 x) ,
2
17 x (8.5)2
26. (a) From the figure in the text we have P
2
0 for
1
8.5 x,
. It follows that RP
L2
or sin
2
2
( x 8.5)2
2 L2
and A ( )
is minimized, then L2 is minimized. Now f ( x)
0 when x
51 ,
8
x2
x 2 )(17 x (8.5)2 )
17 x3
17 x
2
(8.5 x 2 )
x2
x2
4( L2
3
(c) When x
x2
L2
17 x3
17 x (8.5) 2
(b) If f ( x)
L x
x 2 (8.5 x) 2 , RQ
(8.5) 2
L2
11
x
2
2
1
2
sin
L x 2 , PB
L x , QB
HQ 11 CH QB 11
0
6
x, RA
2
DR 11 RA 11
L x2
1) so A ( )
2014 Pearson Education, Inc.
296
Chapter 4 Applications of Derivatives
(b) In this case V ( x)
x 2 (18 x). Solving V ( x) 3 x(12 x ) 0
x 0 or 12; but x 0 would result in
no cylinder. Then V ( x) 6 (6 x) V (12) 0 there is a maximum at x 12. The values of
x 12 cm and y 6 cm give the largest volume.
27. Note that h 2
0
0
r2
3 h2 . Then the volume is given by V
3 and so r
r2
3, and so dV
dh
h
h 1, and
dV
dh
0 for 1 h
(1 r 2 ). The critical point (for h
r 2h
3
3
(3 h 2 )h
h
3
0 ) occurs at h 1. Since dV
dh
h3 for
0 for
3, the critical point corresponds to the maximum volume. The cone of
greatest volume has radius 2 m, height 1 m, and volume 23 m3 .
( x 0) 2
28. Let d
x2
D
2 x
D
x
a
y
b
2
y2
b2 x
a2
ab2
a2 b2
( y 0) 2
x2
b2
a
0
2
2b 2
a2
x2
b
a
ab 2
a2 b2
0
y
b
D
2x 2
2
x b
x
y 2 and ax
1
b
a
y
b
a
x b. We can minimize d by minimizing
b
a
x b
is the critical point
2b 2
a2
2x
b ab 2
a a2 b2
y
b
ab 2
a 2 b2
the critical point is a local minimum
,
x
2b 2
a
.D
a 2b .
a2 b2
a 2b is
a2 b2
0
2
2 2b2
D
a
the point on the line
1 that is closest to the origin.
29. Let S ( x)
1,
x
x
x
0
2
x3
only consider x 1. S ( x )
2
x3
S ( x)
4 x2 , x
1
x
30. Let S ( x)
8
S
0
S (1)
1
x2
S ( x)
1
2
2
(1/2)3
31. The length of the wire b
equilateral triangle P
x2 1 . S
x2
2 0
13
1
x2
S ( x) 1
8
x2 1
x2
0
x2 1 0
0
x
1. Since x
( x)
0
local minimum when x
8 x3 1
x2
1.
2
0
8 x3 1 0
x
1.
2
perimeter of the triangle circumference of the circle. Let x length of a side of the
r b2 3 x .
3 x, and let r radius of the circle C 2 r. Thus b 3 x 2 r
The area of the circle is r 2 and the area of an equilateral triangle whose sides are x is 12 ( x) 23 x
Thus, the total area is given by A
3
2
A
3
2
A
x
3
2
9
2
0
0, we
local minimum when x 1
8 x3 1 . S
x2
8x
0
( x)
3
2
(b 3 x)
x
3 2
x
4
3b
2
r2
9
2
b 3x 2
2
3 2
x
4
x. A
3
2
0
local minimum at the critical point. P
triangular segment and C
b 3x
2
2
b 3x
b
9b
3 9
3b
2
x
9
2
3 2
x
4
b 3x
4
x
x
3b
3 9
3 b
m is
3 9
0
9b
3 9
3
3 2
x .
4
2
3b
3 9
.
m is the length of the
the length of the circular segment.
32. The length of the wire b perimeter of the triangle circumference of the circle. Let x length of a side of the
r b2 4 x . The area of the
square P 4 x, and let r radius of the circle C 2 r. Thus b 4 x 2 r
circle is r 2 and the area of a square whose sides are x is x 2 . Thus, the total area is given by A
b 4x 2
2
x2
b 4x
4
8
0
square segment and C
2
x
4
b
x2
.A
2
2
A
2x
4
2
(b 4 x)
2 x 2b
8
x, A
b 4x
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b 44b
b
4
2x
2b
8
x
r2
0
4b m is the length of the
4 4b
4
m is the length of the circular segment.
local minimum at the critical point. P
b 4x
2
0
x2
2014 Pearson Education, Inc.
Section 4.6 Applied Optimization
x, 43 x be the coordinates of the corner that intersects the line. Then base
33. Let ( x, y )
y
x
4 3
3 2
4 x, thus
3
3. A
2
A
0
A(0)
(3 x) 43 x 4 x 43 x 2 , 0 x 3. A 4 83 x, A 0
local maximum at the critical point. The base 3 32 32 and the height
the area of the rectangle is given by A
4
3
A 23
0
x, 9 x 2 be the coordinates of the corner that intersects the semicircle. Then base
9 x 2 , thus the area of the inscribed rectangle is given by A
y
2 9 x
x
3 x and height
2.
34. Let ( x, y )
height
297
2
(2 x)
2
2(9 x ) 2 x
x
9 x2
2
18 4 x
9 x2
2
4 x2
3. A is continuous on the closed interval 0
0, A(3)
0, and A
3 2
2
3 2
2
3 2
9
,A
x
0
3
18 4 x
2
0
(2 x ) 9 x 2 , 0
3 2
2
x
x
2x and
3. Then
3 2
2
, only x
lies in
A has an absolute maxima and absolute minima.
absolute maxima. Base of rectangle is 3 2 and height
is 3 2 2 .
35. (a) f ( x)
x2
(b) f ( x)
2
x
a
x
a
x
f ( x)
f ( x)
x 2 (2 x3 a), so that f ( x)
2x
3
(x
3
a), so that f ( x )
0 when x
2 implies a 16
0 when x 1 implies a
36. If f ( x) x3 ax 2 bx, then f ( x) 3 x 2 2ax b and f ( x)
(a) A local maximum at x
1 and local minimum at x 3
9.
27 6a b 0 a
3 and b
(b) A local minimum at x 4 and a point inflection at x 1
24.
6 2a 0 a
3 and b
1
6 x 2a.
f ( 1) 0 and f (3)
f (4)
0 and f (1)
0
0
3 2a b
0 and
48 8a b
0 and
37. (a) s (t )
16t 2 96t 112 v(t ) s (t )
32t 96. At t 0, the velocity is v (0) 96 ft/sec.
(b) The maximum height occurs when v(t ) 0, when t 3. The maximum height is s (3) 256 ft and it occurs
at t 3 sec.
(c) Note that s (t )
16t 2 96t 112
16(t 1)(t 7), so s 0 at t
1 or t 7. Choosing the positive value
128 ft/sec.
of t, the velocity when s 0 is v(7)
38.
Let x be the distance from the point on the shoreline nearest Jane s boat to the point where she lands her boat.
Then she needs to row 4 x 2 mi at 2 mph and walk 6 x mi at 5 mph. The total amount of time to reach
the village is f ( x)
f ( x)
0. we have:
4 x2
2
6 x
5
x
1
5
2 4 x2
hours (0
5x
x
6). Then f ( x)
2 4 x2
25 x 2
1
1
2 2 4 x2
4 4 x2
(2 x ) 15
21x 2
16
x
2 4 x2
x
1.
5
4 .
21
Solving
We discard
the negative value of x because it is not in the domain. Checking the endpoints and critical point, we have
f (0)
2.2, f
4
21
2.12, and f (6)
3.16. Jane should land her boat 4
from the point nearest her boat.
Copyright
21
2014 Pearson Education, Inc.
0.87 miles down the shoreline
298
Chapter 4 Applications of Derivatives
39. 8x
h
x 27
2
8 216
x
( x 27) 2 when x
minimized. If f ( x)
( x 27)2
0. Note that L ( x)
2
8 216
x
is minimized when f ( x)
2 8 216
x
h2
8 216
and L ( x)
x
h
( x 27) 2 is
0, then
216
x2
2( x 27)
( x 27) 1 1728
3
0
x
0
x
27 (not acceptable
since distance is never negative) or x 12 . Then
L(12)
2197 46.87 ft.
40. (a) s1
s2
t
sin t
3
or
sin t
sin t
3
4
3
sin t cos 3
sin 3 cos t
(b) The distance between the particles is s (t ) | s1 s2 |
sin t
s (t )
3 cos t cos t
2 sin t
3 sin t
3 cos t
d | x|
since dx
0, 3 , 56 , 43 , 116 , 2 ; then s (0) 23 , s 3
greatest distance between the particles is 1.
sin t
(c) Since s (t )
3 cos t cos t
2 sin t
3 sin t
sin t sin t
x
| x|
1
2
3
sin t
tan t
3
3 cos t
critical times and endpoints are
0, s 56
1, s 43
we can conclude that at t
3 cos t
3
cos t
2
1 sin t
2
sin t
0, s 116
3
3
2
1, s (2 )
the
and 43 , s (t ) has cusps and the
distance between the particles is changing the fastest near these points.
k
d2
41. I
, let x
distance the point is from the stronger light source
6 x
distance the point is from the other
k1
light source. The intensity of illumination at the point from the stronger light is I1
k2
illumination at the point from the weaker light is I 2
the intensity of the second light
16k2
I
x
x
3
4 m. I
8k 2 .
16(6 x )3 k2 2 x3k2
2 k2
(6 x )
48k2
k1
3
x4
3
6 k2
(6 x ) 4
x (6 x )
3
I (4)
8k 2
I1
and I
(6 x) 2
x2
0
6 k2
44
(6 4)4
, and intensity of
. Since the intensity of the first light is eight times
. The total intensity is given by I
16(6 x )3 k2 2 x3k2
0
x3 (6 x )3
48k2
x2
0
I1 I 2
16(6 x)3 k2
8k2
k2
x2
(6 x ) 2
2 x 3 k2
0
local minimum. The point should be 4 m from the
stronger light source.
42. R
v02
g
d 2R
d 2
sin 2
dR
d
4v02
g
2v02
g
cos 2 and ddR
sin 2 4
4v02
g
0
2v02
g
0
cos 2
0
2
. d R2
4
d
4v02
g
local maximum. Thus, the firing angle of
sin 2
4
45
4
will maximize the range R.
43. (a) From the diagram we have d 2 4r 2 w2 . The strength of the beam is S kwd 2 kw (4r 2 w2 ).
When r 6, then S 144kw kw3 . Also, S ( w) 144k 3kw2 3k (48 w2 ) so S ( w) 0 w
4 3;
S 4 3 0 and 4 3 is not acceptable. Therefore S 4 3 is the maximum strength. The dimensions of
the strongest beam are 4 3 by 4 6 inches.
Copyright
2014 Pearson Education, Inc.
Section 4.6 Applied Optimization
(b)
299
(c)
Both graphs indicate the same maximum value and are consistent with each other. Changing k does not
change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce
the strongest beam).
44. (a) From the situation we have w2
12. Also, S (d )
144 d 2 . The stiffness of the beam is S
2
2
4 kd (108 d )
where 0
d
cause S
0. The maximum occurs at d
kwd 3
kd 3 (144 d 2 )1/2 ,
critical points at 0, 12, and 6 3. Both d
144 d 2
0 and d
12
6 3. The dimensions are 6 by 6 3 inches.
(c)
(b)
Both graphs indicate the same maximum value and are consistent with each other. The changing of k has
no effect.
45. (a) s 10 cos( t ) v
10 sin( t ) speed |10 sin( t )| 10 |sin( t ) | the maximum speed is
10
31.42 cm/sec since the maximum value of |sin ( t )| is 1; the cart is moving the fastest at t 0.5 sec,
0 cm and
1.5 sec, 2.5 sec and 3.5 sec when |sin ( t )| is 1. At these times the distance is s 10 cos 2
10 2 cos ( t )
a
| a | 10 2 |cos ( t ) |
| a | 0 cm/sec 2
(b) | a | 10 2 |cos ( t )| is greatest at t 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec, and 4.0 sec, and at these times the
magnitude of the cart s position is | s | 10 cm from the rest position and the speed is 0 cm/sec.
46. (a) 2sin t
sin 2t
2sin t 2sin t cos t
0
(2sin t )(1 cos t )
(b) The vertical distance between the masses is s (t ) | s1 s2 |
0
t
( s1 s2 )
k
where k is a positive integer
2 1/2
((sin 2t 2sin t ) 2 )1/2
1 ((sin 2t 2sin t ) 2 ) 1/2 (2)(sin 2t 2sin t )(2 cos 2t 2 cos t ) 2(cos 2t 2 cos t )(sin 2t 2 sin t )
2
|sin 2t 2 sin t|
4(2 cos t 1)(cos t 1)(sin t )(cos t 1)
2
4
critical times at 0, 3 , , 3 , 2 ; then s (0) 0,
|sin 2t 2sin t |
s (t )
s 23
sin 43
2sin 23
3 3
,
2
the greatest distance is 3 2 3 at t
47. (a) s
(b) ds
dt
2
3
s( )
0, s 43
2sin 43
208t 144
(12 12t 2 ) 64t 2
8 knots
Copyright
3 3
,
2
s (2 )
0
and 43
(12 12t ) 2 (8t )2 ((12 12t )2 64t 2 )1/2
1 ((12 12t ) 2 64t 2 ) 1/2 [2(12 12t )( 12) 128t ]
2
ds
dt t 1
sin 83
2014 Pearson Education, Inc.
ds
dt t 0
12 knots and
300
Chapter 4 Applications of Derivatives
(d) The graph supports the conclusions in parts (b)
and (c).
(c) The graph indicates that the ships did not see
each other because s (t ) 5 for all values of t.
(e) lim
t
ds
dt
lim
t
(208t 144)2
144(1 t )
2
64t
lim
2
t
144 2
t
2
208
144
1
t
1
2082
144 64
64
208
4 13 which equals the square
root of the sums of the squares of the individual speeds.
48. The distance OT TB is minimized when OB is a
straight line. Hence
1
2.
kax kx 2 , then v
49. If v
v
a
2
2k
ka 2kx and v
0. The maximum value of v is
2k , so v
2
ka
4
0
x
a.
2
At x
a
2
there is a maximum since
.
50. (a) According to the graph, y (0) 0.
(b) According to the graph, y ( L) 0.
(c) y (0) 0, so d 0. Now y ( x) 3ax 2 2bx c, so y (0) 0 implies that c 0. Therefore, y ( x) ax3 bx 2
and y ( x) 3ax 2 2bx. then y ( L)
aL3 bL2 H and y ( L) 3aL2 2bL 0, so we have two linear
equations in two unknowns a and b. The second equation gives b 3aL
. Substituting into the first
2
equation, we have aL3
2 H3 x3
L
y ( x)
51. The profit is p
3 H2
L
nx nc
3aL3
2
2
3
H , or aL2
x , or y ( x)
H 2
x 3
L
a (bc 100b) x 100bc bx . Then p ( x)
0
x
c
2
3
x 2
L
L
50. At x
c
2
2x
68 x 2400. Then p ( x)
maximum since p (17)
2b. Solving
50 there is a maximum profit since p ( x)
4 x 68 and p
2b
0 for all x.
(50 x )(200 2 x ) 32(50 x) 6000
4. Solving p ( x )
0. It would take 67 people to maximize the profit.
Copyright
L
a b(100 x)( x c)
bc 100b 2bx and p ( x )
52. Let x represent the number of people over 50. The profit is p ( x )
2
3 H2 and the equation for y is
.
n( x c ) [ a( x c) 1 b(100 x)]( x c)
2
p ( x)
2 H3 . Therefore, b
H , so a
2014 Pearson Education, Inc.
0
x 17. At x 17 there is a
Section 4.6 Applied Optimization
kmq 1 cm h2 q, where q
53. (a) A(q )
2 km , 0, and
h
points are
0
2 km , but
h
kmq 2
A (q )
2km
h
only
hq 2 2 km
h
2
2q 2
2kmq 3 . The critical
and A ( q)
2 km
h
is in the domain. Then A
301
0
2km
h
at q
there
is a minimum average weekly cost.
( k bq ) m
q
(b) A(q )
cm h2 q
2kmq 3
A ( q)
kmq 1 bm cm h2 q, where q
0
A (q )
2km
h
0 so the most economical quantity to order is still q
average weekly cost.
2km
h
0 at q
as in (a). Also
which minimizes the
c ( x)
the average cost of producing x items,
54. We start with c( x) the cost of producing x items, x 0, and x
assumed to be differentiable. If the average cost can be minimized, it will be at a production level at which
d c( x)
dx
x
xc ( x ) c( x)
0
0 (by the quotient rule)
x2
c( x)
0 (multiply both sides by x 2 )
xc ( x ) c( x)
c ( x)
where c ( x) is the marginal cost. This concludes the proof. (Note: The theorem does not assure a
x
production level that will give a minimum cost, but rather, it indicates where to look to see if there is one. Find
the production levels where the average cost equals the marginal cost, then check to see if any of them give a
minimum.)
x3 6 x 2 9 x, where x 0. Then p ( x)
3x 2 12 x 9
55. The profit p( x) r ( x) c( x ) 6 x ( x3 6 x 2 15 x)
3( x 3)( x 1) and p ( x )
6 x 12. The critical points are 1 and 3. Thus p (1) 6 0
at x 1 there is a
local minimum, and p (3)
6 0 at x 3 there is a local maximum. But p (3) 0 the best you can do is
break even.
c( x)
56. The average cost of producing x items is c ( x)
x 2 20 x 20, 000 c ( x) 2 x 20 0 x 10, the
x
only critical value. The average cost is c (10) $19,900 per item is a minimum cost because c (10) 2 0.
57. Let x
the length of a side of the square base of the box and h
6 x2
The total cost is given by C
C
x
0
4
12 x3 768
x2
48
h
3
42
0
4(4 xh)
12 x3 768
and C (4)
0
6(4) 2
x
768
4
6 x 2 16 x 482
6x2
x
12 1536
2
4; C
288
x
x2h
the height of the box. V
768 ,
x
x
0
12 x 768
2
C
C (4) 12 1536
2
4
48
x
0
h
12 x3
x2
48 .
x2
768
local minimum.
the box is 4 ft 4 ft 3 ft, with a minimum cost of $288.
58. Let x the number of $10 increases in the charge per room, then price per room 50 10 x, and the number of
the total revenue is R ( x ) (50 10 x)(800 40 x )
rooms filled each night 800 40x
400 x 2
x
15 ;
2
6000 x
R ( x)
dR
59. We have dM
maximum.
CM
40000, 0
800
R
x
20
15
2
R ( x)
800
2
M 2 . Solving d R2
dM
0
C 2M
Copyright
800 x 6000; R ( x)
0
800 x 6000
0
local maximum. The price per room is 50 10 15
2
0
M
C . Also. d 3 R
2
dM 3
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2
0
at M
C
2
$125.
there is a
302
Chapter 4 Applications of Derivatives
2cr0 r 3cr 2
cr0 r 2 cr 3 , then v
60 . (a) If v
2 r0
,
3
v 0 is r 0 or
there is a maximum.
cr 2r0 3r and v 2cr0 6cr 2c r0 3r . The solution of
2r
2r0
2r0
but 0 is not in the domain. Also, v 0 for r 30 and v 0 for r
at r
3
3
(b) The graph confirms the findings in (a).
61. If x
0, then x 1
2
2
2
2
2
then a a 1 b b 1 c c 1 d d 1
62. (a) f ( x)
x
a
2
x
2. In particular if a, b, c and d are positive integers,
16.
a2 x2
f ( x)
2
x2 1
x
x2 1 2x
0
1/ 2
x 2 a 2 x2
a
2
x
1/ 2
a 2 x2 x 2
2
a2 x
a2
2 3/ 2
a 2 x2
3/ 2
0
f ( x) is an increasing
function of x
(b) g ( x)
d x
b2
b2
b
2
d x
d x
2 3/ 2
g ( x)
2
0
b2
d x
2 1/ 2
b2
d x
2
d x
b2
d x
1/ 2
2
2
b2
d x
b2
2
d x
d x
2
2 3/ 2
g ( x) is a decreasing function of x
dt is an increasing function of x (from part (a)) minus a decreasing function
(c) Since c1 , c2 0, the derivative dx
dt
1 f ( x) 1 g ( x)
d 2t
1 f ( x) 1 g ( x) 0 since f ( x) 0 and
of x (from part (b)): dx
2
c
c
c
c
g ( x)
0
dt
dx
1
2
is an increasing function of x.
dx
1
2
63. At x c, the tangents to the curves are parallel. Justification: The vertical distance between the curves is
D ( x) f ( x) g ( x), so D ( x) f ( x) g ( x). The maximum value of D will occur at a point c where D 0. At
such a point, f (c ) g (c) 0, or f (c) g (c).
64. (a) f ( x) 3 4 cos x cos 2 x is a periodic function with period 2
(b) No, f ( x) 3 4 cos x cos 2 x 3 4cos x (2 cos 2 x 1) 2(1 2 cos x cos 2 x)
f ( x) is never negative.
Copyright
2014 Pearson Education, Inc.
2(1 cos x ) 2
0
Section 4.6 Applied Optimization
65 . (a) If y
cot x
2 csc x where 0
x 4 . For 0 x
value of y
1.
4
x
, then y
we have y
0 and y
(csc x)
2 cot x csc x . Solving y
0 when 4
x
. Therefore, at x
0
4
303
1
2
cos x
there is a maximum
(b)
The graph confirms the findings in (a).
66. (a) If y
tan x 3 cot x where 0
but
x
(b)
3
3
x
x,
2
sec2 x 3csc 2 x. Solving y
then y
2
is not in the domain. Also, y
2
2sec x tan x 6 csc x cot x
there is a minimum value of y
0
0 for all 0
2 3.
tan x
x
2
3
x
3
,
. Therefore at
The graph confirms the findings in (a).
67. (a) The square of the distance is D ( x)
point occurs at x 1. Since D ( x)
2
x 32
x
2
0
0 for x 1 and D ( x)
minimum distance. The minimum distance is D(1)
(b)
x2
2x
9,
4
so D ( x )
0 for x 1, the critical point corresponds to the
5
.
2
The minimum distance is from the point 32 , 0 to the point (1, 1) on the graph of y
the value x 1 where D ( x), the distance squared, has its minimum value.
Copyright
2 x 2 and the critical
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x, and this occurs at
304
Chapter 4 Applications of Derivatives
68. (a) Calculus Method:
The square of the distance from the point 1, 3 to x, 16 x 2 is given by
D ( x)
( x 1)2
Then D ( x)
16 x 2
1
2
2
2
3
2
48 3 x
2
2
x2
( 6 x)
2 x 1 16 x 2
6x
2
48 3 x 2
2
2 48 3 x 2
. Solving D ( x)
2 x 20 2 48 3 x 2 .
3
0 we have:
6 x 2 48 3 x2
36 x
4(48 3x 2 ) 9 x
48 3 x 2 12 x2 48 x
2 We discard x
2 as
an extraneous solution, leaving x 2. Since D ( x) 0 for 4 x 2 and D ( x) 0 for 2 x 4, the critical
point corresponds to the minimum distance. The minimum distance is D(2) 2 .
Geometry Method:
The semicircle is centered at the origin and has radius 4. The distance from the origin to 1, 3 is
12
3
2
2. The shortest distance from the point to the semicircle is the distance along the radius
containing the point 1, 3 . That distance is 4 2
2.
(b)
16 x 2 , and
The minimum distance is from the point 1, 3 to the point 2, 2 3 on the graph of y
this occurs at the value x
4.7
1.
2 where D ( x), the distance squared, has its minimum value.
NEWTON S METHOD
x2
y
x 1
2
3
x2
y
4 6 9
12 9
x3 3 x 1
2. y
1
3
1
90
x4
3. y
6
5
2
2x 1
x 3
2
3
y
29
90
1
21
xn 1
13
21
3x2 3
xn2 xn 1
;
2 xn 1
xn
.61905; x0
xn
xn
1
x0
1
1 1 21 11
x1
xn3 3 xn 1
3 xn2 3
x1 1 121 11
1
; x0
0
x1
2
2
3
x2
4
9
2
3
4
3
0 13
1
3
1
3
x2
1
1
5
3
2 4 42 11
x2
2
3
1
27
1
3
1.66667
1 1
3
0.32222
y
1296 750 1875
4320 625
51
11
31
31
xn4 xn 3
4 x3 1
xn 1
xn
6
5
5763
4945
1.16542; x0
171
4945
4 xn3
1
; x0
1
1
x1
x1 1 1 41 13
6
5
x2
1 1 41 13
2
x2
1.64516
Copyright
2014 Pearson Education, Inc.
6
5
1296 6
625 5
864
1
125
3
2 16322 13
Section 4.7 Newton s Method
2x x2 1
4. y
29
12
1
2
5
12
1
12
2 2x
xn 1
xn
.41667; x0
2
x1
2 xn xn2 1
;
2 2 xn
2 42 44 1
x0
0
x1
0 02 00 1
5
2
x2
5
2
5
25
4
5
4
2
4 x3
y
113
2000
xn 1
2500 113
2000
2387
2000
6. From Exercise 5, xn 1
5
4
625 512
2000
5
4
xn4 2
xn
4 xn3
1
5
2
20 25 4
12
5
4
x2
5
4
625
2
256
125
16
1 14
5
4
2 5
; x0
1
1
x1
x1 1 1 42
xn4 2
xn
4 xn3
113
2000
; x0
1 1 42
f (x )
8. It does matter. If you start too far away from x
Starting with x0
9. If x0
h
0
f ( x0 )
f ( x0 )
x0
1
12
2 h
if x0
h
h
h
1
2 h
x1/3
f ( x)
xn
1
x1
0
xn
h
1
3
x1/3
n
1
3
h
h 2 h
f ( x)
x1
2, x2 4, x3
forth. Since xn 2 xn
n
xn
.
1
x
5
4
625 2
256
125
16
xn
x0 for all n
0. That is all, of
, the calculated values may approach some other root.
2
as the root, not x
2
.
f ( h)
f ( h)
2/3
1
8, and x4 16 and so
we may conclude that
is equivalent to solving x3 3x 1
is equivalent to solving x3 3x 1
is equivalent to solving x3 3x 1
is equivalent to solving x3 3x 1
All four equations are equivalent.
x 1 0.5sin x
x2
h.
2 xn ; x0
xn 2/3
625 512
2000
h;
f ( x0 )
f ( x0 )
x0
5
4
f (h)
f (h )
h
h 2 h
h
1
2
0.5, for instance, leads to x
x1
h
h
12. f ( x)
5
2
1 14 1
2 1
1.1935
0 and f ( x0 ) 0
xn 1 xn f ( xn ) gives x1 x0
x2 x0
n
the approximations in Newton s method will be the root of f ( x ) 0.
11. i)
ii)
iii)
iv)
1
2
x2
1.1935
7. f ( x0 )
10.
1
2
2.41667
x4
5. y
y
305
0.
0.
0.
0.
f ( x) 1 0.5cos x
Copyright
xn
1
xn
xn 1 0.5sin xn
1 0.5 cos xn
; if x0
2014 Pearson Education, Inc.
1.5, then x1 1.49870
306
Chapter 4 Applications of Derivatives
13. f ( x)
tan x 2 x
x2
14. f ( x)
x4
1.155327774
x4
2 x3
sec2 x 2
f ( x)
x16
x2
x17
2x 2
0.630115396; if x0
xn 1
tan( xn ) 2 xn
xn
sec 2 xn
; x0
1
x1 1.2920445
1.165561185
f ( x)
2.5, then x4
4 x3 6 x 2
2x 2
xn 1
xn
2.57327196
xn4 2 xn3 xn2 2 xn 2
4 xn3 6 xn2 2 xn 2
; if x0
0.5, then
15. (a) The graph of f ( x) sin 3 x 0.99 x 2 in the
window 2 x 2, 2 y 3 suggests three
roots. However, when you zoom in on the
x-axis near x 1.2, you can see that the graph
lies above the axis there. There are only two
roots, one near x
1, the other near x 0.4.
(b) f ( x) sin 3 x 0.99 x 2
f ( x) 3cos 3x 2 x
xn 1
xn
sin 3 xn 0.99 xn2
3cos 3 xn 2 xn
and the solutions are approximately
0.35003501505249 and 1.0261731615301
16. (a) Yes, three times as indicted by the graphs
f ( x)
3sin 3 x 1
(b) f ( x) cos 3x x
cos(3 x ) x
n
n
xn 1 xn
; at approximately
3sin(3 xn ) 1
0.979367, 0.887726, and 0.39004 we have
cos 3x x
17. f ( x)
2 x4
4 x2 1
x0
0.5, then x3
even function.
18. f ( x)
tan x
approximate
xn 1
xn
2 xn4 4 xn2 1
8 xn3 8 xn
; if x0
2, then x6
1.30656296; if
0.5411961; the roots are approximately 0.5411961 and 1.30656296 because f ( x) is an
f ( x)
sec2 x
xn 1
tan( xn )
xn
sec2 ( xn )
to be 3.14159.
19. From the graph we let x0
xn 1 xn
x2 .45018
8 x3 8 x
f ( x)
cos( xn ) 2 xn
sin( xn ) 2
at x
0.5 and f ( x)
x1
; x0
3
x1
3.13971
cos x 2 x
.45063
0.45 we have cos x
Copyright
2 x.
2014 Pearson Education, Inc.
x2
3.14159 and we
Section 4.7 Newton s Method
20. From the graph we let x0
f ( x)
cos x x
0.7 and
xn 1
xn cos( xn )
1 sin( xn )
xn
x1
.73944
x2
we have cos x
x.
.73908
at x
0.74
21. The x-coordinate of the point of intersection of y
x3
x2
xn 1
1
x
xn
0
xn3
3 xn2
x 2 ( x 1) and y
x3
The x -coordinate is the root of f ( x)
xn2
1
xn
2 xn 12
xn
x1
0.83333
x2
0.81924
x2
x3
the solution of x 2 ( x 1)
1 is
x
1
x
f ( x)
0.81917
x7
3x2
2x
0.81917
1
x
1 . Let x
0
x2
r
xn
xn 3 xn
1
2 xn
2 x
2
x1 1.4
2 xn
2
23. Graphing e x and x 2
x2
1.35556
x1
f ( xn )
f ( xn )
xn
0.536981, x2
xn
1.35498
x7
1.35498
r
1
1.3550
x 1 shows that there are two places where the curves intersect, one at x = 0 and the
other between x = 0.5 and x = 0.6. Let f ( x)
xn 1
x3
1
0.8192
22. The x-coordinate of the point of intersection of y
x and y 3 x 2 is the solution of x 3 x 2
1
x 3 x 2 0 The x -coordinate is the root of f ( x)
x 3 x2
f ( x)
2 x. Let x0
xn 1
307
e
xn2
xn2 xn 1
1 2 xn 2 xn e
x2
0
0.534856, x3
2
e x
x2
x 1, x0
0.5, and
. Performing iterations on a calculator, spreadsheet, or CAS gives
0.53485, x4
0.53485. (You may get different results depending upon
what you select for f(x) and x0 , and what calculator or computer you may use.) Therefore, the two curves
intersect at x = 0 and x = 0.53485.
24. Graphing ln(1 x 2 ) and x
1 shows that there are two places where the curves intersect, one between x = 1
and x = 0.9, and the other between x = 0.5 and x = 0.6. Let f ( x)
xn 1
x0
x1
f ( xn )
f ( xn )
xn
xn
ln(1 xn2 ) xn 1
2 xn
1 xn2
1
ln(1 x 2 ) x 1, and
. Performing iterations on a calculator, spreadsheet, or CAS with
0.5 gives x1 0.590992, x2 0.583658, x3 0.583597, x4 0.583597 and with x0
0.9 gives
0.928237, x2
0.924247, x3
0.924119, x4
0.924119. (You may get different results
depending upon what you select for f(x) and x0 , and what calculator or computer you may use.) Therefore, the
two curves intersect at x = 0.924119 and x = 0.583597.
25. If f ( x)
x
3
x0
x3 2 x 4, then f (1)
2x 4
1
1 0 and f (2) 8
0
by the Intermediate Value Theorem the equation
0 has a solution between 1 and 2. Consequently, f ( x)
x1 1.2
x2
1.17975
x3
Copyright
1.179509
x4
3x2
1.1795090
2 and xn 1
xn
xn3 2 xn 4
3 xn2 2
. Then
the root is approximately 1.17951.
2014 Pearson Education, Inc.
308
Chapter 4 Applications of Derivatives
26. We wish to solve 8 x 4 14 x3 9 x 2 11x 1 0. Let f x
32 x3 42 x 2 18 x 11
f ( x)
x0
8 xn4 14 xn3 9 xn2 11xn 1
xn
32 xn3 42 xn2 18 xn 11
.
approximation of corresponding root
0.976823589
0.100363332
0.642746671
1.983713587
1.0
0.1
0.6
2.0
27.
xn 1
8 x 4 14 x3 9 x 2 11x 1, then
4 x4
f ( x)
4 x2
f ( x) 16 x3 8 x
xi 1
f ( xi )
f ( xi )
xi
xi3 xi
xi
4 xi2 2
. Iterations are performed using the
procedure in problem 13 in this section.
2 or x0
0.8, xi
1 as i gets large.
(a) For x0
0.5 or x0 0.25, xi
0 as i gets large.
(b) For x0
(c) For x0 0.8 or x0 2, xi 1 as i gets large.
(d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal
21
7
value.) For x0
21
7
between x0
21
,
7
or x0
21
7
or x0
Newton s method does not converge. The values of xi alternate
as i increases.
28. (a) The distance can be represented by
( x 2) 2
D ( x)
x2
1
2
2
, where x
D ( x) is minimized when f ( x)
( x 2) 2
minimized. If f ( x)
0. The distance
( x 2)2
x2
2
1
2
2
x2
1
2
2
is
, then
4( x3 x 1) and f ( x) 4 (3 x 1) 0. Now
1 .
x3 x 1 0 x ( x 2 1) 1 x
0
2
f ( x)
f ( x)
1
x
1
1
(b) Let g ( x)
x0
1
x4
x88
30. Since s
x200
3
r2
sin 23r
1.00282
1
r
xn 1
2x
( x 2 1)2
1
xn 1
xn
xn2 1
xn
;
2 xn
2
2
xn 1
1
1
r
xn
( xn 1)40
40( xn 1)39
39 xn 1
. With x0
40
2, our computer gave
1.11051, coming within 0.11051 of the root x 1.
3 . Bisect
r
r
of length 1. Then sin 2
f (r )
( x 2 1) 2 (2 x) 1
g ( x)
40( x 1)39
f ( x)
x89
r
( x 2 1) 1 x
0.68233 to five decimal places.
( x 1)40
29. f ( x)
x87
x2 1
x
3
r
sin 2
f (r )
r3 1.00282
3
2r 2
the angle
1
r
cos 23r
r
1.0028
Copyright
to obtain a right triangle with hypotenuse r and opposite side
sin 23r
1
r2
; r0
1
r
1
3
1.00282
sin 23r
rn 1
rn
1
r
0. Thus the solution r is a root of
sin
3
2 rn2
2.9916
2014 Pearson Education, Inc.
3
2 rn
cos
1
rn
3
2 rn
1
rn2
r1 1.00280
Section 4.8 Antiderivatives
4.8
309
ANTIDERIVATIVES
1. (a) x 2
(b)
x3
3
3
(c) x3
x2
2. (a) 3x 2
(b)
x8
8
8
(c) x8
3x2 8 x
3. (a) x 3
(b)
x 3
3
(b)
x 2
4
4. (a)
x 2
x 3
3
(c)
x3
3
x
x2 3x
2
(c) x2
x2
2
5. (a)
1
x
(b)
5
x
(c) 2 x
5
x
6. (a)
1
x2
(b)
1
4x 2
4
(c) x4
1
2 x2
x3
(b)
x
8. (a) x 4/3
(b)
9. (a) x 2/3
x
(c) 23 x3
2 x
1 x 2/3
2
(c) 34 x 4/3
3 x 2/3
2
(b)
x1/3
(c) x 1/3
10. (a) x1/2
(b)
x 1/2
(c) x 3/2
11. (a) ln |x|
(b)
7 ln |x|
(c) x
5 ln |x|
(c)
4 ln
3
7. (a)
12. (a)
1 ln
3
x
(b)
2 ln
5
x
x
x
1
x
(c)
cos ( x )
sin 2x
(c)
2
(b)
2 tan 3x
(c)
cot x
(b)
cot 32x
(c) x 4 cot (2 x)
csc x
(b)
1 csc(5 x )
5
(c) 2 csc 2x
18. (a) sec x
(b)
4 sec(3 x )
3
(c) 2 sec 2x
1 e3 x
3
(b)
e x
(c)
(b)
3 e4 x /3
4
(c)
13. (a) cos ( x)
(b)
14. (a) sin ( x)
(b)
15. (a) tan x
16. (a)
17. (a)
19. (a)
20. (a)
1 e 2x
2
Copyright
3cos x
2014 Pearson Education, Inc.
cos (3x )
sin 2x
2 tan 3 x
3
2
2e x /2
5e x /5
sin x
310
Chapter 4 Applications of Derivatives
21. (a)
1
ln 3
3x
1 x 3 1
3 1
22. (a)
23. (a) 2sin 1 x
1 x2
2
24. (a)
25.
27.
29.
x2
2
( x 1) dx
3t 2
1 x
2
1
ln(1/2)
t
2
t2
4
(2 x3 5 x 7)dx
31.
1
x2
32.
1
5
x2
1
3
2
x3
x 2/3
35.
x
3
x dx
36.
x
2
37.
8y
38.
1
7
2
1/ 4
y
1
y
5/ 4
(b)
1
x 1
(c)
(b)
1 tan 1 x
2
(c)
1 tan 1 (2 x )
2
(b)
1 x3
3
1
ln 2
(c)
1
ln
26.
(5 6 x) dx
1 x1/2
2
dx
7x C
x2
3 x 2/3
2
x1/2
1
3
2 x 1 x 3 dx
40.
x 3 ( x 1) dx
41.
t t t
t2
t 3/ 2
t2
t1/ 2
t2
1x
3
2x 2
2
x
x3/ 2
2 x 1/2 dx
x 3 dx
x 4/3
3
2
1
7
4
3
x3/ 2
1
2
8 y2
2
1
x
x3
3
x
3
2 x2
2
C
x
5
1
x2
x 5/4 dx
y
y
x
1
1
y3/ 4
3
4
1/ 4
1
2 x1
x
2
2
y
7
4
y1/ 4
1
x
C
t 1/2 t 3/2 dt
Copyright
4 y2
C
t1/ 2
1
2
x2
t
1/ 2
1
2
x 13 x3
C
4
4
x
4 x1/2 C
C
C
C
C
C
1 x6
2
x2 C
1
4
2 t
2014 Pearson Education, Inc.
1
2
x 2
t4 C
C
1/ 4
1
ln(5/3)
C
C
8 y 3/4
3
2
x
1
2 x2
x
1 x3/2
3
C
C
C
1
4
2 x2
2
3 x 4/3
4
2
2
t3
6
4t 3 dt
C
1/ 2
2 x1
3
2
5 x 3x2
(1 x 2 3 x5 ) dx
2 x 3/2
3
C
2x
t2
2
34.
x1/3 dx
dt
x3
3
C
2 x 2 x 2 dx
x 2
1
1
1
5
2 x dx
y 5/4 dy
1
7
dy
1
30.
x
dx
8 y 2 y 1/4 dy
dy
39.
dt
5 x2
2
2x 3
1
5
C
2
3
2
x
(c)
28.
x 2
2 x dx
x 1/3 dx
2 x
C
1 x4
2
dx
33.
1
ln 2
x C
t3
dt
5 x
3
(b)
2
t
C
C
C
x
ln x
Section 4.8 Antiderivatives
42.
43.
45.
47.
t
4
t
3
t1/ 2
t3
4
t3
dt
2 cos t dt
7 sin 3 d
4t 3 t 5/2 dt
dt
2
4 t2
t
3/ 2
2 sin t C
44.
5sin t dt
21cos 3 C
46.
3cos 5 d
3cot x C
48.
sec2 x dx
3
3csc 2 x dx
49.
csc cot
2
51.
(e 3 x
5e x )dx
53.
(e x
4 x )dx
55.
(4sec x tan x 2sec2 x) dx
4sec x 2 tan x C
56.
1 (csc 2
2
1 cot x
2
57.
(sin 2 x csc2 x) dx
59.
1 cos 4t
2
dt
1
2
1 cos 4t
2
dt
1t
2
1 sin 4t
2
4
C
t
2
sin 4t
8
C
60.
1 cos 6t
2
dt
1
2
1 cos 6t
2
dt
1t
2
1 sin 6t
2
6
C
t
2
sin 6t
12
C
61.
1
x
1 csc
2
d
C
e3 x
3
5e x
e x
4x
ln 4
C
x csc x cot x) dx
5
x2 1
dx
1 cos 2 x
2
3 x 3 dx
65.
(1 tan 2 ) d
sec2 d
66.
(2 tan 2 ) d
(1 1 tan 2 ) d
67.
cot 2 x dx
68.
(1 cot 2 x) dx
69.
cos (tan
3 1
(sin
3 sin 5
5
2 sec
5
(2e x 3e 2 x ) dx
54.
(1.3) x dx
58.
(2 cos 2 x 3sin 3 x) dx sin 2 x cos 3 x C
tan d
(1.3) x
ln(1.3)
2
1
y1/ 4
1 y2
x
2 1
dx
x
dy
2
2
tan
C
cot x x C
(2 csc2 x) dx
Copyright
C
52.
(1 sec2 ) d
1) d
C
tan x
3
C
(1 (csc2 x 1)) dx
sec ) d
5cos t C
2 sec
5
64.
(csc 2 x 1) dx
C
50.
62.
C
tan
2
3t 3/ 2
2e x
C
3 e 2x
2
C
C
C
cot x C
63.
3x
1 csc x
2
5 tan 1 x C
ln x
3 1
C
2
t2
C
3
2
cos
2 x cot x C
C
2014 Pearson Education, Inc.
2sin 1 y
C
4 y 3/4
3
C
311
312
70.
Chapter 4 Applications of Derivatives
csc
csc sin
4
d (7 x 2)
71. dx
28
(3 x 5)
3
d
72. dx
csc
csc sin
d
4(7 x 2)3 (7)
28
C
1
sin
sin
d 1 tan (5 x 1) C
73. dx
5
1 (sec 2 (5 x
5
d
74. dx
3
d (ln
dx
79.
d 1
dx a
80.
d
dx
tan 1 ax
sin 1 ax
81. If y
1
x
dy
x 2
a
tan 1 x
x
ln x 12 ln(1 x 2 )
1
x
tan
1 x2
x
1 x2
d x
dx a
1
C
x
d x
dx a
x 2
a
1
x
2
( x 1)(1) x (1)
d
x
76. dx
C
x 1
1
( x 1)2
1
1
C
csc2 x3 1
1
3
d ( xe x
dx
78.
1
a
C
tan
1))(5) sec2 (5 x 1)
1
x 1
x 1 C)
sec2 d
d
(3 x 5) 2
csc2 x3 1
( 1)( 1)( x 1) 2
d
1 C
75. dx
x 1
77.
C
1
c o s2
d
(7 x 2)3
(3 x 5) 2 (3)
3
C
3cot x3 1
1
1 sin 2
d
1
a2 1
ex
C)
x ex
(1) e x
ex
xe x
1
a2 x 2
x2
a2
1
1
x 2
a
a 1
1
( x 1) 2
( x 1)2
a 2 x2
C , then
1
x
dx
x
1 x2
1
2
x (1 x )
tan 1 x
x2
x(1 x 2 ) x3 x (tan
dx
2
2
x (1 x )
1
x)(1 x2 )
dx
tan 1 x dx,
x2
which verifies the formula
82. If y
dy
x(sin 1 x) 2
2 x 2 1 x 2 sin 1 x C , then
1
2 x (sin
(sin 1 x)2
1 x
x)
2
2x
2
1 x2
sin 1 x 2 1 x 2
1
dx
1 x2
(sin 1 x) 2 dx, which verifies the
formula
2
d x sin x C
83. (a) Wrong: dx
2
d ( x cos x C )
(b) Wrong: dx
(c) Right:
d (
dx
1 tan 2
2
1 sec 2
2
C
C
C
x2
2
cos x
cos x x sin x
x cos x sin x C )
3
84. (a) Wrong: dd sec3
(b) Right: dd
(c) Right: dd
2 x sin x
2
x sin x
x2
2
Copyright
x sin x
x sin x
cos x x sin x cos x
3sec 2 (sec tan
3
1 (2 tan ) sec2
2
1 (2sec )sec
2
cos x
)
x sin x
sec3 tan
tan sec 2
tan sec 2
tan
tan sec2
2014 Pearson Education, Inc.
Section 4.8 Antiderivatives
(2 x 1)3
3
d
85. (a) Wrong: dx
3(2 x 1)2 (2)
3
2
C
2(2 x 1)2
d ((2 x 1)3 C ) 3(2 x 1) (2)
(b) Wrong: dx
d ((2 x 1)3 C ) 6(2 x 1) 2
(c) Right: dx
d ( x2
86. (a) Wrong: dx
x C )1/2
d ( x2
(b) Wrong: dx
d 1
(c) Right: dx
3
d
87. Right: dx
x 3 3
x 2
d
88. Wrong: dx
sin( x 2 )
x
1 ( x2
2
x)1/2 C
3
2x 1
C
y
10 x
y 10 x
dy
1
x2
93. dx
x 1
y
dy
9 x2
94. dx
dy
dx
dy
97. ds
dt
98.
1
2
1 cos t
ds
dt
x
2
1
2
x 2 C. Then y (1)
1 x2
2
x2
2
C ; at x
y
x 1
1
x
or y
2
C
3.
C. Then y ( 1) 1
C
2 and y
x
2
3.
2
22 7(2) C
2
1 we have 1 10(0) 02
0 and y
x2
2
4
0 we have 0
C; at x
2 and y 1 we have 1
C
C
2 1
10
x 2 7 x 10
y
C
1
22
2
x2
2
y 10 x
C
1
1
2
C
1
2
3x3 2 x 2 5 x C ; at x
y
3 x1/3
y
1
3
1 and y
0 we have 0
3( 1)3 2( 1)2 5( 1) C
C
9
9 x1/3 C ; at x
y
1 and y
5 we have
5 9( 1)1/3 C
C
4
4
x 1/2
s
y
x1/2 C ; at x
t sin t C ; at t
cos t sin t
s
99. ddr
9 x1/3
1
2 x
96. dx
2
( x 2)4
y 3 x3 2 x 2 5 x 10
3 x 2/3
y
x
4x 5
C 10
95.
x
x 2 7 x C ; at x
x 2
x
15( x 3) 2
5
( x 2) ( x 2)2
2
x2
x
dy
( x 3) 2
2x 1
x cos( x 2 ) sin( x 2 )
dy
y
1)1/2 (2)
3 (2 x
6
2
y
2x 7
1
2 x 2 cos( x 2 ) sin( x 2 )
2x
dy
2x 1
2
dy
91. dx
3
2x 1
2 x2 x C
2x 1
2x
x cos( x 2 )(2 x ) sin( x 2 ) 1
x
90. Graph (b), because dx
3(2 x 1)2
2 x2 x
1)3/2 C
2 ( x 2) 1 ( x 3) 1
3 xx 32
( x 2) 2
C
89. Graph (b), because dx
92. dx
6(2 x 1)2
x) 1/2 (2 x 1)
d 1 (2 x
dx 3
C
(2 x 1) 2
x C ) 1/2 (2 x 1)
1 ( x2
2
313
s
4 and y
0 and s
4 we have 4
sin t cos t C ; at t
41/2 C
C
0 sin 0 C
C
0 we have 0
and s 1 we have 1 sin
2
4
cos
x1/2
y
s
2
t sin t 4
C
C
0
sin t cos t
sin
r
cos (
) C ; at r
0 and
Copyright
0 we have 0
cos ( 0) C
2014 Pearson Education, Inc.
C
1
r
cos (
) 1
314
Chapter 4 Applications of Derivatives
100. ddr
cos
101. dv
dt
1 sec t tan t
2
102. dv
dt
8t csc2 t
4t 2
v
103. dv
dt
3
d2y
dx 2
d2y
x2
x3
dy
dx
r
2
108. d 2s
dt
d3y
3
4
3x
1
2
C
2
4 2
1
r
cot 2
sin (
1 sec t
2
v
C
3sec 1 2 C
) 1
1
2
C
7
2
C=
y
3t 2
16
dy
dx
2
2x
C1; at ds
dt
6 x C1; at
2
dy
8 x C2 ; at dx
dt
0
1 (0)
2
C2
C3
sin t cos t
cos t
C2
C=1
4
4(0) C2
2
0 and x
y
d2y
dx
0 and x
2 and t
d
dt
C3
y
sin t 6
8 and x
2
1
2
0
C2
1
0 we have
2
t
y
C2
0
t
Copyright
y
8
2
2
r
t 1 2t 2 or r
1
t
ds
dt
C1
8
3t 2
16
s
t3
16
2t 2
C2 ; at
6(0) C1
1t
2
d
dt
C2
C3
0
5
2t C2 ; at ddt
C3 ; at
d2y
2 and t
y
dx 2
dy
dx
3
x
1
2
6x 8
3x 2 8 x
4 x2
and t
5
0 we have
0 we have
2
7 and t
sin t cos t 6t C2 ; at y
1 sin (0) cos (0) 6(0) C2
2
dr
dt
2
0
C1
03 4(0)2 C3
dt
cos t sin t C1 ; at y
2
3(0)2 8(0) C2
2
0 we have d 2
1t
2
C1
t 2
C1
t
16
s
0 we have 5
1
2
2
C2
0 we have
3
0 we have 0
2t
2 x C2 ; at y
(1) 2 C1
3(4) 2
16
4 we have 3
C2
5 and x
2
C1; at d 2
2(0) C2
2
C1
0 we have 1 02 03
1 and t 1 we have 1
3 and t
43
16
4 we have 4
dx
2
2(0) 3(0)2 C1
0 we have 4
0 we have C1
t 2 C1; at dr
dt
dr
dt
d2
dt 2
0
y
4 and x
x3 4 x C2 ; at y 1 and x
2 and x
x3 4 x 2 C3 ; at y
y
111. y (4)
C1; at dx
x2
y
dy
d2y
6
dy
dx
2
dy
2 x 3x2
C1 ; at dx
C2 0
ds
dt
4 and t
dt
1
2
we have 7
t 1 2t C2 ; at r 1 and t 1 we have 1 1 1 2(1) C2
3t
8
3
110. d 3
2
1
4x 1
2t 3
2
t3
dt
dx
7 and t
C
C
8 tan 1 t tan t C ; at t = 0 and v = 1 we have 1 8 tan 1 (0) tan(0) C
v
dy
dx
2(0) C2
2
107. d 2r
109.
cot t C ; at v
1 sec (0)
2
0 we have 1
3sec 1 t C ; at t = 2 and v = 0 we have 0
v
2 x 3x2
0
s
C ; at v 1 and t
sin( 0) C
2
, t>1
2 6x
y
0
4t 2
v
1
0 we have 1
8 tan 1 t tan t 1
dy
dx
dx 2
1 sec t
2
v
sec2 t
8
1 t2
v
106.
) C ; at r 1 and
3sec 1 t
v
105.
sin (
cot t 7
t t2 1
104. dv
dt
1
r
0 we have 7
1 and t
sin t cos t 6t
cos (0) sin (0) C1
0 we have
y
2014 Pearson Education, Inc.
cos t sin t 3t 2 C3 ; at
C1
6
Section 4.8 Antiderivatives
1 and t
y
C4
1
112. y (4)
y
C1
4
y
2
y
cos(0) 12 sin(2(0)) 23 (0)
y
3x1/2
3 x
d2y
3
C4
C4
4
y
3t 2
0
0 we have
y
cos x 2sin(2 x) 4 x C2 ; at
y
cos x 2sin(2 x) 4 x
C4 ; at y
cos x
2 x3
3
1 sin(2 x )
2
y
C
3(0)2 C1
0 and x = 0 we have 0
1
3 and x
2(9)3/2 C
C; at (9, 4) we have 4
3x 2 C1; at y
0 and x
C2
2 x3
3
1 sin(2 x)
2
x3 C2 ; at y = 1 and x = 0 we have C2
y
cos t sin t
0 we have 1 sin(0) cos(2(0)) 2(0)2 C3
1 and x
cos x
2 x3/2
y
dy
dx
6x
dx 2
y
sin x 4 cos(2 x ) 4
0 we have 1 cos(0) 2sin(2(0)) 4(0) C2
sin x cos(2 x ) 2 x
3
0
sin (0) cos (0) 03 C4
0 we have 0
sin x 4 cos (2 x) C1; at y
sin x cos(2 x) 2 x 2 C3 ; at y
y
114. (a)
y
sin(0) 4 cos(2(0)) C1
1 and x
0 and t
C3
sin t cos t t 3 1
y
cos x 8sin(2 x)
0
113. m
cos (0) sin (0) 3(0)2 C3
1
sin t cos t t 3 C4 ; at y
y
y
0 we have
315
C3
0
0 we have
4
50
C1
2 x 3/2 50
y
dy
dx
0
3x 2
x3 1
(b) One, because any other possible function would differ from x3 1 by a constant that must be zero because
of the initial conditions
dy
115. dx
1 43 x1/3
1 43 x1/3 dx
y
0.5 1 14/3 C
dy
116. dx
1
x 1
( 1)
2
2
y
C
x x 4/3
0.5
x2
2
( x 1)dx
( 1) C
1
2
C
x x 4/3 C ; at (1, 0.5) on the curve we have
1
2
x C ; at ( 1, 1) on the curve we have
y
x2
2
x 12
dy
117. dx sin x cos x
y
(sin x cos x)dx
cos x sin x C ; at (
1 = cos( ) sin( ) + C
C = 2 y = cos x sin x 2
dy
118. dx
1
2 x
1
2
sin x
x 1/2
we have 2 11/2 cos (1) C
119. (a)
ds
dt
9.8t 3
(iii) at s
s
4.9t
y
C
y
0
2
s(1) = ((4.9)(9)
x 1/2 sin x dx
x1/2 cos x C ; at (1, 2) on the curve
x cos x
9 + 5)
(4.9
3t 2; displacement = s(3)
s0 and t = 0 we have C
displacement
1
2
4.9t 2 3t C ; (i) at s = 5 and t = 0 we have C = 5
s
displacement = s(3)
C= 2
sin x
, 1) on the curve we have
s (3) s (1)
s0
s
4.9t
4.9t 2 3t 5;
3 + 5) = 33.2 units; (ii) at s = 2 and t = 0 we have
s(1) = ((4.9)(9)
2
s
9
2)
(4.9
3
2) = 33.2 units;
3t s0 ;
((4.9)(9) 9 s0 ) (4.9 3 s0 )
33.2 units
(b) True. Given an antiderivative f(t) of the velocity function, we know that the body s position function is
s = f(t) + C for some constant C. Therefore, the displacement from t = a to t = b is
(f(b) + C) (f(a) + C) = f(b) f(a). Thus we can find the displacement from any antiderivative f as the
numerical difference f(b) f(a) without knowing the exact values of C and s.
Copyright
2014 Pearson Education, Inc.
316
Chapter 4 Applications of Derivatives
120. a (t ) v (t ) 20
v(t) = 20t + C; at (0, 0) we have C = 0
v(60) = 20(60) = 1200 m/sec.
2
121. Step 1: d 2s
C1
0
ds
dt
k (0)
2
2
44 44
k
2
123. (a) v
(b) s
0
ds
dt
dt
ds
dt
a
when t = 0
kt 2
2
968
k
1936
k
242
(88) 2
2k
kt 2
2
44t. Then ds
dt
4(1)3/2 C
4
45
968
45
k
s0
a (0)2
2
C
16
kt 44
0
44
k
t
and
21.5 ft 2 .
sec
v 10t 3/2 6t1/2
0
C
0
4t 5/2
s
0 and t = 0 we have C1
2.6t 2
s
at C ; ds
dt
a dt
4. Then s
v0 when t = 0
v0 (0) C1
s0 when t = 0. Thus ds
dt
Thus s
( gt v0 )dt
Thus s
1
2
(c)
k
0
C1
s0
0
0
C
v0
at 2
2
s
4t 3/2
gt 2
f ( x) dx 1
f ( x) dx
gt C1 ; ds
(0)
dt
g dt
1
2
gt 2
v0t C2 ; s (0)
ds
dt
0
2.6t 2
ds
dt
s
2.6t 2 C2 ; at s = 4
4
2.6
1.24 sec, since t > 0
5.2t
4
t
at v0
s
at 2
2
v0 t C1; s
v0
v0
s0
1 ( g )(0) 2
2
(b)
g ( x) dx
g with Initial Conditions: ds
dt
( g )(0) C1
v0 (0) C2
C1
v0
C2
s0
v0t s0 .
x C1
x C
1
x
C1
x C
(e)
[ f ( x) g ( x)]dx
1
x
( x 2) C1
(f)
[ f ( x) g ( x )]dx
1
x
( x 2) C1
Copyright
(d)
x
g ( x) dx
x 2 C1
x C
( x 2) C1
x C
x
s0
v0 t s0
dt
127 (a)
88t
44t C1; at s = 0 when t = 0 we have
2
126. The appropriate initial value problem is: Differential Equation: d 2s
s
kt 2
2
4t 5/2 4t 3/2 C ;
5.2t C1; at ds
dt
and t = 0 we have C2
2
125. d 2s
(88)2
k
s
s
0
45
4(1)5/2
0
5.2
dt
kt 44
(10t 3/2 6t1/2 )dt
v dt
2
124. d 2s
s
44 when t = 0 we have
4 10(1)3/2 6(1)1/2 C
4
s (1)
0
(15t1/2 3t 1/2 )dt 10t 3/2 6t1/2 C ;
a dt
ds (1)
dt
(88)2
2k
242
C1
88t C2 ; at s = 0 and t = 0 we have C2
88
k
t
ds
dt
C = 44
44 2
k
2
k t2
kt C ; at ds
dt
k dt
44(0) C1
k
s 44
k
88 and t = 0 we have
s
88 88
k
2
44 = k(0) + C
0
kt 88
88 2
k
k
k
dt
kt 88
0
Step 3: 242
kt C1; at ds
dt
ds
dt
88
Step 2: ds
dt
2
122. d 2s
ds
dt
k
dt
v(t) = 20t. When t = 60, then
x C
2014 Pearson Education, Inc.
x C
ds
dt
v0 and
gt v0 .
Section 4.8 Antiderivatives
317
128. Yes. If F ( x) and G ( x) both solve the initial value problem on an interval I then they both have the same
first derivative. Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that
F ( x) G ( x) C for all x. In particular, F ( x0 ) G ( x0 ) C , so C F ( x0 ) G ( x0 ) 0. Hence F ( x) G ( x)
for all x.
129 132. Example CAS commands:
Maple:
with(student):
f : x - cos(x)^2 sin(x);
ic : [x Pi,y 1];
F : unapply( int( f(x), x ) C, x );
eq : eval( y F(x), ic );
solnC : solve( eq, {C} );
Y : unapply( eval( F(x), solnC ), x );
DEplot( diff(y(x),x) f(x), y(x), x 0..2*Pi, [[y(Pi) 1]],
color black, linecolor black, stepsize 0.05, title "Section 4.8 #129" );
Mathematica: (functions and values may vary)
The following commands use the definite integral and the Fundamental Theorem of calculus to construct the
solution of the initial value problems for Exercises 129-132.
Clear x, y, yprime
yprime[x_] Cos[x]2 Sin[x];
initxvalue
; inityvalue 1;
_
y[x ] Integrate[yprime[t], {t, initxvalue, x}] inityvalue
If the solution satisfies the differential equation and initial condition, the following yield True
yprime[x] D[y[x], x]//Simplify
y[initxvalue] inityvalue
Since exercise 132 is a second order differential equation, two integrations will be required.
Clear[x, y, yprime]
y2prime[x_] 3 Exp[x/2] 1;
initxval 0; inityval 4; inityprimeval
1;
_
yprime[x ] Integrate[y2prime[t],{t, initxval, x}] inityprimeval
y[x_]
Integrate[yprime[t], {t, initxval, x}] inityval
Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative
(blue).
y2prime[x] D[y[x], {x, 2}]//Simplify
y[initxval] inityval
yprime[initxval] inityprimeval
Plot[{y[x], yprime[x]}, {x, initxval 3, initxval 3}, PlotStyle {RGBColor[1,0,0], RGBColor[0,0,1]}]
Copyright
2014 Pearson Education, Inc.
318
Chapter 4 Applications of Derivatives
CHAPTER 4
PRACTICE EXERCISES
1. No, since f ( x)
x3 2 x tan x
2. No, since g ( x)
csc x 2cot x
3x 2
f ( x)
2 sec2 x
0
f ( x) is always increasing on its domain
csc x cot x 2 csc 2 x
g ( x)
cos x
sin 2 x
g ( x) is always decreasing on its domain
3. No absolute minimum because lim (7 x)(11 3x)1/3
(11 3 x ) (7 x )
4(1 x )
(11 3 x )2/3
(11 3 x )2/3
x 1 and x
11
3
2)
0
(11 3 x)1/3 (7 x)(11 3x ) 2/3
. Next f ( x)
x
1 (cos x
sin 2 x
2
sin 2 x
are critical points. Since f
0 if x 1 and f
0
if x 1, f (1) 16 is the absolute maximum.
4. f ( x)
ax b
x2 1
a ( x 2 1) 2 x ( ax b )
f ( x)
(x
2
1)
f ( x)
( x 2 1)2
so that f
3a b
|
1
g ( x)
ex
x
ex 1
g ( x)
g
1 (9a
64
0
6b a)
8. Solving both equations yields a
|
|
|
1/3
1
3
positive to negative so there is a local maximum at x
5.
; f (3)
( x 2 1)2
3a b
8
require also that f (3) 1. Thus 1
2(3 x 1)( x 3)
( ax 2 2bx a )
2
|
0
5a 3b
6 and b
. Thus f changes sign at x
0. We
10. Now,
3 from
3 which has a value f (3) 1.
the graph is decreasing on (
, 0), increasing on (0, );
0
an absolute minimum value is 1 at x = 0; x = 0 is the only critical point of g; there is no absolute maximum
value
6.
f ( x)
2e x
1 x2
f ( x)
(1 x 2 ) 2e x 2e x 2 x
2 2
(1 x )
2e x (1 x )2
f
(1 x 2 )2
|
the graph is increasing on (
, );
1
x = 1 is the only critical point of f; there are no absolute maximum values or absolute minimum values.
7. f(x) = x
2 ln x on 1
x
f ( x) 1 2x
3
f ( x)
increasing on (2, 3); an absolute minimum value is 2
8.
f ( x)
4
x
ln x 2 on 1
x
4
f ( x)
4
x2
2
x
|
|
|
1
2
3
the graph is decreasing on (1, 2),
2 ln 2 at x = 2; an absolute maximum value is 1 at x = 1.
2x 4
x2
f
|
|
|
1
2
4
the graph is decreasing on
(1, 2), increasing on (2, 4); an absolute minimum value is 2 + ln 4 at x = 2; an absolute maximum value is 4 at
x = 1.
9. Yes, because at each point of [0, 1) except x 0, the function s value is a local minimum value as well as a
local maximum value. At x 0 the function s value, 0, is not a local minimum value because each open
interval around x 0 on the x-axis contains points to the left of 0 where f equals 1.
10. (a) The first derivative of the function f ( x) x3 is zero at x 0 even though f has no local extreme value at
x 0.
(b) Theorem 2 says only that if f is differentiable and f has a local extreme at x c then f (c) 0. It does not
f has a local extreme at x c.
assert the (false) reverse implication f (c) 0
11. No, because the interval 0 x 1 fails to be closed. The Extreme Value Theorem says that if the function is
continuous throughout a finite closed interval a x b then the existence of absolute extrema is guaranteed on
that interval.
Copyright
2014 Pearson Education, Inc.
Chapter 4 Practice Exercises
319
12. The absolute maximum is | 1| 1 and the absolute minimum is |0| 0. This is not inconsistent with the
Extreme Value Theorem for continuous functions, which says a continuous function on a closed interval attains
its extreme values on that interval. The theorem says nothing about the behavior of a continuous function on an
interval which is half open and half closed, such as [ 1, 1), so there is nothing to contradict.
13. (a) There appear to be local minima at x
1.75
and 1.8. Points of inflection are indicated at
1.
approximately x 0 and x
(b) f ( x)
x 7 3 x5 5 x 4 15 x 2
indicates a local maximum at x
x 2 ( x 2 3)( x3 5). The pattern y
3
|
|
5 and local minima at x
3
0
3
|
|
5
3
3.
(c)
14. (a) The graph does not indicate any local
extremum. Points of inflection are indicated
3 and x 1.
at approximately x
4
(b) f ( x)
x7
2 x 4 5 103
local maximum at x
x
7
x 3 ( x3 2)( x7 5). The pattern f
5 and a local minimum at x
3
2.
(c)
Copyright
)(
0
2014 Pearson Education, Inc.
|
7
5
3
|
2
indicates a
320
Chapter 4 Applications of Derivatives
15. (a) g (t ) sin 2 t 3t
g (t ) 2sin t cos t 3 sin(2t ) 3 g 0 g (t ) is always falling and hence must
decrease on every interval in its domain.
(b) One, since sin 2 t 3t 5 0 and sin 2 t 3t 5 have the same solutions: f (t ) sin 2 t 3t 5 has the same
derivative as g (t ) in part (a) and is always decreasing with f ( 3) 0 and f (0) 0. The Intermediate Value
Theorem guarantees the continuous function f has a root in [ 3, 0].
dy
16. (a) y tan
sec 2
0
y tan is always rising on its domain
y tan
d
interval in its domain
is not in the tangent s domain because tan is undefined at
(b) The interval 4 ,
need not increase on this interval.
increases on every
2
. Thus the tangent
f ( x) 4 x3 4 x. Since f (0)
2 0, f (1) 1 0 and f ( x) 0 for 0 x 1, we
17. (a) f ( x) x 4 2 x 2 2
may conclude from the Intermediate Value Theorem that f ( x) has exactly one solution when 0 x 1.
(b) x 2
18. (a) y
2
4 8
x
x 1
x2
0
2
0
x
.7320508076
0, for all x in the domain of xx 1
1
( x 1) 2
y
3 1 and x
domain.
(b) y x3 2 x
y 3x 2 2 0 for all x
have a local maximum or minimum
x is increasing in every interval in its
x 1
y
the graph of y
.8555996772
x3 2 x is always increasing and can never
19. Let V (t ) represent the volume of the water in the reservoir at time t, in minutes, let V (0) a0 be the initial amount
and V (1440) a0 (1400)(43,560)(7.58) gallons be the amount of water contained in the reservoir after the rain,
where 24 hr 1440 min. Assume that V (t ) is continuous on [0, 1440] and differentiable on (0, 1440). The Mean
V (1400) V (0)
1440 0
Value Theorem says that for some t0 in (0, 1440) we have V (t0 )
456,160,320 gal
1440 min
a0 (1440)(43,560)(7.48) a0
1440
316, 778 gal/min. Therefore at t0 the reservoir s volume was increasing at a rate in excess of
225,000 gal/min.
20. Yes, all differentiable functions g ( x) having 3 as a derivative differ by only a constant. Consequently, the
d (3 x ). Thus g ( x ) 3 x K , the same form as F ( x).
difference 3x g ( x) is a constant K because g ( x ) 3 dx
21. No, xx 1 1 x 11
( x 1) x (1)
d
x
dx x 1
22. f ( x)
x
x 1
( x 1)2
g ( x)
2x
( x 2 1)2
differs from x 11 by the constant 1. Both functions have the same derivative
1
( x 1)2
d
1
dx x 1
f ( x) g ( x)
.
C for some constant C
23. The global minimum value of 12 occurs at x
the graphs differ by a vertical shift.
2.
24. (a) The function is increasing on the intervals [ 3, 2] and [1, 2].
(b) The function is decreasing on the intervals [ 2, 0) and (0, 1].
(c) The local maximum values occur only at x
2, and at x 2; local minimum values occur at x
at x 1 provided f is continuous at x 0.
25. (a) t
0, 6, 12
(b)
t
3, 9
(c)
6
t 12
(d)
0
26. (a) t
4
(b)
at no time
(c)
0
t
(d)
4 t
Copyright
4
2014 Pearson Education, Inc.
t
6, 12
8
3 and
t 14
Chapter 4 Practice Exercises
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
Copyright
2014 Pearson Education, Inc.
321
322
Chapter 4 Applications of Derivatives
37.
38.
39.
40.
41.
42.
43. (a) y
16 x 2
y
a local maximum at x
concave up on (
|
|
4
4
the curve is rising on ( 4, 4), falling on (
4 and a local minimum at x
, 0), concave down on (0, )
4; y
2x
y
a point of inflection at x
, 4) and (4, )
the curve is
|
0
0
(b)
44. (a) y
x2
x 6
falling on ( 2, 3)
( x 3)( x 2)
y
local maximum at x
|
|
2
3
the curve is rising on (
2 and a local minimum at x 3; y
2x 1
, 2) and (3,
y
|
1/2
(b)
concave up on 12 ,
, concave down on
Copyright
, 12
a point of inflection at x
2014 Pearson Education, Inc.
1
2
),
Chapter 4 Practice Exercises
45. (a) y
6 x3 6 x 2 12 x
6 x( x 1)( x 2)
and (2, ), falling on (
x
2; y
18 x
2
y
, 1) and (0, 2)
12 x 12
6 (3x
2
the graph is rising on ( 1, 0)
|
|
|
1
0
2
a local maximum at x
2 x 2)
6 x
1
7
1
x
3
0, local minima at x
7
y
3
1 and
|
1
|
7
1
3
, 1 3 7 and 1 3 7 ,
the curve is concave up on
inflection at x
(b)
x 2 (6 4 x)
46 . (a) y
3,
2
1
323
7
3
, concave down on 1 3 7 , 1 3 7
points of
7
3
6 x2
4 x3
y
a local maximum at x
up on (0, 1), concave down on (
3;y
2
the curve is rising on
|
|
0
3/2
12 x 12 x 2 12 x(1 x)
, 0) and (1, )
y
points of inflection at x
, 32 , falling on
|
|
0
1
concave
0 and x 1
(b)
x4
47. (a) y
2 x2
x2 ( x2
2)
y
|
2
2,
y
, falling on
4x
3
4x
2, 2
|
2
the curve is rising on
a local maximum at x
4 x( x 1)( x 1)
concave down on (
|
0
y
, 1) and (0, 1)
2 and a local minimum at x
|
|
|
1
0
1
,
2 and
2;
concave up on ( 1, 0) and (1, ),
points of inflection at x
0 and x
1
(b)
48. (a) y
4 x2
x4
x 2 (4 x 2 )
y
falling on ( , 2) and (2, )
4 x (2 x 2 )
y
|
2
down on
2, 0 and
2,
|
|
|
2
0
2
the curve is rising on ( 2, 0) and (0, 2),
a local maximum at x 2, a local minimum at x
2; y 8 x 4 x3
|
|
concave up on
,
2 and 0, 2 , concave
0
2
points of inflection at x
Copyright
0 and x
2014 Pearson Education, Inc.
2
324
Chapter 4 Applications of Derivatives
(b)
49. The values of the first derivative indicate that the curve is rising on (0, ) and falling on ( , 0). The slope of
the curve approaches
as x 0 , and approaches as x 0 and x 1. The curve should therefore have a
cusp and local minimum at x 0, and a vertical tangent at x 1.
50. The values of the first derivative indicate that the curve is rising on 0, 12 and (1, ), and falling on ( , 0) and
1 , 1 . The derivative changes from positive to negative at x 1 , indicating a local maximum there. The slope
2
2
as x
of the curve approaches
cusps and local minima at both x
0 and x 1 , and approaches
0 and x 1.
as x
0 and as x
1 , indicating
51. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches
as x 0 and as x 1, indicating vertical tangents at both x 0 and x 1.
Copyright
2014 Pearson Education, Inc.
Chapter 4 Practice Exercises
52. The graph of the first derivative indicates that the curve is rising on 0, 17 16 33 and 17 16 33 ,
325
, falling on
, 0) and 17 16 33 , 17 16 33
a local maximum at x 17 16 33 , a local minimum at x 17 16 33 . The derivative
approaches
as x 0 and x 1, and approaches as x 0 , indicating a cusp and local minimum at
x 0 and a vertical tangent at x 1.
(
53. y
x 1
x 3
55. y
x2 1
x
x
57. y
x3 2
2x
x2
2
1 x4 3
1
x
1
x
Copyright
54. y
2x
x 5
56. y
x2 x 1
x
58. y
x4 1
x2
2
10
x 5
x 1 1x
x2
2014 Pearson Education, Inc.
1
x2
326
Chapter 4 Applications of Derivatives
x2 4
x2 3
59. y
61.
2
lim x x3 1x 4
x 1
63.
lim tanx x
65.
66.
67.
68.
69.
70.
71.
1
x2 3
1
lim 2 x 3
x 1 1
tan
2
lim sin 2x
x
sin( mx )
m cos( mx )
lim
x 0 sin( nx )
x
/2
lim
x
x sec x
0
lim cosxx
0
0 x
x
x2
lim
x
x
lim
x
x2
x3
x
x
x
sin x
lim cos
x
0
x
3sin(3 x )
7 sin(7 x )
lim
/2
0
1
x
x
1
1 1
1
2
2
0 21
1
3
7
x2
x
0
lim (1 x 2 )
x
0
lim
x
0
x2
x 1
2
1
1
x
lim
x
1
x
2
lim
x2 x
x2
x3
1
0
x
lim 14
0x
x
x2 x 1
x
2
x 1
1
x2 x
x2 x
lim
x
2x 1
x2 x 1
x2 x
x 2 for x > 0 so this is equivalent to
x2 x 1
x2
2
x
2 2
2
2 2
0 2 x (2 sec ( x ) tan( x ) 2 x ) 2 sec ( x )
x
lim (1 x 2 ) 14
x
0
2
lim 1seccosxx
0
0
2x 1
x
x
72.
x
x 1
Notice that x
lim
0
1
2
lim 1 x4
1
x2
/2
cos x
lim 1 sin
x
0
lim 14
cos(3 x )
cos(7 x )
lim
x
lim (csc x cot x)
x
x
a
b
2 cos(2 x )
lim
2 x sec 2 ( x 2 )
x
1
1
m
n
lim
x 0 n cos( nx )
lim sec(7 x) cos(3x )
sin(2 x )
lim
0 2 x sec ( x )
x 1 bx
lim tan x
x 0 x sin x
64.
lim 2 sin x2cos2x
0 tan( x )
x
a
lim axb
1
x 1x
4
x2 4
1
a
lim xb 1
62.
5
0
x
x2
x2 4
60. y
1
x
2
1
x2
1
x3 ( x 2 1) x3 ( x 2 1)
(x
2
1)( x
2
1
1
x
1)
1
1
3
lim 24x
x
x
1
x
73. The limit leads to the indeterminate form 00 : lim 10 x 1
x
Copyright
2
lim 6 x 3
4x
x
lim
x
lim 12 x2
x
(ln10)10 x
1
12 x
ln10
2014 Pearson Education, Inc.
12
lim 24
x
x
lim 21x
x
0
Chapter 4 Practice Exercises
74. The limit leads to the indeterminate form 00 : lim 3 1
lim
sin x
75. The limit leads to the indeterminate form 00 : lim 2
e
x
x
1
1
sin x
76. The limit leads to the indeterminate form 00 : lim 2
e
x
x
lim
2
lim
2
e
1
1
(ln 2)( cos x )
ex
sin x
e
lim
4e x
e xe x
t ln(1 2t )
79. The limit leads to the indeterminate form 00 : lim
2
t
lim 5cosx x
1
1
lim
5
e
x
4
x
t
ln 2
ex
x
x
ln 2
(ln 2)( cos x )
x
lim 5sin
x
x 1
xe
t
sin x
x
x
78. The limit leads to the indeterminate form 00 : lim 4 4xe
x
ln 3
x
77. The limit leads to the indeterminate form 00 : lim 5 x5cos x
x
(ln 3)3
1
327
2
1 2t
2t
80. The limit leads to the indeterminate form 00 :
lim
x
sin 2 ( x)
4e
x 4
lim
3 x
x
2 (sin x )(cos x )
e
4
x 4
lim
1
x
4
sin(2 x )
e
x 4
1
lim
x
t
81. The limit leads to the indeterminate form 00 : lim et
t
82. The limit leads to the indeterminate form
83.
84.
lim 1 bx
x
lim 1 2x
x
kx
x
7
x2
x / b bk
1 x1/b
lim
2
2
1
t
cos(2 x )
ex
4
2 2
4
t
lim e t 1
t
: lim e 1/ y ln y
y
t
t
lim e1
ln y
lim
y
e
y 1
1
y
lim
y
e
1
lim
y 1( y 2 )
y
y
ey
1
0
ebk
1 0 0 1
85. (a) Maximize f ( x)
1
2
f ( x)
x
x 1/2
36 x
1 (36
2
x1/2 (36 x)1/2 where 0
x ) 1/2 ( 1)
36 x x
2 x 36 x
critical points at 0, 18 and 36; g (0)
36
derivative fails to exist at 0 and 36; f (0)
f (36) 6 the numbers are 0 and 36
x
36 x x1/2 (36 x)1/2 where 0
(b) Maximize g ( x)
36 x x
2 x 36 x
x
x
6, g (18)
36
2 18
g ( x)
1
2
x 1/2
6 2 and g (36)
1 (36
2
6
6, and
x) 1/2 ( 1)
the numbers
are 18 and 18
86. (a) Maximize f ( x)
x
are
20
3
0 and x
and
x (20 x)
20
3
20x1/2
x3/2 where 0
x
20
f ( x ) 10 x 1/2
are critical points; f (0) f (20) 0 and f 20
3
40 .
3
Copyright
2014 Pearson Education, Inc.
20
3
20 20
3
3 x1/2
2
40 20
3 3
20 3 x
2 x
0
the numbers
328
Chapter 4 Applications of Derivatives
(b) Maximize g ( x)
79 . The
4
79 and 1 .
4
4
x
87. A( x)
x
critical points are x
x 2 ) for 0
1 (2 x )(27
2
x (20 x )1/2 where 0
20 x
x
79
4
and x
x
20
20. Since g 79
4
g ( x)
81
4
2 20 x 1
2 20 x
and g (20)
0
20 x
1
2
20, the numbers must be
27
A ( x) 3(3 x)(3 x) and A ( x)
6 x. The
critical points are 3 and 3, but 3 is not in the
domain. Since A (3)
18 0 and A 27 0, the
maximum occurs at x
is A(3) 54 sq units.
x2h
88. The volume is V
x2
area is S ( x)
S ( x)
the largest area
3
32
x2
4 x 322
2( x 4)( x 2
32 . The surface
x2
128 , where x 0
x
h
x
4 x 16)
the critical points are 0
x2
and 4, but 0 is not in the domain. Now
S (4) 2 256
0 at x 4 there is a minimum.
3
4
The dimensions 4 ft by 4 ft by 2 ft minimize the
surface area.
2
89. From the diagram we have h2
r2
V
r2
12 h 2 . The volume of the
4
12 h 2 h
r2h
(12h
4
4
2 3. Then V (h) 34 (2
3
2
cylinder is
h3 ), where
0 h
h)(2 h) the
critical points are 2 and 2, but 2 is not in the
domain. At h 2 there is a maximum since
V (2)
3
0. The dimensions of the largest
cylinder are radius
2 and height 2.
90. From the diagram we have x radius and y height
12 2x and V ( x) 13 x 2 (12 2 x), where 0 x 6
V ( x) 2 x(4 x) and V (4)
8 . The critical
x 4 gives the
points are 0 and 4; V (0) V (6) 0
maximum. Thus the values of r 4 and h 4 yield
the largest volume for the smaller cone.
91. The profit P
P ( x)
2p
(5 x) 2
2 px
py
x , where p is the profit on grade B tires and 0 x 4. Thus
p 405 10
x
the critical points are 5
5 , 5, and 5
5 , but only 5
5 is in the
2 px
( x 2 10 x 20)
domain. Now P ( x)
0 for 0
maximum. Also P (0)
x
5
5 and P ( x)
8 p, P 5
5
4p 5
5
absolute maximum. The maximum occurs when x
i.e., x
276 tires and y
0 for 5
5
x
11 p, and P (4) 8 p
5
5 and y
2 5
553 tires.
Copyright
2014 Pearson Education, Inc.
4
at x
at x
5
5
5 there is a local
5 there is an
5 , the units are hundreds of tires,
Chapter 4 Practice Exercises
92. (a) The distance between the particles is | f (t )| where f (t )
f (t )
sin t sin t
Alternatively, f (t )
sin t
8
cos 8
4
. Solving f (t )
cos t cos t
0 graphically, we obtain t
4
. Then,
1.178, t
0 may be solved analytically as follows. f (t )
sin t
cos t
sin 8
sin 8
8
critical points occur when cos t
sin t
0, or t
8
cos 8
8
3
8
cos t
8
329
4.320, and so on.
8
8
sin t
2sin 8 cos t
k . At each of these values, f (t )
units, so the maximum distance between the particles is 0.765 units.
(b) Solving cos t cos t 4 graphically, we obtain t 2.749, t 5.890, and so on.
8
cos 38
8
8
so the
0.765
Alternatively, this problem can be solved analytically as follows.
cos t cos t 4
cos t
cos t
cos 8
8
8
cos t
8
sin t
sin 8 cos t
8
2sin t 8 sin 8 0
sin t
The particles collide when t
7
8
0; t
8
8
8
cos 8
8
7
8
sin t
8
sin 8
k
2.749. (Plus multiples of
if they keep going.)
93. The dimensions will be x in. by 10 2x in. by 16 2x in., so V ( x ) x(10 2 x)(16 2 x ) 4 x3 52 x 2 160 x for
0 x 5. Then V ( x) 12 x 2 104 x 160 4( x 2)(3 x 20), so the critical point in the correct domain is x 2.
This critical point corresponds to the maximum possible volume because V ( x ) 0 for 0 x 2 and V ( x) 0
for 2 x 5. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its
volume is 144 in.3
Graphical support:
Copyright
2014 Pearson Education, Inc.
330
Chapter 4 Applications of Derivatives
94. The length of the ladder is d1 d 2 8sec
6 csc .
We wish to maximize I ( ) 8sec
6 csc
I ( ) 8sec tan
8sin
3
6 cos
6csc cot . Then I ( ) 0
3
0
4 4 3 36 and d 2
d1
3
6
2
tan
3
36 4 3 36
length of the ladder is about 4 3 36
3/2
4 3 36
4
g (2)
2
Value Theorem. Then g ( x)
forth to x5
3 3x
x 3 75
g (3)
and so forth to x5
14
xn 1
x4
4
5 x2
2
t2
2
t dt
8t 4
4
t3
6
98.
8t 3
99.
3 t
4
t2
dt
3t1/2
100.
1
3
t4
dt
1 t 1/2
2
4 x3 3x 2
xn
1
( r 5)
t2
2
C
4t 2 dt
3t 4 dt
2t 4
t3
6
1 .
( r 5)2
C
3
2
2
0
; x0
g ( x)
xn4 xn3 75
xn
3t 3/ 2
t2
2
4t
3
2
1
4 xn3 3 xn2
2
x1
2.22
x2
2.196215, and so
0 in the interval [3, 4] by the Intermediate
; x0
3
x1
3.259259
1 t1/ 2
2 1
dr
( r 5) 2
6
C
r
2
3
1
( r 5)
1
( r 5)
r
2
C
t
C
1
t3
C
C. Differentiate the solution to check:
3
2
C. Differentiate the solution to check:
2
3
3
4
t
C.
6 dr
. Thus
2t 3/2
3t 3
( 3)
2
Thus
C
C
1
r
r
3 3 xn2
0 in the interval [2, 3] by the Intermediate
7x C
102. Our trial solution based on the chain rule is
d
dr
g ( x)
3 xn xn3 4
xn 1
101. Our trial solution based on the chain rule is
d
dr
0
3.22857729.
( x3 5 x 7) dx
2 t
2
21 0 and g (4) 117
Value Theorem. Then g ( x)
97.
0 and g (3)
2.195823345.
x4
96. g ( x)
4 3 36
19.7 ft.
3 x x3
95. g ( x)
the
r
2
2
C.
103. Our trial solution based on the chain rule is ( 2 1)3/2 C. Differentiate the solution to check:
d
d
( 2 1)3/2 C
3
2
1. Thus 3
Copyright
2
1d
( 2 1)3/2 C.
2014 Pearson Education, Inc.
x2
3.229050,
Chapter 4 Practice Exercises
104. Our trial solution based on the chain rule is
d
d
2
7
C
2
7
. Thus
2
7
2
7
d
C. Differentiate the solution to check:
2
7
C.
105. Our trial solution based on the chain rule is 13 (1 x 4 )3/4 C. Differentiate the solution to check:
x 4 )3/4 C
d 1 (1
dx 3
x3 (1 x 4 ) 1/4 . Thus
x3 (1 x 4 ) 1/4 dx
5 (2 x)8/5
8
3/5
106. Our trial solution based on the chain rule is
5 (2
8
d
dx
x)8/5 C
(2 x )3/5 . Thus
(2 x)
1 (1
3
x 4 )3/4
C.
C. Differentiate the solution to check:
5 (2
8
dx
x)8/5 C.
s C. Differentiate the solution to check:
107. Our trial solution based on the chain rule is 10 tan 10
d
ds
s . Thus sec 2 s ds 10 tan s C .
sec2 10
10
10
s C
10 tan 10
1
108. Our trial solution based on the chain rule is
d
ds
1
csc2 s. Thus
cot s C
csc2 s ds
1
2
109. Our trial solution based on the chain rule is
d
d
1
2
csc 2
C
cot s C. Differentiate the solution to check:
csc 2 cot 2 . Thus
1
cot s C.
csc 2
C . Differentiate the solution to check:
1
2
csc 2 cot 2 d
csc 2
C.
110. Our trial solution based on the chain rule is 3sec 3 C . Differentiate the solution to check:
d
d
3sec 3 C
sec 3 tan 3 . Thus sec 3 tan 3 cot 2 d
3sec 3 C .
111. Our trial solution based on the chain rule is 2x sin 2x C. Differentiate the solution to check:
d x
dx 2
sin 2x C
sin 2 4x . Thus sin 2 4x dx
1 cos x
2
2
1
2
x
2
sin 2x C.
112. Our trial solution based on the chain rule is 2x
d x
dx 2
113.
3
x
114.
5
x2
115.
1 et
2
116.
(5s
1 sin x
2
x dx
2
x2 1
1 cos x
2
x2
2
3ln x
1 et
2
5s
ln 5
s6
6
the solution to check:
C.
C
5x 2
dx
e t dt
s 5 ) ds
1
2
C
1 sin x C. Differentiate
2
cos 2 2x . Thus cos 2 2x dx 2x 12 sin x
e
2
x2 1
t
1
C
dx
5 x 1 2 tan 1 x C
1 et
2
e t
C
C
117.
Copyright
( 1
)d
2
2
2014 Pearson Education, Inc.
C
331
332
Chapter 4 Applications of Derivatives
r
(2
118.
2 r
ln 2
)dr
d
120.
C
1d
4
d
2
16
y
1
x
( x2
dx
1
3
y 1 when x 1
123. dr
dt
4(1)
4(1)
8(1) C1
3/2
sin t dt
cos t 1
r
xy
for x
128. A
dA
dx
1)
x2
dx
C
1
C
C
x 1x C; y
( x2
2 x 2 ) dx
1
3
C
6t1/2 8
0
sin t C ; r
y
1 when x 1
1 11 C
1
C
1
x3
3
x3
3
x3
3
2x x 1 C
2 x 1x
2 x 1x C ;
1
3
C1
6t1/2 C ; dr
8 when t 1 10(1)3/2 6(1)1/2 C 8
dt
(10t 3/2 6t1/2 8) dt 4t 5/2 4t 3/2 8t C ; r 0 when t 1
r
0. Therefore, r
0 when t
0
cos t C1; r
0 when t
0
(cos t 1) dt
sin t t
4t 5/2
sin 0 C
1 C1
C2 ; r
0
1 when t
4t 3/2 8t
0
2
0. Thus, d 2r
C
sin t
dt
C1
1. Then
0
0 0 C2
1
C2
1. Therefore,
cos 1 x differ by the constant 2
cos 1 x C and y
126. Yes, the derivatives of y
y
3 sec 1 x
2
dx
sin t t 1
125. Yes, sin 1 x and
127. A
1
x x2 1
(15t1/2 3t 1/2 ) dt 10t 3/2
10t 3/2
dr
dt
x x 1 C
2 11
dt
cos t dt
dt
3
2
dx
16
2
8. Thus dr
dt
5/2
2
124. d 2r
r
3
t
15 t
C
3
2 x x2 1
1
2
1
x
x
1
16
sin 1 4
2
(1 x 2 ) dx
dx
x
122. y
2
16 1
x2 1
x2
121. y
dr
dt
119.
e
2
xe x
1
2
1/2
xy
and dA
dx
1
e
2
e x
dA
dx
2
( x)( 2 x)e x
0 for 0
x
cos 1 ( x) C are both
2
e x (1 2 x 2 ). Solving dA
dx
1
2
absolute maximum of 1 e 1/2
ln x
x2
ln x
1
1 x2
1 2x2
0
1
2e
2
.
at x
0
1
2
x
1 ; dA
dx
2
units long by
units high.
x ln2x
x
0 for x < e
ln x
x
dA
dx
absolute
1 ln x . Solving dA 0
1
dx
x2
maximum of lnee 1e at x = e units long
1
x2
Copyright
and y
2014 Pearson Education, Inc.
0
x
1
e2
e; dA
dx
units high.
0 for x > e and
0
Chapter 4 Practice Exercises
129 . y = x ln 2x
y
x
y
0
x
1
2
1;
2
x
x
2
2x
ln(2 x ) 1 ln 2 x; solving
1;
2
x
y
1 and
2
1 at
2
0 for x
relative minimum of
1
f 2e
1
e
1
2
minimum is
and f 2e
1
2
at x
0
y
0 for
absolute
and the absolute
e
2
maximum is 0 at x
130. y = 10x(2
333
ln x)
y
10(2 ln x) 10 x 1x
20 10 ln x 10
10(1 ln x);
solving y 0
x = e; y 0 for x > e and y
for x < e relative maximum at x = e of 10e; y
on (0, e2 ] and y (e2 ) 10e 2 (2 2 ln 3)
e and the absolute
4
e x / x 1 for all x in (
2 x3
x4 1
x4 1 1 x
f ( x)
x
4
1
2
0
2
absolute minimum is 0 at x
maximum is 10e at x = e
131. f ( x)
0
0
, );
4
ex/ x 1
1 x4
x
4
1
3
(1 x 2 )(1 x 2 ) x / x 4 1
e
( x 4 1)3/ 2
4
ex/ x 1
4
; lim e x / x 1
are critical points. Consider the behavior of f as x
x
x
the following table (14 digit precision, 12 digits displayed):
x
x / x4 1
0
4
e x/ x 1
1
0.0000 10000 0000 00000
0.9999 9000 0050
10000
0.0001 0000 0000 000
0.9999 0000 5000
1000
0.0010 0000 0000 00
0.9990 0049 9833
100
0.0099 9999 9950 00
0.9900 4983 3799
10
0.0999 9500 0375 0
0.9048 4194 1895
100000
0
10
0
1
0.0999 9500 0375 0
1.1051 6539 265
Copyright
2014 Pearson Education, Inc.
0
1 x2
4
lim e x / x 1
0
x= 1
1 as suggested by
334
Chapter 4 Applications of Derivatives
100
0.0099 9999 9950 00
1.0100 5016 703
1000
0.0010 0000 0000 00
1.0010 0050 017
10000
0.0001 0000 0000 000
1.0001 0000 500
0.0000 10000 0000 00000
1.0000 1000 005
0
1
100000
Therefore, y = 1 is a horizontal asymptote in both directions. Check the critical points for absolute extreme
values: f ( 1)
e
2 /2
2 /2
e
e 2 /2
0.4931, f (1)
the absolute minimum value of the function is
at x = 1, and the absolute maximum value is e 2 /2 at x = 1.
2
e 3 2 x x ; the domain of g is all x such that 3 2 x x 2
132. f ( x)
down with x-intercepts at x = 3 and x = 1, therefore 3 2 x x 2
1 x
domain of g; g ( x)
g ( 3)
3 2 x x2
2
e0
g (1)
1, g (1)
e 3 2x x
e
2
0
7.3891
1+x=0
0. The parabola y
0 if 3
the absolute minimum value of the function is 1 at x = 3 and x =
1, and the absolute maximum value is e at x = 1.
ln x
x
y
x
solving y
for x
1
x x
5/2
y
ln x
2 x3/ 2
3x
(2
4
5/2 3
ln x
4
0
2 ;
ln x = 2
2
e and y
ln x
on (0, e
e2 ; y
x
0 for x
maximum of 2e ; y
8
x e8/3 ; the
3
8/3
a
2 ln x
2x x
1 x 5/2
2
ln x)
e
0
curve is concave down
there is an inflection point at e8/3 ,
(b) y
2
e x
e
2
0
8
3e4/3
.
2
2
4 x 2e x (4 x 2 2)e x ;
x 0; y 0 for x > 0 and
0 for x < 0
y
); so
2
2 xe x
y
y
2e x
solving y 0
0
2
) and concave up on (e8/3 ,
a maximum at x = 0 or
1; there are points of inflection at
1 ;
2
x
1
2
x
the curve is concave down for
1
2
and concave up otherwise.
Copyright
1, and this interval is the
x = 1 is a critical point;
2
133. (a) y
x
3 2 x x 2 is concave
2014 Pearson Education, Inc.
Chapter 4 Practice Exercises
(c)
y
335
(1 x )e x
e x
y
e x
y
y
(1 x)e x
e x
xe x
0
xe x
( x 1)e x ; solving
0
x = 0; y
0 for x < 0
x > 0 and y
0 for
a maximum at x =
0
0 of (1 0)e 1; there is a point of inflection
at x = 1 and the curve is concave up for x > 1
and concave down for x < 1.
134. y = x ln x
y
y
0
y
0 for x
ln x x 1x
ln x + 1 = 0
ln x 1; solving
ln x = 1
e 1 and y
0 for x
a minimum of e 1 ln e 1
1
e
e 1;
x
e 1
e 1. This
at x
1
x
minimum is an absolute minimum since y
is
positive for all x > 0.
135. In the interval
< x < 2 the function
sin x < 0
(sin x )sin x is not defined for all values
in that interval or its translation by 2 .
136. v
dv
dx
x 2 ln 1x
0
x 2 (ln1 ln x)
2 ln x 1 0
maximum at x
x 2 ln x
ln x
e 1/2 ; hr
1
2
x
x and r = 1
Copyright
e
dv
dx
1/2
h
2 x ln x x 2 1x
; dv
dx
e1/2
0 for x
e
x(2 ln x 1); solving
e 1/2 and dv
dx
1.65 cm
2014 Pearson Education, Inc.
0 for x
e 1/2
a relative
336
Chapter 4 Applications of Derivatives
CHAPTER 4
ADDITIONAL AND ADVANCED EXERCISES
1. If M and m are the maximum and minimum values, respectively, then m
then f is constant on I.
3 x 6, 2 x 0
9 x2 , 0 x 2
2. No, the function f ( x)
maximum value of 9 at x
f ( x)
M for all x
has an absolute minimum value of 0 at x
0, but it is discontinuous at x
I . If m
M
2 and an absolute
0.
3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical point.
On a half-open interval the extreme values of a continuous function may be at a critical point or at the closed
endpoint. Extreme values occur only where f
0, f does not exist, or at the endpoints of the interval. Thus
the extreme points will not be at the ends of an open interval.
4. The pattern f
at x
3.
|
|
|
|
1
2
3
4
indicates a local maximum at x 1 and a local minimum
6( x 1)( x 2) 2 , then y 0 for x 1 and y 0 for x
1. The sign pattern is
f has a local minimum at x
1. Also y 6( x 2) 2 12( x 1)( x 2)
|
|
5. (a) If y
f
1
2
6( x 2)(3 x)
y 0 for x 0 or x 2, while y 0 for 0 x 2. Therefore f has points of inflection at
x 0 and x 2. There is no local maximum.
1 and 0 x 2; y 0 for 1 x 0 and x 2. The sign pattern
(b) If y 6 x ( x 1)( x 2), then y 0 for x
1 and
is y
|
|
|
. Therefore f has a local maximum at x 0 and local minima at x
1
x
0
2. Also, y
18 x
2
1
7
has points of inflection at x
1
x
3
1
7
3
7
f (6) f (0)
f (c) f ( x)
c x
0
f (c )
f ( x)
for all x (c, b] we have
x [ a, b].
8. (a) For all x, ( x 1)2
(b) There exists c
| f (b) f (a )|
0
f ( x)
7
3
and y
0 for all other x
f (c )
2 for some c in (0, 6). Then f (6)
f (c ). Also if f is continuous on (c, b] and f ( x)
0
( x 1) 2
f ( x)
f (c)
(1 x 2 )
( a, b) such that c 2
1
2
1
f (0) 12
0 on [a, c), then by the Mean Value Theorem for all x [a, c ) we have
f ( x ) f (c)
x c
0
x
.
6. The Mean Value Theorem indicates that 6 0
indicates the most that f can increase is 12.
7. If f is continuous on [a, c ) and f ( x)
0 for 1 3 7
, so y
3
1 c
f (b ) f ( a )
b a
2x
0
f ( x)
(1 x 2 )
f (b ) f ( a )
b a
f (c ). Therefore f ( x)
0 on (c, b], then
f (c ) for all
x
1
1.
2 1 x2
2
c
1 , from part (a)
2
1 c2
|b a | .
9. No. Corollary 1 requires that f ( x)
0 for all x in some interval I, not f ( x )
0 at a single point in I.
10. (a) h( x) f ( x) g ( x) h ( x) f ( x) g ( x) f ( x) g ( x) which changes signs at x a since f ( x), g ( x) 0
when x a, f ( x), g ( x) 0 when x a and f ( x), g ( x) 0 for all x. Therefore h( x) does have a local
maximum at x a.
(b) No, let f ( x) g ( x) x3 which have points of inflection at x 0, but h( x) x 6 has no point of inflection
(it has a local minimum at x 0).
Copyright
2014 Pearson Education, Inc.
f
Chapter 4 Additional and Advanced Exercises
1 a
b c 2
11. From (ii), f ( 1)
x
lim f ( x)
c
x
x
dy
3x2
12. dx
1
bx
a 1; from (iii), either 1
x 1
bx 2 cx 2
lim
0, then lim
0
1
x
2
x
2kx 3
x
0
lim
x
bx c
1
2
x
b
4 k 2 36
6
x
lim f ( x). In either case,
0 and c 1. For if b 1, then lim
x
. Thus a 1, b
2
x
1
x
1
0 and if
2
x
x c
0, and c 1.
x has only one value when 4k 2 36
k2
0
9 or k
3.
1 x2
1 (2)
2
13. The area of the ABC is A( x)
1
x
1
1
x
1
2k
x
(1 x 2 )1/2 , where 0
lim
lim f ( x) or 1
x
337
x 1. Thus A ( x)
x
1 x2
0 and 1 are critical points. Also A ( 1) 0 so
A(0) 1 is the maximum. When x 0 the ABC is
isosceles since AC BC
2.
14.
lim
h
f (c h ) f (c )
h
0
f ( c h) f (c )
f
h
f (c) 12 | f (c) |
f (c h )
3 f (c )
2
h
1
2
for
f (c)
1| f
2
f ( c h)
h
1 f (c )
2
(c )
(a) If f (c) 0, then
maximum.
(b) If f (c) 0, then
minimum.
| f (c) | 0 there exists a
(c) | . Then f (c)
1
2
0
f (c h )
h
| f (c ) |
f (c) 12 | f (c) | . If f (c)
0; likewise if f (c )
0 such that 0 | h |
0, then | f (c) |
1
2
0, then 0
| f (c ) |
f (c)
f (c h )
h
f (c )
1
2
f (c )
3
2
f (c).
h
0
f (c h )
0 and 0
h
f (c h )
0. Therefore, f (c) is a local
h
0
f (c h )
0 and 0
h
f (c h )
0. Therefore, f (c) is a local
2y
, where
g
15. The time it would take the water to hit the ground from height y is
g is the acceleration of gravity.
The product of time and exit velocity (rate) yields the distance the water travels:
2y
g
D( y )
64(h
8 g2 (hy
y)
0, D h2
are critical points. Now D (0)
drill the hole is at y
tan
b a
h
1
a
h
tan
y
h
1/2
h 2
2
8 g2 h 2h
h tan a
.
h a tan
)
b a;
n
Solving for tan
tan(
)
0
2h tan
4h g2 and D (h)
gives tan
tan
tan
1 tan tan
bh
h 2 a (b a )
(h 2
Differentiating both sides with respect to h gives 2h tan
d
dh
y 2 ) 1/2 ( h 2 y )
4 g2 ( hy
D ( y)
0
0, h2 and h
the best place to
h.
2
16. From the figure in the text, tan(
a
h
y 2 )1/2 , 0
b
2h
bh
h 2 a (b a )
Copyright
b
2bh 2
bh 2
; and tan
or (h 2
2014 Pearson Education, Inc.
h2
These equations give
a (b a)) tan
a (b a )) sec 2
ab(b a)
a.
h
d
dh
bh.
b. Then
a(b a )
h
a (a b).
338
Chapter 4 Applications of Derivatives
2 r 2 2 rh.
h RHR rH
17. The surface area of the cylinder is S
From the diagram we have Rr HH h
and S (r )
2
2 r (r h)
1 HR r 2
2 r r
r HR
H
2 Hr , where 0
r
R.
Case 1: H R S (r ) is a quadratic equation containing the origin and concave upward S (r ) is maximum
at r R.
Case 2: H R S (r ) is a linear equation containing the origin with a positive slope S (r ) is maximum
at r R.
Case 3: H R S (r ) is a quadratic equation containing the origin and concave downward.
RH . For simplification
4 1 HR r 2 H and dS
0 4 1 HR r 2 H 0 r 2( H
Then dS
R)
dr
dr
RH .
2( H R )
we let r*
(a) If R
H
2 R, then 0
H
2R
H
2( H
the right endpoint R of the interval 0
(b) If H
2R2
2R
2 R, then r*
(c) If H
R
2 R, then 2 R H
a maximum at r
r
R)
S (r ) is maximum at r
2H
H
2( H
R.
H
2( H R )
R)
RH
2( H R )
1
Conclusion: If H (0, 2 R ], then the maximum surface area is at r
RH .
at r r* 2( H
R)
mx 1 1x
18. f ( x)
1
m
If f
19. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
0
2 sin(5 x )
3x
0
lim sin(5 x) cot(3x)
x
x
x
0
x
lim (sec x tan x)
x
lim xx sin
tan x
x
sin( x 2 )
4
0 sin
x
2
10
3
1 10
3
x
2x
x
lim
x
x
0
lim 1 cos2x
1 sec x
x
5 tan x
x
lim
x
2x
0 sin 2 2 x
x
lim
x
(2 x 2 ) sin( x 2 ) 2 cos( x 2 )
x sin x 2 cos x
2
lim x x 2 x2 4
x
Copyright
), then the maximum is
0
1
m
x
yields a minimum.
1
2x
lim
x
5
3
0 cos 2 2 x
1
0 cos 2 2 x 2
lim
2
2x
x
0
tan x
3
2
x sec x
lim sec x tan
2
R. Therefore, S (r ) is
Thus the smallest acceptable value for m is 14 .
lim cos x2 1
x
x
( x 2)( x 2 2 x 4)
( x 2)( x 2)
0. Then f ( x)
lim
cos x
sin x
lim
r*
(2 R,
3sin(5 x ) sin(3 x) 5cos(5 x ) cos(3 x)
3cos(3 x )
0
2sin 2 x cos 2 x
2x
x
R. If H
R
lim
1
lim
sin x
lim 1cos
x
lim sec x2 xtan x
x
x
sin(5 x )
lim 10
0 3 (5 x )
lim x cos x sin x
x
lim sec x2 1
3
lim x2 8
1.
4
2 x cos( x 2 )
lim x sin x
x
x
m
x
lim 1 cos2 x
x
x
lim
0 when x
2 m 1 0
sin(5 x ) cos(3 x )
sin(3 x )
0
x
2
x3
and f ( x)
lim
0
lim x csc 2 2 x
m
2sin(5 x )
3
(5 x )
5
lim
x
1
x2
m
0, then m 1
lim
x
f ( x)
R. Therefore, the maximum occurs at
R because S (r ) is an increasing function of r.
RH .
2( H R )
r*
RH
2( H R )
r*
2
2
1 0
2
1
2
4 4 4
4
3
sin x
2 tan x sec2 x
1
2014 Pearson Education, Inc.
sin x
lim 2sin
x
x
cos3 x
3
lim cos2 x
x
1
2
1
2
Chapter 4 Additional and Advanced Exercises
20. (a)
(b)
lim
x 5
x 5
lim
2x
x 7 x
x
x
x 5
x
x 5
x
lim
x
x
1
x
2x
x
x 7 x
x
lim
1
lim
lim
x
5
x
5
x
1
1
2
1 7
1
2
1 0
1
x
2
21. (a) The profit function is P ( x ) (c ex) x (a bx)
ex 2 (c b) x a. P ( x)
2ex c b
x c2eb . P ( x)
2e 0 if e 0 so that the profit function is maximized at x c2eb .
(b) The price therefore that corresponds to a production level yielding a maximum profit is
p x
c e c2eb
c b
2e
c b
2
339
0
dollars.
2
( c b)
e c2eb
(c b) c2eb a
a.
4e
2
(d) The tax increases cost to the new profit function is F ( x ) (c ex ) x (a bx tx)
ex (c b t ) x a.
Now F ( x)
2ex c b t 0 when x t b2ec c 2be t . Since F ( x)
2e 0 if e 0, F is maximized
2
(c) The weekly profit at this production level is P ( x)
c b t
2e
when x
c b t dollars. Thus, such a tax
c e c 2be t
2
t dollars if units are priced to maximize profit.
2
units per week. Thus the price per unit is p
increases the cost per unit by c 2b t
c b
2
22. (a)
The x-intercept occurs when 1x 3 0
(b) By Newton s method, xn 1
xn 3 xn2
xn
23. x1
x0
f ( x0 )
f ( x0 )
x0q a
x0
qx0q
a
x0q 1
In the case where x0
24. We have that ( x h) 2
dy
2 x 2 y dx
qx0q x0q a
q 1
q
x0
3
1.
3
x
Here f ( xn )
xn 2
qx0q
x0q ( q 1) a
1
qx0q
and m1
1
x0
q 1
q
we have x0q
a and x1
a q 1
x0q 1 q
r 2 and so 2( x h) 2( y h) dx
d2y
dx
2
1
xn
xn
3
xn
1
xn2
1
xn
3 xn2
dy
2
x y dx
1
so that x1 is a weighted average of x0
dy
dx
Copyright
1
q
a
x0q 1
dy
( y h) 2
2y
xn 1
1.
q
dy
dy
xn2
a 1
x0q 1 q
2h 2h dx , by the former. Solving for h, we obtain h
equation yields 2 2 dx
1 . So
xn (2 3xn ).
1
and qa 1 with weights m0
f ( xn )
.
f ( xn )
xn
2 xn 3 xn2
1
x
q 1
a
x0q 1 q
dy
0 and 2 2 dx
x
1
dy
y dx
dy
dx
a .
x0q 1
1
q
2( y h)
d2y
dx 2
0 hold. Thus
. Substituting this into the second
dy
0. Dividing by 2 results in 1 dx
2014 Pearson Education, Inc.
y
d2y
dx
2
dy
x y dx
1
dy
dx
0.
340
Chapter 4 Applications of Derivatives
25. ds
dt
ks
ds
s
k dt
ln s
kt C
s0 ekt
s
the 14th century model of free fall was
exponential; note that the motion starts too slowly at
first and then becomes too fast after about 7 seconds.
26. Two views of the graph of y 1000 1 (.99) x
1
x
are shown below.
At about x = 11 there is a minimum. There is no maximum; however, the curve is asymptotic to y = 1000. The
curve is near 1000 when x 643.
27. (a) a(t )
s (t )
kt 2
2
s (t )
k (k
0)
s (t )
kt C1 , where s (0)
88t C2 where s (0)
Solving for t we obtain t
2
k 88 88k 200k
88
0
2
C2
88 200 k
k
88
kt 2
2
0 so s (t )
(b) The initial condition that s (0)
traveled a distance s 44
k
200k
44 ft/sec implies that s (t )
above. The car is stopped at a time t such that s (t )
k 44
2 k
2
44 44
k
968
k
f 2 ( x) g 2 ( x)
h ( x)
2 f ( x) f ( x ) 2 g ( x) g ( x)
0
968 2002
dy
x satisfies all three conditions since dx
88 200k
k
kt 2
2
3x 2
2 for all x
y
x3
2 x C where 1 13
Copyright
88t 100.
88
0 or
38.72 ft/sec2 .
44t where k is as
this time the car has
25 feet. Thus halving the initial
88
2[ f ( x) f ( x ) g ( x) g ( x)]
5, h( x)
1 everywhere, when x
5 for all x in the
0, y
0, and
everywhere.
30. y
2
2
882
200
44 . At
k
t
2[ f ( x) g ( x) g ( x)( f ( x))] 2 0 0. Thus h( x) c, a constant. Since h(0)
domain of h. Thus h(10) 5.
29. Yes. The curve y
kt 88. So
0, thus k 88
0 so that k
velocity quarters stopping distance.
28. h( x)
s (t )
kt 44 and s (t )
kt 44
442
2k
88
88t. Now s (t ) 100 when kt
2
. At such t we want s (t )
0. In either case we obtain 882
88
C1
21 C
C
4
2014 Pearson Education, Inc.
y
x3
2 x 4.
d2y
dx 2
0
Chapter 4 Additional and Advanced Exercises
31. s (t )
t2
a
v
t3
3
s (t )
C. We seek v0
t4
12
maximum for this t*. Since s (t )
[ (3C )1/3 ]4
12
that t* (3C )1/3 . So
C
s (0)
t1/2 t 1/2
v(t )
32. (a) s (t )
4 t 5/2
15
(b) s (t )
4 t 3/2
3
Ct k and s (0)
C (3C )1/3
(4b )3/ 4
. Thus v0
3
2t1/2
k2 where s (0)
ax 2 bx c with a
33. The graph of f ( x)
0 we have that
we require (2b)
4ac
34. (a) Clearly f ( x)
a12
b
ac
(a1 x b1 ) 2
a12 x 2
f ( x)
0
2
a22
an2 x 2
4 . Thus
15
k2
(an x bn )2
an2 x 2
bn2
a22
an2 b12 b22
bn2
a1b1 a2b2
an bn
2
a12
a22
an2 b12 b22
bn2 .
Now notice that this implies that f ( x)
L
dL
d
0 for all x if f ( x )
2b
0 by Exercise 29. Thus
b2
0 for some real x
(a1 x b1 )
2 a1b1 a2b2
2
(an x bn )
anbn x
4ac
0, by quadratic formula.
2
b12 b22
bn2
0
an bn
2
a12
a22
an2 b12 b22
bn2
a1b1 a2b2
an bn
2
a12
a22
an2 b12 b22
bn2 But now f ( x)
k a b cot
4
b csc
R
0
cos
r
r 4b csc2
r 4 csc
1 5 4
6
ai x
bi
dL
d
4
0 for all i 1, 2,
2
k b csc4
0
cos 1 (0.48225)
0
cos
R 4 cot )
0; but b csc
4
cos 1 r 4 , the critical value of
R
61
Copyright
0
ai x bi
; solving
(b csc )(r 4 csc
r4
R4
0
, n.
b csc cot
r4
R
bR 4 csc cot
R 4 cot
0
(2b )2 4 ac
. Thus
2a
0. Thus
a1b1 a2b2
2
(b)
b12 b22
an bn x
a12
,n
4.
3
2an bn x bn2
2
for all i 1, 2,
4b
3
0 for all x. Expanding we see
2 a1b1 a2 b2
an2 x 2
31/3 C 4/3
v (t )
0 are, by the quadratic equation
an bn
a22
b
0 so
0.
(b) Referring to Exercise 33: It is clear that f ( x)
35. (a)
s (t )
a1b1 a2b2
a12
and also s (t*)
2 t 3/2 2t1/2
3
4 t 3/2 4 t 4 .
3
3
15
4
3
4 t 5/2
15
k
0 is a parabola opening upwards. Thus f ( x )
b12
2a1b1 x
b
k where v (0)
for at most one real value of x. The solutions to f ( x)
2
3C )
12
b for some t* and s is at a
4
s (t ) 12t Ct
(3C )1/3 34C
2 2 3/4
b .
3
2 t 3/2
3
s (t )
C. We know that s (t*)
(3C )1/3 (C
b
(4b)3/ 4
3
4t
3
s (0)
341
2014 Pearson Education, Inc.
0 since
0
342
Chapter 4 Applications of Derivatives
Copyright
2014 Pearson Education, Inc.
CHAPTER 5 INTEGRATION
5.1
AREA AND ESTIMATING WITH FINITE SUMS
x2
1. f ( x)
(a)
(b)
(c)
(d)
x
1 0
2
1
2
and xi
x
1 0
4
1
4
and xi
x
1 0
2
1
2
and xi
1 0
4
1
4
and xi
x
(b)
(c)
(d)
1
i
2
a lower sum is
i x
i
4
a lower sum is
i x
i
2
an upper sum is
i
4
an upper sum is
i x
i x
i 0
3
i 0
2
i 1
4
i 1
x3
2. f ( x)
(a)
Since f is increasing on [0, 1], we use left endpoints to
obtain lower sums and right endpoints to obtain upper
sums.
i 2 1
2
2
1
2
02
1 2
2
i 2
4
1
4
1
4
02
1 2
4
i 2 1
2
2
1
2
1 2
2
i 2 1
4
4
1
4
1 2
4
12
1
8
1 2
2
3 2
4
1 7
4 8
7
32
5
8
3 2
4
1 2
2
12
1
4
30
16
15
32
Since f is increasing on [0, 1], we use left endpoints to
obtain lower sums and right endpoints to obtain upper
sums.
x
1 0
2
1
2
and xi
x
1 0
4
1
4
and xi
x
1 0
2
1
2
and xi
x
1 0
4
1
4
and xi
1
i
2
a lower sum is
i
4
a lower sum is
i x
i
2
an upper sum is
i x
i
4
an upper sum is
i x
i x
i 0
3
i 0
2
i 1
4
Copyright
i 1
i 3 1
2
2
1
2
03
1 3
2
1
16
i 3 1
4
4
1
4
03
1 3
4
1 3
2
i 3 1
2
2
1
2
1 3
2
i 3 1
4
4
1
4
1 3
4
2014 Pearson Education, Inc.
13
1 3
2
1 9
2 8
3 3
4
3 3
4
36
256
9
64
100
256
25
64
9
16
13
343
344
Chapter 5 Integration
1
x
3. f ( x)
(a)
x
Since f is decreasing on [1, 5], we use left endpoints to
obtain upper sums and right endpoints to obtain lower
sums.
5 1
2
2 and xi
1 and xi
(b)
x
5 1
4
(c)
x
5 1
2
2 and xi
1 i x 1 2i
(d)
x
5 1
4
1
1 i x 1 i
4. f ( x)
(a)
(b)
2
1 i x 1 2i
4 x2
x
x
2 ( 2)
2
2 ( 2)
4
x
(d)
x
2 ( 2)
2
2 ( 2)
4
2 and xi
2 i x
1 an xi
2 i x
2 2i
1
3
16
15
1
5
1
4
77
60
1
5
8
3
1
3
1
3
25
12
1
4
a lower sum is 2 (4 ( 2) 2 ) 2 (4 22 )
1
2 i
2 and xi
2 i x
2 2i
1 and xi
2 i x
2 i
a lower sum is
(4
( xi )2 ) 1
i 0
(4 ( xi )2 ) 1
i 3
an upper sum is
(4
( xi )2 ) 1
i 1
1
2
f
1
4
x
3
4
f
1
2
Using 4 rectangles
1
4
1
4
f 81
1
8
2
x
f 83
3
8
2
3
(4 ( xi )2 ) 1
i 2
Using 2 rectangles
Copyright
0
an upper sum is 2 (4 (0)2 ) 2 (4 02 ) 16
(4 12 )) 14
x2
4
6
2
1((4 ( 1)2 ) (4 02 ) (4 02 )
5. f ( x)
2
Since f is increasing on [ 2, 0] and decreasing on
[0, 2], we use left endpoints on [ 2, 0] and right
endpoints on [0, 2] to obtain lower sums and use right
endpoints on [ 2, 0] and left endpoints on [0, 2] to
obtain upper sums.
1((4 ( 2) 2 ) (4 ( 1)2 ) (4 12 ) (4 22 ))
(c)
2
i 1
4
1 1 1 1 1
a lower sum is
xi
2 3
i 1
1
1 2 2 1
an upper sum is
xi
i 0
3
1 1 11 1
an upper sum is
2
xi
i 0
1 i x 1 i
xi
1
xi
a lower sum is
1 0
2
2
1
4
3
4
1 0
4
1
4
f 85
5
8
2014 Pearson Education, Inc.
2
1
2
2
f 87
7
8
2
21
64
10
32
5
16
Section 5.1 Area and Estimating with Finite Sums
6. f ( x)
x3
Using 2 rectangles
1
2
x
f 43
f 14
1
2
Using 4 rectangles
1
4
f 83
f 81
1 13 33 53 73
4
83
7. f ( x)
1
x
x
496
4 83
Using 4 rectangles
1 f 32
8. f ( x)
4 x2
f 52
496
57 9
2( f ( 1)
3
2
16
9. (a) D
(b) D
9
4
28
2 64
7
32
f 87
31
128
124
83
x
5 1
2
2
x
5 1
4
1
2( f (2)
f
2 14 2
f 92
1 23
2 ( 2)
2
x
f (1))
3 2
2
4
1
4
3
f (4))
2
5
2
7
2
9
496
315
2
2(3 3) 12
Using 4 rectangles
1
1 0
4
f 72
Using 2 rectangles
1 f
3
4
3
2
1
4
1488
3579
1
2
4
f 85
Using 2 rectangles
2 12
1 0
2
3
1
345
x
1
2
f
4
2 ( 2)
1
4
1
f 32
2
1 2
2
16 10
2
4
1 2
2
4
3 2
2
11
(0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) 87 inches
(12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) 87 inches
10. (a) D
(1)(300) (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300)
(1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) 5220 meters (NOTE: 5 minutes 300 seconds)
(b) D (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300)
(1.8)(300) (1.5)(300) (1.2)(300) (0)(300) 4920 meters (NOTE: 5 minutes 300 seconds)
11. (a) D
(0)(10) (44)(10) (15)(10) (35)(10)
(35)(10) (44)(10) (30)(10) 3490 feet
(b) D (44)(10) (15)(10) (35)(10) (30)(10)
(44)(10) (30)(10) (35)(10) 3840 feet
(30)(10) (44)(10) (35)(10) (15)(10) (22)(10)
0.66 miles
(44)(10) (35)(10) (15)(10) (22)(10) (35)(10)
0.73 miles
12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the
midpoints of each time interval to approximate this area using rectangles. Thus,
D (20)(0.001) (50)(0.001) (72)(0.001) (90)(0.001) (102)(0.001) (112)(0.001) (120)(0.001)
(128)(0.001) (134)(0.001) (139)(0.001) 0.967 miles
(b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours 22.7 sec.
At 22.7 sec, the velocity was approximately 120 mi/hr.
Copyright
2014 Pearson Education, Inc.
346
Chapter 5 Integration
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left endpoints in summing
acceleration t. Thus, t 1 and speed [32.00 19.41 11.77 7.14 4.33](1) 74.65 ft/sec
(b) Using right endpoints we obtain a lower estimate: speed [19.41 11.77 7.14 4.33 2.63](1)
45.28 ft/sec
(c) Upper estimates for the speed at each second are:
t 0
1
2
3
4
5
v 0 32.00 51.41 63.18 70.32 74.65
Thus, the distance fallen when t
3 seconds is s [32.00 51.41 63.18](1) 146.59 ft.
14. (a) The speed is a decreasing function of time
(distance) attained. Also
t
0
1
right endpoints give a lower estimate for the height
2
3
4
5
v 400 368 336 304 272 240
gives the time-velocity table by subtracting the constant g 32 from the speed at each time increment
t 1sec. Thus, the speed 240 ft/sec after 5 seconds.
(b) A lower estimate for height attained is h [368 336 304 272 240](1) 1520 ft.
15. Partition [0, 2] into the four subintervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2]. The midpoints of these
subintervals are m1 0.25, m2 0.75, m3 1.25, and m4 1.75. The heights of the four approximating
27 , f ( m ) (1.25)3 125 , and f ( m ) (1.75)3 343
1 , f ( m ) (0.75)3
rectangles are f (m1 ) (0.25)3 64
2
3
4
64
64
64
Notice that the average value is approximated by 12
1
length of [0,2]
approximate area under
x3
curve f ( x )
3 3 1
4
2
1 3 1
4
2
5 3 1
4
2
7 3 1
4
2
31
16
. We use this observation in solving the next several exercises.
16. Partition [1,9] into the four subintervals [1, 3], [3, 5], [5, 7], and [7, 9]. The midpoints of these subintervals are
m1
2, m2
1,
4
f (m2 )
Area
4, m3
2 12
6, and m4
1,
2
8. The heights of the four approximating rectangles are f (m1 )
f (m3 )
1,
6
and f (m4 )
2 14
2 16
2 18
1.
8
25
12
The width of each rectangle is x
average value
2. Thus,
25
12
area
length of [1,9]
8
25 .
96
17. Partition [0, 2] into the four subintervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2]. The midpoints of the
subintervals are m1 0.25, m2 0.75, m3 1.25, and m4 1.75. The heights of the four approximating
rectangles are f (m1 ) 12 sin 2 4 12 12 1, f (m2 ) 12 sin 2 34 12 12 1, f (m3 ) 12 sin 2 54
1
2
1
2
2
Thus, Area
1
2
1
2
1, and f (m4 )
(1 1 1 1)
1
2
2
1
2
sin 2
7
4
1
2
2
1
2
area
length of [0, 2]
average value
1. The width of each rectangle is x
2
2
1.
18. Partition [0, 4] into the four subintervals [0, 1], [1, 2], [2, 3], and [3, 4]. The midpoints of the subintervals
are m1 12 , m2 23 , m3 25 , and m4 72 . The heights of the four approximating rectangles are
f (m1 ) 1
1
cos
cos
3
8
4
1
2
4
4
1
cos 8
4
0.97855, f (m3 ) 1
Copyright
0.27145 (to 5 decimal places), f (m2 ) 1
cos
5
2
4
4
1
cos 58
4
2014 Pearson Education, Inc.
cos
0.97855, and
3
2
4
4
1.
2
Section 5.1 Area and Estimating with Finite Sums
f (m4 ) 1
Area
cos
7
2
4
4
1
cos 78
4
347
0.27145. The width of each rectangle is x 1. Thus,
(0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1)
2.5
average value
area
length of [0,4]
2.5
4
5
8
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left endpoints:
(a) upper estimate (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) 758 gal,
lower estimate (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) 543 gal.
(b) upper estimate (70 97 136 190 265 369 516 720) 2363 gal,
lower estimate (50 70 97 136 190 265 369 516) 1693 gal.
(c) worst case: 2363 720t 25, 000 t 31.4 hrs;
best case: 1693 720t 25, 000 t 32.4 hrs
20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate
uses left endpoints;
(a) upper estimate (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) 60.9 tons
lower estimate (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) 46.8 tons
(b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) 126.6 tons,
so near the end of September 125 tons of pollutants will have been released.
21. (a) The diagonal of the square has length 2, so the side length is 2. Area
2
2
2
(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle
measuring 216 8 .
Area 16 12 sin 8 cos 8
4 sin 4 2 2 2.828
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle
measuring 232 16 .
32 12 sin 16 cos 16 8 sin 8 2 2 3.061
(d) Each area is less than the area of the circle, . As n increase, the area approaches .
Area
22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle
1 sin 2 .
measuring 22n n The area of each isosceles triangle is AT 2 12 sin n cos n
2
n
(b) The area of the polygon is AP
nAT
n sin 2
2
n
, so lim n2 sin 2n
n
lim
n
(c) Multiply each area by r 2 .
AT 12 r 2 sin 2n
AP
lim
n
23-26.
n r 2 sin 2
n
2
AP
r2
Example CAS commands:
Maple:
with( Student[Calculus 1] );
f := x -> sin(x);
a := 0;
b := Pi;
Plot( f (x), x a..b, title "#23(a) (Section 5.1)" );
N : [ 100, 200, 1000 ];
# (b)
Copyright
2014 Pearson Education, Inc.
sin 2n
2
n
348
Chapter 5 Integration
for n in N do
Xlist : [ a+1.*(b-a)/n*i $ i 0..n ];
Ylist : map( f, Xlist );
end do:
for n in N do
Avg[n] : evalf(add(y,y Ylist)/nops(Ylist));
# (c)
end do;
avg : FunctionAverage( f (x), x a..b, output value );
evalf( avg );
FunctionAverage(f(x),x a..b, output plot);
# (d)
fsolve( f(x) avg, x 0.5 );
fsolve( f(x) avg, x 2.5 );
fsolve( f(x) Avg[1000], x 0.5 );
fsolve( f(x) Avg[1000], x 2.5 );
Mathematica: (assigned function and values for a and b may vary):
Symbols for , , powers, roots, fractions, etc. are available in Palettes.
Never insert a space between the name of a function and its argument.
Clear[x]
f[x_] : x Sin[1/x]
{a, b} { /4, }
Plot[f[x],{x, a, b}]
The following code computes the value of the function for each interval midpoint and then finds the
average. Each sequence of commands for a different value of n (number of subdivisions) should be
placed in a separate cell.
n 100; dx (b a) /n;
values Table[N[f[x]],{x, a dx/2, b, dx}]
average Sum[values[[i]],{i, 1, Length[values]}] / n
n 200; dx (b a) /n;
values Table[N[f[x]],{x, a dx/2, b, dx}]
average Sum[values[[i]],{i, 1, Length[values]}] / n
n 1000; dx (b a) /n;
values Table[N[f[x]],{x, a dx/2, b, dx}]
average Sum[values[[i]],{i, 1, Length[values]}] / n
FindRoot[f[x]
average,{x, a}]
5.2
1.
2.
3.
SIGMA NOTATION AND LIMITS OF FINITE SUMS
2
k 1
3
k 1
4
k 1
6k
k 1
6(1)
1 1
k 1
k
1 1
1
cos k
6(2)
2 1
2 1
2
6
2
12
3
3 1
3
7
0 12
2
3
7
6
cos(1 ) cos(2 ) cos(3 ) cos(4 )
Copyright
1 1 1 1 0
2014 Pearson Education, Inc.
Section 5.2 Sigma Notation and Limits of Finite Sums
4.
5.
6.
5
k 1
3
sin k
sin(1 ) sin(2 ) sin(3 ) sin(4 ) sin(5 )
0
( 1) k 1 sin k
( 1)1 1 sin 1
( 1) k cos k
( 1)1 cos(1 ) ( 1) 2 cos(2 ) ( 1)3 cos(3 ) ( 1) 4 cos(4 )
k 1
4
0 0 0 0 0
k 1
6
7. (a)
k 1
5
(b)
2k 1
2k
k 0
4
(c)
k 1
( 1) 2 1 sin 2
( 1)3 1 sin 3
0 1
3
2
349
3 2
2
( 1) 1 ( 1) 1 4
21 1 22 1 23 1 24 1 25 1 26 1 1 2 4 8 16 32
20 21 22 23 24
25 1 2 4 8 16 32
2k 1 2 1 1 20 1 21 1 22 1
23 1 24 1 1 2 4 8 16 32
All of them represent 1 2 4 8 16 32
6
8. (a)
( 2)k 1 ( 2)1 1 ( 2)2 1 ( 2)3 1 ( 2)4 1 ( 2)5 1 ( 2)6 1 1 2 4 8 16 32
k 1
5
(b)
k 0
3
(c)
( 1)k 2k
k
2
( 1)0 20 ( 1)1 21 ( 1)2 22 ( 1)3 23 ( 1) 4 24
( 1)k 1 2k 2
( 1) 2 1 2 2 2
( 1)5 25 1 2 4 8 16 32
( 1) 1 1 2 1 2 ( 1) 0 1 20 2 ( 1)1 1 21 2
( 1)2 1 22 2 ( 1)3 1 23 2
1 2 4 8 16 32;
(a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two
4
9. (a)
k 2
2
(b)
k 0
1
(c)
( 1) k 1
k 1
( 1)k
k 1
1
k
( 1)2 1
2 1
( 1)0
0 1
( 1)k
k 2
( 1) 1
1 2
( 1)3 1
3 1
( 1)1
1 1
( 1)0
0 2
( 1)4 1
4 1
( 1) 2
2 1
1 12
( 1)1
1 2
1 12
1
3
1
3
1 12
1
3
(a) and (c) are equivalent; (b) is not equivalent to the other two.
4
10. (a)
(k 1)2
k 1
3
(b)
k
(c)
k
1
(k 1)2
1
3
k2
(1 1) 2 (2 1)2 (3 1)2 (4 1) 2
0 1 4 9
( 1 1)2 (0 1) 2 (1 1)2 (2 1) 2 (3 1) 2
( 3)2 ( 2) 2 ( 1) 2
0 1 4 9 16
9 4 1
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
11.
14.
6
k 1
5
k 1
k
12.
2k
15.
4
k 1
5
k2
13.
( 1)k 1 k1
k 1
Copyright
16.
2014 Pearson Education, Inc.
4
1
k 12
5
k
( 1)k k5
k 1
350
Chapter 5 Integration
n
17. (a)
10
k 1
10
k 1
13
k 1
13
(c)
27.
28.
k 1
7
n
k 1
2
7
k 1
k
6
6
6
(k 2 5)
k 1
k 1
5
k 1
k 1
5
k 1
k 1
2
k
n
ak
7
k 1
k3
4
5 6
n
k 1
ak
11
6 2( 5) 16
(b)
1 0 n
n
k 1
(d)
(b)
552
(b)
k2
k 1
5
k 1
k
3
2
6(6 1)(2(6) 1)
6
5
k 1
7
k 1
k2
k2
k 1
1
4
7
k 1
5
5(6)
5
k 1
7
k 1
3
5
3
2
k
22.
6(6 1)(2(6) 1)
6
5
(2k 2 k )
k 1
56
3(6)
(3k 2 5k )
7
k 1
n
250bk
250
n
(bk 1)
k 1
10
k 1
k 1
n
k 1
bk
bk
250(1)
n
1 1 n
k 1
k2
10(10 1)(2(10) 1)
6
385
13
k2
13(13 1)(2(13) 1)
6
k 1
819
912 8281
6
1
225
n
3025
7(7 1)
2
3
7
bk
0
k 1
k
5 6 1
91
k 1
k2
bk
55
6
3
2
k 1
k (2k 1)
7
bk
8(0)
2
5
k (3k 5)
k3
225
k 1
13(13 1) 2
2
k3
k 1
7
k 1
n
ak
ak
1
n
ak
k 1
k 1
n
k 1
k 1
k 1
n
13(13 1)
2
k
(3 k 2 )
5
n
10(10 1) 2
2
k3
15
1 (6)
6
10(10 1)
2
k
2k
k 1
3( 5)
k 1
n
k 1
20. (a)
26.
8
(ak 1)
(c)
25.
2ak )
k 1
n
19. (a)
24.
(bk
8ak
ak
bk
bk )
n
(c)
k 1
(ak
k 1
18. (a)
k 1
n
bk )
k 1
n
(e)
n
(ak
k 1
n
(d)
6
1
6
k
6
k 1
n
(c)
23.
3
k 1
n b
(b)
21.
3ak
7(7 1) 2
2
Copyright
15
k 1
k
15
5(5 1)
2
61
3
2
k 1
5
k
15
73
5(5 1)(2(5) 1)
6
7(7 1)(2(7) 1)
6
2
1 5(5 1)
225
2
k
k3
k
k
5
5
7(7 1)
2
5(5 1) 3
2
2
1 7(7 1)
4
2
5(5 1)
2
308
3376
588
2014 Pearson Education, Inc.
240
250
Section 5.2 Sigma Notation and Limits of Finite Sums
29. (a)
7
k 1
3 3(7)
21
500
(b)
k 1
7
7(500)
j
262
(c) Let j
k 2
k
j 2; if k
3
j 1 and if k
264
30. (a) Let j
k 8
k
j 8; if k
9
j 1 and if k
36
j
3
j 1 and if k
17
j 15
28(28 1)
2
(b) Let j
15
k 2
( j2
(c) Let j
k
k 17
31. (a)
(c)
32. (a)
(c)
33. (a)
n
k 1
n
4
j 1
1
n
k
2
k 1n
j2
15
j 1
4j
j 17; if k
33
54
18
15
j 1
4
2n
15(15 1)(2(15) 1)
6
j 1 and if k
( j 2 33 j 272)
j 1
54(54 1)
2
272(54)
4n
k 1
k 1
n
15
( j 17)(( j 17) 1)
(k 1)
n
j 2; if k
k
j 1
54(54 1)(2(54) 1)
6
54
j 1
j2
k 1
k
1
n
1 n ( n 1)
2
n2
k 3
k 9
262
10
28
k
10 10(262)
j 1
( j 8)
j 1
n
1
k 1
n ( n 1)
2
n
2n n 1 2n 2
4
71
17
k 3
15(15 1)
2
j
54
28
j 1
j
28
j 1
2620
8
j 1
4(15) 1240 480 60 1780
k 3
54
j 1
n
k 1
c cn
n
k 1
c
n
c
n
n
c
n 1
2n
(b)
Copyright
k (k 1)
272
n2 n
2
(b)
(c)
2014 Pearson Education, Inc.
( j 2) 2
j 1
71
54
33 j
15
k2
53955 49005 14688 117648
(b)
n
264
630
4 j 4)
j 1
54
8(28)
3500
36
28
351
352
Chapter 5 Integration
34. (a)
(b)
(c)
35. (a)
(b)
(c)
36. (a)
(b)
(c)
37. | x1 x0 | |1.2 0|
and | x5
1.2, | x2
x4 | |3 2.6|
x1 | |1.5 1.2| 0.3, | x3
x2 |
2.3 1.5
0.8, | x4
x3 |
2.6 2.3
0.3,
0.4; the largest is || P || 1.2.
38. | x1 x0 | | 1.6 ( 2)| 0.4,| x2 x1 | | 0.5 ( 1.6) | 1.1,| x3 x2 | | 0 ( 0.5) | 0.5,
| x4 x3 | |0.8 0| 0.8, and | x5 x4 | |1 0.8| 0.2; the largest is || P || 1.1.
39. f ( x) 1 x 2
1 0
n
1
n
1 ci2 1n
1
n
Let x
n
i 1
n
1
i2
3
n i 1
n3
n3
1
2
3
n
6
lim 1
n
Copyright
1
n2
and ci
n
i 1
1
i x
i 2
n
1
n ( n 1)(2n 1)
6 n3
n
. Thus, lim
n
2
3
n
1
n2
6
2014 Pearson Education, Inc.
i 1
1 13
i . The right-hand
n
n
1
n2 i 2
n3 i 1
3
2
1 2 n 3n3 n
6n
1 ci2 1n
2
3
sum is
Section 5.2 Sigma Notation and Limits of Finite Sums
40. f ( x)
2x
3 0
n
Let x
is
n
i 1
2ci n3
n
x2 1
i 1
n
3
n
n
ci2 1 n3
i 1
27
n
n
i 1
18
27
n
2
lim
3
n
x x2
Let x
Let x
x(1 x)
n
i 1
1 3n
n
27
n
18
9
n2
3
2
ci
1 0
n
9.
3i . The right-hand
n
n
3
9i 2 1
2
n
i 1 n
27 n( n 1)(2 n 1)
6
n3
n
3. Thus, lim
lim
Let x
n
i 1
3ci
n
sum is
9(2 n3 3n2 n )
3
2 n3
3
ci2 1 n3
i 1
9 3 12.
n
Copyright
1
n
1
n2
and ci
n
i x
i 2
n
i
n
1
n
i 1
n
(
n
1)(2
n
1)
1
6
n3
n
. Thus, lim
n
3
n
2
i 1
1
2
3
n
and ci
n
n
3
n
2
2014 Pearson Education, Inc.
3
ci
sum is
2 n3 3n 2 n
6 n3
ci2 n1
2
6
5.
6
right-hand sum is
n
n
3
i 23 i 2
2
n i 1
n i 1
3n 2 3n 2 n 2 3n 1
2n 2
3n 2
2
1
2 ni
n
3i
n
. Thus, lim
2
n2 n
2 n2
i . The
n
i x
i 1
2 n ( n 1)(2 n 1)
6
n3
1
n
2
n
i . The right-hand
n
n
n
1
1
i
i2
n2 i 1
n3 i 1
1
n2
6
1 0 1
n
n
2 1
2ci n
3
i x
and ci
2
3 n ( n 1)
2
n2
3
3
3 n 2 n
2
3
lim
1
n
1
n
1
n
3x 2 x 2
n
1 0
n
2 1
ci n
1 n ( n 1)
2
n2
3
1 1n 2 n
2
6
44. f ( x)
3i 2
n
n
i . The right-hand sum is
n
n
2
n( n 1)(2 n 1)
3
1
3ci2 1n
3 ni
i 2 33
3
n
6
n
n
i 1
i 1
i 1
3 1
2
n
n n2
2 n3 3n 2 n
. Thus, lim 3ci2 1n
2
2n3
n
i 1
2 3n 12
n
2 1.
lim
2
2
n
3x2
n
43. f ( x)
i x
9
n2
n
42. f ( x)
and ci
9 n2 9 n .
n2
lim 9 9n
n
n
The right-hand sum
18 n( n 1)
2
n2
18
i
n2 i 1
2
lim 9n 2 9n
i 1
i2
3i .
n
i x
n
6i 3
n n
6i 3
n n
3 0
n
Let x
and ci
i 1
n
Thus, lim
41. f ( x)
3
n
n
353
i 1
3ci
2ci2 1n
1
n2
3
2
2
3
13 .
6
354
Chapter 5 Integration
45. f ( x)
2 x3
46. f ( x)
x2
i . The right-hand sum is
n
n
n
n
2
3
3
2
2 n ( n 1)
1
2ci3 1n
2 ni
i
4
4
2
n n
n
i 1
i 1
i 1
2 1
1
n
2 2
n n2
2 n ( n 2 n 1)
n2 2 n 1
.
Thus,
lim
2ci3 1n
4
2
2
4n
2n
n
i 1
1 n2 12
n
1.
lim
2
2
n
x3
Let x
1 0
n
Let x
0 ( 1)
n
1
n
and ci
1
n
and ci
The right-hand sum is
n
i 1
n
i 1
1
5i
n2
2
n
lim 2
n
5.3
1.
4.
7.
4 n2 6 n 2
3n 2
5
n
5
4
6
n
2
n2 2n 1
4n 2
2
n2
2
n
1
3
2
5
2
6
n
4
5
n
3
2
1
n2
2
n
4i 2
n3
1
4
n2
. Thus, lim
n
n
i3
n4
n
i 1
ci2 ci3 1n
i 1
n
n
i 1
n
5
i
2
n i 1
i 1
4 n ( n 1)(2n 1)
6
n3
ci2
2 52
4
4
3
1
4
2
n
ci3 1n
7.
12
THE DEFINITE INTEGRAL
2 2
0
x dx
41
1 x
dx
5.
(sec x) dx
8.
0
/4
9. (a)
(c)
(d)
(e)
(f )
2.
2
g ( x) dx
2
2
3 f ( x) dx
1
5
2
5
1
5
1
f ( x ) dx
0
1
2x3dx
3 1
21 x
/4
0
dx
5
1
2
f ( x ) dx
2
f ( x ) dx
[ f ( x) g ( x )] dx
[4 f ( x) g ( x)] dx
5
1
4
3( 4)
1
1
5
1
5
g ( x) dx
12
f ( x) dx
f ( x) dx
5
1
f ( x) dx
Copyright
1
6.
(b)
1
7
0
( x 2 3x) dx
4 x 2 dx
(tan x ) dx
0
3
5
3.
6 ( 4) 10
g ( x) dx
5
1
g ( x ) dx
6 8
2
4(6) 8 16
2014 Pearson Education, Inc.
5
1
g ( x) dx
i3 1
n3 n
n
4
i 2 14 i3
3
n i 1
n i 1
2
1 n ( n 1)
4
2
n
2 5ni
1
n2
1 ni .
1 i x
3
1
1 ni
n
5 n( n 1)
2
n2
2 ( n)
n
2 5n2n 5
i 2
n
i x
8
4i 2
n2
n
Section 5.3 The Definite Integral
10. (a)
(b)
(c)
(d)
(e)
(f )
11. (a)
(c)
12. (a)
(c)
13. (a)
(b)
14. (a)
(b)
9
2 f ( x ) dx
1
9
7
1
9
7
2
1
1
9
3
2
3
4
9
7
0
3
0
3
4
f ( z ) dz
4
f (t ) dt
3
3
h(r ) dr
1
3
1
3
1
2)(6)
4
x
2 2
21
16. The area of the trapezoid is A
1 (3
2
1)(1)
2
3/2
1/2
h( x) dx
9
2(5) 3(4)
1 5
2
9
7
6
f ( x) dx
(b)
5
(d)
2
3
0
(b)
2
f ( z ) dz
(d)
7 3
4
6 0
6
h(u ) du
6
9
7
h( x) dx
2
1
2
1
5 4 1
3 f ( z ) dz
0
g (u ) du
3
0 g (r )
3
2
dr
1
1
h(r ) dr
1 (B
2
3 dx
1 (B
2
( 2 x 4) dx
3
1
b) h
21 square units
b) h
2 square units
Copyright
2014 Pearson Education, Inc.
2
3
1
2
[ f ( x )] dx
4
h(u ) du
15. The area of the trapezoid is A
1 (5
2
f ( x) dx
g ( x) dx
h(r ) dr
5 4
5
f (t ) dt
h(u ) du
9
7
[ f ( x) h( x)] dx
f ( z ) dz
0
h( x) dx
( 1) 1
g (t ) dt
[ g ( x)] dx
9
7
2
f ( x) dx 3
7
f (t ) dt
1
9
7
9
f ( x) dx
1
2
g (t ) dt
2
f ( x) dx
f ( x)] dx
2( 1)
f ( x) dx
7
f ( x) dx
1
f (u ) du
0
0
1
1
9
f (t ) dt
2
3
9
f ( x ) dx
[ h( x )
9
f ( x) dx
[2 f ( x) 3h( x)] dx
f ( x) dx
1
7
4
1
[ f ( x) h( x)] dx
7
9
3
3
9
2
1
0
3
1
2
f ( z ) dz
f ( x) dx
g (t ) dt
0
3
g (t ) dt
5 3
5
2
1
2
( 2) 1
355
356
Chapter 5 Integration
1
2
17. The area of the semicircle is A
3
9
2
3
9 x 2 dx
9
2
0
4
16 x 2 dx
4
1
4
2
| x| dx
1
r2
1
4
(4)2
1 bh
2
area of the triangle on the right is
1.
2
Then, the total area is 2.5
2.5 square units
1 bh
2
20. The area of the triangle is A
1
(3)2
square units
19. The area of the triangle on the left is A
1 (2)(2) 2. The
2
A 12 bh 12 (1)(1)
1
1
2
square units
18. The graph of the quarter circle is A
4
r2
1 (2)(1)
2
1
(1 | x|) dx 1 square unit
21. The area of the triangular peak is A 12 bh 12 (2)(1) 1.
The area of the rectangular base is S
w (2)(1) 2.
Then the total area is 3
1
1
(2 | x|) dx
3 square
units
Copyright
2014 Pearson Education, Inc.
Section 5.3 The Definite Integral
1 x2
( y 1)2
22. y 1
x2
357
y 1
1 x2
( y 1) 2 1 x 2
1, a circle with center (0, 1) and radius
of 1 y 1 1 x 2 is the upper semicircle. The area
of this semicircle is A 12 r 2 12 (1) 2 2 . The area
of the rectangular base is A w (2)(1) 2. Then the
1
total area is 2
2
1
1 x 2 dx
1
2
2
square units
23.
25.
bx
0 2
b
a
2 s ds
2
27. (a)
2
0
28. (a)
(b)
29.
31.
2
1
2
dx
1
1
1
x dx
d
b2
4
1 (b )( b )
2
2
24.
1 b(2b)
2
1 a (2a )
2
4 x 2 dx
1[
2
(2) 2 ]
1 x 2 dx
3x
1 x 2 dx
2
2
(1)2
2
(2 )2
2
2
2
a2
26.
2
0
3x
2
b2
1
0
3x dx
1
3 x dx
(b)
0
1
1
0
3 x dx
1
1
32.
Copyright
1 b(4b)
2
3t d t
1 b(3b)
2
b
a
2
0
4 x 2 dx
1 x 2 dx
2
2
4 x dx
1 [(1)(3)]
2
30.
1
2
3
1 x 2 dx
b
0
2.5
0.5
x dx
1[
4
(1) 2 ]
r dr
2014 Pearson Education, Inc.
2
3
2
1 [(1)(3)]
2
(0.5)2
2
2
a2 )
(2) 2 ]
4
(2.5)2
2
5 2
3 (b 2
2
1 a (3a )
2
1 [(1)(3)]
2
5 2
2
1[
4
2b 2
2
2
1[
2
3
2
24
(1) 2 ]
2
358
33.
35.
37.
39.
41.
43.
44.
45.
46.
47.
48.
49.
50.
Chapter 5 Integration
3
x dx
0
1/2 2
t
0
2a
a
3
1
2
x dx
(2 a ) 2
2
a2
2
0
1
3
2
3u 2 du
1
1/2
2
0
0
1
42.
0
2
2
0
3
3
u du
24
x 5) dx
(3x 2
x 5) dx
0
2 z dz
1
1/2
3
3
3 dz
2 2
x
0
1
0
2
3
0
2 2
1 2
0
u du
24
dx
(3 x 2
1 2
u
0
2
0
3
u du
x dx
x 5) dx
2
0
5 x dx
3a
2
a2
2
2
(3b)3
3
5
1 2
2
2 32
03
3
du
3
24 13
03
3
3
3 23
1 2
x dx
0
1
0
x dx
12
2
2
0
a2
9b3
x dx
02
2
2
5 22
10
02
2
1
1 12 32
3[0 3]
03
3
13
3
1 3
2
1
0
3
3 23
13
3
0
3 73
7
7
3
22
2
7
4
9 9
24 38
02
2
5 dx
7
2
Copyright
24
2
2 0
23
3
3
5 dx
3
2
3 dz
1/2 2
u
0
du
2
0
3
dx
2
dz 1[1 2] 12 22
0
z dz
0
u 2 du
3
2
3b 2
x
0
4 6
02
2
1 2z
2 1
1 dz
d
x dx
a
3
2
2
2
1
dz
3(2 0)
2
2 dt
0
1z
22
2 2
1
2
t dt
02
2
2
2 22
3 dt
0
1 dz
(3x 2
5
14
1
24u 2 du
3
2
40.
t dt
3a
0.009
3
0
38.
b
3
2
2 dt
(2 z 3) dz
1
1
2
1 2z dz
2
0
3
7(1 3)
t
3a 2
2
(0.3)3
3
ds
/2 2
36.
3
3b
(2t 3) dt
2
1
3
0.3 2
s
0
34.
7
3
1
24
x dx
0
3
1 3
2
dt
7 dx
7
3
b 2
0
3
3
7 2
2014 Pearson Education, Inc.
7
5[2 0] (8 2) 10
3
3 13
03
3
12
2
02
2
0
5(1 0)
Section 5.3 The Definite Integral
51. Let x b n 0 bn and let x0 0, x1
x, x2 2 x, ,
xn 1 ( n 1) x, xn n x b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
f (c1 ) x f ( x) x 3( x )2 x 3( x)3
f (c2 ) x f (2 x) x 3(2 x)2 x 3(2) 2 ( x)3
f (c3 ) x f (3 x) x 3(3 x) 2 x 3(3) 2 ( x)3
f (cn ) x
Then Sn
3( x )3
3
n
k 1
k2
b
0
b
n
3(n)2 ( x)3
3k 2 ( x )3
k 1
n ( n 1)(2 n 1)
6
b3
n3
3
1
n2
b 0
n
52. Let x
n
f (ck ) x
k 1
2 n3
b
2
3(n x)2 x
f (n x) x
n
3
lim b2 2 n3
n
3x 2 dx
and let x0
0, x1
b3 .
1
n2
x, x2 2 x, . . . ,
xn 1 (n 1) x, xn n x b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
( x)2 x
( x )3
f (c1 ) x
f ( x) x
f (c2 ) x
f (2 x ) x
(2 x) 2 x
(2)2 ( x)3
f (c3 ) x
f (3 x) x
(3 x)2
x
(3)2 ( x)3
f (cn ) x
f (n x) x
(n x) 2 x
( n ) 2 ( x )3
n
Then Sn
k 1
b3
n3
n
f (ck ) x
k 1
b3
6
n ( n 1)(2 n 1)
6
b
x 2 dx
0
b 0
n
53. Let x
lim
n
k 2 ( x )3
b3
6
2 n3
2 3n
( x)3
n
k 1
k2
1
n2
1
n2
b3 .
3
b
n
and let x0 0, x1 x, x2 2 x, ,
xn 1 ( n 1) x, xn n x b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
f (c1 ) x f ( x) x 2( x)( x) 2( x) 2
f (c2 ) x f (2 x) x 2(2 x)( x) 2(2)( x)2
f (c3 ) x f (3 x) x 2(3 x)( x) 2(3)( x )2
f (cn ) x
Then Sn
2( x)
b
0
2
f (n x) x
n
k 1
n
k 1
2 x dx
f (ck ) x
k
2
b2
n2
2(n x)( x)
n
2k ( x ) 2
k 1
n ( n 1)
2
lim b 2 1 1n
n
2(n)( x)2
b 2 1 1n
b2 .
Copyright
2014 Pearson Education, Inc.
359
360
Chapter 5 Integration
b 0
n
54. Let x
b
n
and let x0 0, x1
x, x2 2 x, ,
xn 1 ( n 1) x, xn n x b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
x 1 ( x) 1 ( x )2
f (c1 ) x f ( x) x
x
2
2
f (c2 ) x
2 x
2
3 x
2
f (2 x) x
f (c3 ) x
f (3 x) x
f (cn ) x
f (n x) x
n
Then Sn
1 b2
4
k 1
1 1n
1
3
56. av( f )
3
3
3 13
3
3 13
1 (2)(
2
1 (3)(
2
x) 2
1 dx
lim
n
x
x)
x)2
x)2
x
1(
2
1 b2
4
1 1n
3 2
1
x dx
3 0
( x 2 1) dx
x
2
1 ( n)(
2
1 ( x)
1 k(
2
x) 2
x
n
k 1
k
b
x
1 b2
4
n
1
k 1
1 b2
2 n2
b.
3
1
1dx
3 0
3
1
3
3
1
3 0
1 33
6 3
57. av( f )
3
0
1 ( x)
n x
2
k 1
b x
0 2
b
1
3 0
55. av( f )
58. av( f )
f (ck ) x
n
1 ( x)
3
3 0
x2
2
1 1 0.
1
3
1
2
( 3 x 2 1) dx
3
0
dx
3 2
0
x dx
3.
2
1
1 0
1
0
(1 0)
1
1 0
1
0
1
0
x dx
1 dx
2.
(3 x 2 3) dx
3(1 0)
1 2
0
3
1 2
1
0
0
x dx
3 dx
2.
Copyright
2014 Pearson Education, Inc.
n ( n 1)
2
b
n
( n)
Section 5.3 The Definite Integral
3
1
3 0
59. av( f )
0
(t 1) 2 dt
1 3 t 2 dt
3 0
0) 1.
1 33
3 3
2 32
3 2
02
2
60. av( f )
1
1 ( 2)
1
1 1 t 2 dt
3 0
1 (3
3
3
1 ( 2)
3
3
1 13
3 3
0
1
1
2
2
dt 13 12
( 2) 2
2
dt
1 1
3 2
1 31 dt
3 0
t dt
3.
2
1
2
1
1
(| x| 1) dx
1
( x 1) dx 12 ( x 1) dx
0
0
1
(b) av( g )
1 3x
2 1
1 0 1 dx
2 1
x dx
( 1)2
2
1 02
2 2
1.
2
(c) av( g )
1 1 t 2 dt
3 2
1
1 ( 1)
61. (a) av( g )
1
2
t 2 t dt
2
2 2
1
t
3 0
2 3t
3 0
1 (0
2
3
1
3 1
1
3
1
1
dx 12 1 dx
0
02
2
1 12
2 2
1 3 (x
2 1
1 32
2 2
1
1 1 (| x |
4 1
1 ( 1 2)
4
( 1))
(| x | 1) dx
3
dx 12 1 dx
1
3 1( 1)
1 1x
2 0
12
2
1 (1
2
0)
1) dx
1 (3
2
1)
1.
(| x | 1) dx
3
1) dx 14 (| x | 1) dx
1
1
4
(see parts (a) and (b) above).
Copyright
2014 Pearson Education, Inc.
361
362
Chapter 5 Integration
0
1
0
1
0 ( 1)
62. (a) av( h)
1
1
1 0
(b) av( h)
12
2
(c) av(h)
1
2
0
1
2
1
2
1
( 1) 2
2
02
2
x dx
02
2
1.
2
1
1 ( 1)
1
1
0
1
2
( x) dx
1
0
x dx
| x | dx
1
| x | dx
1
1.
2
| x | dx
0
0
| x | dx
1
| x | dx
1 (see
2
parts (a) and (b) above).
63. Consider the partition P that subdivides the interval [a, b] into n subintervals of width
the right endpoint of each subinterval. So the partition is P
ck
a
As n
k (b a )
. We
n
and P
n
get the Riemann sum
n
f (ck ) x
k 1
b a,
n
a, a
c (b a )
n
b a
n
c
k 1
b
0 this expression remains c(b a). Thus,
a
a
c dx
2(b a )
, ...,
n
a
c (b a )
n
n
n
1
k 1
c(b
We get the Riemann sum
n
n
f (ck ) x
k 1
P
k 1
2 2nk
0 the expression
1 n2
4( n 1)
n
2
n
n
k 1
4k
n
1
8
n2
2 has the value 4 2
Copyright
n
k
k 1
2
n
1
k 1
6. Thus,
2
0
2 , 2 2 , ... ,
n
n
x
2 0
n
n
2 and ck
8 n ( n 1)
2
n2
2
n
n
4( n 1)
n
(2 x 1) dx
6.
2014 Pearson Education, Inc.
be
c(b a).
2
n
0,
n
b a and let c
k
n
n (b a )
and
n
a) .
64. Consider the partition P that subdivides the interval [0, 2] into n subintervals of width
be the right endpoint of each subinterval. So the partition is P
x
2
n
and let ck
k n2
2 . As n
2k .
n
and
Section 5.3 The Definite Integral
65. Consider the partition P that subdivides the interval [a, b] into n subintervals of width
x
2(b a )
the right endpoint of each subinterval. So the partition is P a, a b n a , a
, ... , a
n
n
n
n
k (b a )
k (b a ) 2
2 b a
b a
ck a
.
We
get
the
Riemann
sum
f
(
c
)
x
c
a
k
k
n
n
n
n
k 1
k 1
k 1
n
n
n
n
2
2
(b a ) 2
2 2ak (b a ) k (b a )
2 2a(b a)
b a
b a
a
a
k
k2
n
n
n
n
n2
n2
k 1
k 1
k 1
k 1
2 a (b a )2 n ( n 1)
2
n2
na 2
b a
n
(b a) a 2
(b a )a 2
1
1
b3
3
x dx
1
n
3
3
n
2
(b a )3
6
(b a )
6
a (b a ) 2 1
b 2
a
a(b a )2
(b a )3 n ( n 1)(2 n 1)
6
n3
a (b a) 2
ba 2
As n
a3
and P
ab 2
0 this expression has value
2a 2b a3
1 (b3
3
3b 2 a 3ba 2
1
n
1 kn . We get the Riemann sum
1 k 1n
k 1
2
Thus,
1 kn 1 2nk
3( n 1)
2n
0
1
b3
3
a3 )
1, 1 1n , 1 2 1n ,
k
n
( n 1)(2 n 1)
6n2
2
n
2
n
k 1
. As n
( x x 2 ) dx
3
n2
1
k
k 1
n
k 1
2
n
3 n( n 1)
2
n2
n
1
n
n( n 1)(2n 1)
6
1
n3
0 this expression has value 2 32
1 k n3
n
18k
n
3
k 1
36( n 1)
n
2
Thus,
(3x 2
1
18
5.
6
1
3
5.
6
1 3nk . We get the Riemann sum
27 k 2 2 6k
n
n2
27( n 1)(2 n 1)
1
18
n
n
1
k 1
. As n
2n2
2 x 1)dx
72
n2
n
k
k 1
and || P ||
1, 1
n
n
f (ck ) x
k 1
n
81
n3
k
n( n 1)
n 722
2
n
18
n
k 1
1 2
2
1 3nk
3
k 1
2
3,
n
2 ( 1)
n
3,
n
2
2
n
n
1
k 1
2
n
6k
n
n( n 1)
n 122
2
n
8k 3
n3
1 3nk
and
3
n
2
1 n3
9.
9.
1 2nk . We get the Riemann sum
12 k 2
n2
, 1 n
0 this expression has value 18 36 27
ck be the right endpoint of each subinterval. So the partition is P
1 k n2
3
n
81 n ( n 1)(2 n 1)
6
n3
68. Consider the partition P that subdivides the interval [ 1, 1] into n subintervals of width x
ck
1 and
n
1 0
n
, 1 n
2
1 kn
1 kn
f (ck ) x
k 1
n
1
k2
3
n
k 1
n
and || P ||
n
let ck be the right endpoint of each subinterval. So the partition is P
3
n
Thus,
0 ( 1)
n
x
67. Consider the partition P that subdivides the interval [ 1, 2] into n subintervals of width x
and ck
a3 .
3
a3 .
3
let ck be the right endpoint of each subinterval. So the partition is P
n
be
( b a )3 ( n 1)(2 n 1)
62
n
n 1
n
66. Consider the partition P that subdivides the interval [ 1, 0] into n subintervals of width
and ck
b a and let c
k
n
n (b a )
and
n
1
n2
1
2
(b a ) a 2
363
2
n
n
1
k 1
24 n ( n 1)(2 n 1)
6
n3
6
n
16
n4
Copyright
n
k
k 1
n
1, 1
n
f (ck ) x
k 1
n
12
n2
n ( n 1) 2
2
k
k 1
2
8
n3
2,
n
ck3 n2
2
n
k 1
n
3
1 2
n
k 1
2,
n
1 ( 1)
n
, 1 n
2
n
2
n
and let
1 and
3
1 2nk
k
k 1
2 6 nn 1 4
( n 1)(2 n 1)
n
2014 Pearson Education, Inc.
2
4
( n 1)2
n
2
1
1
2 6 1n
364
Chapter 5 Integration
1
Thus,
1
2
1
1
2 6 1n
4
x3dx
3
n
1
n2
1
4
2
n
1
1
1
n2
. As n
and || P ||
0 this expression has value 2 6 8 4
0.
b a and
n
n (b a )
n
69. Consider the partition P that subdivides the interval [a, b] into n subintervals of width x
k (b a )
. We
n
a
b a
n
n
a
3
k 1
n
f (ck ) x
k 1
3a 2 k ( b a )
n
3ak 2 (b a )2
n
3a 2 (b a ) 2 n ( n 1)
2
n2
na3
b a
n
n
get the Riemann sum
k 3 ( b a )3
2
n
b a
n
3
3a (b a )3 n ( n 1)(2 n 1)
6
n3
(b a) a 3
3a 2 ( b a ) 2 n 1
2
n
a (b a )3 ( n 1)(2n 1)
2
n2
(b a) a 3
1
3a 2 ( b a ) 2 1 n
2
1
a (b a ) 3
2
2
3a (b a )
2
value (b a ) a3
2
2
3
n
1
1
n2
(b a )
4
a (b a )3
k 1
n
ck3 b na
a
n
b a
n
3a 2 (b a )
n
3
k 1
(b a )4
n( n 1) 2
2
k
,a
n
3a ( b a ) 2
n
k 1
2
2
n
1
1 3
n n
n
k
k 1
k.
n
b4
4
a4 .
4
1
. As n
Thus,
n
We get the Riemann sum
and || P ||
b 3
a
1
n3
k3
k 1
3 n ( n 1)
2
n2
n
b4
4
x dx
3 n 1
2 n
1
0 this expression has value 32
1
4
5 . Thus,
4
71. To find where x x 2 0, let x x 2
b 1 maximize the integral.
0
x(1 x) 0
|| P ||
3ck
2
0
73.
f ( x)
1
1 x2
0
(3x x3 ) dx
x
f occurs at 1
1
2
min f
1 1
dx
0 1 x2
f (1)
1
1 12
ck3
1
n
3 1
2 1
Therefore, (1 0) min f
1. That is, an upper bound 1 and a lower bound
Copyright
n
k3
k 1
0 this expression has
,0 n
1
n
1
n
n
k 1
1
4
1
1
n
3 kn
2
n
1 0
n
1
n
and let ck be
1 and
k 3
n
1
n2
1
. As n
and
5.
4
x 1, then 0
0 or x
2, 2
maximum value of f occurs at 0
1.
2
n
3
a4 .
4
1,
n
0 or x 1. If 0
2
is decreasing on [0, 1]
2
2
1 ( n 1)
2
4
n
x 2 ( x 2 2) 0
x
72. To find where x 4 2 x 2 0, let x 4 2 x 2 0
4
2
++++++ 0
0
0 +++++++, we can see that x 2 x
0 on
minimize the integral.
1,0
n
k 1
n ( n 1) 2
2
1
n4
0, 0
f (ck ) x
k 1
n
( b a )3
1
n2
the right endpoint of each subinterval. So the partition is P
0 k 1n
k2
k 1
70. Consider the partition P that subdivides the interval [0, 1] into n subintervals of width x
ck
b and
3
k (b a )
n
a
k 1
n
,
let ck be
n4
(b a )4 ( n 1)2
4
n2
(b a ) 4
4
4
2(b a )
n
a, a b n a , a
the right endpoint of each subinterval. So the partition is P
ck
0.
max f
1 1
0 1 x2
1.
2
2014 Pearson Education, Inc.
dx
x x2
a
0 and
2. By the sign graph,
a
2 and b
2
f (0) 1; minimum value of
(1 0) max f
Section 5.3 The Definite Integral
1
1 02
74. See Exercise 73 above. On [0, 0.5], max f
0.5
(0.5 0) min f
1
1 12
and min f
1
4
Then
f ( x) dx
0
1
1
0.5 1 x 2
dx
1 for all x
1
2
dx
1
0
1
x2
dx
1 1
0 1 x2
13
20
sin x 2 dx
0.8. Therefore
1.
2
dx
0
0.5 1
2
5
1
(1 0)( 1)
0.5 1
1 x2
2
5
(0.5 0) max f
0.5. Therefore (1 0.5) min f
0.5 1
1 x2
2
5
1 sin x 2
75.
0
1
1 (0.5)2
1, min f
365
1
1 (0.5)2
On [0.5, 1], max f
1
1
0.5 1 x 2
1
4
(1 0.5) max f
dx
0.8
2.
5
9 .
10
dx
1
(1 0)(1) or
0
1
sin x 2 dx 1
0
sin x 2 dx cannot
equal 2.
76. f ( x)
x 8 is increasing on [0, 1]
1
(1 0) min f
77. If f ( x)
Then b
78. If f ( x)
x 8 dx
0
b a
0
(b a ) min f
0 on [a, b], then min f
0
(b a) max f
79. sin x
x for x
0
sin x x
1
0
1
sin x dx
0
1
x dx
2
80. sec x 1 x2 on
2
,
0
2
since [0, 1] is contained in
1
0
1
sec x dx
0
1 dx
b
a
0 for x
2
1 x2
1
2
0
1 1 x 2 dx
2 0
1
0
b
a
b
av( f ) dx
a
K dx
a
b
a
b
a
f ( x ) dx
f ( x ) dx
(b a ) max f .
(b a ) max f . Then
2
0
,
1 . Thus
2
1
2
0
2
1 x2 dx
(1 0)
1
0 (see Exercise 78)
sin x dx
1
sec x dx
2 2. Therefore,
0.
0 on
b
1
b a a
b
K (b a)
1
0
sec x dx
81. Yes, for the following reasons: av( f )
0 8
0.
(sin x x) dx
02
2
sec x
,
f ( x) dx
0
f (0)
3.
0. Now, (b a) min f
1
12
2
x 8 dx
0
f ( x) dx
a
0
sin x dx
2
b
0
0
3 and min f
0 on [a, b]. Now, (b a) min f
0 and max f
b a
1 8
1
2 2
0 and max f
a
b
f (1)
(1 0) max f
0 on [a, b], then min f
a
max f
1
0
1
1 13
2 3
0
0
x dx
2
1 x2
dx
1
sec x dx
sec x dx
0
7.
6
0 (see Exercise 77)
2
1 x2 dx
Thus a lower bound is 76 .
f ( x) dx is a constant K. Thus
av( f ) dx
b
(b a) b 1 a
f ( x ) dx
a
(b a ) K
b
a
f ( x) dx.
82. All three rules hold. The reasons: On any interval [a, b] on which f and g are integrable, we have:
(a) av( f
g)
b
1
[ f ( x)
b a a
1
b a
g ( x)]dx
b
a
f ( x ) dx
b
a
b
1
b a a
g ( x ) dx
av( f ) av( g )
(b) av(kf )
b
1
kf ( x) dx
b a a
1
b a
k
b
a
Copyright
f ( x) dx
b
k b 1a
f ( x ) dx
a
0
an upper bound is 12 .
sec x
0
0
1
sin x dx
k av( f )
2014 Pearson Education, Inc.
b
f ( x) dx b 1 a g ( x) dx
a
366
Chapter 5 Integration
b
1
b a a
(c) av( f )
b
1
g ( x)
b a a
f ( x) dx
Therefore, av( f )
b
g ( x) on [a, b], and b 1 a g ( x) dx
a
dx since f ( x)
av( g ).
av( g ).
83. (a) U max1 x max 2 x
max n x where max1 f ( x1 ), max 2 f ( x2 ) , , max n f ( xn ) since f is
min n x where min1 f ( x0 ), min 2 f ( x1 ), ,
increasing on [a, b]; L min1 x min 2 x
min n f ( xn 1 ) since f is increasing on [a, b]. Therefore
U L (max1 min1 ) x (max 2 min 2 ) x
(max n min n ) x
( f ( x1 )
f ( x0 )) x ( f ( x2 )
f ( x1 )) x
( f ( xn )
f ( xn 1 )) x
( f ( xn )
f ( x0 )) x
( f (b) f (a )) x.
(b) U max1 x1 max 2 x2
max n xn where max1 f ( x1 ), max 2 f ( x2 ) , , max n f ( xn ) since f
is increasing on [a, b]; L min1 x1 min 2 x2 ... min n xn where min1 f ( x0 ), min 2 f ( x1 ), ,
min n
f ( xn 1 ) since f is increasing on [a, b]. Therefore
U
(max1 min1 ) x1 (max 2 min 2 ) x2
U
L
( f ( x1 )
f ( x0 )) x1 ( f ( x2 )
( f ( x1 )
f ( x0 )) xmax
L
( f ( xn )
lim (U
P
L)
0
( f ( x2 )
f ( x0 )) xmax
lim ( f (b)
P
f ( x1 )) x2
0
(max n min n ) xn
( f ( xn ) f ( xn 1 )) xn
f ( x1 )) xmax
( f (b)
( f ( xn )
f ( a )) xmax
f (a)) xmax
f (b)
0, since xmax
f ( xn 1 )) xmax . Then
f ( a) xmax since f (b)
f (a ). Thus
P.
84. (a) U max1 x max 2 x
max n x where
max1 f ( x0 ), max 2 f ( x1 ), , max n f ( xn 1 )
since f is decreasing on [a, b];
L min1 x min 2 x
min n x where
min1 f ( x1 ), min 2 f ( x2 ), , min n f ( xn )
since f is decreasing on [a, b]. Therefore
U L (max1 min1 ) x (max 2 min 2 ) x
... (max n min n ) x
( f ( x0 ) f ( x1 )) x ( f ( x1 ) f ( x2 )) x
... ( f ( xn 1 ) f ( xn )) x
( f ( x0 ) f ( xn )) x ( f (a ) f (b))
(b) U max1 x1 max 2 x 2 ... max n
since f is decreasing on [a, b]; L min1
min1 f ( x1 ), min 2 f ( x2 ), , min n
U
L
x.
xn where max1 f ( x0 ), max 2 f ( x1 ), , max n
x1 min 2 x2
min n xn where
f ( xn ) since f is decreasing on [a, b]. Therefore
(max1 min1 ) x1 (max 2 min 2 ) x2
(max n min n ) xn
( f ( x0 ) f ( x1 )) x1 ( f ( x1 ) f ( x2 )) x2
( f ( xn 1 ) f ( xn )) xn
( f ( a) f (b) xmax
f (b) f (a ) xmax since f (b) f (a ). Thus
lim (U L )
lim f (b) f (a) xmax 0, since xmax
P.
P
P
0
85. (a) Partition 0,
x2
2
2 x,... , xn
( f ( x0 )
f ( xn )) xmax
0
into n subintervals, each of length x
n x
2
2n
. Since sin x is increasing on 0,
of the circumscribed rectangles of areas f ( x1 ) x
f ( xn ) x
f ( xn 1 )
with points x0
2
x,
, the upper sum U is the sum of the areas
(sin x ) x, f ( x2 ) x
(sin n x) x.
Copyright
0, x1
2014 Pearson Education, Inc.
(sin 2 x ) x,... ,
Section 5.3 The Definite Integral
Then U
cos
(sin x sin 2 x ... sin n x ) x
cos 4 n
cos
2
cos 4 n
4n
4 n sin 4 n
cos
sin
2
x
2
1
2
cos n
2sin
x
x
2
x
cos 4 n
cos n
367
1
2 2n
2n
2sin 4 n
4n
4n
4n
/2
(b) The area is
0
sin x dx
lim
cos 4 n
cos
n
sin
2
4n
4n
1 cos 2
1
1.
4n
n
86. (a) The area of the shaded region is
xi mi which is equal to L.
i 1
n
xi M i which is equal to U.
(b) The area of the shaded region is
i 1
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part
of the figure and the first part of the figure. Thus this area is U L.
n
87. By Exercise 86, U
L
n
xi M i
i 1
mi
min { f ( x) on ith subinterval}. Thus U
n
i 1,
xi mi where M i
i 1
, n. Since
i 1
n
L
n
(M i
mi ) xi
i 1
n
xi
max { f ( x) on the ith subinterval} and
xi
(b a ) the result, U
xi provided xi
i 1
L
(b a ) follows.
i 1
88. The car drove the first 150 miles in 5 hours and the second
150 miles in 3 hours, which means it drove 300 miles in
mi/hr 37.5 mi/hr. In
8 hours, for an average value of 300
8
terms of average value of functions, the function whose
average value we seek is v(t )
average value is
89-94.
(30)(5) (50)(3)
8
30, 0 t 5
50, 5 t 8 ,
and the
37.5.
Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
f : x -> 1-x;
a : 0;
b : 1;
N : [4, 10, 20, 50];
P : [seq( RiemannSum( f(x), x a..b, partition n, method random, output plot ), n N )]:
display( P, insequence true);
Copyright
2014 Pearson Education, Inc.
for each
368
Chapter 5 Integration
95-102. Example CAS commands:
Maple:
with( Student[Calculus1] );
f : x - sin(x);
a : 0;
b : Pi;
plot( f(x), x a..b, title "#95(a) (Section 5.3)" );
N : [ 100, 200, 1000 ];
# (b)
for n in N do
Xlist : [ a 1.*(b-a)/n*i $ i 0..n ];
Ylist : map( f, Xlist );
end do:
for n in N do
Avg[n] : evalf(add(y,y Ylist)/nops(Ylist));
# (c)
end do;
avg : FunctionAverage( f(x), x a..b, output value );
evalf( avg );
FunctionAverage(f(x),x a..b, output plot);
fsolve( f(x) avg, x 0.5 );
fsolve( f(x) avg, x 2.5 );
fsolve( f(x) Avg[1000], x 0.5 );
fsolve( f(x) Avg[1000], x 2.5 );
# (d)
89-102. Example CAS commands:
Mathematica: (assigned function and values for a, b, and n may vary)
Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands
Clear[x, f, a, b, n]
{a, b} {0, }; n 10; dx (b a)/n;
f Sin[x]2 ;
xvals Table[N[x],{x, a, b dx, dx}];
yvals f /.x
xvals;
boxes MapThread[Line[{{#1, 0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals dx, yvals}];
Plot[f, {x, a, b}, Epilog boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of
commands.
Clear[x, f, a, b, n]
{a, b} {0, }; n 10; dx (b a)/n;
f Sin[x]2 ;
xvals Table[N[x], {x, a dx, b, dx}];
yvals f /.x
xvals;
boxes MapThread[Line[{{#1, 0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, dx,xvals, yvals}];
Plot[f, {x, a, b}, Epilog boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
Copyright
2014 Pearson Education, Inc.
Section 5.4 The Fundamental Theorem of Calculus
Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands.
Clear[x, f, a, b, n]
{a, b} {0, }; n 10; dx (b a)/n;
f Sin[x]2 ;
xvals Table[N[x], {x, a dx/2, b dx/2, dx}];
yvals f /.x
xvals;
boxes MapThread[Line[{{#1, 0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, dx/2, xvals dx/2, yvals}];
Plot[f, {x, a, b},Epilog boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
5.4
1.
2.
THE FUNDAMENTAL THEOREM OF CALCULUS
2
0
1
x2
1
2
3.
4.
x3
3
3
1
2
( x 3)
x
299
1
1
5.
3x
3
4
8.
9.
1
x2
0
32
1
1
x
x dx
3
(1)3
3
1
2
(5)
3(2)2
2
(1) 2 3(1)
1
2
x4
16
3
1
3
4
1
( 1)3
3
1
1
125
1
1 1
300
0
(1)
44
16
43
(0)3
3
3
13
14
16
3(0) 2
2
124
125
64 16 1
x 2 3x
1
( 2)
4
4
( 2)2 3( 2)
1
2 x3/2
3
0
32
5 x 1/5
1
2sec 2 x dx [2 tan x]0 /3
1
3
5
2
2
3
( 5)
2 tan 3
Copyright
81
105
6
4
4
0 1
5
2
(2 tan 0)
2 3 0
10
3
( 1)2 3( 1)
4
x4
4
x3
3
1
(2)3
3
1
(1)300 ( 1)300
300
32 3(3)
x 6/5 dx
/3
0
1
x 3 2 x 3 dx
2
x 2 3x
( x 3)
x3
dx
4
2
4
7.
dx
4
x 300
300
dx
1
3
0
2
3x2
2 0
x3
3
( x 2 3 x) dx
2 x 3 dx
4
6.
2
x( x 3) dx
2 3
2014 Pearson Education, Inc.
1
16
753
16
20
3
369
370
10.
11.
Chapter 5 Integration
(1 cos x ) dx [ x sin x ]0
0
3 /4
/4
4
0
13.
0
/3
14.
sin u
/3
/3
/3
16.
/4
0
/6
0
0
dt
1
/2 2
/3
csc 34
4
(1/2)
1 cos 2t
2
/8
0
/3
1 cos 2t
dt
2
/3
sin 2 t dt
1
4
3
2
1
4
6
/4
tan 2 x dx
0
t
2
4sec 2 t
4 tan
19.
20.
1
1
t
/6
0
3
4
4
3
3
0
1 sin 2(0)
4
1
2 2
1 sin 2
4
2
1
1
/3
/3
(r 2
4) dt
2r 1) dr
3
3
(t 3 t 2
3
2( 3) 2
2 sec 6
tan 4
4 3
(tan(0) 0) 1 4
4
/6
0
(2sec2 x 2sec x tan x 1) dx
(2 tan 0 2sec 0 0)
6
2
1 cos 2(0)
2
cos 2
(4sec2 t
1 cos(2t )
.
2
3
4
3
1
2
4 tan 3
4
(t 1)(t 2
3
1 (0)
2
(sec2 x 2sec x tan x tan 2 x) dx
4
dt
2
4
(r 1)2 dr
3
2
/3
sin 2t
4
3
2
1 cos 2 x /8
2
0
sin 2 x dx
3
1 sin 2t 0
4
2
(sec2 x 1) dx [tan x x]0 /4
(sec x tan x ) 2 dx
4
18.
2
4
1t
2
dt
[2 tan x 2sec x x ]0 /6 2 tan 6
17.
4
1
csc 4
Use the double angle formula cos 2t 1 2sin 2 t which implies that sin 2 t
sin 2 t dt
6
15.
4
cos u 0
du
cos2 u
1 cos 2t
2
2
sin ) (0 sin 0)
[ csc ]3 /4/4
csc cot d
/3
12.
(
t 2 ) dt
6
2
2
4
4
4 tan t
(4( 1) 4)
2 3
t
4
3
3
3
4 3 3
3
r3
3
r2
r
1
1
t4
4
4t 4) dt
3
4
Copyright
4
( 1)3
3
3
3
t3
3
( 1)2
2t 2
4t
( 1)
13
3
12 1
3
3
3
2(
3)2
4(
3)
2014 Pearson Education, Inc.
10 3
8
3
4
Section 5.4 The Fundamental Theorem of Calculus
21.
1
22.
23.
24.
u7
2 2
1
u5
1 y5 2 y
3 y3
dy
2 s2
s
1
3
1
1/3
2/3
8 x 1 2 x
x
2(8) 53 (8)5/3 3(8)
sin 2 x
/2 2sin x
25.
26.
/3
0
0
dx
/2
28.
4
4
1
0 2
30.
31.
32.
33.
4
cos x
sin 2
29.
0
| x | dx
2 1
x
1/2
4
0
0
1
( 1)3
3
3
2
s 1
2
8
dx
1
2
2
sin x
/2
5
2
1 sin 2 x
4
0
4
/2 1
(cos x
2
4
x dx
0
x1/3 ) dx
2
3
4
4
2
( 3)
22
3
dx
1 4 x2
2
x 1dx
1 e3ln 2
3
2
(ln x e x )
1
1/2
4sin 1 x
0
1/ 3
0
x
4
2
dx
1 22 x 2
1
(4
1 e0
3
1 e ln 8
3
2 x 53 x5/3 3 x 2/3
137
20
(sin( ))
sin 2
0
1
2 sec2 x dx
4
x2
2 0
4
1 (cos x
/2 2
1
3
(ln 2 e 2 ) (ln1 e 1 )
4 sin 1 12
4sin 1 0
1/ 3
1 tan 1 (2 x )
2
0
4 6
1 tan 1
2
( 4)2
2
02
2
cos x) dx
/2
0
2
3
ln 2
4(0)
1
e2
1
e
2
3
1 tan 1 (0)
2
2014 Pearson Education, Inc.
1 tan 1
2
42
2
02
2
16
cos x dx [sin x]0 /2
7
3
2 )
Copyright
3
3 x 4/3
4
1
9 3
8
5
6
x2
2
8
3
8 1
/3
x dx
1
3
4
2
0
tan(0)
cos x) dx
4
3 (1) 4/3
4
/3 cos 2 x 1
2
0
x tan x
5 (0)
2
1 sin 2(0)
4
1
2 23/4 1
2
1
1
2
(2 x 2/3 2 x 1/3
cos x dx
( 3)3
3
2
( 1)
2(1) 35 (1)5/3 3(1)2/3
/2
| x | dx
cos x dx
dx
1/ 3
2
4
2
( 2)8
16
1
4(1)4
(cos2 x 2 sec2 x) dx
tan 3
| x | dx
e x dx
1 x
4
0
ln 2
1 e3 x
3
0
0
0
/3
18
16
sin 0 1
ln 2 3 x
e dx
0
1
dx
sec2 x dx
5
2 3
1 sin 2
4
3
27.
8 2 x1/3 x 2 x 2/3
x1/3
3 (8)4/3
4
1
1
4u 4
2y 1
s
1
2/3
2sin x cos x
2sin x
5
2
y3
3
2 y 2 ) dy
(cos x sec x)2 dx
/3 1
cos 2 x
2
u8
16
u 5 du
(1 s 3 2 ) ds
dx
1/3
1
u7
2 2
( y2
2
ds
s2
1
1
du
371
2
3
372
34.
35.
36.
37.
38.
Chapter 5 Integration
0
x 1
1
0
2
xe x dx
1 ex
2
x
2 1 x
/3
0
39. (a)
2
40. (a)
(b)
41. (a)
(b)
42. (a)
x
0
43. (a)
(b)
1 e1
2
0
x
0
sin x
1
t4
/3
tan
0
d
d
x3
0
d
dx
3t 2 dt
t 4 1/2
u
0
x
0
3
e t dt
26
5
1 (sin x )3
3
0
sin x sin 0
du
1 sin 3
3
3
x
d
dx
sin x
0
(cos x ) 12 x 1/2
sin x
d
dx
t4
2 u 3/2
3
0
[tan y ]0tan
/3
1 sin 3 (0)
3
cos t dt
1
3t 2 dt
cos x
2 x
d (sin 3
dx
x 1)
3sin 2 x cos x
3sin 2 x cos x
2 (t 4 )3/2
3
t 2 (4t 3 )
2 t6
3
d
dt
tan (tan )
d
d
0
4
t
u du
0
tan (tan ) 0
0
(sec 2 (tan )) dd (tan )
3
e x 1
3
d ( x3 )
e x dx
d
dx
tan
0
sec2 y dy
x3
0
e t dt
(sec2 (tan )) sec2
d
dx
3
e x 1
3
3x 2e x
3
3x 2e x
Copyright
d 2 t6
dt 3
4t 5
(sec 2 (tan )) sec 2
x3
e t
3
8
cos x
2 x
sin 3 x 1
d (t 4 )
t 4 dt
sec2 y dy
e t dt
5
d (sin x )
(3sin 2 x) dx
(tan(tan ))
0
1 x2
1 (ln 2) 2
2
(sin x)2 cos x dx
[t 3 ]1sin x
sec 2 y dy
tan
1)
d ( x)
(cos x ) dx
u du
0
)
1 (ln1) 2
2
cos x 12 x 1/2
3t 2 dt
t4
d
dt
0
[sin t ]0 x
u du
0
1 (e
2
1 (ln 2) 2
2
cos t dt
1
1 e0
2
2
2
x)
sin x
d
dx
1
(
x(1 x 2 )1/2 dx
cos t dt
d
dx
d
d
(b)
1
sin 2 x cos x dx
d (sin
dx
(b)
2
5
dx
2
1
ln
1
2
1 (ln x ) 2
2
1
2 ln x
dx
1 x
5
x 10
1
ln
dx
1
2014 Pearson Education, Inc.
4t 5
Section 5.4 The Fundamental Theorem of Calculus
t
44. (a)
x4
0
3
1 x2
t
d
dt 0
(b)
45. y
47.
y
48.
y
t
d
dt 0
x
x4
x4
0
x
y
50.
y
51.
y
52.
y
53.
y
54.
y
55.
y
56.
y
1 x
3
x2
x
2
x
t2
1t
x
0
x
2
4
sin x
dt
1 t
2
, x
tan x dt
1 t2
ex
2
0
1
t
1 3
t
2x
sin 1 t
0
x1/
1
dt
dy
dx
x2
2
1 5 t 3/2
5 2
t2
t
sin( x )2
sin t 3 dt
x2
1
2
e
x2
2 x 1/3
t dt
1
x
2
x 3
(t
0
x2
4
x
2
3
1 t
d (
dx
1 t 3/2
2
1
2 t
x1
1 t
x)
sin t 3 dt
1 t 3/2
2
3
1
1 t 2 t
3
2 t t2
dy
dx
dt
(sin x)
3
2 t t2
1
2
x
1,
x
x
0
1/2
d ( x2 )
x sin ( x 2 )3 dx
sin x
2 x
x2
2
sin t 3dt
0
4
1)10 dt
2
d
dx
d (sin x)
1
1 sin 2 x dx
1
1 tan 2 x
1
x2
3
dy
dx
2
dy
dx
dt
d
dt
dy
dx
dy
dx
dt
3
dy
dx
0
3sin 1 t
sin t 3dt
(t 3 1)10 dt
0
3
1 t
3sin 1 t
46. y
d
dx
x
x t2
3 t2 4
dt
t2
sin t 2 dt
dy
dx
2
1 t 5/2
5
0
1 x2
0
x2
t
d 1 t 5/2
dt 5
dx
2
dy
dx
sin t 3 dt
3sin 1 x
dx
1 x2
sin t 2 dt
2 x 2 sin x6
49.
3
1 t 2 dt
0
x5
5
dx
d
dx
d (tan x)
dx
ex
2
dy
dx
1
sec 2 x
2 xe x
1
1 x2
e2
2
d (2 x )
(2 x )1/3 dx
cos t dt
dy
dx
d (sin 1 x )
cos(sin 1 x) dx
sin 1 t dt
dy
dx
1
sin 1 x
d
dx
Copyright
x
1
x
0
(t 3 1)10 dt
3( x3 1)10
1
cos x
cos x
cos2 x
(cos x )
sec 2 x
cos x
cos x
1
1 x2
2 xe 2
2 x /3 2 x ln 2
1 x2
1
sin 1 x
24 x /3 ln 2
1
1
1 x2
1
x
1
1
2014 Pearson Education, Inc.
x
0
2
(t 3 1)10 dt
1 since x
2
373
374
57.
Chapter 5 Integration
x2
2x
0
2
Area
3
x( x 2)
( x2
2
0
x3
3
( x2
2
x3
3
( 2)
0 or x
2;
( x2
2 x)dx
2
x3
3
2 x)dx
3
( 2)3
3
x
0
2 x)dx
2
x2
0
0
x2
( 3)3
3
2
03
3
02
( 2)3
3
23
3
22
03
3
( 3)
2
x2
0
2
( 2)2
02
28
3
58. 3 x 2 3 0
x2 1 x
1;
because of symmetry about the y -axis,
Area
1
2
0
2
(3 x 2 3)dx
1
(3x 2 3) dx
[ x3 3x]10 [ x3 3x]12
2
2[ ((13 3(1)) (03 3(0))) ((23 3(2)) (13 3(1))]
2(6) 12
59. x3 3 x 2 2 x 0
x( x 2 3 x 2) 0
x( x 2)( x 1) 0 x 0, 1, or 2;
1
Area
0
x4
4
( x3 3 x 2
x3
x2
24
4
60.
x1/3
x
or 1
x 2/3
0
1
x3
0 or 1
0
x2
2
2
14
4
1
2
13 1
x2
1
x) dx
( x3 3x 2
x2
14
4
22
( x1/3
3 (0) 4/3
4
1
4
0
x
3 x 4/3
4
1
4
x4
4
1
x1/3 1 x 2/3
0
Area
1
23
2
2 x)dx
0
x1/3
x
0 or x
x) dx
1
3(
4
12
2
3 (0) 4/3
4
02
2
3 (8)4/3
4
82
2
3 (1) 4/3
4
12
2
3
4
1)
2
0 or 1 x 2/3
03 0 2
0
x
0
1;
8
1
1
x2
3 x 4/3
2 0
4
2
( 1)
1) 4/3
2
3 x 4/3
4
02
2
04
4
1
2
3 (1) 4/3
4
( 20
13 12
0
( x1/3
2 x) dx
( x1/3
x) dx
8
x2
2 1
83
4
Copyright
2014 Pearson Education, Inc.
Section 5.4 The Fundamental Theorem of Calculus
61. The area of the rectangle bounded by the lines y
y 1 cos x on [0, ] is
0
shaded region is 2
2, y
0, x
(1 cos x) dx [ x sin x]0
(
, and x
6
3
63. On
4
2
4
3
2
cos 6
3
2
. The area between the curve y
On 0,
sin x on
4
sec
2
sec tan and y
/4
sec tan is
4
2
2 1
4
is
0
2
4
sec2 t on
under the curve y
curve y 1 t 2 on [0, 1] is
y
66.
y
67.
y
68.
y
69.
y
, 1 is 1
2
x1
t
x
x
x
0
1
x
0
, 0 is
(1 t 2 ) dt
t
1 dt
t
and y ( )
sec t dt 4
dy
dx
sec x and y (0)
dy
dx
/4
sec tan
1
x
2, and y
sec
/4
0
sec 4 sec 0
2, y
1
t2
3 0
1
0, t
4
d
and y (1)
4
4
sec
0 is 2
.
is 2
/4
( 2 1).
4
2
4
4
0
2
, 0 is
4
. The area
2 1. Therefore the area of
, and t 1 is 2 1
[tan t ]0
4
13
3
03
3
0
tan 0 tan
the area of the shaded region is 2
sec x and y ( 1)
dt 3
0
0, and
0, y
sec 2 t dt
4
dy
dx
x1
1 t
2
dy
dx
0,
3
.
sec t dt 4
1
0
dt 3
1
4
5 . Therefore
3
2
3
2, y
0 is
0 is
( 2 1). Thus, the area of the total shaded region is
2
2 1
4
sin 56 , and y
1
2
6
sin x dx [ cos x]5 /6/6
/6
,
4
sec tan d
64. The area of the rectangle bounded by the lines y
65.
is
sin
3. Therefore the area of the shaded region is 3
: The area of the rectangle bounded by
the shaded region on 0,
4
, 56
,y
2 1. Therefore the area of the shaded region on
4
under the curve y
on
6
5
6
5 /6
,x
6
, 0 : The area of the rectangle bounded by the lines y
( sec 0)
. Therefore the area of the
.
. The area under the curve y
cos 56
0 is 2 . The area under the curve
sin ) (0 sin 0)
62. The area of the rectangle bounded by the lines by the lines x
1 5
2 6
375
11
1t
3 0 3
1
1
0
0
sec t dt 4
sec t dt 4
dt 3 0 3
sec t dt 3
Copyright
3
2.
3
4
2
2
. The area
1. The area under the
Thus, the total area under the curves
5
3
2
4
1
3
2
.
(d) is a solution to this problem.
0 4
0 4
4
4
(c) is a solution to this problem.
(b) is a solution to this problem.
3
(a) is a solution to this problem.
70.
y
x
1
1 t 2 dt 2
2014 Pearson Education, Inc.
376
Chapter 5 Integration
b /2
71. Area
b /2
bh
2
72. k
bh
6
0
bh
6
k
3
2
( x 1)2
2 3 14 1
75. (a) t
t
1/2
1
2
2
x
dx
2
2 2 14
78.
79.
x
0
0
8
3
0
1)3/2
t
1/2
[t1/2 ]0x
dt
1
( x 1)2
dx
x; c(100) c(1)
2 x
x cos x
x 1 9
1 t
2
3( x 1)
3
1
x 1
2 3
0
dt
f (1)
T
85 3 25 16
25
1
1
85 3 25 t dt 25
25 0 0
1 85(0) 2(25 0)3/2
25)3/2 25
f ( x)
f ( x)
f ( x)
4
0
100
1
(3 1)
1 cos kx / k
k
0
sin kx dx
1 $9.00
1
(0 1)
0
d x
dx 1
d x
dx 0
f (t ) dt
d ( x2
dx
cos x
9
x 2
85t 2(25 t )3/2
f (1)
5 53 4
10.17 ft;
8
15 t 4/3
4
0
29
3
9.67 ft
2 x 1)
2x 2
x sin x
3; f (1)
f (4)
2
3x 5
Copyright
25
0
75 F
1)3/2
15 (0)4/3
4
f (t ) dt
9
1 ( x 1)
3( x 1) 2
1 2 (t
8 3
1)3/2
1 2 (0
8 3
76 F;
4 1 5(4)1/3
H
t 1 5t1/3 dt
15 (8) 4/3
4
f (t ) dt
2
/k
the area
k
2
k
1
8
1
8 0 0
x2 2 x 1
L( x)
bh
0 1 5(0)1/3 1 ft; t
8 1 5(8)1/3 13 ft
H
H
f (t ) dt
f ( x)
2
3
4.5 or $4500
85(25) 2(25
1 2 (8
8 3
x
bh
3
x1
0 2
c
(b) average height
1
b /2
b 2
2
2
0 T 85 3 25 0 70 F; t 16
25 T 85 3 25 25 85 F
1
25
77.
b /2
sin kx will occur over the interval 0,
(b) average temperature
76. (a) t
t
3b
1 cos (0)
k
k
1
2 x
0
4h
bh
one arch of y
dc
dx
74. r
bh
2
4 hx3
3b 2
hx
b
2
h
3b
1 cos
k
73.
b 2
2
2
4h
b
2
h
x 2 dx
4h
b2
h
2014 Pearson Education, Inc.
cos (4)
1 1 9
1 t
2
dt
(4) sin (4) 1
2 0
2;
Section 5.4 The Fundamental Theorem of Calculus
80. g ( x)
x2
3
1
sec(t 1) dt
( 1)2
g ( 1)
3
L( x)
2( x ( 1)) g ( 1)
81. (a)
(b)
(c)
(d)
(e)
(f )
(g)
sec(t 1) dt
1
3
1
sec(t 1) dt
2( x 1) 3
2 x sec( x 2 1)
3 0
g ( 1)
2( 1) sec (( 1)2 1)
2;
3;
2x 1
True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus.
True: g is continuous because it is differentiable.
True: since g (1) f (1) 0.
False, since g (1) f (1) 0.
True, since g (1) 0 and g (1) f (1) 0.
False: g ( x) f ( x) 0, so g never changes sign.
True, since g (1) f (1) 0 and g ( x) f ( x) is an increasing function of x (because f ( x) 0).
82. Let a
(a)
1
(sec( x2 1))(2 x)
g ( x)
377
n
x0
x1
x2
xn
b be any partition of [a, b] and left F be any antiderivative of f.
[ F ( xi ) F ( xi 1 )]
i 1
[ F ( x1 ) F ( x0 )] [ F ( x2 ) F ( x1 )] [ F ( x3 ) F ( x2 )]
[ F ( xn 1 ) F ( xn 2 )] [ F ( xn ) F ( xn 1 )]
F ( x0 ) F ( x1 ) F ( x1 ) F ( x2 ) F ( x2 )
F ( xn 1 ) F ( xn 1 ) F ( xn )
F ( xn ) F ( x0 ) F (b) F (a)
(b) Since F is any antiderivative of f on [a, b]
F is differentiable of [a, b] F is continuous on [a, b].
Consider any subinterval [ xi 1, xi ] in [a, b], then by the Mean Value Theorem there is at least one
number ci in ( xi 1, xi ) such that [ F ( xi ) F ( xi 1 )] F (ci )( xi xi 1 ) f (ci )( xi xi 1 ) f (ci ) xi .
n
Thus F (b) F (a )
[ F ( xi ) F ( xi 1 )]
i 1
n
(c) Taking the limit of F (b) F (a )
F (b) F (a )
83. (a) v
ds
dt
(b) a
df
dt
3
(c) s
0
d t
dt 0
b
a
i 1
n
i 1
f (ci ) xi .
f (ci ) xi we obtain lim ( F (b) F (a))
P
0
lim
P
n
0 i 1
f (ci ) xi
f ( x ) dx
f ( x) dx
f (t )
v(5)
f (5)
2 m/sec
is negative since the slope of the tangent line at t = 5 is negative
f ( x) dx
9
2
1 (3)(3)
2
m since the integral is the area of the triangle formed by y = f(x), the x-axis
and x = 3
(d) t = 6 since from t = 6 to t = 9, the region lies below the x-axis
(e) At t = 4 and t = 7, since there are horizontal tangents there
(f) Toward the origin between t = 6 and t = 9 since the velocity is negative on this interval. Away from the
origin between t = 0 and t = 6 since the velocity is positive there.
(g) Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the
x-axis than below it.
84.
lim
x
1 x dt
x 1 t
lim
x
1
x
2 t
x
1
lim
x
1
x
Copyright
2 x 2
lim 2
x
2
x
2 0
2014 Pearson Education, Inc.
2
378
Chapter 5 Integration
85 88.
Example CAS commands:
Maple:
with( plots );
f : x - x^3-4*x^2 3*x;
a : 0;
b : 4;
F : unapply( int(f(t),t a..x), x );
p1: plot( [f(x),F(x)], x a..b, legend ["y
# (a)
f(x)","y
F(x)"], title "#85(a) (Section 5.4)" ):
p1;
dF : D(F);
# (b)
q1: solve( dF(x) 0, x );
pts1: [ seq( [x,f(x)], x remove(has,evalf([q1]),I) ) ];
p2 : plot( pts1, style point, color blue, symbolsize 18, symbol diamond, legend "(x,f(x))
where F'(x) 0" ):
display( [p1, p2], title "85(b) (Section 5.4)" );
incr : solve( dF(x)>0, x );
decr : solve( dF(x)<0, x );
# (c)
df : D(f );
# (d)
p3 : plot( [df(x),F(x)], x a..b, legend ["y f '(x)","y
F(x)"], title "#85(d) (Section 5.4)" ):
p3;
q2 : solve( df(x) 0, x );
pts2 : [ seq( [x,F(x)], x remove(has,evalf([q2]),I) ) ];
p4 : plot( pts2, style point, color blue, symbolsize 18, symbol diamond, legend "(x,f(x))
where f '(x) 0" ):
display( [p3,p4], title "85(d) (Section 5.4)" );
89-92.
Example CAS commands:
Maple:
a:
u:
f:
F:
1;
x - x^2;
x - sqrt(1-x^2);
unapply( int( f(t),t a..u(x) ), x );
dF : D(F);
cp : solve( dF(x) 0, x );
solve( dF(x)>0, x );
solve( dF(x)<0, x );
# (b)
d2F : D(dF);
solve( d2F(x) 0, x );
# (c)
plot( F(x), x -1..1, title "#89(d) (Section 5.4)" );
Copyright
2014 Pearson Education, Inc.
Section 5.5 Indefinite Integrals and the Substitution Method
93.
Example CAS commands:
Maple:
f : `f `;
q1: Diff( Int( f(t), t a..u(x) ), x );
d1: value( q1 );
94.
Example CAS commands:
Maple:
f : `f `;
q2 : Diff( Int( f(t), t a..u(x) ), x,x );
value( q2 );
85-94.
Example CAS commands:
Mathematica: (assigned function and values for a, and b may vary)
For transcendental functions the FindRoot is needed instead of the Solve command.
The Map command executes FindRoot over a set of initial guesses
Initial guesses will vary as the functions vary.
Clear[x, f, F]
{a, b} {0, 2 }; f[x_] Sin[2x] Cos[x/3]
F[x_] Integrate[f[t],{t, a, x}]
Plot[{f[x], F[x]},{x, a, b}]
x/.Map[FindRoot[F'[x] 0, {x, #}] &, {2, 3, 5, 6}]
x/.Map[FindRoot[f '[x] 0, {x, #}] &, {1, 2, 4, 5, 6}]
Slightly alter above commands for 89 94.
Clear[x, f, F, u]
a 0; f[x_]
x2
2x 3
2
u[x_] 1 x
F[x_] Integrate[f[t], {t, a, u(x)}]
x/.Map[FindRoot[F'[x] 0, {x, #}] &, {1, 2, 3, 4}]
x/.Map[FindRoot[F"[x] 0, {x, #}] &, {1, 2, 3, 4}]
After determining an appropriate value for b, the following can be entered
b 4;
Plot[{F[x],{x, a, b}]
5.5
INDEFINITE INTEGRALS AND THE SUBSTITUTION METHOD
1. Let u
2x 4
5
2(2 x 4) dx
2. Let u
7x 1
7 7 x 1 dx
du
2 dx
2u 5 12 du
du
7 dx
1/2
7(7 x 1)
1 du
2
5
u du
1 du
7
dx
dx
1 u6
6
C
1 (2 x
6
4)6 C
dx
7u1/2
Copyright
1
7
du
u1/2 du
2 u 3/2
3
C
2 (7 x
3
2014 Pearson Education, Inc.
1)3/2 C
379
380
Chapter 5 Integration
x2 5
3. Let u
du
4
2 x( x 2 5) dx
x4 1
4. Let u
3
4x
( x 4 1)2
4x
(3 x 2)(3 x
x)
x
7. Let u
4
4 x) dx
x
du
dx
(1
3x
du
2 x2
8. Let u
1
4
2t
du
1
2
10. Let u 1 cos 2t
1 cos 2t
2
9 r dr
1 r3
y4
12. Let u
2
3u du
13. Let u
x sin ( x
1
x
1 cos 2 1
x
x2
14. Let u
15. (a) Let u
1 u
10
u du
1
x
1/3
u
5)
3
C
5
1
u
1 du
2
C
1
x4 1
C
(3 x 2)dx
1 (3 x 2
10
C
4 x)5 C
dx
2 u1/3 du
2 du
1 cos 3x
3
2 34 u 4/3 C
x )4/3 C
3 (1
2
C
x dx
1
4
1
4
cos u C
cos 2 x 2 C
dt
1
2
sec u tan u du
1 sin t dt
2
2
2
2u du 32
du
sec u C
2 du
1
2
sec 2t C
sin 2t dt
u3 C
1 cos 2t
2
3
3
C
3r 2 dr
3du 9r 2 dr
3u 1/2 du
3(2)u1/2 C
6(1 r 3 )1/2 C
du
4 y2 1
u
3
C
x3/2 1
2
1 du
2
sin 2t dt
11. Let u 1 r 3
2
sin u du
2 dt
sec 2t tan 2t dt
1
2
2 du
1 du
4
4 x dx
x sin (2 x ) dx
9. Let u
du
du
2
4
1 du dx
3
1 cos u C
3
3 dx
1 sin u
3
sin 3 x dx
2(3 x 2)dx
dx
1 ( x2
3
C
u 2 du
21
du
4
4u
u 4 12 du
1 dx
2 x
x )1/3 1
x
1u 3
3
x3 dx
1 du
4
(6 x 4)dx
2
6. Let u 1
(1
4 x3 dx
du
x dx
u 4 du
4 x3 ( x 4 1) 2 dx
dx
1/3
41
du
2
2u
du
3x 2
5. Let u
1 du
2
2 x dx
32
du
(y
1) dx
dx
1
x2
4y
2
csc2 2 cot 2 d
3
3 du 12 ( y 3
2 y ) dy 12( y 4
4 y 2 1) 2 ( y 3
2 y ) dy
C
2 du
3
du 23 u2
x dx
1 sin 2u
4
C
1
3
u
2
sin 2u
( x3/2 1) 16 sin (2 x 3/2
2) C
dx
2
cos ( u ) du
cot 2
1)
3 x1/2 dx
2
2 sin 2 u
3
du
du
4
(4 y 3 8 y ) dy
du
cos 2 (u ) du
2 csc2 2 d
1
2
u du
1 u2
2 2
Copyright
1 du
2
C
1
4
C
1
2x
1 sin
4
csc 2 2 d
u2
4
C
1 cot 2
4
2
2014 Pearson Education, Inc.
C
2
x
C
1
2x
1 sin 2
x
4
C
Section 5.5 Indefinite Integrals and the Substitution Method
(b) Let u
csc 2
du
csc2 2 cot 2 d
16. (a) Let u
5x 8
dx
5x 8
(b) Let u
17. Let u
dx
5x 8
2
5
2u
5
3 2s
du
2
2
1
u
1 du
2
du
5 ds
d
du
2 d
4
1 du
2
u
7 3 y2
20. Let u
du
3 y 7 3 y 2 dy
21. Let u 1
1
x (1
22. Let u
x )2
du
2 du
dx
1
2 x
3x 2
sin 3x
3dx
2
(sec u )
1 du
3
1 du
3
du
1 cos x
3
3
5
2
5
C
5x 8 C
2 u 3/2
3
1
2
2 s )3/2 C
1 (3
3
C
(2u1/2 ) C
1
x
2
5
5s 4 C
d
4 u 5/4
5
1
2
2 (1
5
C
2 5/4
)
C
dy
1
2
2 u 3/2
3
1 (7
3
C
3 y 2 )3/2 C
dx
C
x
u (3 du )
tan 2x
1 sec2 x
2
2
7
du
tan 7 2x sec2 2x dx
1
3
sec2u du
1 tan 3
3
3 du
3 61 u 6
dx
u (2 du )
2 u 3/2
3
2 u 7/2
7
C
2 sin 3/2 x
3
u C
1 tan(3 x
3
2 sin 7/2
7
dx
dx
sin 5 3x cos 3x dx
26. Let u
2 du
5
u1/2 u 5/2 du
24. Let u tan x du sec 2 x dx
tan 2 x sec2 x dx
u 2 du 13 u 3 C
25. Let u
2 u1/2
5
dx
5x 8
cos x dx
du
sec (3 x 2) dx
1
5
du
2
sin x 1 sin x cos x dx
2
C
ds
1
2
23. Let u
2
5x 8 C
2 du
C
1 csc 2
4
C
C
8) 1/2 (5) dx
1 du 3 y
2
1 u1/2 du
2
dx
2
u
1 (2u1/2 )
5
1 du
2
1 u1/4 du
2
1 du
2
u2
du
u
csc 2 cot 2 d
dx
du
1 du
5
1/2
6 y dy
u
x
sin x
1
5
C
1 du ds
2
1 u1/2 du
2
2 ds
1 1 du
u 5
ds
19. Let u 1
4
du
1 u
2 2
1 du
5
1/2
1 u
5
1 (5 x
2
C 52
du
du
5s 4
1
5s 4
1
u
2
du
5 dx
5x 8
3 2s ds
18. Let u
1u
2
du
1
5
1 du
2
u2
4
2 csc 2 cot 2 d
2 du
2 18 u 8
Copyright
1 tan
3
x C
cos 3x dx
1 sin 6 x
2
3
C
C
sec2 2x dx
C
1
4
tan 8 2x
C
2014 Pearson Education, Inc.
2) C
x C
381
382
Chapter 5 Integration
r3
18
27. Let u
1
5
r 3 1 dr
r 2 18
3
5
r
r 4 7 10
x1/2 sin( x3/2 1) dx
csc v 2
csc v 2 cot v 2
30. Let u
31. Let u
32. Let u
sec z
dz
1
t2
cos
34. Let u
1
t
2
cos( t
3) dt
1
d
dx
x
u 1
u
38. Let u 1 1x
x 1 dx
x5
39. Let u
1
x2
cos 1
du
37. Let u 1 x
x
1 x
du
2 1x
2 1x dx
1
x2
2
dx
u
2 cos( x3/2
3
cot v 2
1
x2
u du
1
t2
sin u C
1
t
1
d
sin 1t 1
1
2
C
dt
2sin u C
2sin( t
3) C
cos 1 d
1 sin 2 1
2
C
d
2du
2 du
2u C
2
C
dt
du
cot
dv
sin(2t 1) dt
cos u du
C
1) C
1
cot
csc
2 csc
C
d
2
sin
C
du
u 1/2 du
2 u 3/2
3
2u1/2 C
2 (1
3
x )3/2 2(1 x )1/2 C
dx
x 1 dx
x
du
C
cos u ) C
2 sec z
2du
d
csc
1/2
1
x2
2
1 u2
2
cot
u 1
du
4
r5
7 10
C
du
1 t 1/2 dt
2
csc
du
1 du
2
cos u du
1
u du
csc
cos
sin 2
t 2 dt
du
sin 1 cos 1 d
2(
3
sin u du
2u1/2 C
(cos u )(2 du )
du
2
3
1
2 cos(2t 1)
C
(cos u )( du )
sin 1
36. Let u
1
2u
3 t1/2 3
t
35. Let u
1
1 dt
1
2
C
x1 2 dx
2sin(2t 1) dt
sec z tan z dz
du
u 1/2 du
1
u
1 t 1 1
1
t
1
t
33. Let u
du
du
C
csc v 2 cot v 2 dv
2du csc v 2
2du
2u C
2 csc v 2
C
1 du
2 u2
sec z tan z
sec z
6
1
1
2
dv
dt
cos2 (2t 1)
(sin u ) 23 du
du
cos(2t 1)
sin(2t 1)
4
2 u4
2 u 3 du
2 du
3
r3
18
C
r 4 dr
2 du
3 x1/2 dx
2
du
6
6 u6
6 u 5 du
u 3 ( 2 du )
dr
r 2 dr
6 du
1 r 4 dr
2
du
x3/2 1
29. Let u
dr
u 5 (6 du )
r5
7 10
28. Let u
r2
6
du
1
x2
1 1x dx
u1/2 du
u du
2 u 3/2
3
C
dx
u1/2 du
2 u 3/2
3
Copyright
C
2
3
3/2
2 1x
C
2014 Pearson Education, Inc.
2
3
3/2
1 1x
C
Section 5.5 Indefinite Integrals and the Substitution Method
1
x2
du
x 2 1 dx
x2
1
x3
3
x3
du
40. Let u 1
1
x3
41. Let u 1
x3 3
x11
x3 1
42. Let u
x4
3
x 1
43. Let u
9
x4
x2
44. Let u
1 u11
11
dx
4 x. Then du
1/2
(4 u )( u
) du
(u 3/2 u1/2 ) du
1
2
48. Let u
x3 1
2 u 5/2
5
49. Let u
2 u 3/2
3
C
4
du
x2
1u 2
4
50. Let u
2x 1
1 3 u 4/3
4 4
(cos x )e
52. Let u
3u1/3
2 ( x3
5
4u 5 ) du
2 ( x3
3
dx
sin 2
du
2
d
eu
x dx and x 2
1 u 5/2
5
C
2 ( x3
3
1 du.
2
Thus
3 (2 x
4
esin x
2sin cos d
eu du
eu
C
C
C
2 (4
5
x)
5/2
(4 u ) u ( 1) du
x)3/2 C
8 (4
3
4 u7
7
2 u6
3
(u 10)u1/3 du
Copyright
4 (1
7
x )7
2 (1
3
x)6 C
(u 4/3 10u1/3 ) du
u 1. Thus x3 x 2 1 dx
1 u 3/2
3
(u 1) 12 u du
1 ( x 2 1)5/2 1 ( x 2 1)3/2 C
5
3
C
(u 3/2 u1/2 ) du
(u 1) u du
x
( x 2 4)3
x
(2 x 1)2/3
dx
( x2
1 ( u 1)
2
2/3
u
4) 3 x dx
1
2
du
1)1/3 C
C
sin 2 d
esin
x)8
1 (1
8
C
1)3/2 C
x dx. Thus
1)4/3
C
3/2
3
x3
(u11 u10 )du
4 u. Thus x 4 xdx
8 u 3/2
3
1 u8
8
4) 2 C
eu du
1
1)3/2 C
(u 1) u10 du
u 1. So 3x5 x3 1 dx
1)5/2
3 (2 x
16
2
27
C
5)4/3 C
15 ( x
2
2 x dx and 12 du
C
C
u 5. Thus ( x 5)( x 5)1/3dx
dx
C
dx and x 1 u. Thus ( x 1) 2 (1 x)5 dx
2 u 3/2
3
1)
3/2
2 u 3/2
27
du = cos x dx
sin x
(sin 2 )esin
) du
4u 6
1 2 u 5/2
2 5
1 (u
2
x
51. Let u = sin x
dx and x
2 u 5/2
5
1/2
2 x dx and 12 du
1 (x2
4
C
2 u1/2
3
3/2
5)7/3
1
x2
1
1)11 C
1 (x
11
4u
1
3
C
u1/2 du
1
9
u 1. Thus x( x 1)10 dx
3 x 2 dx and x3
du
u 91 du
u 1/2 du
1
3
du
1)12
3 (x
7
C
1 u 3/2
3
x 2 dx
1 du
3
dx and x
x 2 1. Then du
47. Let u
dx
( u7
x 5. Then du
15 u 4/3
2
3
x3
1 dx and ( 1)du
(2 u ) 2 u 5 ( 1) du
3 u 7/3
7
1
x4
1 dx and ( 1)du
(u
45. Let u 1 x. Then du
46. Let u
dx
1
u1/2 du
1
2
1
x4
dx and x
1 (x
12
C
dx
1 du
9
1 1
u 3
dx
x3 1
x 1. Then du
1 u12
12
dx
1
x3
u 12 du
dx
3x 2 dx
du
dx
1 du
2
dx
1
x2
1
x3 3
x3
1
x4
dx
2
x3
383
2
C
2014 Pearson Education, Inc.
1
4
u 3 12 du
1
2
u 3 du
u1/3 u 2/3 du
384
Chapter 5 Integration
e x
53. Let u
1
xe
1
sec2 e x 1 dx
x
1
54. Let u 1 e x
1
x2
e sec 1 e
55. Let u = ln x
1 dx
x ln x
1
x
dz
1 ez
du
1 dt
t
x2
z
z
e
e
x x
2r
3
59. Let u
5
9 4r 2
e
1
2
61.
esin
62.
ecos
63.
(sin
1
1x
1 x2
2
1
e
dx
x )2
1 x2
2r
3
ln ln x
1
2
dt
dx
1
C
1
x
1
sec u C
sec 1 e x
C
z
z
1
dr
1 u2
2 2
u du
1 (ln t ) 2
4
C
C
e z dz
1 du
u
dz
ln u
z
ln 1 ez
ln(e z 1) C
C
e
x dx
1 sec 1 u
2
du
u u2 1
1 sec 1 ( x 2 )
2
C
C
dr
3
2
C
du
5
2
9 1 u2
du
5 tan 1 u
6
5 tan 1 2 r
6
3
C
C
e d
e d
d
dx
1x
1 x
du
dx
x
2 tan e x
sec u tan u du
3 du
2
2 dr
3
1
1
dx
1 du
2
1
2
( x2 )2 1
du
5
9
dr
60. Let u
e
x
1
2
1
x
1
xe
z ln(1 e z ) C
2 x dx
x dx
4
e x 12 dx
du
e z dz
dz
du
dx
1 dx
2
ln t
t
1
2
dt
(ln(1 e ) ln e z ) C
58. Let u
2 tan u C
C
du
e
1
1 ez e
z
e x dx
1 dx
x
ln u
e z 1
57. Let u
x
1
x
2 du
2 sec2 u du
tan 1 e
1 du
u
ln t1/ 2
t
dt
ex
du
56. Let u = ln t
ln t
t
1
du
1
x
e x dx
1
2 x
du
(e )
du
2
1
u u
eu du, where u
2
1
sec 1 u C
sin 1 x and du
eu du , where u
cos 1 x and du
u 2 du , where u
sin 1 x and du
Copyright
sec 1 (e ) C
eu
dx
1 x2
dx
1 x
2
dx
1 x2
1
esin x
C
eu
u3
3
C
C
C
1
ecos x
(sin
1
3
2014 Pearson Education, Inc.
x )3
C
C
C
Section 5.5 Indefinite Integrals and the Substitution Method
64.
tan 1 x
1 x2
u1/2 du , where u
dx
tan 1 x and du
2 u 3/2
3
dx
1 x2
2 (tan 1 x)3/2
3
C
C
2
3
385
(tan 1 x)3
C
1
65.
1 y2
1
1
dy
(tan 1 y )(1 y 2 )
tan
y
1 du ,
u
dy
dy
tan 1 y and du
where u
ln u
1 y2
ln tan 1 y
C
C
1
66.
1
(sin
1
y) 1 y2
67. (a) Let u
1 y2
dy
tan x
2
1
sin
1 du ,
u
dy
y
sec2 x dx; v
du
2
2
18 tan x sec x dx
18u
du
(2 tan 3 x )2
(2 u 3 )2
6
6
C
C
2 u3
2 tan 3 x
3
2
(b) Let u
tan x
2
2
2 tan x
2
2
x 1
6 du
6 dv
(2 u )2
v2
u3
3u 2 du
dv
6 dv
6 dw
(2 v )2
w2
6 du
du
6
v
u2
6
u
dx; v
sin u
1 w3/2
3
w dw
(b) Let u
sin( x 1)
1 (1
3
C
du
1 2
2 3
v3/2
2
(c) Let u 1 sin ( x 1)
1 (1
3
C
3(2r 1) 2
6
(2 r 1) cos 3(2 r 1)2 6
3(2 r 1)2 6
du
1 sin
6
70. Let u
sin
cos3
71. Let u
cot x
3(2r 1) 2
cos
du
du
2
d
2 du
u 3/ 2
d
cos3
dx
1 du
12
1
sin
sin
d
cos x
sin x
6(2r 1)(2) dr
2 u
C
dv
du
dw
1 (1
3
1
2
u du
1 du
12
1
2
v dv
sin 2 ( x 1))
1 du
2
1 u1/2 du
2
1))3/2 C
C
sin( x 1) cos( x 1) dx
1 2 u 3/2
2 3
(2r 1) dr ; v
(cos v) 16 dv
v dv
v 1 v 2 dv
1 (1 sin 2 ( x
3
1 dv u du
2
1 v1/2 dv
2
3/2
C
2u du
u 1 u 2 du
1 dw
2
2v dv
1 sin v
6
u
C
1 (1
3
C
1
2 u
dv
1 sin
6
u
du
sin 2 ( x 1))3/2 C
1 dv
6
C
6 C
du
sin x
dv
2sin( x 1) cos( x 1) dx
cos u
u
dr
sin 2 u )3/2
u 2 )3/2 C
1 sin 2 ( x 1) sin( x 1) cos( x 1) dx
69. Let u
1 (1
3
2
cos( x 1) dx; v 1 u
1 v3/2
3
C
6
2 v
dv
C
1 sin 2 u sin u cos u du
v 2 )3/2 C
1 sin 2 ( x 1) sin( x 1) cos( x 1) dx
dw
2du
sin
2 u 3/2 du
d
2( 2u 1/2 ) C
cos x dx.
du
u
ln u
C
ln sin x
Copyright
C
C
cos u du; w 1 v 2
dv
2 v
6w 1 C
6
2 tan 3 x
2
2
C
ln sin 1 y
C
6 du 18 tan x sec x dx
1 sin 2 ( x 1) sin( x 1) cos( x 1) dx
1
2
ln u
6dv 18u 2 du; w
6 w 2 dw
6
2 u
C
6
2 tan 3 x
C
dy
1 y2
6 du 18 tan 2 x sec2 x dx; v
3 tan 2 x sec2 x dx
du
18 tan x sec x dx
(2 tan 3 x )2
68. (a) Let u
sin 1 y and du
3 tan x sec2 x dx
du
18 tan x sec x dx
(2 tan 3 x )2
3
(c) Let u
where u
C
2014 Pearson Education, Inc.
4
u
C
4
cos
C
1
12 u
du
386
Chapter 5 Integration
72.
csc x dx
du
u
csc x dx
73. Let u
3t 2 1
s
2
12t (3t
s
du
2
y
4 x( x
8)
y
0 when x
2 14 u 4
2 x dx
dx
0
1) 4 C
1 (3
2
u
(csc 2 x csc x cot x ) dx.
du
C
2 du
1/3
1 u4
2
C
3 8 C
(2du )
3(8) 2/3 C
0
csc x cot x
2du 12t dt
u (2 du )
du
1/3
ln csc x cot x
3
3
x2 8
C
6t dt
1) dt
cot x
csc x csc x cot x dx. Let u
ln u
3
3 when t 1
74. Let u
csc x
csc x (1) dx
1 (3t 2
2
C
C
5
s
1) 4 C ;
1 (3t 2
2
1) 4 5
4 x dx
2 32 u 2/3
C
C
y
12
3u 2/3 C
3( x 2 8) 2/3 C ;
3( x 2 8)2/3 12
75. Let u
t 12
du dt
2
s
8sin t 12 dt
8sin 2 u du 8 u2 14 sin 2u C 4 t 12
s 8 when t 0 8 4 12 2sin 6 C C 8 3 1 9 3
s 4 t 12 2sin 2t 6 9 3 4t 2sin 2t 6 9
76. Let u
r
4
3cos
r
8
r
77. Let u
ds
dt
du
2
when
3
2
0
2t
0 and
s
at t
0 and s
s
78. Let u
dy
dx
sin 2t
2 dt
2 du
2
tan 2 x
at x
2 cos u C1
C1
2
2
50
(1 25 )
u (2 du )
dy
0 and y
1 we have 1
79. Let u 2t
du
s
6 sin 2t dt
at t
0 and s
3
4
2du
s
1 (0)
2
2 cos 2t
C1 100
C2
C2
sin 2t
2
C;
2
2
3
4
u 2 C1
C1;
2
ds
dt
2 cos 2t
sin 2t
2
100
2
50 2t
2
C2 ;
1 25
100t 1
2
4sec2 2 x dx; v
2x
dv
2dx
1 dv
2
dx
tan 2 2 x C1;
dy
dx
4
0 C2
tan 2 2 x 4
1 tan v
2
3v
2
C2
C2
1
y
2 dt 3 du 6dt
(sin u )(3 du )
3 cos u C
0 we have 0
r
3 sin
4
2
3 sin
4
2
sin u 50u C2
2
0 and dx 4 we have 4 0 C1 C1
y
(sec2 2 x 3) dx
(sec2 v 3) 12 dv
at x
3
2 4
3
2 4
C
4 dt
2sec2 2 x dx
4sec2 2 x tan 2 x dx
1 sin 2u
4
C 2 34
3 cos 2
4
8
(cos u 50) du
sin
100t 25
du
3
2
2 cos
100 dt
0 we have 0
2
C
(sin u )( 2 du )
100 we have 100
2cos 2t
3 u2
3 sin
4
2
3
r
4
8
dt
2
ds
dt
3
8
2
du
2
4sin 2t
at t
3cos 2 u du
8
3 sin
4
2
C;
6
d
d
4
2 sin 2t
(sec2 2 x 1) 4
sec 2 2 x 3
1 tan 2 x
2
3 x C2 ;
1 tan 2 x 3 x 1
2
3 cos 2t C ;
3cos 0 C
C
3
s
3 3cos 2t
s 2
Copyright
2014 Pearson Education, Inc.
3 3cos( )
6m
Section 5.6 Substitution and Area Between Curves
80. Let u
v
at t
t
2
du
dt
cos t dt
0 and v
2
du
dt
(cos u )( du )
8 we have 8
sin u du 8t C2
s 8t cos ( t ) 1
387
sin u C1
(0) C1
C1
sin( t ) C1;
8
ds
dt
v
cos( t ) 8t C2 ; at t 0 and s
s (1) 8 cos
1 10 m
sin( t ) 8
0 we have 0
s
( sin( t ) 8) dt
1 C2
C2
1
81. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on the
left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
cos 2 x 1 C
sin 2 x C1 1 cos2 x C1 C2 1 C1 ; also cos 2 x C2
C3 C2 12 C1 12 .
2
2
2
82. (a)
1/60
1
1
60
0
0
Vmax
[1
2
(b) Vmax
(c)
Vmax
0
1
Vmax 120
60
cos(120 t )
1/60
V max [cos 2
2
0
2
2
t
2(240)
339 volts
sin 2 120 t dt
1
240
1/60
sin 240 t
1/60 1 cos 240 t
2
0
2
Vmax
2
Vmax
2
0
1
60
1
240
2
Vmax
2
dt
1/60
0
sin(4 )
(1 cos 240 t ) dt
1
240
0
sin(0)
SUBSTITUTION AND AREA BETWEEN CURVES
1. (a) Let u
3
y 1
y 1 dy
0
du
dy; y
4 1/2
u
1
0
u 1, y
4
2 u 3/2
3
1
du
3
u
y 1 dy
1
1 1/2
0
u
2. (a) Let u 1 r 2
du
1
0
0
r 1 r 2 dr
du
1
2
1
1
2 u 3/2
3
0
(1)3/2
2
3
2r dr
1 du
2
u du
0
1 u 3/2
3
1
1
0
r 1 r 2 dr
3. (a) Let u
/4
0
tan x
0
du
tan x sec2 x dx
1
2
u du
1
0
u du
/4
4. (a) Let u
0
0
tan x sec2 x dx
cos x
du
3cos 2 x sin x dx
1
u du
sin x dx
1
1
0, y
0
u 1
u 1, r
1
u
(1)3/2
1
3
0
2
3
(8)
(1)
2
3
0
r dr ; r
0
u
1
1
u2
2
0, x
12
2
0
u
0
1
3
0, r
1
u
0
3u 2 du
u2
2
du
0
1
2
u
4
1, x
0 12
1
2
sin x dx; x
0
1
[ u3 ] 1
Copyright
u 1
4
0
(b) Use the same substitution as in part (a); x
0
u
2
3
0
sec2 x dx; x
0
(1)3/2
2
3
1
(b) Use the same substitution for u as in part (a); r
1
4
(4)3/2
2
3
(b) Use the same substitution for u as in part (a); y
0
cos 0]
1] 0
2Vrms
1/60
Vmax
2
5.6
Vmax sin 120 t dt
1
0
u
u 1, x
( 1)3 ( (1)3 )
2
2014 Pearson Education, Inc.
0
u
1
14
3
Vmax
120
2
388
Chapter 5 Integration
(b) Use the same substitution as in part (a); x 2
3
3cos 2 x sin x dx
2
5. (a) u 1 t 4
21
1 4
t (1 t 4 )3 dt
t 3 dt ; t
1 du
4
u 3 du
0
2
4
u 1, t
24
16
u
16 1
14
16
(b) Use the same substitution as in part (a); t
1 3
t (1
1
7
0
t 4 )3 dt
t2 1
6. (a) Let u
21
2 4
du
u 3 du
8 1 1/3
u
1 2
du
t dt; t
4 r2
7. (a) Let u
1
5r
1 (4 r 2 )2
dr
du
5
0
1 du
2
du
5r
0 (4 r 2 ) 2
dr
8. (a) Let u 1 v3/2
1 10 v
0 (1 v3/ 2 ) 2
dv
5
51 2
u
4 2
du
dv
x2 1
9. (a) Let u
3
4x
0
dx
x2 1
du
1u
2
3 v1/2
2
dv
20 du
3
20 2 u 2
3 1
du
du
4x
3
x2 1
dx
x4 9
10. (a) Let u
1
x
0
x4 9
3
dx
4 2
u
4
du
10 1
9 4
du
3
du
2 du
4
1
x
1 x4 9
dx
2
7
u
8
(8)4/3
3
8
(1) 4/3
0
45
8
u 1
5, r 1
u
4, r
u
5
1 (5) 1
2
u
0
u
2, v
20 1
3 9
20 1
3 2
3
20
3
10
3
1
1
7
18
3
4(4)1/2
u
u
u 1 43/2
4
u 1, x
[4u1/2 ] 14
1
8
u 1, v 1
1
2
0
5
1 (4) 1
2
5
20 1 2
3 u 1
du
4 x dx; x
2u 1/2 du
1
4, x
70
27
u
4(1)1/2
3
9
u
4
4
4
0
4 x3 dx
u 1/2 du
9 1 1/2
u
10 4
u
10 v dv; v
1 du
4
x3 dx; x
10
1 (2)u1/2
4
9
(b) Use the same substitution as in part (a); x
0
u
8, t
1
5
(b) Use the same substitution as in part (a); x
3
1
45
8
du
20 1 9
3 u 2
2 x dx
4 2
1 u
u
0
1 5
4
5
9 1 20
2 u2 3
3
8
1
(b) Use the same substitution as in part (a); v 1
4 10 v
1 (1 v3/ 2 )2
2
0
du
2 1 20
1 u2 3
u
2, t
8
rdr ; r
(b) Use the same substitution as in part (a); r
1
1
u 1, t
7
8 1 1/3
u
1 2
du
2r dr
51 2
u
5 2
u
u 4/3
3
4
1
2
du
1 1 1/3
u
82
t (t 2 1)1/3dt
7
1
15
16
1
(b) Use the same substitution as in part (a); t
0
u
0
1
2
2t dt
t (t 2 1)1/3 dt
u 1, x 3
3u 2 du 2
1
4t 3 dt
du
1 3
0
1
du
10 1 1/2
u
9 4
Copyright
0
u
1 (10)1/2
2
1
u 10, x
du
3
9, x 1
u 10
1 (9)1/2
2
10 3
2
0
u
10
2
2014 Pearson Education, Inc.
9
2
Section 5.6 Substitution and Area Between Curves
11. (a) Let u
1
0
4 5t
t
1
(u 4), dt
5
1
du; t
5
1 9
( u 4) u du
25 4
t 4 5t dt
1 2 5/2
u
25 5
9
8 3/2
u
3
4
1
25
0
2
8
(243)
(27)
5
3
1
1
25
2
8
(16,807)
(343)
5
3
12. (a) Let u 1 cos 3t
/6
0
u
1 49 3/2
u
4u1/2 du
25 9
t 4 5t dt
du
11
03
/3
1
3
du
13. (a) Let u
2
4 3sin z
cos z
4 3sin z
0
dz
du
4 1 1
du
4 u 3
1
14. (a) Let u
0
/2
dz
2 tan 2t
du
2
1 du
3
/2
u
2 du
15. Let u
1
0
t5
t 5 2t (5t 4
16. Let u 1
4
2t
y
dy
1 2 y (1
y)
2
du
(5t 4
2) dt
du
3 1
2 u2
dy
2 y
du
3
1
2
u du
2) dt ; t
3 1/2
u
0
3
2
0
2
2 3
[u ] 1
u
0, t
3
2 u 3/2
3
0
du
;y 1
u
0, t
1 (0) 2
6
6
1 (1) 2
6
0
u 1 cos 2
1
1
6
u 1 cos
3
2
1
2
u
4, z
4 3sin(
sec2 2t dt ; t
u (2 du ) [u 2 ] 1
1
2
0
2
u
4
)
4, z
u
2 tan 4
u
4
0
1 sec 2 t dt
2
2
2
(2 tan 2t ) sec2 2t dt
86,744
375
cos z dz; z
u
u 2 du
2, y
4
[ u 1 ]32
Copyright
2
22 12
(b) Use the same substitution as in part (a); t
/2
49.
49
1 (2) 2
6
1
506
375
0
du
2 tan 2t sec2 2t dt
9.
8 3/2
u
3
9
u 1, t
6
(b) Use the same substitution as in part (a); z
cos z
4 3sin z
u
1 (1)2
6
0
1 u2
3 2
du
3cos z dz
4 1 1
4 u 3
9
sin 3t dt ; t
1 u2
3 2
u du
21
u
1 3
(1 cos 3t ) sin 3t dt
9, t
1 2 5/2
u
25 5
(b) Use the same substitution as in part (a); t
/6
u
2
8
(32)
8)
5
3
2
8
(243)
27
5
3
3sin 3t dt
(1 cos 3t ) sin 3t dt
4, t 1
1 9 3/2
4u1/2 du
u
25 4
(b) Use the same substitution as in (a); t 1
9
u
389
3
u 1, t
32 12
8
1
3
u
2 (3)3/2
3
u
u
2
2 (0)3/2
3
3
2 3
3
1
3
1
2
1
6
2014 Pearson Education, Inc.
1, t
0
u
2
390
Chapter 5 Integration
17. Let u
/6
0
cos 2
du
tan 6
3 /2
du
20. Let u 1 sin 2t
/4
0
4y
(4 y
22. Let u
1
9
du
1
y2
y2
0
3
2
0
4 y3 1) 2/3 (12 y 2
3/2
du
du
1
2
1 sin 2
4
/4
0
tan
du
1 e tan
sec2
(1 0) ( e 1)
26. Let u
/2
/4
cot
1 ecot
(1 0) (1 e)
1
d ;
1
u
2 u
3 2
1
2
0, t
sin 2 u du
sec 2 d ;
/4
0
0
u
2
u
sec 2 d
( y2
u
1
/4
u 1;
e du
tan 4
1
12
4
5 4 cos
9
4(1)3 1 8
3
0
2 (4)1/2
3
2
1
5
u
9, y 1
2 (9)1/2
3
2 (2
3
u
3)
4
2
3
u
1 sin 2
4
2 (0)
3
3
1
0
/4 u
u
0
3(8)1/3 3(1)1/3
2
3 2
0
1
3
4(1) (1)2
4 y 4) dy; y
3
0,
1 sin 2u
4
u
u
u
1 (2u1/2 )
3
9
du
u
3
2
1 (1)5/2
5
4
1 sin 2u
4
0,
0
8
1 du
3
0
3
2(1) 4
4
u 1, y 1
[3u1/3 ] 1
3
2
1 ,
3
1 (0)5/2
5
1
3
4
1
4(1)2
2
95/4 1 35/2 1
1
u 1, t
0
u 2/3 du
4 1 1/2
u
9 3
9
1
2
cos 2 6
5 4 cos 0 1, t
0
0
1
2
tan 6
5 4 u 5/4
4 5
du
(3 y 2 12 y 12) dy
1
0
d
8
cos 2 u 23 du
0
u
1 2 u 5/2
2 5
1
4
1
1
3
2u 4 1/ 3
cos 2t dt ; t
u 3/2 du
2 du
3
t 2 dt ; t
t 2 sin 2 1 1t dt
25. Let u
1 du
2
4 y 4) dy
3 1/2 d
2
1
5 9 u1/4
4 1
1/2
u
0
u
6
1 u 2
2
2
du
1/ 3
sin t dt ; t
2 y 4) dy
du
cos 2 ( 3/2 ) d
24. Let u 1 1t
1/2
1
2
u 1,
1
4
6 u4
(4 2 y 12 y 2 ) dy; y
du
( y 3 6 y 2 12 y 9) 1/2 ( y 2
23. Let u
1 du
4
0
sec 2 6 d ;
5u1/4 14 du
1
y 3 6 y 2 12 y 9
1 1/2 u 3
2 1
u 5 (6 du )
2 cos 2t dt
4 y3 1
sin 2 d ;
6 du
4sin t dt
(1 sin 2t )3/2 cos 2t dt
21. Let u
0
1/ 3
du
1 du
2
d
1
5(5 4 cos t )1/4 sin t dt
0
0
1 sec 2
6
6
du
5 4 cos t
u 3
1
cot 5 6 sec 2 6 d
19. Let u
1
1/2
cos 3 2 sin 2 d
18. Let u
1
2
2sin 2 d
1
2
/4
tan
0
1 sin(
4
[eu ] 1
0
0
2
2)
tan
/4
1 sin 0
4
tan(0)
( e1 e0 )
e
du
csc2 d ;
csc 2 d
/2
/4
/4
csc2 d
u 1,
0 u
1
e du
/2
cot
u
0;
/2
/4
[eu ]10
e
Copyright
2014 Pearson Education, Inc.
cot
/2
cot( /4)
( e0 e1)
Section 5.6 Substitution and Area Between Curves
sin t
0 2 cos t
27.
t
28.
dt
u
0
sin t
0 2 cos t
3
/3 4sin
1 4cos
0
ln 2 cos t
d
ln 3 ln1 ln 3; or let u
0
31
1 u
dt
ln 1 4cos
0
du
/3
u
ln u
3
1
1
ln 3 ln 13 ; or let u 1 4cos
/3 4sin
1 4cos
3 and
29. Let u
ln x
du
1
x
dx; x 1
u
0 and x
30. Let u
ln x
du
1
x
dx; x
u
ln 2 and x
1
ln 4
32. Let u
33. Let u
/2
0
du
1
ln 2
1
ln 22
ln x
du
ln 2
cos 2x
ln u
dx; x
1
x
4 ln 2
1
1/ 2
cos t dt; t
(sec2
1)cos
0
sec
(sec 2
sec
sec
tan
tan
0
/3 sec
sec tan
sec
tan
3
2
ln 2
1
2
0
2
1
4
du
ln u
2 2ln x
x
dx
ln 4;
u
ln 4;
4sin d
with
ln 3 ln 13
ln 2
du
ln 2
u2
2u du
ln 4 1
ln 2 u
4
ln 4
dx
2 x (ln x )2
ln 2
16 dx
2 2 x ln x
ln16;
0
1/ 2
1
2 ln u
/3
0
u 1 and x
2ln 1
2
/2
u 1;
2
cos
0
(ln 2) 2
ln 4
ln u ln 2
1 ln 4
u ln 2
u 2 du
1 ln16 u 1/2
2 ln 2
du
u1/2
ln16
ln 2
/4
2
2ln 2
/3
tan 2 cos d
ln 2
/2 cos t
/4 sin t
cot t dt
0
1 ;
2
u
sec d
1
du
1/ 2 u
dt
/3
0
cos d
Rewrite the first integral as
ln u
d . Let u
/3
2
1
3
ln 2
sec
2
tan
du
(sec tan
sec 2 ) d
3;
3 ; thus the original definite integral is equal to
3 .
Copyright
1
3
0
4 dx
2 x ln x
du
ln 2
u 1 0 1;
31
u
du
1
u
u
sin 2x dx; x
and t
3
.
2
0
u 1 and
1
ln 4
/3 sec 2
sec tan
sec tan
0
d
d
u
/3
sin
sec tan ) d ;
2
1/ 2 du
u
1
0
ln 2
ln 2
2 du
sec
ln 2;
ln 2 and x 16
2
4
ln 2
1
2ln 2
2 ln 2
dx
dx
ln 1
The second integral is
/3
u
ln 2
1 sin x
2
2
2
35. tan 2 cos
2
u
4
ln 2 and x
1
ln 2
11
3u
d
2
2
ln 2ln
ln 2
u
1
2ln 2
dx; x
du
sin t
2
1
ln 2
/2 sin( /2)
0 cos( /2)
tan 2x
34. Let u
1
x
ln x
ln16
2
ln lnln22
4
ln ln
ln 2
ln(ln 4) ln(ln 2)
31. Let u
2
0
sin t dt with t
du
ln 3 ln1 ln 3
ln1 ln 3
u
3
2 cos t
391
2014 Pearson Education, Inc.
392
Chapter 5 Integration
36. Let u
cos3x
/12
0
du
6 tan 3x dx
/2
2cos d
/2 1 (sin )2
37.
3sin 3 x dx; 2 du
/12 6sin 3 x
cos3 x
0
1
du ,
1 1 u2
2
/4 csc2 x dx
/6 1 (cot x )2
1
du ,
3 1 u2
1
tan 1 u
39.
ln 3 e x dx
0
1 e2 x
3
1
du ,
1 u2
40.
/4
e
4 dt
4
2
1
t 1 ln t
/4
0
du ,
1 u2
4 tan 1 u
41.
42.
1
1
4sin 1 2s
4 ds
0
4
s
3 2 /4
0
2
ds
9 s2
where u
2
2 sec sec
2
x x
2
1
x
1 3 2 /4
2 0
1
dx
/3
/4
2
2 3
cos sec
x x
2
1
x
1
dx
/3
/6
cot x
/4
ln t and du
1
t
1
2
sec 2 u du, where u
/3
/4
tan 3
cos u du, where u
sin u
/3
/6
sin 3 sin 6
Copyright
1,
u
3, x
u 1,
2
6
u 1,
4
12
0
u 1, x
ln 3
u
3,
u
e /4
0, t
u
/4,
4 tan 1 4
0
sin 1 0
2
3
2 ds; s
0
1
2 4
0
sec 1 x and du
2
tan 4
u
4
dt; t 1
2 s and du
sin 1 22
2
ln 2
12
tan 1 0
4 6
u 1/ 2;
ln1 2ln 2
2
3
e x dx; x
/12
2ln 1
2 4
4
4
where u
1/ 2
1
csc 2 x dx; x
3
4 tan 1 4
0
u 1, x
cos d ;
du
e x and du
where u
du ,
9 u2
tan u
44.
du
tan 1 1 tan 1 3
3
3 2 /2
1 sin 1 u
2
3 0
43.
sin
4 sin 1 12 sin 1 0
0
2 ln u
tan 1 3 tan 1 1
1
0
2 tan 1 1 tan 1 ( 1)
1
where u
3
tan 1 u
2
where u
1
2 tan 1 u
38.
dx
6sin 3 x dx; x
1/ 2 du
u
1
u
3 2
4
0, s
3 2
,
2
u
8
dx
;
x x 1
x
2
u
4
,x
2
u
3
3 1
sec 1 x and du2
dx
;
x x 1
x
3 1
2
2014 Pearson Education, Inc.
2
3
u
6
,x
2
u
3
,
,
Section 5.6 Substitution and Area Between Curves
2 /2
45.
2
dy
1
y 4y
2
2
1
du ,
u u2 1
where u
2
sec 1
sec 1 u
46.
3
1 16 u 1
25 1 u1/2
y dy
5y 1
0
4 x2
0
A
2
2 u 3/2
3
du
2
4
0
0
2 (4)3/2
3
(1 cos x) sin x dx
49. Let u 1 cos x
0
A
0
du
sin x
sin x dx
sin u du
2
51. For the sketch given, a
A
(1 cos 2 x )
2
0
/3
1 sec 2 t
/3 2
1
2
/3
/3
53. For the sketch given, a
A
2
2
(4 x 2
,b
3
4sin 2 t dt
sec 2t dt 2
x 4 ) dx
/3
/3
1
,A
du
4 x2
3
0
2
3
u
1 u1/2
2
3
2,
4, x
du
u 16,
36
25
2
2
u
4 1 1/2
u
0 2
2
0
4 1/2
du
u
0
du
0
u
/2 2
1
2
/3
/3
2
(cos x)(sin(
x
2
1 [tan
2
Copyright
2(2)3/2
u
sin
sin x)) dx
32
5
sin 2 x
sin 2 x
2
0
1
2
2
0
u 1 cos 0
2(0)3/2
0, x
2
0 2
25/2
0
u
(sin u ) 1 du
1 [(
2
1 cos 2 x
;
2
0) (0 0)]
sec 2 t ( 4 sin 2 t )
sec 2 t dt 4
32
3
2
0, x
2
2; f ( x) g ( x)
x5
5
)
2
2u 3/2
0
du
cos x dx; x
; f (t ) g (t )
1
2
2
u 1 cos (
2 1/2
3
0
2
(1 cos 2t ) dt
2, b
0, x
sin x dx; x
(1 cos 2 x) dx
3
u 1, y
64 8
4
u
2
; f ( x) g ( x) 1 cos2 x
0, b
52. For the sketch given, a
A
2
0
2
2
2
( cos ) ( cos 0)
1
2 0
dx
du
2
2
2, y
12
2
3
0
u
2
3u1/2 ( du )
cos x dx
[ cos u ]0
u
0, x
2
du
0
Because of symmetry about x
0
u
2
u
2 0
u du
du
2
1 u1/2
2
0
0
2
3(sin x) 1 cos x dx
50. Let u
4
3
0
1
25
1
u
16
3
sin x dx; x
2
16
1
4
5 dy; y
2u1/2
x dx; x
x 4 x dx
2 (0)3/2
3
du
5 y 1 and du
1 du
2
2
2 dy; y
sec 1 2
2
1 2 u 3/2
25 3
2 x dx
x 4 x 2 dx
48. Let u 1 cos x
0
du, where u
u1/2 u 1/2 du
1
25
47. Let u
2
2 y and du
393
sin 2t dt
t ] /3/3
2 t
2 x2 ( x4
32
3
sin 2t
2
1 sec 2 t
2
/3
2
1
2
/3
/3
4 x2
32
5
64
3
4sin 2 t ;
/3 (1 cos 2t )
2
/3
sec t dt 4
3 4 3
/3
2 x2 )
2
3
x4 ;
64
5
2014 Pearson Education, Inc.
320 192
15
128
15
4
3
dt
2
394
Chapter 5 Integration
54. For the sketch given, c
1
A
0
( y2
0, d
1 2
y
0
y 3 ) dy
55. For the sketch given, c
1
A
0
10
3
0, d
1
1
( x2
1
2
y
x3
3
1
1
1
0
0
2 x5
5
10 y 2 dy
12 y 3 dy
2 23
x2 ( 2 x4 )
( x2
1
1
3
1
2
5
1
3
x
1
2
3
(2 x 2
2
16
3
1 4
Therefore, AREA
0
2
2
0
4 8
y3
1
12
4
0
y4
1
0
2
2
y2
1
0
2x4 ;
2
3
10 12
15
4
5
x2
4
2, and the curve y
2
0
22
15
, minus the area of a triangle
2
1 x4 dx 12 (1)(1)
3
2
x
x 12
0
x2 , 0
1
2
x 1 plus the area of a triangle (formed by
1 2
x
0
1 (1)(1)
2
dx
1
x3
3 0
1
2
2x2
2 x)
2x 4
9 16
3
( 18 9 12)
2x
3
3
x2
4x
A1
( x2
1
(2 x 2
2 x 4) dx
4)
2 x2
11 ;
3
2 x) ( x 2
2
3
2
2
3
1 4
16
3
2x 4
4 8
9;
11
3
A1 A2
9
2 and b
2 x4
4
(2 x3 8 x) dx
0 and b
8 x2
2
(8 x 2 x3 ) dx
Therefore, AREA
4 8
2 x 4) dx
A2: For the sketch given, a
A2
2
5
2 and b 1: f ( x) g ( x)
60. AREA A1 A2
A1: For the sketch given, a
A1
4) ( x 2
16
3
3
A2: For the sketch given, a
A2
1
12
1
3
3 and we find b by solving the equations y x 2 4 and y
x2 2 x
4
x 2 2 x 2 x 2 2 x 4 0 2( x 2)( x 1)
x
2 or x 1 so
2
4x
1
4
2 y ) 10 y 2 12 y 3 2 y;
10
3
x2
2, and the x-axis) with base 1 and height 1. Thus, A
2 x2
2
1
3
5
6
1
2
2: f ( x) g ( x )
2 x3
3
(1 0)
4
2 y dy
x and y 1) with base 1 and height 1. Thus, A
59. AREA A1 A2
A1: For the sketch given, a
simultaneously for x: x 2
b
0
0
58. We want the area between the x -axis and the curve y
x 1, x
(1 0)
3
(12 y 2 12 y 3 ) (2 y 2
1; f ( y ) g ( y )
57. We want the area between the line y 1, 0
8
2 12
1
y4
4
0
1, b 1; f ( x) g ( x)
2 x 4 ) dx
(formed by y
1
y3
3
dy
y3 ;
4
3
(3 0) (1 0)
56. For the sketch given, a
A
1 3
y
0
dy
(10 y 2 12 y 3 2 y) dy
0
y2
1; f ( y ) g ( y )
38
3
0: f ( x) g ( x)
8x2
2
0
2
x 2 5 x) ( x 2 3x)
(16 8)
( x2
3 x) (2 x3
8;
A1 A2 16
Copyright
2 x3 8 x
0 (8 16) 8;
2: f ( x) g ( x)
2
2 x4
4 0
(2 x3
2014 Pearson Education, Inc.
x 2 5 x)
8 x 2 x3
1
2
5
6
Section 5.6 Substitution and Area Between Curves
61. AREA A1 A2 A3
A1: For the sketch given, a
1
A1
2
( x2
2 and b
A2: For the sketch given, a
2
A2
1
( x2
3
2
( x2
62. AREA
2 and b
x3
3
1
3
2
x2
2
2
2x
1
27
3
9
2
9 92
8
3
2
9
2
8
3
2
8
3
4
2
4
2
8
3
6
9 65
4
2
1
2
x2
14 3
6
1
2
11 ;
6
x 2)
3 8 12
2
9 92
4
7
3
( x2
1
3
4
x 2
4
( x 2) (4 x 2 )
3: f ( x) g ( x)
3
1
2
x2
(4 x 2 ) ( x 2)
2: f ( x) g ( x )
2x
11
6
A1 A2 A3
9;
2
x 2
8;
3
49
6
A1 A2 A3
A1: For the sketch given, a
0
1
3
A1
2
2 and b
3
x
3
for x:
x3
3
x
3
x
x
3
4
3
1 ( x3
3
x
1 3 ( x3
3 2
Therefore, AREA
2
2
2 24
3
2)
1 x4
3 4
3
1
27
3
4x
4
3
0
1 2 ( x3
3 0
4 x) dx
3: f ( x) g ( x)
2x2
3
2
25
12
4
3
1
3
x
81
4
32 25
12
x3
3
2 9
x
x3
3
2
2
8 83
x3
3
1
1
3
3
11
1
3
x3
3
3x
0, or x
1 2 (4 x
3 0
x3 )
x
3
16
4
1 ( x3
3
8
19
4
8 83
3
1
32
3
Copyright
x and y
2, x
2 x x2 3
x2
1 ( x3
3
x
4x)
4;
3
32
3
(2 x x 2 3)dx
9
4
3
4 x2
64. a
1, b 3;
f ( x) g ( x ) (2 x x 2 ) ( 3)
A
0 13 (4 8)
2
2)( x 2)
A2
2 and b
4 x) dx
(4 x 2 ) dx
8
3
0
A1 A2 A3
63. a
2, b 2;
f ( x) g ( x ) 2 ( x 2
A
x
x (x
3
4 x)
A3: For the sketch given, a
A3
0
2x2
x3
3
x
3
x
0 and we find b by solving the equations y
3
x
3
x3
3
0: f ( x) g ( x)
1 x4
3 4
( x3 4 x ) dx
A2: For the sketch given, a
9
x2
2
1
2x
x3
3
x 2) dx
x 2) dx
Therefore, AREA
x2
2
1 and b
A3: For the sketch given, a
A3
x3
3
x 2) dx
( x 2) (4 x 2 )
1: f ( x) g ( x)
395
2014 Pearson Education, Inc.
2 so b
1
3
x
3
simultaneously
2: f ( x) g ( x)
2 x2
2
x4
4 0
14
25 ;
12
4 x)
1 81
3 4
1 (8
3
4)
4;
3
396
Chapter 5 Integration
65. a
0, b
2;
f ( x) g ( x ) 8 x x 4
2
x5
5 0
8 x2
2
2
A
16 32
5
0
(8 x x 4 )dx
80 32
5
48
5
66. Limits of integration: x 2 2 x x
x2 3x
x( x 3) 0 a 0 and b 3;
f ( x) g ( x ) x ( x 2 2 x) 3 x x 2
3
A
0
3
x3
3 0
3 x2
2
(3x x 2 ) dx
27
2
0
27 18
2
9
67. Limits of integration: x 2
x 2 4 x 2 x2
2 x( x 2) 0 a 0 and b 2;
f ( x) g ( x ) ( x 2 4 x) x 2
2 x2 4 x
2
A
( 2 x2
0
16
2
16
3
32 48
6
0
2
4 x2
2 0
2 x3
3
4 x)dx
4x
9
2
8
3
68. Limits of integration: 7 2 x 2 x 2 4 3 x 2 3 0
3( x 1)( x 1) 0 a
1 and b 1;
2
2
f ( x) g ( x ) (7 2 x ) ( x 4) 3 3 x 2
1
A
1
6 23
(3 3 x 2 )dx
1
x3
3
3 x
3 1 13
1
1 13
4
69. Limits of integration: x 4 4 x 2 4 x 2
x4 5x2 4
2
2
( x 4)( x 1) 0
( x 2)( x 2)( x 1)( x 1) 0
x
x 4 5 x 2 4 and
g (x ) f ( x) x 2 ( x 4
1
A
2
2
1
x5
5
1
5
( x4
2, 1,1, 2; f ( x) g ( x )
5
3
x4 5x2
( x4
5x 2
5 x3
3
4x
4
32
5
1
2
40
3
4 x2
x4
4)
1
4 dx
4 x2
1
0
4) x 2
5x 2
( x4 5x2
4
4) dx
4) dx
x5
5
5 x3
3
8
1
5
4x
5
3
1
1
4
Copyright
x5
5
1
5
5 x3
3
5
3
4
4x
2
1
32
5
40
3
8
2014 Pearson Education, Inc.
1
5
5
3
4
60
5
60
3
300 180
15
8
Section 5.6 Substitution and Area Between Curves
70. Limits of integration: x a 2
a2
x2
0
0
A
1 2 (a 2
2 3
a
x 2 dx
x 2 )3/2
0
0
a
1 ( a 2 )3/2
3
1 ( a 2 )3/2
3
x
5
x 6 or y
x
x2
0
x
x a2
0 or
a,0, a;
x 2 dx
1 2 (a 2
2 3
a
x 2 )3/2
0
2 a3
3
71. Limits of integration: y
5y
0
0 or a 2
x
x a2
a
x2
x, x
x
6 ; for
5
x
0
and
x, x 0
x 5x 65
0;
2
5 x x 6 25( x) x 12 x 36
x 2 37 x 36 0 ( x 1)( x 36) 0
x
1, 36 (but x
36 is not a solution);
for x
0:5 x
x 6
x 2 12 x 36
25 x
x 2 13x 36 0 ( x 4)( x 9) 0
x 4,9; there are three intersection points and
0
A
x 6
5
1
( x 6)2
10
2
3
36
10
2
3
25
10
4 x 6
5
x dx
x
3/2
0
1
2
3
100
10
2 and x
2x
2
8
x
2
4 x2
x
2
2
2: x
8
2
x
8 2x2
4
0
x2
4|
16
3/2
225
10
9
dx
( x 6)2
10
2 x 3/2
3
2
3
4, x
2 or x
2
x
4 x , 2
x2
2
4
2
0
2
3
x 6
5
x
4
4
2 x 3/2
3
36
10
4
| x2
72. Limits of integration: y
for x
( x 6)2
10
3/2
9
x dx
0
9
4
3/2
4
100
10
2
2
4
x 4; for
x2 8
x2
2
0
x
2:
x
0; by
symmetry of the graph,
A
2
2
2
2
8
2
x2
2
0
0
4
2
(4 x 2 ) dx
4
x2
2
2 32
4
64
6
x2
4 dx
16
8
6
40
3
2 x2
56
3
Copyright
2
0
64
3
2 8x
4
x3
6 2
2014 Pearson Education, Inc.
50
10
20
3
5
3
397
398
Chapter 5 Integration
73. Limits of integration: c 0 and d
f ( y) g ( y) 2 y 2 0 2 y 2
3
A
0
3
2 y3
3
2 y 2 dy
3;
2 9 18
0
74. Limits of integration: y 2 y 2
c
1 and d 2; f ( y ) g ( y )
2
A
4
2
1
4
y2
2
( y 2 y 2 ) dy
8
3
1
2
1
3
2
( y 1)( y 2)
( y 2) y 2
8
3
6
2
y3
3
2y
1
2
1
9
2
1
3
2
75. Limits of integration: 4 x y 2 4 and
4 x 16 y
y 2 4 16 y
y 2 y 20
( y 5)( y 4) 0 c
4 and d 5;
A
1
4
5
4
125
3
189
3
1
4
1
4
y2 4
4
16 y
4
f ( y) g ( y)
( y2
25
2
9
2
1 64
4 3
243
8
180
76. Limits of integration: x
y2
c
3 2 y2
1 and d
3 3y
2
3 y
y3
3
3(1 y )
1
1
3 1 13
77. Limits of integration: x
y2 2 3y2
2( y 1)( y 1)
f ( y) g( y)
A 2
2 1 13
1
1
0
2
(3 2 y )
1
1
(1 y ) dy
(2 3 y 2 ) ( y 2 )
1 13
4 23
y
0
2
2
3 2 1 13
y 2 and x
2 y
4
3( y 1)( y 1)
1 13
3
20 y
3 2 y2
2 y2 2 0
0 c
1 and d
(1 y 2 ) dy
2
y 2 and x
A 3
5
y2
2
80
1; f ( y ) g ( y )
2
y3
3
1
4
16
2
3 y2 3
0
y 2 y 20
4
y 20) dy
100
0
2 3 y2
y
3
1
y
2(1 y 2 )
x
y
2
0
1
8
3
2014 Pearson Education, Inc.
2
0
1
1
Copyright
x 3y2
1
1;
2 2 y2
3
4
1
2
x
Section 5.6 Substitution and Area Between Curves
78. Limits of integration: x y 2/3 and
x 2 y4
y 2/3 2 y 4
c
1 and d
(2 y 4 ) y 2/3
f ( y) g ( y)
y5
5
2y
3 y 5/3
5
3
5
2 2 15
1
A
1;
(2 y 4
1
3
5
2 15
1
12
5
1
y 2/3 ) dy
3
5
2 15
y 2 1 and x | y | 1 y 2
79. Limits of integration: x
y2 1 | y | 1 y2
y 4 2 y 2 1 y 2 (1 y 2 )
4
2
2
4
y 2y 1 y
y
2 y4 3y2 1 0
(2 y 2 1)( y 2 1) 0 2 y 2 1 0 or y 2 1 0
or y 2
1
2
2
y
or y
y2
1
2
1.
Substitution shows that 2 2 are not solutions
y
1;
2
2
for 1 y 0, f ( x) g ( x)
y 1 y
( y 1)
2
2 1/2
1 y
y (1 y ) , and by symmetry of the graph,
A
0
2
1 y2
1
0
y3
3
2 y
y (1 y 2 )1/2 dy
2 12
1
0
2
1
(1 y 2 ) dy 2
0
2(1 y 2 )3/ 2
3
2 (0 0)
1
1 13
0
1
y (1 y 2 )1/2 dy
2
3
0
2
80. AREA A1 A2
Limits of integration: x 2 y and
x y3
y( y
y2
2
y 2)
for 1
y
1
1
4
for 0
y
4
y2
1
5 ;
12
0
y3
(2 y
8
3
0
8;
3
y
3
2 y ) dy
2, f ( y ) g ( y )
2
16
4
0
y ( y 1)( y 2)
( y3
1
3
A2
y2 2 y
0, f ( y ) g ( y )
0
A1
0
y3
2y
y 2 ) dy
0
y
2
2y
y
y4
4
y3
y2
y3
3
1, 0, 2:
y2
0
1
y2
y4
4
Therefore, A1 A2
Copyright
y3
3
2
0
5 8
12 3
37
12
2014 Pearson Education, Inc.
399
400
Chapter 5 Integration
81. Limits of integration: y
4 x 2 4 and y x 4 1
4
2
4
x 1
4x 4
x 4 x2 5 0
2
( x 5)( x 1)( x 1) 0 a
1 and b 1;
2
4
f ( x) g ( x )
4x 4 x 1
4 x2 x4 5
1
A
1
4
3
1
5
( 4 x2
4
3
5
4 x3
3
x 4 5) dx
1
5
5
4
3
2
1
5
x5
5
5x
5
104
15
1
1
82. Limits of integration: y x3 and y 3x 2 4
x3 3 x 2 4 0 ( x 2 x 2)( x 2) 0
( x 1)( x 2)2 0 a
1 and b 2;
3
2
f ( x) g ( x ) x (3x 4) x3 3 x 2 4
2
A
16
4
1
24
3
( x3 3 x 2
1
4
8
x4
4
4) dx
3 x3
3
2
4x
1
27
4
1 4
83. Limits of integration: x 4 4 y 2 and x 1 y 4
4 4 y2 1 y4
y4 4 y2 3 0
y
3 y
3 ( y 1)( y 1) 0 c
1 and d
since x
0; f ( y ) g ( y )
3 4y
3y
2
2
y
4 y3
3
4
y5
5
1
(4 4 y ) (1 y )
1
A
1
y2
4
c
85. a
3 1
y2
4
3 2
8
12
0, b
A
3 y2
4
2 and d
3 (y
4
2
y2
4
2
2
1
8
12
y2
4
3 4
y3
3 y 12
0
y
4
2
2
2
12 4 8
2sin x sin 2 x
(2sin x sin 2 x)dx
2 1 12
dy
16
12
2)( y 2)
(3 y 2 )
2; f ( y ) g ( y )
A 3
2( 1) 12
56
15
1
5
3 0
; f ( x) g ( x)
0
y 4 ) dy
3 y 2 and x
84. Limits of integration: x
3 y2
(3 4 y 2
2 3 34
1
1
4
2 cos x
cos 2 x
2
0
4
Copyright
2014 Pearson Education, Inc.
Section 5.6 Substitution and Area Between Curves
86. a
3
,b
3
/3
A
/3
8 23
87. a
; f ( x ) g ( x) 8cos x sec2 x
(8cos x sec2 x)dx [8 sin x tan x ] /3/3
8 23
3
3
(1 x 2 ) cos 2x
1, b 1; f ( x) g ( x)
1
A
x
2
1
x3
3
2 2
3
1 x 2 cos 2x
2 sin
4
3
x
2
4
6 3
1
dx
1 13
1
2
1 13
2
88. A A1 A2
a1
1, b1 0 and a2 0, b2 1;
f1 ( x) g1 ( x) x sin 2x and f 2 ( x) g 2 ( x )
sin 2x
A1
A2
2 2
89. a
4
/4
A
/4
/4
/4
/4
90. c
2 42
,b
/4
4
2
A
cos 2x
1
2
4
by symmetry about the origin,
2 A1
2
2
x
1
2
sin 2x
0
1
x2
2 0
2
2
x dx
0 12
2
1 0
4
; f ( x ) g ( x) sec2 x tan 2 x
(sec2 x tan 2 x ) dx
[sec 2 x (sec2 x 1)] dx
1 dx [ x] /4/4
,d
4
2 1 4
/4
/4
4
2
; f ( y) g ( y)
tan y ( tan 2 y )
A
4
2 tan 2 y
2 sec2 y 1 dy
1 4
2(sec2 y 1)
2[(tan y
4 1 4
y )] /4/4
4
Copyright
2014 Pearson Education, Inc.
401
402
Chapter 5 Integration
91. c
0, d
; f ( y) g ( y)
2
3 sin y cos y 0
A
/2
3
0
2(0 1)
92. a
3sin y cos y
3
1
sec2 3x
1
3
4
3
0
sec2 3x
x1/3
x1/3 dx
3 x 4/3
4
tan 3x
3
/2
2
1, b 1; f ( x) g ( x)
A
3 23 (cos y)3/2
sin y cos y dy
3
1
1
6 3
3
4
3
93. A A1 A2
Limits of integration: x y 3 and x y
y y3
3
y
y 0 y ( y 1)( y 1) 0 c1
1, d1
and c2 0, d 2 1; f1 ( y ) g1 ( y ) y3 y and
f 2 ( y ) g 2 ( y ) y y3
by symmetry
A
about the origin, A1 A2 2 A2
2
1
0
y2
y 3 ) dy
(y
1
y4
4
2 2
2 12
0
1
4
0
1
2
94. A A1 A2
Limits of integration: y x3 and y x5 x3 x5
x5 x3 0 x3 ( x 1)( x 1) 0 a1
1, b1 0
3
5
and a2 0, b2 1; f1 ( x) g1 ( x) x x and
f 2 ( x) g 2 ( x ) x5 x3
by symmetry about the
origin, A1
1
x6
6 0
4
2 x4
95. A
A2
2 A2
2 14
A
1
6
A1 A2
Limits of integration: y
x
3
A1
x 2
A
A1
1
2
1
0
( x3 x5 ) dx
1
6
x 1, f1 ( x) g1 ( x )
1
0
1
2
A2
1
1
2
1 ; f ( x)
2 2
x 2 dx
1 2
x 1
1
2
x
x 0
x2
2 0
x dx
A2
1
x2
x and y
,x
0
x
g 2 ( x)
1
2
1
x2
1
1
x2
0
1;
2
1
Copyright
2014 Pearson Education, Inc.
Section 5.6 Substitution and Area Between Curves
96. Limits of integration: sin x
b
; f ( x) g ( x )
4
/4
A
0
2
2
97.
98.
5
1
(0 1)
101.
102.
ln 3
0
2x
2 1 x2
1
dx
e2 x
2
2(1 x ) dx
2
2 2x
0 1 x2
2
51
1 u
2
0 and
sin x
/4 cos x
ln 3
e2ln 3
2
0
dx
0
dx; [u 1 x 2
5
1
2 ln u
dx
2
1x
2
ln 12
dx
ln 24
(ln 2)(5 1)
/3 sin x
cos x
e0
2
eln 3
dx
ln16
0
ln cos x
du
e0
9
2
ln cos x
/4
2 x dx; x
2(ln 5 ln1)
0
1
2
3
2eln 2 2e ln 2
0
du
1
/3
0
3 ln 2
2
ln 2 ln 2
ex
5
(ln 2)
2ln 2
2e x /2 2e x /2
1 x
1 2
1
0
tan x dx
e x /2 e x /2 dx
2
1
/3
0
e2 x e x dx
2ln 2
0
1
( ln x ln 2 ln x ) dx
ln 12 ln1
2
100.
5
tan x dx
ln1 ln 1
99.
a
4
2 1
(ln 2 x ln x ) dx
/4
x
(cos x sin x ) dx [sin x cos x ]0 /4
2
2
0
cos x
cos x sin x
403
8
2
1
2 e0 2 e0
u 1, x
2
u
2
2
(4 1) (2 2) 5 4 1
5]
2ln 5
1
2 1
ln 2 2
1
2
3
2
2
ln 2
3
ln 2
103. (a) The coordinates of the points of intersection of the
line and parabola are c x 2
x
c and y c
(b) f ( y ) g ( y )
y
y
2 y
the area of
c
the lower section is, AL
2
c
0
y dy
0
c
2 23 y3/2
0
[ f ( y ) g ( y )] dy
4 c3/2 .
3
The area of
the entire shaded region can be found by setting c
the region into subsections of equal area we have A
Copyright
4: A
2 AL
4
3
43/2
32
3
2
4 8 32 . Since we
3
3
4 c3/2
c 4 2/3
3
2014 Pearson Education, Inc.
want c to divide
404
Chapter 5 Integration
4 c3/2 . Again,
3
condition A
c
c
cx
x3
3
the area of the whole shaded region can be found by setting c
4
c x2
(c) f ( x) g ( x )
AL
c
[ f ( x) g ( x)] dx
4 c 3/2
3
2 AL , we get
32
3
c
4
2/3
c
(c x 2 ) dx
c
c
A
2 c3/2
32 . From the
3
as in part (b).
1
104. (a) Limits of integration: y 3 x 2 and y
3 x2
1 x2 4 a
2 and b 2;
f ( x) g ( x ) (3 x 2 ) ( 1) 4 x 2
2
A
2
(4 x 2 ) dx
4x
x3
3
8
3
16
3
32
3
8
3
8
8
16
2
2
(b) Limits of integration: let x 0 in y 3 x 2
y 3; f ( y ) g ( y )
3 y
3 y
2(3 y )1/2
4
3
A
0 (3 1)
3
2
1
3/2
4
3
105. Limits of integration: y 1
1
2
x
x
,x
(3 y )1/2 dy
0
(8)
x
1
(3 y )1/2 ( 1) dy
2(3 y )3/ 2
3
( 2)
3
1
32
3
2
x
x and y
x
3
2
2
(2 x)2
x
x 4 4x x2
x2 5 x 4 0
( x 4)( x 1) 0 x 1, 4 (but x 4 does not
x
2
satisfy the equation); y 2 and y 4x
4
x
1
A1
0
A2
1
AREA
1 x1/2
x
4
dx
x
2 x 1/2
x
4
dx
4 x1/2
4
A1
A2
37
24
17
8
y2
2y 1 3 y
( y 2)( y 1)
0
y
2 x 3/2
3
x
4
2 1
x
8 0
4
x2
8 1
88 11
24
3
37 51
24
106. Limits of integration: ( y 1) 2
y2
x
64 x3
x 4. Therefore,
A2 : f1 ( x) g1 ( x) 1 x1/2
8 x x
AREA A1
1 23
1
8
4 2 16
8
0
37 ; f ( x )
2
24
4 15
8
4 81
g 2 ( x)
2 x 1/2
17 ; Therefore,
8
3 y
y 2
0
2 since y
2
2
2 y1/2
A1
0; also,
2 y 3 y
4y 9 6y y
y 10 y 9 0
( y 9)( y 1) 0
y 1 since y 9 does not satisfy
the equation;
AREA A1 A2
f1 ( y ) g1 ( y )
2 y 0
f2 ( y) g2 ( y)
(3 y ) ( y 1) 2
6 2 13
3 12 0
1 13
2
2
A2
1
2
7.
6
1 1/2
y
0
1
dy
2
1
0
4;
3
[3 y ( y 1)2 ] dy
Therefore, A1
Copyright
2 y 3/ 2
3
A2
4
3
3y
7
6
1
2
15
6
2014 Pearson Education, Inc.
y2
5
2
1 (y
3
1)3
2
1
c3/ 2
3
x
4
Section 5.6 Substitution and Area Between Curves
a2 : A
107. Area between parabola and y
Area of triangle AOC: 12 (2a )(a 2 )
b
108. A
a
b
2 f ( x) dx
f ( x) dx
a
a
2
0
(a 2
a3 ; limit of ratio
b
2
3
4
4 a3
3
a 0
b
f ( x) dx
a
4a3 ;
3
0
which is independent of a.
f ( x) dx
a
a3
3
2 a3
0
a3
lim
b
f ( x) dx
a
a
2 a 2 x 13 x3
x 2 ) dx
405
4
109. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the region s
upper and lower bounding curves at x 0. The area of the shaded region is actually
0
A
1
x ( x ) dx
1
0
0
x ( x ) dx
1
2 x dx
1
0
2 x dx
2.
110. It is sometimes true. It is true if f ( x) g ( x) for all x between a and b. Otherwise it is false. If the graph of f lies
below the graph of g for a portion of the interval of integration, the integral over that portion will be negative
and the integral over [a, b] will be less than the area between the curves (see Exercise 71).
111. Let u
2x
3 sin 2 x
1 x
du
6 sin u
dx
1
0
du
1
1
x
f odd
x
0
a
a
/2
115. Let u
du
0
a
u
a x
0
1
du
f ( x) dx
0 f ( x) f (a x)
1
I
0
u
6
F (6) F (2)
u
u
0
0
1
f (u ) du
0
f ( x)dx
1
0
f ( x) dx
0
0
f ( u )( du )
1
u 1, x
0
f ( x). Then
1
0
1
0
f (u ) ( du )
1
f (u ) du
0
a
a
a
f ( x) dx
dx; x
0
0
f ( x) dx
0
0
a
u
x
a
f ( x ) dx
a, x
( du )
a
Copyright
du
0
cos
a
u
0
a
du
0 0
a
0
1
f (u )du
0
dx and x
a
f (u ) du
f ( x) dx
2
1
dx
0
f ( u ) du
cos 2
f (a u )
a f ( a u ) f (u )
0
f ( u )( du )
1
f ( x) dx when f is odd. Let u
0
u
0
f ( x)dx
a f ( a x ) dx
f ( x) dx
f
(
x
)
f
(
a
x
)
0
0 f ( x) f ( a x)
a
Therefore, 2 I a
I 2.
I
1
u 1, x
0
3
u 1, x 1
f (u ) du
1
sin x dx [ cos x] /2/2
a
I
0
dx; x
0. Thus
f ( x) dx
/2
(b)
dx; x
f ( x ). Then
f ( x)
114. (a) Consider
du
2, x
[ F (u )]62
du
2
u
3
x
f even
6 sin u
u
du
dx; x
f ( x)
(b) Let u
dx; x 1
f (u )( du )
du
f (u ) du
0
du
dx
0
f (1 x) dx
113. (a) Let u
1
2
u
2
2
112. Let u 1 x
1
2
2 dx
0
f (u ) du
a
a
f (u ) du
f ( x ) dx
0
0.
0.
0
a
f ( a u ) du
0 f (u ) f ( a u )
a f ( x ) f ( a x)
0 f ( x ) f ( a x)
dx
a f ( a x ) dx
0 f ( x) f (a x)
a
0
dx [ x]0a
2014 Pearson Education, Inc.
a 0
u
a.
3
a and
f ( x) dx. Thus
406
Chapter 5 Integration
xy
t
116. Let u
xy 1
x t
117. Let u
b c
a c
dt
xy
du
1
y
1
u
x c
t2
11
yu
du
du
f ( x c) dx
t
xy
dt
dx; x
b
a
118. (a)
y1
1 u
du
a c
f (u ) du
1
t
du
u
b
a
1 du
u
dt
y1
1 t
du
a, x
1 dt ; t
t
x
u
xy
u 1. Therefore,
dt
b c
u
b
f ( x ) dx
(b)
119-122.
y, t
(c)
Example CAS commands:
Maple:
f : x - x^3/3-x^2/2-2*x 1/3;
g : x - x-1;
plot( [f(x),g(x)], x -5..5, legend ["y
f(x)","y
g(x)"], title "#119(a) (Section 5.6)" );
q1: [ -5, -2, 1, 4 ];
# (b)
q2 : [seq( fsolve( f(x) g(x), x q1[i]..q1[i 1] ), i 1..nops(q1)-1 )];
for i from 1 to nops(q2)-1 do
# (c)
area[i] : int( abs(f(x)-g(x)),x q2[i]..q2[i 1] );
end do;
add( area[i], i 1..nops(q2)-1 );
# (d)
Mathematica: (assigned functions may vary)
Clear[x, f, g]
f[x_]
x 2 Cos[x]
g[x_]
x3 x
Plot[{f[x], g[x]}, {x, 2, 2}]
After examining the plots, the initial guesses for FindRoot can be determined.
pts x/.Map[FindRoot[f[x] g[x],{x, #}]&, { 1, 0, 1}]
i1 NIntegrate[f[x] g[x], {x, pts[[1]], pts[[2]]}]
i2 NIntegrate [f[x] g[x], {x, pts[[2]], pts[[3]]}]
i1 i2
Copyright
2014 Pearson Education, Inc.
Chapter 5 Practice Exercises
CHAPTER 5
407
PRACTICE EXERCISES
1. (a) Each time subinterval is of length t 0.4 sec. The distance traveled over each subinterval, using the
midpoint rule, is h 12 (vi vi 1 ) t , where vi is the velocity at the left endpoint and vi 1 the velocity at
the right endpoint of the subinterval. We then add h to the height attained so far at the left endpoint vi to
arrive at the height associated with velocity vi 1 at the right endpoint. Using this methodology we build the
following table based on the figure in the text:
t (sec) 0 0.4
v (fps) 0 10
h (ft) 0 2
t (sec)
v (fps)
h (ft)
6.4
50
643.2
0.8
25
9
1.2
55
25
6.8
37
660.6
1.6
100
56
7.2
25
672
2.0
190
114
7.6
12
679.4
2.4
180
188
2.8
165
257
3.2
150
320
3.6
140
378
4.0
130
432
4.4
115
481
4.8
105
525
5.2
90
564
5.6
76
592
6.0
65
620.2
8.0
0
681.8
NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph.
Remember that some shifting of the graph occurs in the printing process.
The total height attained is about 680 ft.
(b) The graph is based on the table in part (a).
2. (a) Each time subinterval is of length t 1 sec. The distance traveled over each subinterval, using the
midpoint rule, is s 12 (vi vi 1 ) t , where vi is the velocity at the left, and vi 1 the velocity at the right,
endpoint of the subinterval. We then add s to the distance attained so far at the left endpoint vi to arrive at
the distance associated with velocity vi 1 at the right endpoint. Using this methodology we build the table
given below based on the figure in the text, obtaining approximately 26 m for the total distance traveled:
t (sec)
v (m/sec)
s (m)
0
0
0
1
0.5
0.25
2
1.2
1.1
3
2
2.7
4
3.4
5.4
5
4.5
9.35
6
4.8
14
7
4.5
18.65
8
3.5
22.65
(b) The graph shows the distance traveled by the
moving body as a function of time for 0 t 10.
Copyright
2014 Pearson Education, Inc.
9
2
25.4
10
0
26.4
408
Chapter 5 Integration
10
3. (a)
k 1
10
(b)
10
ak
4
1
4
(bk
3ak )
ak
k 1
k 1
10
k 1
3
(ak
k 1
20
(d)
(ak
(c)
(d)
(e)
2
7
sin x
9
1
5
2
f ( x ) dx
2
5
5
2
5
2
1
2
8 1/3 1
u
2
du
u
1
0
0
0
5
2
u2
2 0
1 (12)
3
f ( x) dx
2
2
g ( x ) dx
5
2
1
5
5
2
0, x
1
u du
f ( x) dx
dx
,x
u
f ( x) dx
u
u
0, x
3 (16
8
(cos u )(2 du ) [2sin u ]0
1 2 3
3 2
2
2
5
9
3 1 2
8
3 u 4/3
8
0
cos x dx; x
5
u 1, x
x dx; x 1
du
dx; x
g ( x)) dx
f ( x) g ( x)
5
8
40
u 1/2 12 du
0
g ( x) dx
(
0 2(20)
9
u1/2
1
/2
f ( x) dx
2
1
2
2 x dx
du
2 (7)
7
dx; x 1
(sin x)(cos x) dx
2
1 (20)
2
2
7
du
0
cos 2x dx
bk
k 1
20
2 dx
du
2 du
0 7
k 1
du
x( x 2 1)1/3 dx
x
2
bk
k 1
20
ak
x2 1
7. Let u
(b)
0
20
k 1
(2 x 1) 1/2 dx
/2
25
0
ak
2)
2x 1
6. Let u
9. (a)
20
k 1
20
1
2
k 1
20
k 1
0
2 25 (1)(10) 13
k 1
5 (10)
2
k 1
3(0)
bk )
2bk
7
1
2
(c)
8. Let u
1
k 1
10
31
10
bk
bk
ak
k 1
k 1
20
0
5
2
25 3( 2)
20
3ak
(b)
3
ak
k 1
10
k 1
k 1
k 1
20
1
10
ak
bk
20
4. (a)
5
3
bk 1)
10
5
2
(d)
1
bk
1
2
k 1
10
(ak
5. Let u
2)
10
k 1
10
(c)
1(
4
0)
u
3
u
8
6
0
2 sin 0 2sin
/2
2
u 1
2
1
2
4
6 4
2
2
g ( x ) dx
(2)
2
5
f ( x) dx 15
g ( x) dx
2
Copyright
1 (6)
5
1 (2)
5
8
5
2014 Pearson Education, Inc.
2(0 ( 1))
2
Chapter 5 Practice Exercises
2
10. (a)
0
2
(b)
1
2
2
0
11. x 2
2
0
1
Area
0
( x2
0
2
2
2
x3
12
2
4
3
Area
x
1 x4 dx
0
1
1
5 x 3 x5/3
2
0
2/3
1
5(1) 3(1)5/3
3(1)
2
1 x4 dx
13
4
3
4
2/3
1
2(1)2
33
3 12
1 x
(5 5 x
1
2;
3
x3
12 2
( 2)3
12
3
4
4
3
3
8
3
3
2
3 or x 1;
x
( x 2 4 x 3) dx
13
3
2
2
f ( x ) dx 1 3
0
0
1
x
1
2
2
g ( x ) dx 3
2 x 2 3x
4 x2 0
23
12
13. 5 5 x 2/3
1
2( )
0
3
x3
3
0
1
3
0
Area
x
2
0
2(32 ) 3(3)
1
2
12. 1 x4
1
g ( x) dx 1 2
f ( x) dx
0
4 x 3) dx
2(1)2 3(1)
33
3
1
3
2
2
( x 3)( x 1)
2 x 2 3x
13
3
0
[ g ( x ) 3 f ( x )] dx
4x 3 0
x3
3
1
1
f ( x ) dx
2 f ( x) dx
0
(e)
0
1 (7)
7
g ( x ) dx
g ( x) dx
f ( x ) dx
2
(d)
2
g ( x) dx
0
(c)
1 27
7 0
g ( x) dx
x
8
) dx
1
5 x 3 x5/3
23
2 12
1;
(5 5 x 2/3 ) dx
8
1
5( 1) 3( 1)5/3
5(8) 3(8)5/3
5(1) 3(1)5/3
[2 ( 2)] [(40 96) 2] 62
14. 1
x
Area
x
1
1
3
0
1
0
x 1;
(1
4
x ) dx
1
2 x3/2
x
3
0
2 (1)3/2
0
3
1
4 16
3
3
1
(1
x ) dx
4
2 x3/2
3
1
4 23 (4)3/2
1
2 (1)3/2
3
2
Copyright
2014 Pearson Education, Inc.
409
410
15.
Chapter 5 Integration
f ( x)
b
A
2
1
16.
f ( x)
a
1
0
2
1
x
x
1
1
2
A
d
c
3 2
y
0
20. f ( y )
A
4y
d
c
y3
3
0, a
2
1
0, d
0
A
c
d
c
A
b
a
1
2
1 (6
6
8 3)
[ f ( x) g ( x )] dx
0
(1
0, c
2, d
2
2
1
0
(1 x3 )2 dx
2
(4 y 2 ) dy
32
3
2; f ( y )
[ f ( y ) g ( y )] dy
y 2
4
y
1 or
y 2
y2
, g ( y) 4
4
2 y 2 y2
dy
4
1 4
Copyright
1
x ) 2 dx
0
(1 2 x
x) dx
1
6
18
2 8 83
1, d
1 43
1
(2 y 2 0) dy
y2
2
a
[ f ( x) g ( x )] dx
3
3
21. Let us find the intersection points: 4
y 2 y 2 0 ( y 2)( y 1) 0
y
b
A
1
x2
2 0
0, b 1
[ f ( y ) g ( y )] dy
2
1
9
14
1
7
0, c
2 [ y 3 ]3
0
3
2
1
[ f ( x) g ( x )] dx
0, b 1
0, a
1 12
4 y 2 , g ( y)
a
2 x
[ f ( y ) g ( y )] dy
dy
b
x 43 x3/2
x) dx
2 y 2 , g ( y)
19. f ( y )
1
2
7 4 2
2
2
1
x7
7 0
1
2
2
x2
2
dx
(1 x3 )2 , g ( x)
x4
2
4
2
, a 1, b
x )2 , g ( x)
(1 2 x1/2
x
2
1
x
x, g ( x )
(1
18. f ( x)
2
2
1
x 1
x2
2
dx
2 2
17. f ( x)
, a 1, b
[ f ( x) g ( x )] dx
1
x2
x
A
4
2
1
x2
x, g ( x )
2014 Pearson Education, Inc.
1
0
(1 2 x3
x 6 ) dx
Chapter 5 Practice Exercises
1
4
2
1
4
4
2
1
( y 2 y 2 ) dy
8
3
4
1
2
y2
2
1
4
y2 4
4
22. Let us find the intersection points:
y2
y
y 20
0
1
( y 5)( y 4)
0
y 16
4
y
4 or
y 16
y2 4
5 c
4, d 5; f ( y )
,
g
(
y
)
4
4
d
5 y 16 y 2 4
A
[ f ( y ) g ( y )] dy
dy
4
4
c
4
5
2
y3
1 5 ( y 20 y 2 ) dy 1 y
20
y
4 4
4 2
3
4
16 80 64
1 25 100 125
4
2
3
2
3
1 9 180 63
1 9 117
1 (9 234) 243
4 2
4 2
8
8
23. f ( x)
x, g ( x )
b
A
a
x2
2
sin x, a
0, b
[ f ( x) g ( x )] dx
/4
cos x
2
2
2
32
0
b
A
a
0
/2
2
/2
0
(1 sin x) dx
2 2 1
0, b
0
3
8 23
/2
/2
/2
2
(1 |sin x |) dx
(1 sin x ) dx
2[ x cos x ]0 /2
2sin x sin 2 x
(2sin x sin 2 x) dx
,b
A
,b
2
2 ( 1) 12
26. a
2
0
, f ( x) g ( x)
A
( x sin x) dx
1
[ f ( x) g ( x )] dx
(1 sin x) dx
4
/4
0
24. f ( x) 1, g ( x) |sin x |, a
25. a
2
9
8
1
3
2
y3
3
2y
3
/3
/3
3
2 1 12
2 cos x
cos 2 x
2
0
4
, f ( x ) g ( x) 8cos x sec 2 x
(8 cos x sec2 x) dx [8sin x tan x ] /3/3
8 23
3
6 3
Copyright
2014 Pearson Education, Inc.
411
412
Chapter 5 Integration
27. f ( y )
y , g( y)
d
A
c
2
y
1
4
3
2 y, c
2
3
d
A
c
6y
y2
2
4 73
1
2
2
1
24 14 3
6
x3 3x 2
29. f ( x)
2
y 2 , c 1, d
2
a
A
(a1/2
0
2
a 2 1 43
1
y2
2
0
A2
1
1
( y 2/3
a
32.
A
/4
0
5 /4
x) dx
1
0
( y 2/3
ax
4
3
ax3/2
x4
4
|
|
0
2
x3
a
x2
2 0
y ) dy
area below the x-axis is
3 y 5/3
5
y2
2
0
6
5
5 /4
(cos x sin x) dx
3 /2
0
f
( x3 3 x 2 ) dx
a2
6
(6 8 3)
A2
3 x( x 2)
3
(a 2 ax1/2
0
y ) dy
total area is A1
1
3
4 is a minimum. A
1 ; the
10
0
1
8 2 7
6
3x 2 6 x
f ( x)
31. The area above the x-axis is A1
3 y 5/3
5
7
6
2
13
6
x1/2 ) 2 dx
a2
6
y2
2
6 12
x 2 ( x 3)
1
2
(2 y )] dy
(6 y y 2 ) dy
12 2 83
maximum and f (2)
30.
[ y
4
3
[ f ( y ) g ( y )] dy
y3
3
2
2y
1
2
2
y, g ( y )
1
2 y 3/2
3
2 y dy
6
2
[ f ( y ) g ( y )] dy
2 4 2
28. f ( y )
1, d
/4
11
10
1
the
(sin x cos x ) dx
(cos x sin x) dx
[sin x cos x]0 /4 [ cos x sin x]5 /4/4
[sin x cos x]35 /2
/4
2
2
2
2
2
2
(0 1)
( 1 0)
2
2
2
2
2
2
8 2
2
2
2
2
Copyright
2
2
4 2 2
2014 Pearson Education, Inc.
3
0
a2
f (0)
81
4
4
3
27
a a a
27
4
a2
2
0 is a
Chapter 5 Practice Exercises
33.
e 2ln x
dx
x
A
34. (a)
A1
(b)
A1
35.
y
36.
y
x
1
1
x1
1 t
0
0
y
x sin t
t
37.
y
38.
y
39.
dy
dx
1 x
40.
dy
dx
1
x2 1
5
x
0
dy
dx
dy
dx
ln 2, and A2
ln x
kb
ka
ln kb ln ka
kb
ln ka
ln ab
d2y
1
x
2x
dy
dx
sin x ;
x
x
5
dx
dy
2
1
1 x
1
1 x2
dy
1
sec
1
1 x2
2
cos x
dy
1 x2
C=1
y
sec 1 x
2 C
C
dx
x x2 1
1
1
1 x2
2
1 x2
du
sin x dx
2(cos x) 1/2 sin x dx
du
2u 1/2 ( du )
y
tan x
du
(tan x) 3/2 sec2 x dx
u
b1
dx
a x
ln x
b
a
ln b ln a
2 1 3
sec x (tan x);
3
1
1
y
2 sin 2 t dt 2
1
0
sin 1 0 C
2
C=0
y
sin 1 x
dx
tan 1 ( x) x 1
y
1
2
2
3
3
sec 1 ( x)
y
2
3
, x>1
tan 1 x 2sin 1 x C ;
C=2
tan 1 x 2 sin 1 x 2
y
sin x dx
2 u 1/2 du
1/ 2
2 u1
C
4u1/2 C
sec 2 x dx
u 3/2 du
ln 2 ln1 ln 2
3
2
44. Let u
2
1
C;
sec
tan 1 0 2 sin 1 0 C
2
ln x
u=1
tan 1 ( x) x C ;
y
1 tan 1 0 0 C
dy
x x2 1
dt 3
5
2 sin 2 x; x
1 dx
21
dx
1 x
1 and y (1)
1 2 sec 0
5 sin t
t
y
11
1t
sin 1 x C ; x = 0 and y = 0
y
2
dy
dx
0
u = 0, x = e
2 12 (sec x ) 1/2 (sec x tan x )
dx 2
0 and x
x=1
ln b ln a, and A2
; y (1) 1
d2y
1 2 sec x
dy
1
1
x2
2
dx 2
2 sin 2 t dt 2 so that dx
1
x = 0 and y = 2
43. Let u
20
ln 10
dy
dx
dt 3
x = 2 and y =
42.
ln 20 ln10
(1 2 sec t ) dt
x = 0 and y = 1
41.
20
10
(1 2 sec t ) dt
0
1 dx;
x
1, where u = ln x and du
ln x
dy
dx
dt
[u 2 ]10
2u du
20 1
dx
10 x
kb 1
dx
ka x
x2
x
0
1/ 2
1
2
Copyright
C
413
2u 1/2 C
2
(tan x )1/ 2
C
2014 Pearson Education, Inc.
4(cos x)1/2
C
414
Chapter 5 Integration
45. Let u
2
[2
1
du
u
48.
tan u C
2
t
(t 1)2 1
t
2
t
t
49. Let u
t2
dt
t 2 2t
t4
2t 3/2
du
51.
52.
du
e x sec2 (e x
7)dx
tan u C
tan(e x
54.
C
55.
1 dx
1 3x 4
ln u
e ln x
1 x
dx
1
2 u 3/2
3
0
sin (2
4
(t 2
1
3
1 du
t dt
3
1 cos u C
3
sec tan
ex
1 u1/ 2
1
2
C
t 1
( 1)
t3
3
2
2 t2
4
t
1 sec d
C
1
t
1
t2
u1/2 du
2 u 3/2
3
e x dx
7 and du
e y 1 and du
e y dy
sec 2 x dx
eu du , where u = cot x and du
1 1/2
u du,
0
2 13/2
3
csc2 x dx
C
1 [0
3
4, du = 3 dx; x = 1
ln 7]
where u = ln x, du
2 03/2
3
1 dx;
x
u = 7, x = 1
ln 7
3
x=1
u = 0, x = e
2
3
Copyright
C
C
C
where u = 3x
1 (2 tan u )
2
C
(2t 3/2 ) C
1 cos
3
csc u cot u du, where u
ln 1 ln 7
1) C1
2
1
4 t1
2t 3 ) dt
2sec2 u ) du
7) C
11
du,
7u
1
3
1
7
dt
(u 1/2
1
2
t3
3
eu du, where u = tan x and du
ecot x
1
1
3
56.
e tan x
C
4t 2 ) dt
sec2 u du , where u
(csc2 x)ecot x dx
eu
1)2
(2
csc(e y 1) C
(sec 2 x)e tan x dx
eu
1 du
2
(t 2
sec tan d
e y csc(e y 1) cot(e y 1)dy
csc u C
53.
sin u du
2sec 2 u
) C
2
t3
sin u C1
is still an arbitrary constant
tan (2
dt
1
t2
dt
1
3
50. Let u 1 sec
4
t2
3 t dt
t sin 2t 3/2 dt
u2
4
d
1
u
)1/2
(2
1
4
C1
du
) d
dt
4
1
2
2d
2sec2 (2
2
1/2
t
1) C , where C
2
1
d
(u 2 cos u ) 12 du
1)] d
sin (2
46. Let u
1 du
2
2d
1 2 cos (2
2
47.
du
2014 Pearson Education, Inc.
u=1
u= 1
C
2 (1
3
sec )3/2
C
Chapter 5 Practice Exercises
57.
58.
4 2t
dt
0 t 2 25
9
ln u
25
9 1
du,
25 u
ln 9
tan(ln v )
dv
v
59.
(ln x )
x
u
60.
2
3
C
1 csc 2 (1
r
2
x3x dx
63.
64.
66.
3
2
3 sin 1 u
2
C
6 dr
6
dx
2 ( x 1) 2
2
67.
du
C
, where u = 2(r
3 sin 1 2( r
2
du
4 u2
1
2
dx
C
C
1 and du = dx
tan 1 x 1
2
C
where u = 3x + 1 and du = 3 dx
1 tan 1 (3 x
3
(2 x 1) (2 x 1)2 4
1) and du = 2 dr
, where u = r + 1 and du = dr
, where u = x
C
C
sec 2 x dx
1) C
6sin 1 r 21
du ,
1
3 1 u2
1 1 sec 1 u
2 2
2
2
C
1 u2
du
2 u2
tan 1 u
1 tan 1 u
3
1 dr
r
x 2 and du = 2x dx
2u du , where u = tan x and du
2tan x
ln 2
C
4 ( r 1)2
3x
1
2ln 3
C
3 dr
dx
1 (3 x 1) 2
1 dx
x
C
3u du , where u
1
2
1 4( r 1)2
1
2
C
cot(1 ln r ) C
6sin 1 u2 C
65.
1 dv
v
csc 2 u du , where u = 1 + ln r and du
2 tan x sec 2 x dx
1 (2u )
ln 2
u = 25, t = 4
9
ln 25
du , where u = ln v and du
ln cos(ln v)
ln r ) dr
1 (3u )
2ln 3
62.
ln 9 ln 25
sin u
cos u
1 (ln x) 2
2
cot u C
61.
25, du = 2t dt; t = 0
u 3du, where u = ln x and du
C
2
ln 25
tan u du
ln cos u
t2
where u
1
2
1) C
du
u u2 4
, where u = 2x
1 sec 1 2 x 1
4
2
1 and du = 2 dx
C
Copyright
2014 Pearson Education, Inc.
u= 9
415
416
68.
Chapter 5 Integration
dx
du
( x 3) ( x 3)2 25
u u 2 25
1 sec 1 u
5
5
69.
esin
1 x
2 x x
eu
1
sin
70.
x
1 x2
2 u 3/2
3
C
eu du, where u
1
esin
C
1 sec 1 x 3
5
5
C
dx
2
, where u = x + 3 and du = dx
x
sin 1 x and du
C
u1/2 du , where u
dx
2 (sin 1 x)3/2
3
C
dx
2 x x2
sin 1 x and du
dx
1 x2
C
1
1
71.
1
tan
72.
73.
74.
75.
76.
77.
(tan
1
1 x
1 u3
3
1
1
1
0
y (1 y )
x )2
2
tan
y
1 (tan 1 x )3
3
C
2
1
[2 s 4
4 (1
1
79. Let u
u )1/ 2
u
4
1
u
dy
1 y2
du
1 u 1/2
2
3 1/2
2
du
3
1
x
[13 2(1)2
[2(1) 4
4
2
4
1
[ 2t 1/2 ]14
du
2 dx
2 23 x
(2 dx)
2 dx
18u 3 du
dx
1 x2
7(1)] [( 1)3 2( 1)2
4(1)3 5(1)] 0
18 du
18u
2
2
3
1
( 2)
1
2
4
du ; u
u
3/2 3
2
36 dx; x
3
9
u2 1
Copyright
7( 1)]
6 ( 10) 16
3
2
3(27) 1/3 ( 3(1) 1/3 )
t 3/2 dt
dx
2x 1
1 36 dx
0 (2 x 1)3
7 x]1 1
4v 2 dv [ 4v 1 ]12
4 dt
1 t 3/ 2
78. Let x 1
tan 1 x and du
4 s3 5s]10
x 4/3 dx [ 3 x 1/3 ]127
4 dt
1 t t
tan 1 y and du
C
4 x 7) dx [ x3 2 x 2
dv
u 1/2 du , where u
dy
u 2 du, where u
dx
(8s 3 12s 2 5) ds
27
1
2 tan 1 y C
C
(3x 2
2 4
1 v2
1
dy
2
2u1/2
1 y2
1
3 13
9
32
2
4
x
1
x
4 (33/2 )
3
0
3(1)
2, u
4 (23/2 )
3
u 1, x 1
9
12
3
4 3 83 2
u
3
8
2014 Pearson Education, Inc.
4 (3
3
3 2 2)
Chapter 5 Practice Exercises
80. Let u
7 5r
du
1
dr
0 3 (7 5r )2
1
0
1
1/2
83. Let u
5r
du
/4
0
4t
/3
0
1
16
sec2
3 /4
/4
1/2
88. Let u
dr ; r
(sin 2 u ) 15 du
3 /4
/4
1 dx
6
/2
/6
6
2
x 1/3 dx; x
u 5/ 2
3
2
x3 dx; x
5
2
1
8
0
1
8
u 1
0
/3
2
2/3
3,
4
u 1 12/3
x 1
0
0
3 u 5/2
5
3/4
1
2
u 1, x
1/ 2
1
2
25/16
u 1 9 12
4
25
16
25/16
1 u 1/2
18
1
1
0
u
0, r
u
sin 2u 5
4
0
dt ; t
0
2
u
4
,t
5
sin10
20
2
3
4
u
4
sin 2u 3 /4
4
/4
1 u
4 2
0
0 sin
20
1 3
4 8
tan 3
du
1
3
sin
3
2
4
1
4
sin
8
2
4
cot 4
dx; x
6 cot u du
u
/2
6
/6
6
2
,x
3
2
(csc u 1) du
u
2
[6( cot u u )] /2
/6
6 3 2
6
3du
d ;
0
/3
2
0
3
cot 34
2
d
tan 0
sec 3 1 d
0
u
0,
u
2
3(sec u 1) du
3
[3tan u 3u ]0 /3
3 3
89.
3
3 37
5
3/4
0
1 u
36
(cos 2 u ) 14 du
6 du
cot 6
2
tan 3 d
2
1 [3u1/3 ]2
7
5
du
du
1 u
5 2
1 du
4
4 dt
[tan ]0 /3
dx
3
u
8
du
cot 2
7, r 1
1
90
csc2 x dx [ cot x]3 /4/4
x
6
3
cot 2 6x
3
2
u
1 du
u 3/2 36
1 du
5
dt
4
d
87. Let u
0
1
(1)
du
4
cos 2 4t
1
16
8
6
5
0
1
5
3 du
2
1 du
36
25/16
5 dr
sin 2 5r dr
84. Let u
86.
1
18
u 2/3
u 3/2
36 x3 dx
du
0
27 3
160
9 x 4 ) 3/2 dx
1 25
18 16
85.
0
3/4
3 5/2
4
3
5
82. Let u 1 9 x 4
1/2 3
x (1
0
2
7
dr ; r
x 1/3 dx
x 1/3 (1 x 2/3 )3/2 dx
3 (0)
5
0
2
3
du
5/2
du
(7 5r ) 2/3 du
81. Let u 1 x 2/3
1/8
1
5
5 dr
417
sec x tan x dx [sec x]0 /3
sec 0 sec
Copyright
3
1 2
1
2014 Pearson Education, Inc.
3 tan 3
3 3
(3 tan 0 0)
418
90.
Chapter 5 Integration
3 /4
/4
91. Let u
sin x
/2
0
[ csc z ]3 /4/4
csc z cot z dz
du
cos x dx; x
92. Let u
sin 3x
/2
/2
du
/2 3sin x cos x
1 3sin x
/4
96.
sec 2 x
(1 7 tan x ) 2/3
4 x
1 8
1
2x
4 1 1
u 2
15
16
ln 4
g
2
3x
1
97.
8
x2
dx
2
3
ln 8 21
2
1
e ( x 1) dx
2
[eu ]10
98.
15
16
0
ln 2
2 g 1
3 1 x
2 [u 3/2 ]8
0
3
1 u1/ 2
1
2
du
2
sec 2 x dx; x
ln x
4
4
1
0
sin 32
2
3
8
ln x 12 x 1
u 1 3sin 2 2
2
41/2 11/2
ln(82/3 ) 7
ln 4
3 (8)1/3
7
1
8
15
16
ln1
u 1 7 tan 4
4
3 (1)1/3
7
3
7
1 ln 4
2
ln 8 12
8
1 [e0
2
2 (83/2
3
2
3
ln 8 32 12
ln 4 7
u = 1, x = 1
where u = 2w, du = 2 dw; w = ln 2
eln(1/4) ]
2 (16
3
1)1/2 d
(ln1 12)
u=0
e 1
1 0
eu du ,
2 ln(1/4)
1
2
1 16 u 3/2 du ,
3 4
1/2
1/2
8 1/2
0
u
03/2 )
3e r 1, du
where u
2
3
1
4
1
2
du, where u
e
1, du
4
)
2 (29/2
3
0)
Copyright
u
ln 14 , w = 0
u=0
3
8
1 14
211/ 2
3
2
3
4
1
8
3 u1/3
7
1
1
2
3
1
1
2
u 1 7 tan 0 1, x
3
16
8
u
2
u 1, x
[u1/2 ]14
8
1 u1/3
1
7
1
2
1, x
( 1)5 (1)5
0
2
1
e du, where u = (x + 1), du = dx; x = 2
(e0 e1 )
e 2w dw
e (e
[u 5 ]1 1
2(0)5/2
3
2
sin
3sin x cos x dx; x
8 1 2/3
u
du
1 7
12 x 2 dx
7
e (3er 1) 3/2 dr
ln 9
0
5u 4 du
2(1)5/2
u
2
0 u
1
2 [u 1/2 ]16
4
3
100.
1 du
2
1 1 x2
2 8
dx
0
ln 2
ln 5 r
1
1
x
x
2 (ln 8)
3
1 [eu ]0
ln(1/4)
2
99.
1
1
1 du
7
8 1 1
du
1 u 2/3 7
dx
[2u 5/2 ]10
cos 3x dx; x
4 1 1/2
u
1 2
du
2
u 1
2
15u 4 13 du
7sec 2 x dx
du
1 4 1
2 1 4
dx
5
1 du
3
2
1
2 u 5/2
5
0
6 sin x cos x dx
1
94. Let u 1 7 tan x
0, x
5u 3/2 du
1
du
dx
2
u
csc 4
1
15 sin 4 3 x cos 3x dx
0
0
0
0
3cos 3 x dx
93. Let u 1 3sin 2 x
95.
1
5(sin x)3/2 cos x dx
csc 34
3er dr; r = 0
1
4
e d ;
u = 4, r = ln 5
1
6
=0
u = 0,
32 2
3
2014 Pearson Education, Inc.
= ln 9
u=8
u = 16
8
Chapter 5 Practice Exercises
101.
102.
e1
(1 7 ln x) 1/3 dx
1 x
3 [u 2/3 ]8
3 2/3
1 14 (8
14
3 [ln(v 1)]2
v 1
1
3
dv
1
1 g u 1/3 du , where u =
7 1
3 (4 1)
9
12/3 ) 14
14
ln 4 2
[ln(v 1)]2 v11 dv
u du , where u
ln 2
7 dx,
x
1 + 7 ln x, du
x=1
u = 1, x = e
1 dv;
v 1
ln(v 1), du
v=1
419
u=8
u = ln 2, v = 3
u = ln 4;
1 [u 3 ]ln 4
ln 2
3
103.
g log 4
1 [(ln 4)
3
1 [u 2 ]ln 8
0
2ln 4
104.
3
e 8(ln 3)(log 3 )
(3ln 2)2
4 ln 2
02 ]
e 8(ln 3)(ln )
d
(ln 3)
d
3
8
1
e
1
(ln 2)3
3
3
(ln 2) ]
1 ln 8 u du ,
ln 4 0
) 1 d
1 [(ln 8) 2
ln16
1
1 [(2 ln 2)
3
(ln 2) ]
1 g (ln
ln 4 1
d
1
3
7 (ln 2)3
3
(8 1)
1
where u = ln , du
d ,
=1
u = 0,
105.
2
4(1
3/4
6
3/4 9 4 x 2
dx
2
0 )
1
(ln ) 1 d
2
3/4 32 (2 x ) 2
0
dx
3/2
1
3/2 32 u 2
3
106.
1/5
6
1/5 4 25 x 2
3 sin 1 12
sin 1
6 1/5
5
5 1/5 22 (5 x)2
dx
107.
1
sin 1 u2
1
2
3 dt
2 4 3t 2
3
1
2
3 6
108.
109.
3
1
3 3 t2
6
5
sin 1 12
2
3
2 22
6 1
1
5 1 22 u 2
dx
sin 1
dt
3t
2 3
3
1
1
dy
1/ 3 y 4 y 2 1
1
sec 1 u
1/ 3
3
2
2 3
3
dt
1
3
u=1
3
2 3
1 du ,
2 3 22 u 2
2
t
2
tan 1
1
3
dt
tan 1
3
tan 1 t
3
3
3
1
5
6
5 3
6
where u
1
3
1/ 3
u
3
6
u 1
3
3 dt;
2 3, t = 2
6
2014 Pearson Education, Inc.
u
2 3
3
1
3 3
where u = 2y and du = 2 dy
3
1
5
1, x
2
5
tan 1 3 tan 1 1
sec 1 2 sec 1 2/ 3
Copyright
u
3t , du
3
2
3
1
1
2
1
dy
du,
1/ 3 (2 y ) (2 y )2 1
1/ 3 u u 2 1
1
sec 1 2 y
u
du , where u = 5x, du = 5 dx;
6
5 6
1
2
3
4
3 3
6
t= 2
3 12 tan 1 u2
=1
3
2
x
6
5
d ;
=e
du, where u = 2x, du = 2 dx; x
u
3/2
3 sin 1 u3
3/2
1
8 u du , where u = ln , du
4
3/4
3
u = ln 8
9 ln 2
4
u = 0,
4[u 2 ]10
=8
4
3
36
3,
2
x
3
4
420
110.
Chapter 5 Integration
8
24
4 2 y y 2 16
6 3
111.
4
dy
24
6 12
2
2/3
8
1
4 2 y y 2 42
8
2/3
2
3
dy
2 /3 3 y (3 y )2 1
1
dy
2 /3 y 9 y 2 1
8
y
24 14 sec 1 4
4 2
dy
1
du,
2 u u2 1
y
6sec 1 4
4 2
where u = 3y, du = 3 dy;
2
3
y
[sec 1 u ]2
112.
6/ 5
2/ 5
sec 1 2 sec 1 2
2
1
y 5 y2 3
3
6/ 5
dy
4
5y
5y
2
3
6
114. (a)
yav
(b)
yav
115. f av
1
1
(2bk )
1
bx
1
bx
3 x1/2 dx
3
3
1
2
k
k
2
3
2
u
2
du ,
2
5
5 dy; y
1
3 4
mx 2
2
1
2k
(mx b) dx
2
3
2, y
1 3
3 12
6
m (1)2
2
1
2k
2
12
u
6
3
36
1 (2b)
2
b( 1)
m( k )2
2
b( k )
6
5
2, y
12 3
m ( 1)2
2
b(1)
m ( k )2
2
u
b
b( k )
b
1
3
3 0
1
a 0
0
b
1
b a a
k
mx 2
2
u u
5 y, du
3
1
2
1
2
sec 1 2 sec 1 2
(mx b) dx
k
1
k ( k)
(b) av( f )
1
2k
2
1
1 ( 1)
113. (a) av( f )
1
3
6
dy
2
where u
1 sec 1 u
3
3
u
12
5
2/ 5
6 sec 1 2 sec 1 2
a
0
f ( x) dx
3
3x dx
1
3
ax dx
1
a
0
a
0
1 [ f ( x )]b
a
b a
3
2 x3/2
3
0
a
2 x3/2
3
0
a x1/2 dx
a
a
1
b a
f (a )]
[ f (b)
3 2
(3)3/2
3 3
2 (0)3/2
3
a 2
(a)3/2
a 3
f (b ) f ( a )
b a
2 (0)3/2
3
3
3
(2 3)
a
a
2a
3
2
a
2a
3
so the average value of f over [a, b] is
the slope of the secant line joining the points (a, f (a )) and (b, f (b)), which is the average rate of change
of f over [a, b].
b
116. Yes, because the average value of f on [a, b] is b 1 a
f ( x) dx. If the length of the interval is 2, then b a
a
b
and the average value of the function is 12
f ( x) dx.
a
117. (a)
d ( x ln x
dx
x C)
(b) average value
118. average value
x 1x ln x 1 0
1 e ln x dx
e 1 1
1 2 1 dx
2 1 1 x
ln x
ln x
1 [ x ln x
e 1
2
1
x]1e
1 [(e ln e
e 1
e) (1ln1 1)]
ln 2 ln1 ln 2
Copyright
2014 Pearson Education, Inc.
1 (e
e 1
e 1)
1
e 1
2
Chapter 5 Practice Exercises
421
119. We want to evaluate
365
1
365 0 0
365
1
365
f ( x) dx
2 ( x 101)
37 sin 365
0
2 ( x 101) is 2
sin 365
2
Notice that the period of y
365
37
365
25 dx
0
365
2 ( x 101) dx 25
sin 365
365
0
dx
365 and that we are integrating this function over an
365
37 365 sin 2 ( x 101) dx 25 365 dx is 37 0 25 365
interval of length 365. Thus the value of 365
365
365 0
365
365
0
675
120. 6751 20
(8.27 10 5 (26T 1.87T 2 )) dT
20
1
655
26(675)2
8.27(675)
1 (3724.44
655
2 10
165.40)
8.27(20)
5
3 10
5.43
interval [20, 675], so T
26
dy
d (7 x 2 ) 14 x 2 cos3 (7 x 2 )
2 cos3 (7 x 2 ). dx
dy
d
dx
dy
d
dx
124. dx
125. y
126. y
127. y
128. y
x 6
1 3 t4
2
ln x
e
2
d
dx
dt
cos t
dt
2e
x
1
sin
x
0
dt
1 2t 2
/4
1
x
e t dt
sec x 1
dt
2
t2 1
ln x 2 cos t
e
dt
0
dy
dx
ln(t 2 1) dt
1
tan
1
382.82 or T
26T
396.72. Only T
284000
0
396.72 lies in the
6
3 x4
dt
1
sec x t
0
3105
2 10
396.72 C.
2 cos3 x
123. dx
1.87(20)3
5
2124996
. So T
3.74
dy
122. dx
26(20)2
the average value of Cv on [20, 675]. To find the temperature T at
(26)2 4(1.87)( 284000)
2(1.87)
26
121. dx
675
1.87T 3
3105 20
26T 2
2 105
8.27T
5.43, solve 5.43 8.27 10 5 (26T 1.87T 2 ) for T. We obtain 1.87T 2
which Cv
T
1.87(675)3
5
1
655
25.
dy
dx
tan
/4
2
d (ln x 2 )
ecos(ln x ) dx
dy
dx
ln e2 x
d e x
1 dx
1
1 2(sin
1
x
e t dt
1
d (sec x )
1
sec 2 x 1 dx
x )2
d (sin 1 x )
dx
dy
dx
e x
2 x
2
2 ecos(ln x )
x
ln e2 x
1
1 2(sin
sec x tan x
1 sec2 x
1
1
1
x)2
1
d (tan 1 x )
e tan x dx
1 x2
e tan
1 x2
1x
129. Yes. The function f, being differentiable on [a, b], is then continuous on [a, b]. The Fundamental Theorem of
Calculus says that every continuous function on [a, b] is the derivative of a function on [a, b].
Copyright
2014 Pearson Education, Inc.
422
Chapter 5 Integration
130. The second part of the Fundamental Theorem of Calculus states that if F ( x) is an antiderivative of f ( x)
b
on [a, b], then
1
0
131. y
132. y
1 x 4 dx
1
x
F (b) F (a ). In particular, if F ( x) is an antiderivative of 1 x 4 on [0, 1], then
f ( x) dx
a
F (1) F (0).
x
1 t 2 dt
0
1
cos x 1 t 2
cos x 1
0
1 t2
dt
d (cos x )
dx
1
1 cos 2 x
dy
dx
1 t 2 dt
1
dy
dx
dt
1
sin 2 x
x
d
dx
1 t 2 dt
1
cos x 1
0
1 t2
d
dx
1
sin x
( sin x)
x
d
dx
cos x 1
0
1 t2
d
dx
dt
1 x2
1 t 2 dt
1
dt
csc x
133. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of
the parking lot on each interval by averaging the widths at top and bottom. This gives the estimate
A 15 0 236 36 254 542 51 51 249.5 49.52 54 54 264.4 64.4 2 67.5 67.52 42 5961 ft 2 . The cost is
Area ($2.10/ft 2 ) (5961 ft 2 )($2.10/ft 2 ) $12,518.10 the job cannot be done for $11,000.
134. (a) Before the chute opens for A, a
32 ft/sec 2 . Since the helicopter is hovering v0 0 ft/sec
v
32 dt
32t v0
32t. Then s0 6400 ft
s
32t dt
16t 2 s0
16t 2 6400.
At t 4sec, s
16(4)2 6400 6144 ft when A s chute opens;
(b) For B, s0 7000 ft, v0 0, a
32 ft/sec2
v
32 dt
32t v0
32t dt
16t
2
s0
16t
2
7000. At t 13 sec, s
16(13)
2
32t
7000
s
4296 ft when B s chute opens;
16 ft/sec s
16 dt
16t s0 . For A, s0 6144 ft and for B, s0 4296 ft.
(c) After the chutes open, v
Therefore, for A, s
16t 6144 and for B, s
16t 4296. When they hit the ground, s 0
for A,
4296 268.5 seconds to hit the
0
16t 6144 t 6144
384
seconds,
and
for
B,
0
16
t
4296
t
16
16
B hits the ground first.
ground after the chutes open, Since B s chutes opens 58 seconds after A s opens
CHAPTER 5
ADDITIONAL AND ADVANCED EXERCISES
1
1. (a) Yes, because
0
(b) No. For example,
2. (a) True:
(b) True:
2
5
2
0
5
2
dx
8 x dx [4 x 2 ]10
5
2
f ( x ) dx
5
[ f ( x) g ( x)] dx
4 3 2
(c) False:
1
f ( x) dx
5
1 1 7 f ( x)
7 0
f ( x) dx
2
1 (7)
7
1
4, but
0
1
3/ 2
2 2 x3
8 x dx
2
1
13/2 03/2
4 2
3
0
4 2
3
4
3
f ( x) dx
5
2
2
g ( x) dx
2
f ( x) dx
5
2
f ( x ) dx
5
2
g ( x) dx
9
f ( x)dx
4 3 7
the other hand, f ( x)
g ( x)
2
5
[ g ( x)
Copyright
2
g ( x) dx
f ( x )] 0
5
2
[ f ( x) g ( x )] dx
5
2
[ g ( x)
f ( x)] dx
2014 Pearson Education, Inc.
0
5
2
[ g ( x)
f ( x )] dx
0. On
0 which is a contradiction.
Chapter 5 Additional and Advanced Exercises
3.
1 x
a 0
y
sin ax x
a
0
dy
dx
x
x
x
cos ax
x
0
x
x
y
x
0
1
1 4t
2
1
1
1 4y
1
2
1 4y
0
f (t ) sin at dt
a sin ax
2
2
dt
d ( x)
dx
dy
dx
dy
dx
1/2
(8 y )
y (0)
x
0
f (t ) sin at dt
( x ) sin ax )
dx 2
x
0
x
0
d x f (t ) sin at dt
(sin ax) dx
f (t ) sin at dt
0
f (t ) sin at dt (sin ax) f ( x) sin ax
a2 y
x
f ( x) a 2 sinaax
f (t ) cos at dt
0
cos ax x
a
0
f (t ) sin at dt
0.
dy
dx
y
d
dy
dt
d2y
1 4 y 2 . Then
4y
cos ax ( f
a
f ( x). Therefore, y
f (t ) cos at dt
d y
1
dx 0 1 4t 2
dy
dx
a cos ax
cos ax d x
a
dx 0
f (t ) sin at dt
d2y
a cos ax
f (t ) sin at dt
0
0
f (t ) sin at dt
f (t ) sin at dt. Next,
0
x
f (t ) cos at dt a cos ax
0
f ( x). Note also that y (0)
0
x
f (t ) cos at dt sin ax
x
sin ax
x
( f ( x ) cos ax) sin ax
f (t ) cos at dt (cos ax) f ( x) cos ax
0
a sin ax
x
f (t ) cos at dt
0
a sin ax
4.
d x
dx 0
f (t ) cos ax sin at dt
dy
dx
f (t )sin at dt
d x f (t ) cos at dt
f (t ) cos at dt (cos ax) dx
0
a cos ax
sin ax
a
f (t ) cos at dt
0
1 x
a 0
f (t ) sin ax cos at dt
sin ax
a
f (t ) cos at dt
0
a sin ax
cos ax x
a
0
f (t ) cos at dt
cos ax
cos ax
1 x
a 0
f (t ) sin a ( x t ) dt
423
2
dt
dy
dx
1 4 y2
1 4 y2
1 4 y2
1 4t
d
dx
dx 2
4y
1
0
4 y. Thus
1 4 y2
from the chain rule
1 4 y2
d
dy
d2y
dy
dx
4 y, and the constant of proportionality
dx 2
is 4.
5. (a)
x2
0
f (t ) dt
f ( x2 )
(b)
6.
a
0
b
1
a
2
sin a
F (a )
a
1
2
f ( x ) dx
3
3 3(4) cos 4
a2
2
f ( x) dx
f (a )
7.
x sin x . Thus, x
2x
f ( x)
t3
1 f ( x) 3
3 0
3
t dt
f (4)
f (t ) dt
cos x
f ( x) 2
0
2
d x
dx 0
x cos x
b2 1
2
sin a
2
cos x
2
1
3
cos 2
f (4)
f ( x)
3
f ( x 2 )(2 x)
x sin x
2 sin 2
4
x cos x
cos x
x sin x
1
4
f ( x)
3
3 x cos x
3 3x
f ( x)
cos
x
12
cos a. Let F (a )
a cos
2
f (b )
a
2
d b
db 1
Copyright
a
0
sin a
f ( x) dx
f (t ) dt
f 2
1
2
f (a )
2
a2
2
a
2
sin 2
2
F (a ). Now F (a )
1 sin
2
2
2
2
(b 2 1) 1/2 (2b)
2014 Pearson Education, Inc.
cos 2
b
b2 1
2
f ( x)
x
x2 1
sin a
1
2
2
2
cos a
1
2
424
Chapter 5 Integration
d
8. The derivative of the left side of the equation is: dx
x
d
right side of the equation is: dx
d
dx
x
x
x
0
dy
f (u ) du
x
0
0
f (t ) dt ; the derivative of the
d xu
dx 0
f (u ) x du
d x f (u ) du
x dx
f (u ) du
x
f (t ) dt du
f (u ) du
x
x f ( x)
0
f (u ) du
0
x f ( x ) x f ( x)
f (u ) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0
0, the constant must be 0. Therefore,
when x
3x2
9. dx
d xu
dx 0
f (u ) du
0
u
0
d x
dx 0
f (u )( x u ) du
0
x
0
C
2
(3 x 2
y
4
y
x
3
x3
2) dx
x
u
0
0
f (t ) dt du
v
0
f (u )( x u ) du.
2 x C . Then (1, 1) lies on the curve
13
2(1) C
1
2x 4
10. The acceleration due to gravity downward is 32 ft/sec 2
velocity
x
32t 32
s
( 32t 32) dt
16t
2
2
2
v
32 dt
32t v0 , where v0 is the initial
32t C. If the release point, at t
2
0, is s
0, then
C 0 s
16t 32t. Then s 17 17
16t 32t 16t 32t 17 0. The discriminant of this
quadratic equation is 64 which says there is no real time when s 17 ft. You had better duck.
11.
3
8
3
4
2
0
8)5/3
14.
2
0
1
(1 z )3/2
2 (1
3
2
1
1)3/2
36
5
4) dx
0
0
16
3
3
1
2
2
7
3
sin t dt
1
1 z dz
0
(x2
4(3)
cos 2
1
12
2
cos t
0
0
3
4x
33
3
t dt
1
h( z ) dz
2
3
2
3
1
0
x3
3
4
3/2
1
1
t2
2 0
1
2
0
0
96
5
3
x dx
4
x)3/2
g (t ) dt
4 dx
0
( 4(3) 0)
0
2 (4)
3
0
3
x 2/3dx
[ 4 x]30
8
f ( x) dx
2(
3
13.
8
0
3 x5/3
5
0 53 (
12.
0
f ( x ) dx
3 (7(2) 6) 2/3
14
6
3
55
7 14
42
cos
2
(7 z 6) 1/3 dz
1
3 (7 z
14
2 (1
3
1
6)2/3
2
1
0)3/2
3 (7(1)
14
6)2/3
Copyright
2014 Pearson Education, Inc.
Chapter 5 Additional and Advanced Exercises
2
15.
2
1
f ( x) dx
[ x] 12
x3
3
x
2
16.
1
2
3
0
1
1
( 1) 2
2
1
2
12
2
22
2
0
19.
20.
lim
b 1
b
1
0
1 x2
lim n1 1
1
n 2
n
0
lim 1n [e1/ n
2
1
2 0 0
f ( x) dx
12
2
lim
2
3
1
3
1
3 0 0
0
2
x dx
1
0
dx
lim (sin 1 b sin 1 0)
b 1
1
2
1
2n
1
n
lim
n
1
1
1
n
x
n
1
1 2
lim n1 1
n
1
n
x
1
n
1
n
e(1/ n)
1
n
lim
2
1
lim
n
j 1
j 5 1
n
n
lim 1n
n
2
1 [1
3
dx
lim (sin 1 b 0)
b 1
1
n
1
n 2
1
n
e 2/ n
1
n
5
Copyright
2
n
5
n 5
n
1
1 n
1 x2
2 2
x
2
1
0 0 3 2]
lim sin 1 b
b 1
2
3
2
en (1/ n )
[ln(1 x)]10
0
ln 2
which can be interpreted as a
1 x
e]
e dx [e x ]10
1 0
1 . Then 1 , 2 ,
n
n
n n
j 5 1
is the upper
n
n
j 1
5
5
lim 1 2
n
which can be interpreted as a
1
n
1 1
dx
01 x
1
2n
e 2(1/ n )
lim 1n [e1/ n
n
1
n
1
n
endpoints of the subintervals. Since f is increasing on [0, 1], U
[0, 1]
3
0 dx
x5 on [0, 1]. Partition [0, 1] into n subintervals with x
23. Let f ( x)
1
1 x2
2 2 0
( x 1) dx
form
1
lim tan1 x
Riemann sum with partitioning
1
3
f ( x) dx
x
e]
1
1
2
f ( x) dx
x
tan 1 t dt
0
x
7
6
1
1
2
f ( x) dx
e2/ n
n
dr
1
2
(2 1)
Riemann sum with partitioning
22.
1
[r ]12
x
21.
2
(1 r 2 ) dr
b 1
0
2 dx
2(2) 2(1)
lim [sin 1 x ]b0
dx
x
lim 1x tan 1 t dt
x
1
0
2
b
1
b a a
18. Ave. value
( 1)3
3
1
r dr
b
1
b a a
17. Ave. value
1
13
3
1
r3
3 0
3
1 13
r
2
(1 x 2 ) dx
[2 x]12
1
0
0
1
4 2
h(r ) dr
r2
2
1
dx
3
1 13
( 1 ( 2))
1 23
2
1
425
n6
2014 Pearson Education, Inc.
n5
e 1
, nn are the right-hand
x5 on
sum for f ( x)
1 5
x dx
0
1
x6
6 0
1
6
426
Chapter 5 Integration
x3 on [0, 1]. Partition [0, 1] into n subintervals with x
24. Let f ( x)
endpoints of the subintervals. Since f is increasing on [0, 1], U
[0, 1]
j 3 1
n
n
lim
n
25. Let y
j 1
lim 1n
3
1
n
n
3
2
n
n 3
n
1 0
1 . Then 1 , 2 ,
n
n
n n
j 3 1
is the upper
n
n
j 1
3
3
lim 1 2
n
n
1 0
1 . Then 1 , 2 ,
n
n
n n
j 1
is a Riemann
n n
f ( x) on [0, 1]. Partition [0, 1] into n subintervals with x
endpoints of the subintervals. Since f is continuous on [0, 1],
f
j 1
[0, 1]
26. (a)
j
lim
n
j 1
1
n
f n
lim 1n f 1n
lim 12 [2 4 6
lim n1 n2
2n]
n
n
0
1
2n
n
6
n
4
n
n
1
f nn
f n2
n
0
x3 on
sum for f ( x)
1
x4
4 0
1 3
x dx
0
n3
4
, nn are the right-hand
1
4
, nn are the right-hand
sum of y
f ( x) on
f ( x) dx
2 x dx [ x 2 ]10
1, where f ( x)
2 x on [0, 1]
(see Exercise 21)
(b) lim 116 [115
n
n
f ( x)
215
n
n
n15
(see part (b) above)
lim 115 115 215
1
0
2
n
15
n 15
n
1
sin n dx
lim 116 [115
lim n
n
n
n
lim 116 [115
lim 1n
n
n
[115
215
n
n15
n
n
15
1
n
1 15
x
0
1
x16
16 0
dx
cos x
1
1
0
1
cos
1 ,
16
where
2,
cos 0
where
sin x on [0, 1] (see Exercise 21)
(d) lim 117 115
(e)
lim 1n
sin nn
sin 2n
lim n1 sin n
n
n
n15 ]
x15 on [0, 1] (see Exercise 21)
f ( x)
(c)
215
n
n16
lim
n
215
n15 ]
215
n15 ]
lim 1n
n
1 15
0
x dx
1
0 16
0
n15 ]
1 15
lim n
0
n
x dx
(see part (b) above)
27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and
the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal
to r, the radius of the circle) and a vertex angle of n where n 2n . The area of each triangle is
An
(b)
1 r2
2
sin n
the area of the polygon is A
2
lim nr2 sin 2n
n
lim A
n
2
lim n2 r sin 2n
n
nr 2
2
sin n
r2
sin
nAn
lim
n
1
n
28. Partition [0, 1] into n subintervals, each of length x
Sn
02
lim Sn
n
2
2
n
1
n
2
lim
n
x,
2
n
12
n3
n 1 2
n
, f ( xn 1 )
2
n 1 2
n
22
n3
n
3
Copyright
sin 2n .
r2
2
n
0, x1
(0) 2 x, f ( x1 ) x
lim
2 /n
sin
2
n
12
n2
1 2
0
x dx
( n 1)2
22
n2
n
13
3
2
1
n
12
n3
22
n3
1, x
2,
n 2
n
2
1
x,
n
1.
3
2014 Pearson Education, Inc.
( n 1)2
n3
r2
2
n
x. The sum of these areas is
x
( n 1)2
2
n
with the points x0
The inscribed rectangles so determined have areas f ( x0 ) x
f ( x2 ) x
nr 2
2
. Then
, xn
n
n
1.
Chapter 5 Additional and Advanced Exercises
29. (a)
1
g (1)
3
(b) g (3)
(c)
f (t ) dt
1
1
g ( 1)
1
(d) g ( x )
0
1 (2)(1)
2
f (t ) dt
1
1
f (t ) dt
f ( x)
0
22 )
1(
4
f (t ) dt
1
x
1
3, 1, 3 and the sign chart for g ( x)
f ( x) is |
3
relative maximum at x 1.
(e) g ( 1)
f ( 1)
427
1
2 is the slope and g ( 1)
f (t )dt
1
. So g has a
|
|
1
3
, by (c). Thus the equation is y
2( x 1)
y 2x 2 .
(f ) g ( x) f ( x ) 0 at x
1 and g ( x) f ( x) is negative on ( 3, 1) and positive on ( 1, 1) so there is an
inflection point for g at x
1. We notice that g ( x) f ( x) 0 for x on ( 1, 2) and g ( x) f ( x) 0 for x
on (2, 4), even though g (2) does not exist, g has a tangent line at x 2, so there is an inflection point at
x 2.
(g) g is continuous on [ 3, 4] and so it attains its absolute maximum and minimum values on this interval. We
saw in (d) that g ( x)
1
g (1)
f (t ) dt
1
3
g (3)
1
4
g (4)
1
x
0
3
3, 1, 3. We have that g ( 3)
1
1
f (t ) dt
3
22
2
f (t ) dt
0
f (t ) dt
1
f (t ) dt
1 12 1 1
1
2
Thus, the absolute minimum is 2 and the absolute maximum is 0. Thus, the range is [ 2 , 0].
30.
y
sin x
y
cos
x
x
cos 2t dt 1 sin x
cos(2 )
x
cos 2t dt 1
2. And y
1 1
y
cos x cos(2 x ); when x
sin x 2sin(2 x); when x
,y
we have
sin
cos 2t dt 1 0 0 1 1.
31.
f ( x)
x 1
1/ x t
32.
f ( x)
sin x 1
cos x 1 t 2
33.
g ( y)
34.
g ( y)
35.
y
2 y
y
y 2 et
y t
x2
x 2 /2
dt
1 dx
x dx
f ( x)
dt
dt
ln t dt
g ( y)
dy
dx
1
x
d 1
dx x
1
1 sin 2 x
f ( x)
sin t 2 dt
1
g ( y)
2
ey
y2
ln x 2
e
d ( x2 )
dx
Copyright
y
y
2
1
1 cos 2 x
d
dy
d
dy
1
x2
x
d (sin x)
dx
sin 2 y
d ( y2 )
dy
1
x
2 y
y
2
ln x2
1
x
1
x
2
(2 y ) e
d x2
dx 2
cos x
cos2 x
d (cos x )
dx
sin
ey
y2
2
x
y
y
y
2
d
dy
2
1
2 y
2 x ln x
2014 Pearson Education, Inc.
y
x ln
4e y
2y
x
2
sin x
sin 2 x
1
cos x
sin 4 y
y
e y
2y
4e y
1
sin x
sin y
2 y
2
2y
e
y
2
428
36.
Chapter 5 Integration
3
y
x
ln t dt
x
dy
dx
37.
38.
39.
ln x
0
ln 3 x
sin et dt
32 x
y
e4
x
x
x
y
41.
6 6 x. Thus f ( x)
x ln x x
x2
A2
1
e 2log 4 x
dx
4
2 e ln x dx
ln 4 1 x
(ln x )2
2 ln 2
(c)
e g ( x)
f ( x)
ln 3 x
x 1/2
3
3 x
2
ln x 3
2 x
x
6 6x
2ln x
e
1
e
4 x e4 x
d
dx
4 x
4 xe2 x 8e4 x
0
x ln x
(ln x )2
ln 2
2 ln e x e x
ex
1 2 ln t
f (0)
dt
1 t
df
2x
f ( x)
dx
1
2
(2 x)(2e2 x )
0
x(5 x) dx
dx
( x 3)(2 x) x(5 x)
x 1. Also, f ( x )
x 2 ln x; then x x ln x
2 e ln x dx
ln 2 1 x
df
dx
e4 x
d
dx
d ( x 3)
( x 3)(5 ( x 3)) dx
f ( x)
e 2 log 2 x
dx
x
1
(b)
43.
2
x
4 xe 4 x
A1
42. (a)
ln x
sin x
x
4 xe 2 x
x = 1; x x
1 x 2/3
3
ln 3 x
x
ln e4 x
x x ln x and ln( x x ) x
ln x = 0. ln x = 0
d
dx
d (e 2 x )
(ln e 2 x ) dx
t (5 t ) dt
6 x x2 5 x x2
a maximum.
x
40. ln x ( x )
ln x
d (ln x)
(sin eln x ) dx
y
ln t dt
x 3
f ( x)
d 3
dx
x 2 ln x
(xx
0
x 2 ) ln x
x
2. Therefore, x ( x )
x
6
x 1 gives
0
xx
x 2 or
( x x ) x when x = 2 or x = 1.
1 ;
ln 2
1
2 ln 2
1
A1 : A2
2 :1
2x
0
x 2 C ; f(0) = 0
f ( x)
C=0
e g ( x ) g ( x), where g ( x)
1
44. The area of the blue shaded region is
0
x2
the graph of f(x) is a parabola.
f (2)
e0 1 216
f ( x)
x
1 x4
1
sin 1 x dx
0
2
17
sin 1 y dy, which is the same as the area of the region to
the left of the curve y = sin x (and part of the rectangle formed by the coordinate axes and dashed lines y = 1,
x
1
2
1
. The area of the rectangle is 2
2
0
sin 1 x dx
45. (a) slope of L3
/2
0
/2
sin x dx
0
slope of L2
0
0
1
sin x dx
slope of L1
/2
sin 1 y dy
2
1
b
0
sin x dx, so we have
sin 1 x dx.
ln b ln a
b a
1
a
(b) area of small (shaded) rectangle < area under curve < area of large rectangle
1 (b
b
a)
b1
dx
a x
1 (b
a
a)
1
b
Copyright
ln b ln a
b a
1
a
2014 Pearson Education, Inc.
Chapter 5 Additional and Advanced Exercises
46.
(a) If f is continuously differentiable on a, b , then so is the function g ( x )
integrals on the right side exist and
c
a
b
( x c ) f ( x ) dx
b
a
c
b
x f ( x ) dx
a
( x c ) f ( x ) dx
b
c f ( x ) dx
a
b
a
b
x f ( x ) dx c(0)
first integral above, use the substitution t
(b a ) / 2, x
0
a
x f ( x ) dx
c
x
x c,
c t
(b a ) / 2 ( a b ) / 2
b
a
xf ( x ) dx
0
c
a
t f ( c t ) dt
dx ; when t
c t , dt
0
t f ( c t ) dt
c
a
c
dx; when t
0, x
each t, we have f ( q)
b
a
x f ( x ) dx
0
0, x
c and when t
( x c ) f ( x ) dx
0
t f (c t ) dt
0
c and when
c
a
( x c ) f ( x ) dx.
b
c
(b a ) / 2,
( x c ) f ( x ) dx and
t f ( c t ) dt
f ( c t )) dt.
(c) According to the mean value theorem of Section 4.2, for every t in 0,
c t , c t at which
t f (c t ) dt. For the
( x c ) f ( x ) ( dx )
b. Thus the second integral above is equal to
b
0
( x c ) f ( x ) dx. For the second integral above, use the
x and dt
( x c) f ( x ) dx
t ( f (c t )
a
x
0
a. (Note that when x is in a, c , c x is positive and thus
t f ( c t ) dt
Thus the first integral above is equal to
substitution t
c x,
( a b ) / 2 (b a ) / 2
c x agrees in sign with t.)
( x c ) f ( x ). So the two
( x c ) f ( x ) dx
(b) Split the right side in part (a) into two integrals and write it as
t
429
f (c t ) f (c t )
(c t ) ( c t )
M , f (c t )
t ( f (c t )
f (c t )
f ( q). Since for all these q belonging to
2tM for all t in 0,
f (c t )
f ( c t )) dt
Copyright
f (c t)
2t
0
(t )(2tM )
, there is a point in q in
2
M t3
3
. Thus
( b a )/2
0
2014 Pearson Education, Inc.
(b a ) 3
M.
12
430
Chapter 5 Integration
Copyright
2014 Pearson Education, Inc.
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
6.1
1.
2.
VOLUMES USING CROSS-SECTIONS
A( x)
3.
b
a
b
4.
a
V
2 x2
b
a
2 x; a
1 x2
1
1
4 1 x 2 dx
1 x2
2
A( x ) dx
1
1 (side)
2
5. (a) STEP 1) A( x)
STEP 2) a
1
2 x
1
2 1 x2
2
1
2
x2
2 x dx
4
0
16
1
2
(side) sin 3
1, b 1;
1 23
1
5
4 1 x2 ; a
16
15
1, b 1;
16
3
2
2 1 x2 ; a
1, b 1;
8
3
4 1 13
1
2
1
8 1 13
2
x3
3
0
x4 ; a
1
x5
5
2 1 x2
x3
3
2
1 x2
1 x 2 dx
1
2
4 x
2
2
1 2x2
x 23 x3
1 x2
4
A( x) dx
a
2
2 1 x2
x 4 dx
b
4; V
4
1 2 x2
1
A( x ) dx
2
x2
0, b
4
1
A( x ) dx
(diagonal)
2
A( x)
2
2
(edge)2
A( x)
V
x
(diameter)2
4
A( x)
V
x
(diagonal)2
2
2 sin x
2 sin x sin 3
3 sin x
0, b
b
STEP 3) V
a
3
(side)2
(b) STEP 1) A( x)
STEP 2) a
A( x ) dx
0
sin x dx
2 sin x
3 cos x
2 sin x
0
3(1 1)
2 3
4 sin x
0, b
b
STEP 3) V
a
STEP 2) a
sec2 x
,b
3
b
STEP 3) V
a
4
0
(diameter)2
4
6. (a) STEP 1) A( x)
4
A( x ) dx
4
sec 2 x 1
4 cos x 0
(sec x tan x) 2
4
8
sec2 x tan 2 x 2sec x tan x
2 sin2x
cos x
3
/3
2sec2 x 1 2 sin2 x dx
/3 4
cos x
A( x ) dx
2 3
4 sin x dx
3
2
1
1
2
Copyright
2 3
3
2
1
1
2
2 tan x x 2
4
4
4 3
2014 Pearson Education, Inc.
1
cos x
/3
/3
2
3
431
432
Chapter 6 Applications of Definite Integrals
(edge) 2
(b) STEP 1) A( x)
STEP 2) a
3
b
STEP 3) V
a
7. (a) STEP 1) A( x)
STEP 2) a
b
a
(b) STEP 1) A( x)
a
0, b
b
a
c
4
0, d
25
0 4
10.
2; V
y 4 dy
d
d
c
A( y ) dy
0
2
3
60 30 x
60 x 15 x 2
(60 30 x) dx
2
(120 60) 0
0
20 2(6 3 x )
2
(6 3x )
24 6 x 9 x 2 dx
(height)
4
x
2
x
(6 3 x)(4 3 x)
24 x 3 x 2 3 x3
(6)
0
6 x1/2 3 x dx
diameter 2
2
60
24 6 x 9 x 2
2
0
(48 12 24) 0
6 x 3x
x
2
1
2
4 x3/2
2
x
2
2
4
3 x2
2
0
(32 24) 0
x x3/ 2
4
1 x2
4
1 x2
2
2 x 5/2
5
8
x x3/2
8
1 x2
4
4
8 0
5 y2 0
4
x x3/2
2
5
4
1 x2
4
dx
4
1 x3
12
0
8
16
8 64
5
3
y4 ;
A( y ) dy
c
y5
5
5
4
1
2
1
1
36
4
A( x ) dx
(diameter) 2
1 (leg)(leg)
2
A( y )
V
a
2
A( x ) dx
0, b
b
4 3
4
1
2
STEP 3) V
A( y )
0
A( x )dx
(b) STEP 1) A( x)
9.
(6 3 x) (10)
3
2
1 (base)
2
STEP 3) V
STEP 2) a
2
A( x ) dx
8. (a) STEP 1) A( x)
2 2 3
cos x
2
0, b
b
cos x
2sec 2 x 1 2sin2 x dx
/3
(length) (height)
STEP 3) V
STEP 2) a
/3
A( x ) dx
0, b
2sec2 x 1 2 sin2x
3
(length) (height)
STEP 3) V
STEP 2) a
,b
(sec x tan x) 2
2
4
0
1 y2
25 0
8
2
1 y2
2 1 y 2 dy
2 y
Copyright
y3
3
1
1
1
2
2 1 y2
4 1 13
2
2 1 y2 ; c
8
3
2014 Pearson Education, Inc.
1, d
1;
8
(0)
15
Section 6.1 Volumes Using Cross-Sections
11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have bh
4
5
The equation of the line through (5, 0) and (0, 4) is y
3
4
the height
6 x2
25
12
5
4
5
3x
5
x 4
b
x 6 and V
a
1 (base)
2
3. Thus A( x)
5 6 2
x
0 25
A( x) dx
12
5
x 4, thus the length of the base
(height)
1
2
4
5
2 x3
25
x 6 dx
3x
5
x 4
6 x2
5
5
6x
4
3
433
h
4
5
3 b.
4
x 4 and
3
(10 30 30) 0 10
0
12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar
triangles we have
5
3 y3
25
0
b
h
3
5
3 h.
5
b
(base)2
Thus A( y )
3
5
2
y
9 y2
25
d
V
c
A( y ) dy
5 9 2
y dy
0 25
15 0 15
13. (a) It follows from Cavalieri s Principle that the volume of a column is the same as the volume of a right
prism with a square base of side length s and altitude h. Thus,
STEP 1) A( x) (sidelength)2
STEP 2) a 0, b h;
b
STEP 3) V
a
A( x ) dx
s2 ;
h 2
s
0
s2h
dx
(b) From Cavalieri s Principle we conclude that the volume of the column is the same as the volume of the
V
s2h
1 x
x2
4
prism described above, regardless of the number of turns
14. 1)
The solid and the cone have the same altitude
of 12.
The cross sections of the solid are disks of
2)
x
2
diameter x
x.
2
If we place the vertex of
the cone at the origin of the coordinate system
and make its axis or symmetry coincide with
the x-axis then the cone s cross sections will
be circular disks of diameter 4x
y 1 2x
R( x)
4
2
2
16.
x
2
(see accompanying figure).
The solid and the cone have equal altitudes and
identical parallel cross sections. From
Cavalier s Principle we conclude that the solid
and the cone have the same volume.
3)
15.
x
4
R( y)
x
17. R( y )
4
3y
2
V
2
R( y ) dy
0
4
1 0
0
2
1 2x dx
2
2
R ( x) dx
0
2
0
dx
x2
2
x
2
x3
12 0
2
3
tan 4 y ; u
1
V
8
12
2
V
2
0
4
y
1
2 3y 2
dy
0 2
2
R( y ) dy
du
4
tan 4 y
0
dy
2
dy
4 du
4
dy; y
/4
0
29
0 4
tan 2 u du
y 2 dy
0
4
u
/4
0
2
3 y3
4
0
0, y 1
u
1 sec2 u du
4
Copyright
3
4
2014 Pearson Education, Inc.
4
8
6
;
4 u tan u 0
/4
434
18.
Chapter 6 Applications of Definite Integrals
R( x)
sin x cos x; R ( x)
/2
V
0
u
0, x
19. R( x)
x2
V
2
x2
0
20.
0
2
2
9(3)
2 x4
4
23. R( x)
cos x
/2
cos x dx
2
x5
5 0
32
5
2
x7
7 0
128
7
2
9x
x3
3
18
36
3
3
2
R ( x ) dx
0
1
0
1
x5
5 0
(10 15 6)
1 sin 2u
4
0
R ( x) dx
3
1
V
x x 2 dx
x3
3
0
3
2
2
1
0
u
2
2x
2
V
x x2
R( x)
8
u
R( x) dx
2 6
x dx
0
27
3
du
dx;
2
0
dx
/2 (sin 2 x )2
4
0
R ( x ) dx
9 x 2 dx
3
30
2
are the limits of integration;
1 sin 2 u
0 8
V
2 4
x dx
0
9 x2
R( x)
3
22.
2
2
(sin x cos x)2 dx
u
0
V
x3
0
2
dx
x3
R( x)
2
21.
2
0 and b
a
/2
2
R( x) dx
0
x
0
x2
1
3
2 x3
1
2
x 4 dx
1
5
30
V
/2
0
sin x 0
R( x)
2
/2
(1 0)
dx
Copyright
2014 Pearson Education, Inc.
du
8
du
8
2 dx
2
0
0
dx ;
4
2
16
Section 6.1 Volumes Using Cross-Sections
24.
R( x)
sec x
/4
/4
25.
1
0
R( x)
2
R( x)
28.
R( x)
29.
R( x)
1
ex 1
/2
V
(e 2 1)
/6
/2
[ R( x)]2 dx
cot x
/6
2
/2
dx
/6
cot x dx
V
4
1/4
3
1
4
[ R( x)]2 dx
/4
V
0
1
1/4 2 x
3
[ R( x)]2 dx
2 sec x tan x
1
2
dx
(e x 1 )2 dx
4 1
dx
4 1/4 x
3 2x 2
1
e
dx
4
4
[ln x]1/4
2
[e2 x 2 ]13
2
R( x) dx
2
/4
2 sec x tan x dx
0
/4
2 2 2 sec x tan x sec2 x tan 2 x dx
0
/4
0
2 dx 2 2
/4
0
2x 0
0
/4
0
sec x tan x dx
(tan x)2 sec2 x dx
/4
/2 cos x
dx
/6 sin x
[ln(sin x )] /2
/6
ln 4 ln 14
2
ln 2
V
2 x
2
(e x )2 dx
2e 2
cot x
2
2
0
0
2
(e2 1)
ln1 ln 12
27.
1
e 2x
[1 ( 1)]
1
[ R ( x)]2 dx
0
e 2 x dx
/4
/4
tan x
1
V
1 12
2
e
26.
2
R ( x ) dx
/4
sec2 x dx
e x
R( x)
/4
V
435
2 2 sec x 0
2 2
2 1
tan 3 x
3
0
/4
1
3
/4
(13 0)
2 2 11
3
Copyright
2014 Pearson Education, Inc.
4
2
(e4 1)
ln 4
84.19
436
30.
Chapter 6 Applications of Definite Integrals
R( x) 2 2 sin x
/2
0
0
4
3
4
4
cos 2 x
2
x sin42 x
3
2
4
2
2
R( y)
1
V
(3
8)
1
dy
1
2
2 3
y dy
0
2
R( y ) dy
0
5 y 4 dy
2
4
y
4
4
0
33. R( y )
2sin 2 y
/2
0
1
[1 ( 1)]
1
y 3/2
32. R( y )
/2
0
(0 0 2)
V
1
y5
1 sin 2 x 2sin x dx
0
2sin x
2 cos x
0 0
5 y2
31. R( y )
2
R ( x ) dx
0
1 12 (1 cos 2 x) 2sin x dx
/2 3
2
0
/2
V
/2
4(1 sin x) 2 dx 4
/2
4
2(1 sin x )
/2
V
2
R( y ) dy
0
2sin 2 y dy
cos 2 y 0
/2
[1 ( 1)] 2
y
34. R( y )
cos 4
0
y
cos 4 dy
2
35. R( y )
4
2
y 1
3
1
y 1 0
0
V
y 0
4 sin 4
3
V
4[0 ( 1)]
2
2
R( y ) dy
0
4
2
R ( y ) dy
2
1
4
( 1)
4
3 1
0 ( y 1) 2
4
dy
3
Copyright
2014 Pearson Education, Inc.
Section 6.1 Volumes Using Cross-Sections
36.
2y
R( y)
1
0
y
2
2 y y2 1
0
1
V
y2 1
V
1
y2 1
dy; [u
u 1, y 1
2
2
R ( y ) dy
0
u
2
1
u 1
u 2 du
/4
39. r ( x)
40. r ( x)
1
0
(4 4 x ) dx
2
V
R( x)
1
2
1
( x 3) 2
2
1
2
1
6x 9
x4
x2
32
5
33
5
8
3
24
2
1
R( x)
0
2
r ( x)
x sin x 0
tan y; V
/2
2
0
2
3
2
2
d
c
2 y tan y 0
r ( x)
R( x)
a
R( y)
2
2
2
2
r ( y)
2
/4
2
dx
2
dx
2 x2 1
1
5
dx
x5
5
1
3
6
2
5 30 33
5
x3
3
6 x2
2
8x
2
1
8
117
5
Copyright
r ( x)
2
dx
r ( x)
2
1
2
dx
6 x 8 dx
3 28 3 8
1 12
4
b
cos x; V
x 3
x4
16
2
1 13
1
x2
2 0
x2 1
x2
R ( x)
x
2
2
; R ( y ) 1, r ( y)
0
V
x 2 1 and R( x )
41. r ( x)
(1 cos x) dx
1
2
4
2
; R ( x) 1, r ( x)
1
x3
3 0
x
( 1)
2 sec2 y dy
0
V
2 x and R( x)
2
4
/4
1 tan 2 y dy
1 x 2 dx
0
0
0, d
x and R( x) 1
1
,b
/2
2
38. For the sketch given, c
0
1
2
2
(1 cos x) dx
/2
2 y dy;
2]
37. For the sketch given, a
/2
du
2014 Pearson Education, Inc.
1
2
2
2
dy
dx
437
438
Chapter 6 Applications of Definite Integrals
42. r ( x)
4 x2
2 x and R( x)
2
V
R( x)
1
2
4 x2
1
2
2
16 8 x 2
1
2
r ( x)
x4
x 4 dx
/4
/4
/4
/4
1
0
2
sec x and r ( x)
tan x
0
1
0
1
1
y 2 dy
0
y 2 dy
0
0
R( x)
1
y3
3
1
0
1 dx
0
y3
3
2
r ( x)
2
2
dy
2
dy
dx
1
0
R( y )
2
r ( y)
y 2 1 dy
1
0
0
1
x
1
1
V
y2
1
0
1 2y
y2
1 (1 y) 2 dy
2y
108
5
/4
/4
1
1
46. R( y ) 1 and r ( y ) 1 y
0
1
33
5
dx
V
V
(1 y )2 1 dy
2y
15
2
2)
sec2 x tan 2 x dx
45. r ( y ) 1 and R( y) 1 y
1
1
5
2 x tan x
(
2
2
r ( x)
2 sec2 x dx
1
x5
5
12 x 2 x 2 3 x3
2
1
2
44. R( x)
R ( x)
dx
12 2 3
sec x and R( x)
V
dx
4 4 x x2
32
5
24 8 24
2
(2 x )2 dx
12 4 x 9 x 2
1
43. r ( x)
2
R( y )
1 2y
1
1
0
Copyright
4
3
1
3
2
r ( y)
y2
dy
1
3
2
3
2014 Pearson Education, Inc.
Section 6.1 Volumes Using Cross-Sections
47. R( y )
2 and r ( y )
4
0
(4 y ) dy
48. R( y )
3
3
1
0
3
50. R( y )
1
0
1
0
y2
2
2
3 y2
3
0
1
4y
1
3 2 y
4
3
y
3 4 y1/3
3
5
3 3
51. (a) r ( x)
0
0
(b) r ( y )
0
y2
2
4 y 3/2
3
r ( y)
2
dy
y dy
1
0
7
6
1
V
1
1 dy
0
y 2/3 dy
R( x)
R( y)
0
4 4 y1/3
3 y 3 y 4/3
2
r ( y)
2
dy
y 2/3 1 dy
3 y 5/3
5
1
0
2
2
r ( x)
(4 x) dx
0 and R( x)
4
2
R( y )
0
3y
4 4 x
y2
2
x dx
2
dx
4
x2
2 0
4x
0 and R( y )
(c) r ( x)
1
3
0
3
5
4
4
dy
3
y3
3
y dy
1 2 y
x and R ( x)
V
2
dy
dy
2 y1/3 and r ( y ) 1
2
2
V
18 8 3
6
2 y1/3
r ( y)
(16 8) 8
0
y dy
1
2
2
0
3 2
dy
2
y
R( y )
4
r ( y)
2 and r ( y ) 1
4
4
0
3 y2
R( y )
0
0
V
3 and r ( y )
V
49. R( y )
y
2
V
x
(16 8) 8
R( y )
0
V
4x
4
0
8 x3/ 2
3
Copyright
2
R( x)
4
x2
2 0
2
dy
r ( x)
2
r ( y)
2
16
64
3
2 4
y dy
0
dx
16
2
4
0
2
8
3
2014 Pearson Education, Inc.
y5
5
2
2
0
x dx
32
5
439
440
Chapter 6 Applications of Definite Integrals
4 y 2 and R ( y )
(d) r ( y )
2
16 16 8 y 2
0
52. (a) r ( y )
2
2
0
y2
2
y
2
0
2
V
0
2
0
8
12
2
2
r ( y)
dy
2
1
1
1
4 4 x2
4 1 2 x2
2
15
(45 10 3)
54. (a) r ( x)
0 and R( x)
b
V
0
b
0
h2
x h
x3
3b2
x2
b
2
y5
5
0
8
12
6 4
0
2
2
1 dy
2
R( x)
1
2
64
3
0
32
5
dy
224
15
r ( x)
2
y2
4
4 2y
8
3
2
3
2
1 dy
dx
x 4 dx
1
5
15 10 3
15
2
1
R ( x)
1
1
1
2
3 4x2
1
V
2
16
15
r ( x)
2
x 4 dx
R( x )
1
1
x 4 dx
1
dx
3x
2 x2
1
x5
5
4 x3
3
1
3 2 x2
2
r ( x)
x4 dx
2
1
dx
1
1
2
2
3x
1
2
3
x3
h
b
2
x5
5
dx
x
x h
r ( x)
b
0
2
dx
b h2
0 b2
h2
b
3
x2
2 h2
b
b b
Copyright
1 x2
4
64
15
R ( x)
h
b
0
2
4 y2
16
1 dx
3
4
3
1
5
56
15
(c) r ( x) 1 x 2 and R( x)
1
y3
12
1
V
x 4 1 dx
(45 20 3)
1
8 y3
3
y 2
2
2
0
y2
2
3
1
2 x2
(b) r ( x) 1 and R ( x)
1
2
1
2
1 2 x2
1
1
x5
5
2
dy
2
3
V
1
1 x 2 dx
2 x3
3
x
y 4 dy
2
dy
dy
3y
0 and R( x) 1 x 2
53. (a) r ( x)
r ( y)
y
2
2
y2
4
3 2y
4
2
0
R( y )
2
dy
y2
4
2
(b) r ( y ) 1 and R ( y )
8 y2
0
1 y
2
y3
12
0
2
2
r ( y)
y 2
dy
2
1
R( y )
y
2
2
R( y)
0
2
V
y 4 dy
0 and R( y ) 1
V
2
15
4
x h 2 dx
h 2b
3
2014 Pearson Education, Inc.
1
1
2
2
3
dx
2
3
1
5
Section 6.1 Volumes Using Cross-Sections
(b) r ( y )
0 and R( y )
b2
h
0
a2
55. R( y ) b
a
V
a
a
a
a
a
4b
b
y2
2y
h
1
y
h
b 1
y 2 and r ( y )
R( y )
2
a2
y2
4b a 2
r ( y)
2
y 2 dy
2
For h
y 2 dy
a
a2
a2h
(b) Given
dV
dh
dV
dt
y2
1
3
8 , so
h a
V
a
0.2 m3 /sec and a
h2
a2
2
4b
A(h). Therefore
h3 3h 2 a 3ha 2
10 h
b2 h h
h
0
y 2
h
1
dy
b2h
3
h
3
dy
2 y and area
4, the area is 2 (4)
57. (a) R( y )
2
a2
a
dV
dh
h
b2
dy
y2
y2
56. (a) A cross section has radius r
A(h)dh, so
2
r ( y)
3h 2 0
a2
area of semicircle of radius a
(b) V (h)
y3
2
dy
b
4b
y2
h
a2
b
R( y )
0
b2 y
dy
h2
h
V
441
dV
dt
a2
a3
dh
dt
2a 2 b
r2
dV
dt
1
8
3 units
sec
5 m, find
dV dh
dh dt
A(h)
3
3
8
y3
3
a2 y
a2h
a
3
dh
.
dt h 4
5
2 y. The volume is
dV dh
dh dt
y 2 dy
3
2
3
h
3
0
dh , so dh
dt
dt
dV
1
A( h) dt
5
25 .
0
.
units3 .
sec
h a
( h a )3
3
a 2 h a3
a
dh
dt h 4
a3
3
a3
h2 (3a h)
3
h 2 a ha 2
h 2 (15 h)
3
From part (a), V (h)
h(10 h) dh
dt
y2
2 ydy
0.2
4 (10 4)
1
(20 )(6)
5 h2
h3
3
1
120
m/sec.
58. Suppose the solid is produced by revolving
y 2 x about the y -axis. Cast a shadow of
the solid on a plane parallel to the xy -plane.
Use an approximation such as the Trapezoid Rule,
to estimate
b
a
dk 2
2
n
2
R( y ) dy
k 1
y.
59. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a
disk of radius h has been removed. Thus its area is A1
hemisphere is a disk of radius
We can see that A1
R2
R2
h2
h 2 . Therefore its area is A2
(R2
h 2 ). The cross section of the
R2
h2
2
R2
h2 .
A2 . The altitudes of both solids are R. Applying Cavalieri s Principle we find
Volume of Hemisphere
(Volume of Cylinder)
Copyright
(Volume of Cone)
2014 Pearson Education, Inc.
R 2 R 13
R2 R
2
3
R3 .
442
60.
Chapter 6 Applications of Definite Integrals
x
12
R( x)
36 x 2
V
6
x5
5 0
144
12 x3
144
6
R ( x)
0
2
65
5
12 63
63
144
256 y 2
R( y )
(256)( 7)
62. (a)
7
V
73
3
16
c2
73
3
2
c2
(2c
4)
x 4 dx
cm3 .
163
3
2
0
1053 cm
0
c
c2
16
3
3308 cm3
2c sin x sin 2 x dx
c2
c2
4c . Let V (c)
2
7
y3
3
2c sin x cos22 x dx
1
2
x 2c cos x sin42 x
1
2
2
3
2
and V (1)
2
2
Now we see that the function s absolute minimum value is 2
(See also the accompanying graph.)
0
4c . We find the
2
is a critical point, and V 2
4; Evaluate V at the endpoints: V (0)
2
36
5
256 y
(c sin x )2 dx
c2
(0 2c 0)
extreme values of V (c) : dV
dc
4
256(16 7)
0
0
60 36
5
256 y 2 dy
16
2
2c 0
2
7
R( x) dx
0
196
144
36 x 2
192 gm, to the nearest gram.
dy
2 x dx
2c sin x 1 cos
2
c2
2
2
163
3
(256)( 16)
R( x) | c sin x |, so V
0
R( y )
6
144 0
36 x2 dx
12 36
5
(8.5) 365
The plumb bob will weigh about W
61.
6 x2
0 144
dx
4
2
4
2
2
8
(4
) .
4, taken on at the critical point c
2.
2
(b) From the discussion in part (a) we conclude that the function s absolute maximum value is 2 , taken on
at the endpoint c 0.
(c) The graph of the solid s volume as a function
of c for 0 c 1 is given at the right. As c
moves away from [0, 1] the volume of the solid
increases without bound. If we approximate the
solid as a set of solid disks, we can see that the
radius of a typical disk increases without
bounds as c moves away from [0, 1].
63. Volume of the solid generated by rotating the region bounded by the x-axis and y
x
b about the x-axis is V
region about the line y
b
a
b
a
f ( x)
f ( x) dx
2
[ f ( x)]2 dx
b
1 is V
2 f ( x) 1
1 (b
2
b
a
a)
2
f ( x)
b
a
2
dx
4
f ( x) dx
b
a
8 . Thus
b
(2 f ( x) 1) dx
about the x-axis is V
region about the line y
a
2
f ( x ) dx
2 is V
b
2
f ( x) 1 dx
a
4
2
b
a
2
f ( x ) dx
a
f ( x) dx
b
a
dx
8
4
f ( x ) from x
a to x
6 , and the volume of the solid generated by rotating the same
b
a
Copyright
4
4 b a
2
64. Volume of the solid generated by rotating the region bounded by the x-axis and y
b
a to
4 , and the volume of the solid generated by rotating the same
[ f ( x ) 1]2 dx
a
f ( x ) from x
2
f ( x ) 2 dx 10 . Thus
2014 Pearson Education, Inc.
b
Section 6.2 Volumes Using Cylindrical Shells
b
b
a
6.2
b
2
f ( x ) 2 dx
a
2
f ( x) dx 10
a
(4 f ( x) 4) dx
4
4
b
a
b
6
b
f ( x) dx 4
a
dx
2
f ( x)
a
b
4
a
4 f ( x) 4
2
f ( x)
b
f ( x) dx (b a ) 1
a
dx
443
4
f ( x) dx 1 b a
VOLUMES USING CYLINDRICAL SHELLS
1. For the sketch given, a
b
V
a
2
shell
radius
2
0, b
shell
height
2
dx
2
2 x 1 x4 dx
0
2
2
x3
4
x
0
dx
x2
2
2
2
x4
16 0
4
2
2
16
16
3 6
2. For the sketch given, a
b
V
a
shell
radius
2
0, b
shell
height
d
V
c
shell
radius
2
d
c
shell
radius
2
shell
height
b
a
shell
radius
2
x2 1
u
du
b
a
shell
radius
2
x3 9
[u
V
2
0, b
b
V
a
2
0
2
du
36
9
shell
height
dx
3x 2 dx
3u 1/2 du
3
0
6
y 2 dy
2
2
2x
0
x3
4
dx
2 y
2 3
y dy
0
y4
4
2
3 y2
2 y 3
dy
2
3 3
0
y dy
3
0
2 x 2 32 dx
shell
height
2
0
dx
3 x 2 dx
2
2
2
0
2
0
x 2 1 dx;
2 x
u 1, x
4
u
43/2 1
2
3
1
3
4
2
3
(8 1)
14
3
3;
3
0
2 x
9x
dx;
x3 9
9 x 2 dx; x
3 du
36
2u1/2
9
12
0
36
u
9
9, x
3
u
36]
36
2;
shell
radius
2
x4
16 0
x2
2
3;
u 3/2
0, b
2
3;
dx
2
3
6. For the sketch given, a
V
0
0, b
2 x dx; x
4 1/2
u du
1
V
2
dy
shell
height
dx
2;
0, d
shell
height
x2
4
2 x 2
0
dy
5. For the sketch given, a
V
2
0, d
4. For the sketch given, c
V
2;
dx
3. For the sketch given, c
7. a
2;
2
0
x
2
2 x x
x3
2
0
Copyright
dx
8
2014 Pearson Education, Inc.
y4
4
3
0
9
2
4 1
6
444
Chapter 6 Applications of Definite Integrals
8. a
0, b 1;
b
V
1
0
9. a
shell
radius
2
a
shell
height
1
2
2 32x dx
0
1
dx
0
2 x 2 x 2x dx
3 x 2 dx
1
x3
0
0, b 1;
b
V
a
2
2
10. a
shell
radius
2
1
2 x x2
0
1 13
1
4
shell
height
dx
x3 dx
2
1
0
x2
12 4 3
12
2
2 x (2 x) x 2 dx
1
x4
4 0
x3
3
10
12
5
6
0, b 1;
b
V
a
2
1
0
shell
height
x 2 2 x 2 dx
1
dx
0
1
4
0
1
x4
4 0
4
1
2
1
4
shell
radius
shell
height
dx
x2
2
4
11. a
shell
radius
2
2 x2
2 x
x 2 dx
x x3 dx
0, b 1;
b
V
a
2
2
1
0
x3/2
2
5
2
3
12. a 1, b
4;
2
b
V
a
3
2
4 1/2
x
1
2x2
1
2
x dx
2
shell
radius
dx
2 x
2
5
2
12 20 15
30
shell
height
3
1
0
2
3
x3/2
x5/2
2
3
x3
1
2
x2
1
0
7
15
4
dx
x (2 x 1) dx
1
4
2 x 32 x 1/2 dx
2
1
43/2 1
2 (8 1) 14
13. (a)
x f ( x)
x f ( x)
x sinx x , 0
x
x,
x
sin x, 0
x
sin x,
x
0
0
x f ( x)
x f ( x)
Copyright
sin x, 0
0,
sin x, 0
x
x
0
; since sin 0
x
2014 Pearson Education, Inc.
0 we have
Section 6.2 Volumes Using Cylindrical Shells
b
(b) V
a
shell
radius
2
V
2
0
sin x dx
x 0,
2
tan x,
b
a
V
15. c
0, d
d
V
2
2
shell
radius
y 3/2
0
5
2
5
16
15
3 2 5
0, d
d
V
c
17. c
d
2
32
2
shell
height
0
2
3
/4
0
/4
x
x
tan 2 x, 0
x
/4
0
2
2 y
2 y 5/ 2
5
2
8 2
5
y3
3
8
3
y
; since tan 0
tan 2 x, 0
tan x x 0
/4
( y ) dy
2
0
2
5
16
shell
height
2
2
dy
0
y4
4
1
3
2 y y 2 ( y ) dy
y3
3
2
2
4
16
0
1
3
40
3
2;
2
2
shell
radius
2 y2
0
1
3
1
4
shell
height
y 3 dy
32
12
2
2
dy
0
2 y3
3
2 y 2y
y4
4
2
2
0
y 2 dy
16
3
16
4
8
3
Copyright
by part (a)
0 we have
/4
2 x g ( x) dx and x g ( x)
sec2 x 1 dx
0
2
dy
2
y 2 dy
5
6
c
x g ( x)
0
tan 2 x dx
3
shell
radius
y3
0, d
V
2
2
0
16
/4
x
4
2;
2
2
0,
0
x
cos 0)
tan 2 x, 0
x g ( x)
4
dx
y 2 dy
2
16. c
2 ( cos
sin x, 0
2;
2
c
0
x
shell
height
/4
2
cos x 0
x
shell
radius
2
2 x f ( x) dx and x f ( x)
0
2
x
tan 2 x, 0
x g ( x)
(b) V
dx
2
x tanx x , 0
x g ( x)
14. (a)
shell
height
2014 Pearson Education, Inc.
2
x
1 4
/4 by part (a)
4
2
2
445
446
Chapter 6 Applications of Definite Integrals
18. c
0, d
d
V
2
c
1
shell
radius
2
y3
3
d
V
2
20. c
1
0
V
2 y 2 dy
2
21. c
d
2
22. c
0
d
2
2
0
1
1
3
1
0
y y ( y ) dy
4
3
2
2
dy
0
8
3
shell
height
dy
0
6
2
0
2 y y
y
2
2 y (2
y ) y 2 dy
y2
y3
3
(48 32 48)
16
3
2
shell
height
dy
y2
y 3 dy
2
1
4
b
a
6
2
shell
radius
2y
a
(b) V
dy
dy
2
y4
4
0
1;
b
23. (a) V
y dy
y 3 dy
6
0
y 3 dy
16
4
2
1
y3
3
y2
8
3
4
c
y2
1
4
1
shell
height
shell
radius
2y
0, d
V
y2
2 y 2y
2;
2
2
y3
4
3
dy
2
c
shell
height
shell
radius
2 y2
0 2
0, d
V
1
0
0
2;
2
c
0
shell
radius
0, d
d
1
3
2
1
dy
1;
2
c
2
1
y4
4
0, d
shell
height
y 2 dy
y y
0
2
19. c
1;
6
1
0
y2
y3
3
y4
4
y 2 dy
1
0
5
6
(12 4 3)
2
shell
radius
shell
height
dx
2
shell
radius
shell
height
dx
8 83
2 y (2 y )
2
0
2
0
2 x (3 x)dx
6
2 (4 x) (3 x)dx
2 2
0
x dx
6
2
0
2
x3
2
0
4 x x 2 dx
32
Copyright
2014 Pearson Education, Inc.
16
6
2 x2
2
1 x3
3
0
Section 6.2 Volumes Using Cylindrical Shells
b
(c) V
a
8
3
6
d
(d) V
c
shell
radius
2
2
2
(e) V
c
2
shell
radius
c
2
4y
b
24. (a) V
a
2
a
b
a
2
c
d
(e) V
c
2
c
2
25. (a) V
shell
radius
shell
radius
x2
x dx
2
1 x2
2
0
1 x3
3
6
2 y 13 y 2 dy
6
1 y3
9
0
y2
2
shell
height
dx
3 x4
4
2
1 x5
5
0
shell
height
dx
14 13
y 13 y 2 dy
3
0
60
2 ( y 2) 2 13 y dy
0
0
6
2
6
2
dx
2
0
6
2
4 34 y 13 y 2 dy
0
48
2 x 8 x 3 dx
2
2
8 x x 4 dx
0
2 (3 x) 8 x3 dx
2
2
0
2
shell
radius
shell
height
dy
2
8
0
8
0
48 16 12 32
5
2
1 x5
5
0
4 x2
2
2
24 8 x 3x3
x 4 dx
16 8 x 2 x3
x 4 dx
0
264
5
2
2
32 16 8 32
5
2 y y1/3dy
2
2
2 ( x 2) 8 x3 dx
2
1 x5
5
0
dy
0
336
5
8 4/3
0
2 (8 y ) y1/3dy
2
2 ( y 1) y1/3dx
2
y
8
y 7/3
6
7
dy
8 y1/3
0
8
0
6
7
y 4/3 dy
2
(128)
6 y 4/3
768
7
8
3 y 7/3
7
0
576
7
shell
radius
4 x x3
6
0
2 (7 y ) 2 13 y dy
0
2 (24 24 24)
shell
height
shell
radius
b
2
2
0
6
shell
height
1 x4
2
b
a
2 y 2 13 y dy
6
0
dy
shell
radius
2
6
2 (84 78 24)
2
2
2 ( x 1) (3 x)dx
96
5
12
2
2
2
dy
shell
height
shell
radius
384
7
a
(b) V
2
shell
height
6
1 y3
9
0
y2
96 384
7
d
(f ) V
2
3
16 x 4 x 2
d
(d) V
2
dy
6
1 y3
9
0
shell
radius
24 x 4 x 2
2
(c) V
shell
radius
16 32
5
b
(b) V
2
shell
height
0
24
14 y 13
y2
6
d
(f ) V
2
dx
28
2 (36 24)
d
2
shell
height
shell
height
dy
8
0
8
y 4/3
0
y1/3 dy
2
3 y 7/3
7
936
7
shell
height
2
1 x4
4
1
2
shell
radius
shell
height
2x
3 x2
2
2
1 x4
4
1
dx
2
1
2 (2 x) x 2 x 2 dx
2 (8 8 4) 2
dx
2
1
Copyright
2 32
2
1
4 3x 2
x3 dx
27
2
4 1 14
2 ( x 1) x 2 x 2 dx
2 (4 6 4) 2
2
1
4
2
2
1
2 3 x x3 dx
27
2
2014 Pearson Education, Inc.
8
3 y 4/3
4
0
447
448
Chapter 6 Applications of Definite Integrals
d
(c) V
4
8
5
1 3/2
y dy
0
c
1
shell
height
4
2
1
64
5
(1) 2
d
(d) V
shell
radius
2
c
64
3
shell
radius
2
4
8 y 3/2
3
4
8
3
1
2 y5/2
2
5
0
64
2 64
3
5
b
26. (a) V
a
d
c
4
3
d
c
24
d
(b) V
c
c
d
24
18
5
8 3
1
5
2
shell
radius
2
1
13 y 3
5
0
2
15
3
20
1
5
1
dy
0
4
dy
8 y 3/2
3
1
3
2
1
( y 2) dy
1 y3
3
4
y2
1
2 (4 y )
y ( y 2) dy
1
6
4
4 y
(1) 4
3
4
1
5
108
5
y
4 u
.
1
2
x5
1
1 2 4
4
dy
3 0
1
x 4 dx
y
3
4
8y
3 83 8
2
5
2
2 y 4y
1
x 4 3x3 3x 2
1
6
2
4 y
3
2 y
du
4 y
3
du; y 1
4 u u 3/2 du
3
4
1
5
16
9
4
4 x 4 dx
y 3 dy
1
24
y3
dy
u
3, y
8 u 3/2
3 3
0
y 4 dy
2
5
y2
24
60
y3
3
1
dy
24
12
2 (1 y ) 12 y 2
0
2
y4
2
y5
5
8 y3
15
24
dy
24
1
3
1
0
y 12 y 2
8
5
y3
13
20
y4
y3
y5
5
1
24
1
2
dy
0
4
u
3
2 u 5/2
5
0
y4
4
24
(1 y ) y 2
1
5
24
y5
5
1
0
24
1 8
0 5
1
8
15
24
0
y 3 dy
4
5
1
30
y
13
20
1
5
y
2
5
y2
y3 dy
y2
y 3 dy
2
dy
2 y3
5
1
2
5
2
y
dy
24
0
(8 9 12)
Copyright
12 y 2
1 2
0 5
24
12
y2
y3
3 y3
5
dy
1
24
y 4 dy
0
24
2
2014 Pearson Education, Inc.
2 y3
15
56
5
1 2 4
872
45
2 y 12 y 2
24
y 4 dy
shell
height
y4
0
y 4 dy
shell
height
shell
radius
1
0
y
2 y 5/2
5
2
y
3 y2
1
16
9
2 y
6 y 4 y 8 dy
1
4x
88
5
dy
1
4
1
6
5
2 y3
y3
16
9
shell
height
shell
radius
1
0
y 5/2
2 (1 x) 4 3 x 2
(4 u ) u du
3
24
20
1
dy
y
2 y 5/2
5
y 4 ydy [u
shell
height
(32 39 12)
c
24
3 3
1
4
2
0
4
0
y2
4
3 1
1
1 8 2
y
0 5
24
(d) V
dy
shell
radius
0
d
24
60
shell
height
2
1
24
(c) V
x3 2 x 2
4
1
48 64
32
3
1
y
72
5
y 3/2
1 y3
3
3 x4
4
dy
1
3
y2
1 x5
5
shell
radius
4
1
dx
y 9/4
16
9
4
8
5
2 (4 y )
shell
height
2
y
1
0
shell
radius
1 5/4
0
16
9
27. (a) V
2
1 x6
6
2
(b) V
2
5
2
5
dy
y
2 y dy
2
y 3/2 dy 2
0
y2
16
shell
height
2 y
0
y 3/2
4 y
4
1
dy
3
20
y4
y5
5
1
0
0]
Section 6.2 Volumes Using Cylindrical Shells
d
28. (a) V
2
shell
radius
y4
4
y6
24
2
shell
radius
c
2
d
(b) V
c
2
2
0
d
(c) V
2
c
2
2
0
d
(d) V
2
c
2
2
0
shell
height
2
24
4
2
0
shell
radius
5 y2
shell
height
y4
5
4
shell
radius
y5
4
y3
1
0
2 y
y
shell
height
dy
5 y2
8
5
32
d
c
1
1 y3
3
0
b
About y -axis: V
1
0
a
2 x x x 2 dx
x3
3
2
1
x4
4 0
x3
3
1
x5
5 0
30. (a) V
b
a
4
0
y3
3
R ( x)
3 x2
4
2
y 3/2
y2
2
y4
4
y6
24
y4
4
y6
24
y2
2
y4
4
y2
2
40
3
2
0
2
dy
16
4
64
24
0
2
160
20
16
4
2
0
16
4
y4
4
64
24
dy
64
24
40
24
8
y4
4
y2
y 58
2
dy
8
5
2 (5 y ) y 2
0
2
5 y5
160
32
10
2
dy
y3
0
y4
4
2 (2 y ) y 2
16
3
2
0
y2
2
5 y3
24
0
2
y4
4
y6
24
2
dy
2
2
8
3
2
24
32
dy
160
160
dy
4
dy
y 2 dy
2
15
1
3
x2
1
4
shell
height
dx
x3 dx
6
x2
b
V
R ( x)
a
2
r ( x)
2
dx
1
x2
0
x 4 dx
2
15
1
5
1
3
r ( x)
2 x 4 dx
16 16 16
shell
height
1
6
y4
4
2 y y2
0
5 y5
20
y4
4
2
y and r ( y )
1
2
0
1
y 85
y 4 dy
x and r ( x)
1
3
1
0
2
5
0
5 y3
3
2
2
shell
radius
2
About y -axis: R( y )
y2
2
2
0
2
1
3
(b) About x-axis: R( x)
dy
1
4
y4
4
y2
2
2
dy
32
y5
10
2 (5 y )
0
1
2
2
2
2
y2
2
y2
2
2 y3
3
2
shell
radius
2
y dy
2 y 5/2
5
2
dy
y5
4
4
24
2 (2 y )
0
dy
y3
29. (a) About x-axis: V
2
y5
4
1
4
32
dy
y3
y4
4
2 y 2
0
26
24
shell
height
y4
2
2 y2
y2
2
dy
y
d
V
c
R( y)
2
r ( y)
6
2
4
dx
0
x3
4
x2
x
2
2
4x
2
x 2 dx
4
0
16
Copyright
2014 Pearson Education, Inc.
2
dy
1
0
y
y 2 dy
449
y5
4
dy
450
Chapter 6 Applications of Definite Integrals
b
(b) V
2
shell
radius
shell
height
x2
4
x3
6 0
2
shell
radius
shell
height
a
2
b
(c) V
a
2
b
(d) V
4 3 2
x
0 4
d
31. (a) V
c
8
3
2
7
3
2
b
(b) V
a
2
b
(c) V
2
20
3
x 83 x 2
c
d
32. (a) V
c
2
2
b
(b) V
a
4
0
2
2
x2
2
16
b
a
2
shell
height
2 (4 x) 2 2x dx
4
2
0
2
6 2x
4
4
dx
64 16 x x 2
0
[16 (5) (16) (7) (16)]
0
2
1 x3
3
1
2
2
0
shell
height
x dx
36 6 x
(3) (16)
48
shell
height
x2
10
3
40
3
2
2
2
1
4
3
2
8
3
20
3
2 ( y 1)( y 1) dy
1
2
2 20
3
2
8
3
2
x2
2
x3
3 1
4
3
2
3
x (2 x) dx
32
3
2 x x 2 dx
1
3
2
2
2
16
3
1
1
3
3
( y 1)2
x x 2 dx
2
2
( y 1)3
3
2
1
2
3
2 y y 2 0 dy
0
24
4
8
dx
2
4
2 x5/2
2
5
0
64
2 (80
5
5
3 1
3
2
2
dy
y4
4
2 x(2 x) dx
1
4
0
2 x x 3/2 dx
5
16 2 52
64)
dx
32
5
4
0
4
2 x5/2
5
0
2 (4 x) 2
2
Copyright
x2
2
8 4x
1
2
dx
shell
height
8 x 83 x3/2
28 x
12 8
3
shell
radius
shell
radius
0
5
3
1
2
dy
2
dx
2
y2
2
2
dx
shell
height
shell
radius
2 x 2
shell
height
shell
radius
2 3
y dy
0
2
x2
2
2x
2 y ( y 1) dy
1
(14 12 3)
1 13
shell
radius
d
(d) V
2
shell
radius
0
1
2
3
2
a
y3
3
1
3
2 12
4 83
2
(c) V
4
2
5x2
2
2
4
2
dx
64
3
(8 x )2
0
x3
4
dy
y dy
4
dx
shell
height
4
2 (4 x) 2x 2 x dx
32 32 64
6
2
r ( x)
shell
radius
y2
1
4
0
10 x 28 dx
2
2
2
2
R( x )
a
2 x 2 2x dx
0
32
3
dx
2
4
2 x 2x 2 x dx
0
16 64
6
4
x3
6 0
8x 2 x2
2
4
dx
x dx
32 64
16 64
3
5
2
4
0
2
15
8 4 x1/2
2 x x3/2 dx
(240 320 192)
2014 Pearson Education, Inc.
2
15
(112)
224
15
x2
4
dx
Section 6.2 Volumes Using Cylindrical Shells
d
(d) V
c
16
3
2
d
33. (a) V
c
1
0
1
2
2
d
c
1
3
1
2
0
1
2
2
1
3
0
2 y2
y3 dy
y4
4
2 y3
3
2
2
0
0
0
y 3 dy
2 (1 y ) y
y2
2
(30 20 15 12)
2
1
dy
0
2 y 1
y2
2
2
y3
3
1
y5
5
0
y3
y
y5
5
4
15
1
5
7
30
2
60
2
30
1
1
3
2
y4
4
1
5
1
5
1
y3
3
y 4 dy
y 4 dy
2
2
y 3 dy
2 y y
y5
5
dy
shell
height
y2
y
0
y3
3
2
y3
shell
radius
2
1
dy
shell
height
1
4
0
2 (2 y ) y 2 dy
8
3
(4 3)
y 4 dy
2
dy
shell
height
y2
y
0
32
12
shell
radius
2
1
2
shell
height
shell
radius
2
y2
c
34. (a) V
16
4
2
d
(b) V
shell
radius
2
451
dy
1
0
11
15
(15 10 6)
(b) Use the washer method:
d
V
2
R( y)
c
y3
3
y
y7
7
2
r ( y)
2 y5
5
1
dy
1
0
1 13
0
12
1
7
y
y3
2
5
105
1
y
y3
y
y3
3
2
1
dy
0
1 y2
y6
2 y 4 dy
97
105
(105 35 15 42)
(c) Use the washer method:
d
V
R( y)
c
1
1 y2
0
1
3
1
d
(d) V
c
2
2
1
0
2
1
7
1
1
3
2
r ( y)
y6
shell
radius
1 y
1 1
2
1
dy
0
2 y 2 y 3 2 y 4 dy
1
2
2
5
210
shell
height
y3
y
1
4
1
5
dy
y2
1
0
1
0
(20 15 12)
Copyright
0
y4
2
y2
1 2 y
y3
y
y3
2
dy
1
2 y5
5
0
121
210
1
y
y3
dy
2
1 2y
y2
y3
y 4 dy
2 (1 y ) 1
2
1
0 dy
y7
7
(70 30 105 2 42)
y 4 dy
2
60
2
0
y 3 dy
(1 y ) 1 y
23
30
2014 Pearson Education, Inc.
2
y
y2
y3
3
y4
4
y5
5
1
0
452
Chapter 6 Applications of Definite Integrals
d
35. (a) V
2
2
4 2
5
b
a
1
2
2
shell
radius
shell
height
dx
x3 dx
2
R( x )
a
2 x1/2
x
7
1 16
d
(b) V
c
2
2
0
r ( x)
1
shell
radius
0
x
x2
8
dx
24 3
5
2 3
160
(32 20)
x3/2
0
9
2 x 2 x x2
x2
4
2
x3
8
2
1
8
1
32
1
2
2
y 2
y 3 16
1
(
2
6
16
3)
48
r ( y)
1
3
2 x 5/2
5
x dx
x3 dx
1
0
2
2x x2
1 x
x2
2
1
x4
4 0
2 x3
3
1
dx
2
x dx
1
2
2
3
2
1
4
1
0
2
12
x 1/2 1 dx
1/16
1
2 14 16
1
dy
0
1
2
2 y 14
y
1
16
dy
2
y2
32
y 2
1
1
32
2
2
2
48
5
9
R( y)
dx
6
2
shell
height
2
c
2
9
16
y
y 3 16 dy
1
d
0
(2 1)
1/16
2
(8 1)
1
1
4
2
2
38. (a) V
1
1
3
0
2
160
dx
2
x 2 x2
7
2
32
shell
height
2 x
0
8
2
5
2
shell
radius
b
37. (a) V
6
1
x4
4 0
1
y4
4
44
4
4
dx
x3
3
a
y 5/2
4 2
5
y 2 dy
8y
24
5
(8 5)
2
b
(b) V
2 y
4 23
5
x x x 2 dx
0
2
4
32
2
0
2
2
shell
height
4
22
5
2
8
5
shell
radius
2
a
y 3 dy
24
4
4 85 1
b
2
2
1
dy
5
5
2
36. (a) V
shell
height
2 2 y 3/2
0
2
(b) V
shell
radius
2
c
2
dy
1
24
1
1
8
1
y4
1
3
1
16
dy
1
16
11
48
Copyright
2014 Pearson Education, Inc.
1 x x x 2 dx
(6 8 3)
6
4
x4
32 0
Section 6.2 Volumes Using Cylindrical Shells
b
(b) V
a
2
shell
radius
2
3
1
2
2
39. (a) Disk: V
2 1
3 8
1
32
1
x
2
4
3
1 dx
1
1 16 16
b2
2
a1
2, b1 1; a2 0, b2
(b) Washer: V V1 V2
b1
2
R1 ( x )
a1
b2
V2
1/4
R1 ( x) dx and V2
a1
V1
1
dx
1
2
48
1/4
x1/2
x dx
1
x2
2 1/4
2 x3/2
3
2
11
48
(4 16 48 8 3)
V1 V2
b1
V1
shell
height
453
2
R2 ( x )
a2
1
x 2
3
and R2 ( x)
x,
two integrals are required
2
r1 ( x)
2
R2 ( x) dx with R1 ( x)
a2
x 2
3
dx with R1 ( x)
2
r2 ( x )
and r1 ( x )
x 2
3
dx with R2 ( x)
0; a1
and r2 ( x)
2 and b1
x ; a2
0;
0 and b2
1
y2
2
two integrals are required
d
(c) Shell: V
c
shell
radius
2
shell
height
d
dy
c
shell
2 y height
dy where shell height
3y2
2 2 y2 ;
c 0 and d 1. Only one integral is required. It is, therefore preferable to use the shell method.
However, whichever method you use, you will get V
.
40. (a) Disk: V
V1 V2 V3
di
Vi
2
Ri ( y ) dy, i 1, 2, 3 with R1 ( y ) 1 and c1
ci
( y )1/4 and c3
R3 ( y )
(b) Washer: V
0
Ri ( y )
ci
2
b
(c) Shell: V
a
and b 1
ri ( y )
2
shell
radius
2
shell
height
1 and d 2
b
dx
a
0
2
4
16 x 13 x3
r ( x)
x 1
b
a
shell
radius
2
u
0, x
shell
height
1
2
4
dx
4
64 64
3
4
(b) Volume of sphere
42. V
y , c1
0 and d1 1;
shell
2 x height
dx , where shell height
x2
x4
x2
x4 , a
0
only one integral is required. It is, therefore preferable to use the shell method.
R( x )
a
1;
two integrals are required
However, whichever method you use, you will get V
b
0 and d 2
three integrals are required
dy, i 1, 2 with R1 ( y ) 1, r1 ( y )
( y )1/4 , c2
R2 ( y ) 1, r2 ( y )
41. (a) V
y and c2
V1 V2
di
Vi
1, d3
1, d1 1; R2 ( y )
(5)3
4
3
dx
u
64 64
3
500
3
1
1
]
25 x 2
2
5
6
.
4
(3) 2 dx
4
256
3
500
3
Volume of portion removed
2 x sin x 2 1 dx; [u
0
Copyright
sin u du
25 x 2 9 dx
x2 1
cos u 0
du
2 x dx;
( 1 1)
2014 Pearson Education, Inc.
256
3
2
244
3
4
4
16 x 2 dx
454
Chapter 6 Applications of Definite Integrals
b
43. V
shell
radius
2
a
2
2
r h
3
2
d
44. V
c
4
3
r
r h
2
r
dx
0
h
r
2 y
r2
r2 , y
x h dx
r
2
0
h x2
r
shell
height
r
dy
0
y2
du
0
u
f (a )
[( f 1 ( y )) 2 a 2 ]dy
0
f (a)
2 y dy; y
[ f 1 ( f (t )))2
W (t )
S (t )
2 f (t )
t
a
a 2 ] f (t )
t
x dx 2
y2
r2
r
/3
46. V
0
b
u
a
shell
shell
radius
height
(t 2
2 x[ f (a )
y2
0]
dy
4
0
2
r
2
[4 y tan y ]0 /3
1
dx
0
f ( x)]dx
(R2
r2 )
ln 3
f (t ) a 2 ] 2
2
2 xe x dx
(t 2
(e x 1)dx
0
1 3
3 2
3
L
0
3
2
r 2 1/2
u du
0
W (t )
4
3
u3 2
r2
0
[e x
x]
0
x2
1
(e 1 e0 )
0
ln 3
0
(2 ln 3)
1/2
x2
1
1 x2
27
3
2
2
dx
2x
2 x 2 dx
3
0
x2
2
3
0
1 x 2 dx
x
1 2 x2
x
x 4 dx
3
x3
3 0
12
Copyright
S (t ). Therefore, W(t) = S(t)
3
ARC LENGTH
3
y 2 dy
xf ( x)dx
a 2 ) f (t )
4
3
2
e x
t
a
(e x 1). The volume is
(3 ln 3 1)
dy
dx
y r2
r
h x2
2
0
S ( a);
e x /2 , and area
r = 1, outer radius R
1.
r
0
u du
48. Use washer cross sections. A washer has inner radius
6.3
h x3
3r
a 2 ) f (t ); also
a 2 f (t ) 2 tf (t )
[22 (sec y )2 ]dy
2
a
a
xf ( x ) dx [ f (t )t 2
a
S (t )
t 2 f (t ) 2 tf (t )
for all t [a, b].
V
2
h x dx
3
45. W (a)
47. V
2 x
r 2h
1
3
shell
radius
2
r2
[u
shell
height
2014 Pearson Education, Inc.
1 1e
Section 6.3 Arc Length
2.
dy
dx
3
2
x
4
L
1 94 x dx;
0
4 du
u 1 94 x du 94 dx
9
x 0 u 1; x 4 u 10]
3.
y2
dx
dy
3
y2
1
y3
3
4.
1 y4
1
9
L
1
2
1
5.
L
2
1
16
4
1
12
1
11
dx
dy
1
4 y3
2
1
y3
1 y6
y 3
4
1
(16)(2)
128 1 8 4
32
1
y
dy
dy
1
3
y 2
1
y
1
4
y 2
9
1
y1/2
y1/2
1
4
1
y
dy
y 1/2 dy
9
1
y6
1
2
4
1
16 y 6
2
dy
1
y 3
4
y3
1
1
8
1
16 y 4
32
3
1
3
2
dy
1
2
1
9 12
1
4
dy
1
16 y 6
2
1
4
1
4
1 9
2 1
2
1
2
y4
1
4 y2
y 3/ 2
3
1
1
16 y 4
1
2
dx
dy
9
10 10 1
53
6
dy
2 y1/2
8
27
3
1
3
2
1
y
1
3
3
y3
dx
dy
y2
y 1/2
y
1 2 y 3/2
2 3
3
3
3
1
1 14 y 2
91
1 2
3
dy
( 2)
12
9
1
2
1
16 y 4
dy
27
3
1
( 1 4 3)
12
y4
2
3
1 y1/2
2
dx
dy
1
2
1
4 y2
y 1
4
2
dx
dy
1
4 y2
3
L
9
10
4 2 u 3/2
9 3
1
10 1/2 4
u
du
9
1
L
dx;
1
32
1
4
y6
dy
1
2
1
16 y 6
y4
4
y 2
8
dy
2
1
1
8
123
32
Copyright
2014 Pearson Education, Inc.
455
456
6.
Chapter 6 Applications of Definite Integrals
y2
2
dx
dy
3
L
1 3
2 2
y2
2
y 2 dy
8
3
x1/3
8
x 2/3
8
x1/3
1
3 (32
8
8.
dy
dx
0
2
0
2
0
u3
3
x 2/3
2
0
8
3
dy 2
dx
x
1
2
2/3
8
2/3
8
1
2
(1 x)2
(1 x) 2
3
1u 1
4
1
1
2
(1 x )
16
(1 x )
4
(1 x )
4
2
22
x2
1 (1 x) 4
4
1
4
2
x 1/3 dx
8
3 x 2/3
8
1
2 24
dy 2
dx
(1 x)4
x1/3
3 x 4/3
4
1 1
4 (1 x ) 2
x 2/3
16
1
2
dx
1
3
8
1
99
8
1
2
x 2/3
16
dx
16
y 2 dy
13
4
4
(4 x 4) 2
2x 1
(1 x)2
2
6 12
y2
1
3
x 1/3 dx
4 3)
x2
L
x
1
2
1
4
2 x 4/3
3
8
1
2
1 x 2/3
1
1
27
3
x 1/3
1
4
8
L
1
2
1 3
2 2
1
2
2
1 26
2 3
dy
dx
y 4 dy
3
y 1
2 y 4
2 y 4 dy
2
y3
3
1
2
y4
1
4
y4
1
4
2
7.
1 14 y 4
2
3
2
dx
dy
1
2 y2
(2 1)
2 x 1 14 1 2
(1 x )
(1 x)4
(1 x )
16
4
1
2
1
16(1 x )4
dx
dx
2
2
dx
dx; [u 1 x
1
9 12
1
3
1
4
du
dx; x
108 1 4 3
12
Copyright
106
12
0
u 1, x
2
u
53
6
2014 Pearson Education, Inc.
3]
L
3
1
u2
1u 2
4
du
Section 6.3 Arc Length
9.
dx
dy
1
x
x
4
2
L
1
4
8
ln 2
10.
x 2
4
1
x
dy
dx
x
L
x
1
3
1
9
2
11.
1
4x
dy 2
dx
x
3
2
2
dy
dx
x2
3
1
2
3
1
dy
dx
x2
1
4 x2
dx
x4
1
4 x4
1
1
1/2
1
5
x8
x4
1
2
1
12
dx
1
160
2
3
sec 4 y 1
dx
dy
L
tan y
/4
/4
/4
/4
8
1
16 x8
1
4 x4
1
x
1
2
1 2
4x
2
1
4 x2
5
y
x2
ln x
4
1
1
16 x8
1
x4
1
2
1
2
x
3
y
1
16 x 4
10
y
6
2
1
4 x2
3
1
4x 1
1/2
1
4 x4
2
2
1
3
x8
1
2
1
3
0
53
6
1
16 x8
dx
x4
1
4x
4
dx
1
9 12
x3
3
1
2
y
1
1
4 x4
2
y
dx
x
3
x5
5
1
12 x3
0.5
1
x5
5
1
12 x 3 1/2
0
0.5
373
480
2
sec4 y 1
sec4 y 1 dy
1 ( 1)
x2
2
2
dx
8
3
x4
dx
dy
x
2
y
1
16 x 2
0
dx
dx
1
1
1
1
2
x2
8
ln x
1
3
1
16 x 4
dy 2
dx
0
3
x
1
x3
3
1 x
1/2
dx
x
8
y
4
x2
1
16 x 4
1
3
1 ln
4
1
2
0.5
x 2 dx
16
2
2
dx
dy 2
dx
1
2
L
1/2
1 x
1
1
2
x2
4 14 ln 3
4
1
x2
y
0.5
1 2
4x
dx
x2
16
ln x
1 ln1
4
1
4 x2
3
dx
1
2
3
8
1
16 x 2
x2
2
dx
x4
1
13.
1
16 x 2
1 ln 3
4
L
12.
1
2
1
2
1
x
4
ln 2
1 x
1
x2
2
dx
2 1
1 x
dx
ln1
1
4x
x
x2
16
1
2
1
8
1
3
x 2
4
1
x
1
x2
1
1
2
2
dx
dy
/4
/4
sec 2 y dy
2
Copyright
2014 Pearson Education, Inc.
1
1.5
x
457
458
14.
Chapter 6 Applications of Definite Integrals
dy
dx
1
L
1
dy 2
dx
2x
2
L
1
(c)
16. (a)
L
17. (a)
18. (a)
0
/3
(b)
sec4 x
1 sec4 x dx
dx
dy
cos y
2
(b)
cos2 y
1 cos 2 y dy
0
3.82
y
dx
dy
1 y
1/2
1/2
dx
dy
2
1/2
L
(c)
(b)
2.06
dx
dy
L
7 3
3
1 8)
4x2
dy 2
dx
sec2 x
L
(c)
3
(
3
6.13
dy
dx
L
3 x 2 dx
1 4x 2 dx
L
(c)
2
dy 2
dx
dx
1
1
2
1
1 ( 2)3
3
3
2
dy
dx
3x 4 1
3x 4 1 dx
1
2
3
3 x3
15. (a)
dy 2
dx
3x4 1
1/2
1 y2
1
2
1 y
y2
1 y
1/2
2
(b)
y2
dy
2
1/2
1
1/2 1 y 2
dy
dy
L 1.05
Copyright
2014 Pearson Education, Inc.
Section 6.3 Arc Length
2 dx
dy
19. (a) 2 y 2
3
L
(c)
20. (a)
L
21. (a)
dy
dx
L
4.70
dy
dx
dy 2
dx
tan x
/6
L
0
/6 dx
cos x
/6
/6
0
sin 2 x cos 2 x
cos 2 x
dx
sec x dx
sec2 y 1
dx
dy
/4
/3
/3
L
1 tan 2 x dx
0
(b)
tan 2 x
0.55
/4
23. (a)
(b)
x 2 sin 2 x
1 x 2 sin 2 x dx
0
L
(c)
dy 2
dx
cos x cos x x sin x
0
22. (a)
(b)
( y 1)2
9.29
L
(c)
2
1 ( y 1)2 dy
1
L
(c)
dx
dy
459
1
dx
dy
2
(b)
sec2 y 1
sec2 y 1 dy
/4
/3
| sec y | dy
sec y dy
2.20
dy 2
dx
1 here, so take dy as
corresponds to 4x
dx
1 .
2 x
Then y
x C and since (1, 1) lies on the curve,
x from (1, 1) to (4, 2).
C 0. So y
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
24. (a)
dx
dy
C
2
dy
corresponds to 14 here, so take dx as 12 . Then x
y
1. So y
y
1
y
C and, since (0, 1) lies on the curve,
1 .
1 x
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
Copyright
2014 Pearson Education, Inc.
460
25.
Chapter 6 Applications of Definite Integrals
x
y
0
/4
2 cos x dx
0
26.
y
1 x
1
2/3 3/2
2
4
1
x
(2 0)2
2
r2
4
x2 , 0
r
0
29. 9 x 2
r
4r
y ( y 3) 2
d
dy
r
2
x
2
( y 3)( y 1)
6x
( y 3)2 ( y 1)2
4 y ( y 3)
ds 2
y2
dx 2
2x
dy 2
64
20 x 2 64
y2
31.
2
x
0
dy 2
L
4
y2
1
dy 2
dt
/4
1 cos 2 x dx
0
2 cos 2 x dx
0
2
3
x
1
x
1/ 2
8 34
1
1
dx
1/3
2 /4 x
1
2 /4
2 /4
1/ 2
1 x 2/3
r
dx
0
r 2 x2
9x2
dx
x1/3
x 1/3 dx
3
2
1
x 2/3
2 /4
5 dx
0
5x
2
2 5.
0
x
r 2 x2
d
dy
y ( y 3)2
( y 3)( y 1)
dy;
6x
( y 1)2
4y
y2
4 x dx
y
d
dx
2
64
dx 2
dy
1
r
4
0
1
2 y ( y 3) ( y 3)2
dy 2
( y 1) 2
4y
dy
dx
0
16 x 2
y2
dx 2
x2
r 2 x2
dx
4
r
0
r2
r 2 x2
dx
3( y 3)( y 1)
2
( y 3)( y 1)
dy
6x
dy 2
8 x 2 y dx
dx 2
dx
r 2 x2
y2 2 y 1 4 y
4y
16 x 2
y2
dx 2
0
2
x
1
18 x dx
dy
ds 2
1 dy 2
r
4
L
dy 2
( y 3)2 ( y 1)2
36 x 2
dy 2
dy 2
dy 2
4x
y
4 x dx;
y
dy
y 2 16 x 2
y
2
dx 2
4 x 2 64 16 x 2
y2
dx 2
(5 x 2 16) dx 2
dt , x
2
6
2
1 ( 2) 2 dx
1
dx
2/3
1
L
x1/3
1
2 /4
1 x 2/3
1/3
r 2 , we will find the length of the portion in the first quadrant, and multiply our
dy
dx
4x2
2/3 1/2
total length
2
/4
dx
2 sin(0) 1
1 dx
0
2
cos 2 x
2 5
y2
dx 2
dx 2
x
2
dy 2
d
dx
1
dy
dx
dx
1 x
2/3
3
4
r
dx
dx
dy
30. 4 x 2
x
3
2
3 1
2 2
(3 ( 1))2
0
1
2 sin 4
3
2
28. Consider the circle x 2
result by 4.
y
/4
1
2 /4
/4
L
dy
dx
x 1
2/3
2
4
3
2
3 2 x, 0
d
2 sin x 0
x
3 (1) 2/3
2
y
,
cos 2 x
2/3
1 1 x2/3 dx
2 /4
27.
dy
dx
cos 2t dt
0
2
1
dy 2
dx
dy
dx
1
y
f ( x)
number.
Copyright
2014 Pearson Education, Inc.
x C where C is any real
Section 6.3 Arc Length
461
32. (a) From the accompanying figure and definition of the
differential (change along the tangent line) we see
length of kth tangent fin is
that dy f ( xk 1 ) xk
2
xk
(dy )2
(b) Length of curve
n
lim
n
33.
x2
n
1
2
0
(length of kth tangent fin)
2
b
xk
1 x2 ; P
15
4
2
1
.
a
1 2
4
2
xk
n
k 1
xi
xi 1
2
f ( xk 1 ) xk
2
1
f ( x ) dx
4
0, 14 , 12 , 43 , 1
1
2
n
lim
k 1
f ( xk 1 )
y
2
f ( xk 1 ) xk
k 1
y2
1
4
n
lim
1
2
xk
3
2
L
2
yi
2
yi 1
k 1
2
15
4
3
4
1
2
2
7
4
2
3
2
3 2
4
1
7
4
0
2
1.55225
34. Let ( x1 , y1 ) and ( x2 , y2 ), with x2
x2
L
1 m 2 dx
x1
x2 x1
2
35.
2
3x1/2 ; L( x)
[u 1 9t
du
9dt ; t
2 (10)3 2
27
x3
3
y
x2
x
L( x)
0
x
1
2
27
1
4x 4
(t 1) 2
x
0
u 1, t
dy
dx
x2
1
4(t 1) 2
0
16(t 1) 4
(t 1) 2
1 u3
3
1
4(t 1)2
x 1
1u 1
4
1
x2
x
2
x1
x
dt
x2
x2
2
x1
dy
dx
then
m
x1
y2
y1
2
.
1 9t dt ;
0
1 1 9x
9 1
u 1 9 x]
1 9x
u 3/2
2
27
u du
2 (1
27
1
9 x )3/2
2(10 10 1)
27
16( t 1) 4 16(t 1)8 8(t 1)4 1
x
3t1/2
1
0
y2 y1 2
x2 x1
1
x1
2
y2 y1
x2 x1
x
0
2
y2 y1
,
x2 x1
mx b, where m
1 m 2 x2
x2 x1
x1
dy
dx
L(1)
36.
x2
2 x3/2
y
x2
x1
1 m2 x
2
y2 y1
x2 x1
x1 , lie on y
dt ; [u
1 (x
3
1)3
2x 1
2
x
dt
dt
0
( x 1)2
1
4( x 1) 2
4(t 1)4 1
1
4(t 1)
2
dt
2
x
16( t 1)8 8(t 1)4 1
0
16(t 1)4
t 1
du
1
4( x 1)
Copyright
dt ; t
1
3
1
4
0
1 (x
3
1
;
4( x 1) 2
dt
x
[4(t 1) 4 1]2
1
0
16(t 1) 4
x
[4(t 1)4 1]2
0
16(t 1)4
u 1, t
x
1)3
1
4( x 1)
u
2014 Pearson Education, Inc.
x 4( t 1) 4 1
0 4(t 1)2
dt
x 1]
1 ;
12
dt
L(1)
x 1
1
8
3
1
8
u2
1
12
dt
1u 2
4
59
24
du
2 ;
27
462
Chapter 6 Applications of Definite Integrals
37-42.
Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
with( student );
f : x - sqrt(1-x^2);a : -1;
b : 1;
N : [2, 4, 8];
for n in N do
xx : [seq( a i*(b-a)/n, i 0..n )];
pts : [seq([x, f (x)], x xx)];
L : simplify(add( distance(pts[i 1], pts[i]), i 1..n ));
T : sprintf("#37(a) (Section 6.3)\nn %3d L %8.5f \n", n, L );
# (b)
P[n] : plot( [f (x), pts], x a..b, title T ):
# (a)
end do:
display( [seq(P[n], n N)], insequence true, scaling constrained );
L : ArcLength( f(x), x a..b, output integral ):
L
evalf ( L );
# (c)
Mathematica: (assigned function and values for a, b, and n may vary)
Clear[x, f ]
{a, b} { 1, 1}; f[x_ ] Sqrt[1 x 2 ]
p1 Plot[f[x], {x, a, b}]
n 8;
pts Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / N
Show[p1,Graphics[{Line[pts]}]}]
Sum[ Sqrt[ (pts[[i 1, 1]] pts[[i, 1]])2
(pts[[i 1, 2]] pts[[i, 2]]) 2 ], {i, 1, n}]
NIntegrate[Sqrt[ 1 f '[ x]2 ], {x, a, b}]
6.4
AREAS OF SURFACES OF REVOLUTION
1. (a)
dy
dx
S
(c)
S
dy 2
dx
sec2 x
2
/4
0
sec4 x
(b)
(tan x) 1 sec4 x dx
3.84
Copyright
2014 Pearson Education, Inc.
Section 6.4 Areas of Surfaces of Revolution
2. (a)
dy
dx
S
(c)
3. (a)
S
4. (a)
5. (a)
xy 1
S
21
1 y
S
2
x1/2
0
y1/2
1
y4
(b)
2
(b)
cos2 y
(sin y ) 1 cos2 y dy
3
3 x1/2
y
2 3 x1/2
2
1
2
(b)
2
x 1/2
2
1 3x 1/2
4
1
2
3 x1/2
1
2
1 3 x 1/2 dx
63.37
1 y 1/2
dx
dy
S
S
2
14.42
S
(c)
dx
dy
1
y2
1 y 4 dy
dx
dy
cos y
dy 2
dx
6. (a)
dx
dy
5.02
dx
dy
S
1 4 x 2 dx
1
y
x
2
dy
dx
(c)
(b)
4x2
53.23
S
(c)
2 2
x
0
2
S
(c)
dy 2
dx
2x
2
2
1
dx
dy
2
y 2 y
(b)
2
1 y 1/2
1
2
1 y 1/2 dx
51.33
Copyright
2014 Pearson Education, Inc.
463
464
Chapter 6 Applications of Definite Integrals
dx
dy
7. (a)
S
(c)
0
(b)
tan 2 y
y
1 tan 2 y dy
tan t dt
0
y
0
1
0.5
tan t dt sec y dy
S
dy 2
dx
5
2
x
1
5
2
S
0
x2 1
x
1
x 2 1 dx
1
3
dy
dx
1;
2
b
S
1
a
dy 2
dx
dx
2 y 1
1 (4
2
Lateral surface area
x
2
y
2
x
dx
dy
2y
5 4
8
dx
dy
1;
2
S
5
9
2
2
b
a
2 y 1
1
2
3
(2 1) 2
c
x
2
2
0
1 14 dx
4
1 (8
2
dx
dy
2 x 1
5
1
(3 1)2
) 2 5
dy 2
dx
dx
2
5
3
1
2
(4 2)
42
8
4
5
2
22
0
x
3
2
4
x2
2 0
4
5;
y dy
2
5 y2
2
2
2
2 5
5
x dx
2 5
2
dy
2
0
2
2 y 1 22 dy
4
5
2
0
2 (4), slant height
4
2
0
5 in agreement with the integral value
( x 1)
2
3
2
5 in agreement with the integral value
5; Geometry formula: base circumference
Lateral surface area
11.
d
4
S
1
2 (2), slant height
) 2 5
2; S
x
0.8
t 2 1 dt
1
2
Geometry formula: base circumference
10.
0.6
8.55
x
2
y
x
y
0
9.
0.4
y
t 2 1 dt x dx
1
0.2
tan t dt
0
(b)
x2 1
t 2 1 dt
1
y
x
2.08
dy
dx
(c)
y
0
/3
S
8. (a)
/3
2
2
2
dx
dy
tan y
1
1 2 dx
2
5 3
(x
2 1
5; Geometry formula: r1
Frustum surface area
r1 r2
1
2
x2
2
5
1) dx
2
1
2
3
2
1, r2
slant height
x
3
1
1
2
2, slant height
(1 2) 5
3
5 in
agreement with the integral value
12.
x
2
y
2
1
2
5 y2
x
2y 1
y
2
1
2
slant height
(2 1) 2
the integral value
dx
dy
d
2; S
c
2 x 1
5 (4 2) (1 1)
(3 1)2
5
Copyright
4
dx
dy
2
dy
2
1
2 (2 y 1) 1 4 dy
5; Geometry formula: r1 1, r2
Frustum surface area
(1 3) 5
2014 Pearson Education, Inc.
4
2
5
2
1
(2 y 1) dy
3,
5 in agreement with
Section 6.4 Areas of Surfaces of Revolution
13.
dy 2
dx
x2
3
dy
dx
4
u 1 x9
x
0
x4
9
u 1, x
2
dy
dx
15/4
S
3/4
15.
1.5
S
0.5
1.5
2
dy
dx
17.
dx
dy
dx
5
1
5
2
1
2
(1 x )2
2 x x2
x
98
81
15/4
3/2
1
4
1
4
x
3/4
3
4
dx
1
4
3/2
(1 x )2
2 x x2
dx
5
4
x
dx
3/2
5 45
4
3
53
23
33
23
y2
2
1
4( x 1)
x 1 1 4( x1 1) dx
2
4
3
0
dy 2
dx
2
1.5
0.5
dy 2
dx
6
dx
dy
u 1 y4
y
125 27
27
du
28
3
2x x2 1
1
2 x 1
S
1
2x x
1.5
0.5
15
4
1 x
2
2
4
3
(8 1)
2x x
25/9 1/2 1
u
4
2
2
2
2 x x 2 2 x x 1 22 x x dx
0.5
2
16.
4
3
1
1 (2 2 x )
2 2 x x2
dy
dx
3
dx;
1
4x
3/4
4 3
2
4
3
1
3/2 15/4
x 14
2
3
2
S
x 1 41x dx
2
x3
9
du
25
9
125
27
dy 2
dx
x 1/2
1
2
1
4
u
3
4
1 x9 dx;
0
4 x3 dx
9
du
25/9
2 u 3/2
2 3
1
14.
2 2 x3
9
S
4
3
(125 27)
98
6
2
y4
u
1
4
du
2
2
2 u 3/2
6 3
1
( x 1) 14 dx
S
9
3/2
25
4
9
4
3/2
49
3
1 2 y3
0 3
S
du 4 y3 dy
1
5
5 3/2
4
1
x
3/2
1 54
u 1, y 1
2 12
u du
6 1
2
3
2
5
2
1 y 4 dy;
y3 dy;
2
1
2
1
3
u1/2 14 du
( 8 1)
Copyright
2014 Pearson Education, Inc.
465
466
18.
Chapter 6 Applications of Definite Integrals
1 y 3/2
3
x
y1/2
0, when 1
1 y 3/2
3
area, we take x
dx
dy
y1/2
1
2
3
S
1
3 1 3/2
y
1 3
y1/2
2
3 1 3/2
y
1 3
1/2
9
19.
4 y
15/4
4
0
8
3
20.
5 5
dx
dy
21. S
5 5
8
ln 2 e y e
2
0
1 e2 y
2 2
1
2 2
1
3
y
2
4 2ln 2
x2
2
0
2
x
3/2
y3
9
1
3
y3
3
4
2 (5
3
4
2
3
1
ey e
2
1
y1/2
y 1
27
9
y
9
3
1
3 1
1 3
dy
1/ 2
y
1
9
3
y
dy
x 2 1 dx
2
dy
2
1
3
y 1 ( y 1) dy
3 19
1
0
1
3
1
0
(4 y ) 1 dy
0
8
3
3/2
5
4
53/2
5
3
1
5/8
5
8
2
dy
2
2 y 1 1 2 y1 1 dy
2
3/2
4
2
3
1 5 5
y
0
ln 2 e y e
2
0
3
2
dy
x x 2 1 dx
15
16
ln 2
1
2 x2
2
2
0
2 82 2 5 5
82 2
1
2
x 4 dx
x dx
12
1
2
5/8
2 y1/2 dy
16 2 5 5
2 e 2 y ) dy
ln 2 2 y
(e
2 0
x3
(2 y 1) 1 dy
5/8
1 14 (e2 y
2 ln 2 12 e 2 ln 2
ds
Copyright
y
1
2
4
8 8
ln 2 e y e
2
2
1 e 2 ln 2
2
2 dx
2
15/4
4
3/2
5 15
53/2
4
8
3
2 81 2 ln 2
x x2
2
15/4
35
13/2
y
2 4 y 1 4 1 y dy
2
y )3/2
S
ey e
2
ln 2
1 e 2y
2
0
1 1
2 4
2
2
0
40 5 5 5
8
1
2y 1
y
15/4
S
2
y
2y
3 1/2 1
y
3
dy
1
4 y
8
3
dx
dy
ln 2 e y e
2
0
2
22.
2
5/8
2
1/ 2 2
16
9
1
2 23 y 3/2
2
y 1 dy
y 1 dy
y 2
y1/ 2 y
5 y dy
1
2y 1
y 1
y 2
1 14 y 2
y 1 dy
dx
dy
1
1
4
2
( 18 1 3)
dx
dy
1
4
y
2
3
2
dx
dy
y1/2
2
3 1 2
y
1 3
y1/2
y 1/2
1 y 3/2
3
2
3. To get positive
y
2 e 2 y )dy
0 12
S
2
2
x4
4
2014 Pearson Education, Inc.
2
0
x 1 2x2
x2
2 0
2
2
x 4 dx
4
4
2
2
4
Section 6.4 Areas of Surfaces of Revolution
dx 2
23. ds
y3
24.
y
cos x
25.
y
a2
S
2
1
4 y3
y5
5
2
dy 2
dy
y3
1
4 y3
1
dy
dx
x2
a
a2
a
a2
1
2
1
8
x
a
x
r
h
y
dy
dx
x
2
r 2 x2
h2
2 r
h2
dy 2
dx
r
h
h
r2
h2
S
y
16
7
2
V
16
S 0.5 mm
r2
y
2
29.
y
2
16 dy
5000 V
28.
2
a h
a
R2
y
2
a h
a
2
dx
dy
32
hr
x
0 h
2
2
r 2 h2
y2
16
9
r2
x2
R2
2
y
x
2
2
1
dy
y4
1
4
y 2 dy
253
20
(8 31 5)
dy 2
dx
2
x2
a x2
2
x 2 dx
2
2
1 r 2 dx
d
c
7
; S
16
dy
dx
1
2
x2
x 2 dx
2x
2
a
2
a
a
a dx
2 a x a
2 x 1
1
2
x2
x 2 dx
2
r
2 r
2
x
a h
a
2x
2
dx
dy
162
2
hr
0 h
2
2
x h 2r dx
h2 r 2 h x
0
h2
2 r
h
h
dx
r2
2
dy. Now, x 2
y2 1
y2
16
2
y2
7
dy
y2
162
16
2
162
x
162
y2
y2
y 2 dy
dx
dy
2
dx
x
x
2
2 R
R
2
x
a h
a
Copyright
45.24 cm3 . For 5000 woks, we need
226.2 L
x
x
dy
dx
2
h
r h2
(5)(45.24) L
R
dy
1
16 y 6
(cos x) 1 sin 2 x dx
x2
(904.78)(0.05) cm3
S 0.05 cm
r
/2
1
4 y3
1
2
904.78 cm 2 . The enamel needed to cover one surface of one wok is
288
5000 45.24 cm3
x2
/2
a2
a
27. The area of the surface of one work is S
dx
dy
a
2
y y3
2
40
x
a
2
1
1
8
2
y6
1 dy
2
2
31
5
2
( 2 x)
S
h2
2 r
h2
0
1
4
dx
2
1
16 y 6
4 a2
2 a [ a ( a )] (2 a)(2a )
26.
1/2
1
2
2 y ds
sin 2 x
2
2
1
1
5
x2
x2 1
2
dy; S
dy 2
dx
sin x
y6
1 dy
32
5
2
dy
dx
2
1
4 y3
2
y 1
1
4
2
y3
2
dx
2
226.2 liters of each color are needed.
x2
r x2
2
; S
a h
2
r2
a
x2 1
x2
r x2
2
dx
2 rh, which is independent of a.
dx
dy
2
x2
R2 x2
; S
2
2 Rh
2014 Pearson Education, Inc.
a h
a
R2
x2 1
x2
R2 x2
dx
467
468
Chapter 6 Applications of Definite Integrals
30. (a)
x2
y2
452
45
S
22.5
452
y2
y2 1
2
x
452
2
y2
45
(2 )(45)(67.5) 6075
(b) 19,085 square feet
y
dx
dy
y
2
45
dy
2
y
45
2
2
dx
dy
2
45
452
22.5
y2
2
y2
y2
;
y 2 dy
2
45
45
22.5
dy
square feet
31. (a) An equation of the tangent line segment is
(see figure) y f (mk ) f (mk ) x mk . When
x
xk 1 we have
f (mk ) f (mk )( xk 1 mk )
r1
f (mk )
when x
r2
x
f (mk ) 2k ;
f (mk )
xk we have
f ( mk )
f (mk ) xk
mk
x
f (mk ) 2k ;
f (mk )
(b) L2k
xk
2
f (mk )
2
xk
Lk
r2
xk
r1
2
xk
2
x
2
2
x
f (mk ) 2k
f (mk ) 2k
2
xk
f (mk ) xk
2
2
f (mk ) xk , as claimed
(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent line
segment about the x-axis is given by
parts (a) and (b) above. Thus,
n
(d) S
32.
S
Sk
3/2
1
2 2
0
u 1 x 2/3
S
6.5
4
lim
n
k 1
1 x 2/3
y
r1 r2 Lk
2 f (mk ) 1
n
lim
n
Sk
Sk
3
2
1 x 2/3
3/2
0 3/2
u
1
2
3
1/2
1 x 2/3
1
1
x
2/3
x 1/3dx
3 du
2
f ( mk )
f (mk )
2
6
1 dx
3
2
2
5
4
a
0
6
1
1/ 2
x
1
0
1 x 2/3
0 52
3/2
0
xk
2
f (mk ) xk
2 f ( x) 1
x 2/3 dx
u 1, x 1
2
dx
1
x 2/3
1
f ( x)
1 x 2/3
x 2/3
4
u
1
0
1 x 2/3
3/2
x 1/3 dx;
0
12
5
WORK AND FLUID FORCES
1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F ( x )
The work done by F is W
9k
2
1800
k
3
0
F ( x) dx
k
3
0
x dx
3
k x2
2
0
9k .
2
kx.
This work is equal to 1800 J
400 N/m
Copyright
2
xk .
dy 2
dx
1/3
x 1/3 dx; x
du
u 5/2
1 x 2/3
x 1/3
2
3
2
b
xk
k 1
dy
dx
du
2 f (mk ) 1
2 f (mk )
2014 Pearson Education, Inc.
using
Section 6.5 Work and Fluid Forces
2. (a) We find the force constant from Hooke s Law: F
kx
F
x
k
k
800
4
200 lb/in.
2
(b) The work done to stretch the spring 2 inches beyond its natural length is W
2
2
200 x2
200(2 0)
0
(c) We substitute F
400 in-lb
469
0
kx dx
2
200
0
x dx
33.3 ft-lb
1600 into the equation F
200 x to find 1600
200 x
x
8 in.
3. We find the force constant from Hooke s law: F kx. A force of 2 N stretches the spring to 0.02 m
2
k (0.02)
y
4N
N
100 m
100
0.04
0
N . The force of 4 N will stretch the rubber band y m, where F
100 m
k
y
0.04 m
2
0.04
(100)(0.04)2
2
kx
F
x
k
kx dx
90
k
5
stretch the spring 5 m beyond its natural length is W
0
5. (a) We find the spring s constant from Hooke s law: F
kx
90
1
5
0
F
x
k
(b) The work done to compress the assembly the first half inch is W
0
(0.5)2
(7238) 2
0
F
k
kx dx
0.08 J
4. We find the force constant from Hooke s law: F
0.5
y
0.04
4 cm. The work done to stretch the rubber band 0.04 m is W
x dx 100 x2
0
2
7238 x2
ky
(7238)(0.25)
2
k
N . The work done to
90 m
21,714
8 5
0.5
0
5
2
90 x2
x dx
(90) 25
2
0
21,714
3
kx dx
7238
lb
7238 in
k
0.5
0
1125 J
x dx
905 in-lb. The work done to compress the assembly the
second half inch is:
W
1.0
0.5
kx dx
7238
1.0
0.5
x dx
2
7238 x2
1.0
0.5
6. First, we find the force constant from Hooke s law: F
compresses the scale x
scale this far is W
1/8
0
1
8
in, he/she must weigh F
kx dx
2
2400 x2
1/8
2400
2 64
0
(7238)(0.75)
2
1 (0.5) 2
7238
2
kx
kx
k
150
F
x
1
16
2, 400 18
18.75 lb in.
2714 in-lb
16 150
lb . If someone
2, 400 in
300 lb. The work done to compress the
2.5 ft-lb
16
7. The force required to haul up the rope is equal to the rope s weight, which varies steadily and is proportional
to x, the length of the rope still hanging: F ( x )
2
0.624 x2
50
0.624 x. The work done is: W
50
0
F ( x) dx
50
0
780 J
0
8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the
bag is x ft off the ground is F ( x) 144 4 x. The work done is: W
144 x 2 x 2
18
0
b
a
F ( x ) dx
1944 ft-lb
Copyright
2014 Pearson Education, Inc.
18
0
(144 4 x) dx
0.624x dx
470
Chapter 6 Applications of Definite Integrals
9. The force required to lift the cable is equal to the weight of the cable paid out: F ( x )
180
where x is the position of the car off the first floor. The work done is: W
180
x2
2 0
4.5 180 x
2
4.5 1802 180
2
4.51802
2
0
(4.5)(180 x )
F ( x) dx
4.5
180
0
(180 x ) dx
72,900 ft-lb
10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F ( x)
b
The work done is W
a
k
k2
dx
k
b
1
x2
a
dx
b
k 1x
k b1
a
1
a
k .
x2
k (a b)
ab
11. Let r the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a
constant rate, the amount of water in the bucket is proportional to (20 x), the distance the bucket is being
raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is
F
0.8(20 x). So: W
20
0
0.8 (20 x )dx
0.8 20 x
20
x2
2 0
160 ft-lb.
12. Let r the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a
constant rate, the amount of water in the bucket is proportional to (20 x), the distance the bucket is being
raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is F
20
So: W
0
2(20 x ) dx
2 20 x
20
x2
2 0
2(20 x ).
400 ft-lb.
Note that since the force in Exercise 12 is 2.5 times the force in Exercise 11 at each elevation, the total work is
also 2.5 times as great.
13. We will use the coordinate system given.
(a) The typical slab between the planes at y and y
y
3
has a volume of V (10)(12) y 120 y ft . The
force F required to lift the slab is equal to its weight:
F 62.4 V 62.4 120 y lb. The distance through
which F must act is about y ft, so the work done
lifting the slab is about W force distance
62.4 120 y y ft-lb The work it takes to lift all
the water is approximately
20
W
20
W
0
62.4 120 y
y ft-lb.
0
This is a Riemann sum for the function 62.4 120 y over the interval 0
tank empty is the limit of these sums:
W
20
0
62.4 120 y dy
y2
(62.4)(120) 2
20
0
(62.4)(120) 400
2
5
(b) The time t it takes to empty the full tank with 11
y
20. The work of pumping the
(62.4)(120)(200) 1,497,600 ft-lb
hp motor is t
W
250 ft-lb
sec
1,497,600 ft lb
250 ft-lb
sec
5990.4 sec
1.664 hr t 1 hr and 40 min
(c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is
W
10
0
62.4 120 y dy
1497.6 sec
0.416 hr
y2
(62.4)(120) 2
10
0
(62.4)(120) 100
2
374,400 ft-lb and the time is t
25 min
Copyright
2014 Pearson Education, Inc.
W
250 ft-lb
sec
Section 6.5 Work and Fluid Forces
471
(d) In a location where water weighs 62.26 lb3 :
ft
a) W
(62.26)(24,000) 1,494,240 ft-lb .
1,494,240
250
b) t
5976.96 sec 1.660 hr
t
1 hr and 40 min
t
1 hr and 40.1 min
In a location where water weighs 62.59 lb3
ft
a) W
(62.59)(24,000) 1,502,160 ft-lb
1,502,160
250
b) t
6008.64 sec 1.669 hr
14. We will use the coordinate system given.
(a) The typical slab between the planes at y and y
y has
3
a volume of V (20)(12) y 240 y ft . The force F
required to lift the slab is equal to its weight:
F 62.4 V 62.4 240 y lb. The distance through
which F must act is about y ft, so the work done lifting
the slab is about W force distance
20
62.4 240 y
y ft-lb. The work it takes to lift all the water is approximately W
W
10
20
62.4 240 y
y ft-lb. This is a Riemann sum for the function 62.4 240 y over the interval
10
10
20
W
(b) t
20. The work it takes to empty the cistern is the limit of these sums:
y
10
62.4 240 y dy
W
275 ft-lb
sec
20
y2
(62.4)(240) 2
2,246,400 ft-lb
275
(62.4)(240)(200 50)
(62.4)(240)(150)
2,246,400 ft-lb
10
8168.73 sec
2.27 hours
2 hr and 16.1 min
(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is
W
15
10
62.4 240 y dy
Then the time is t
15
y2
(62.4)(240) 2
W
275 ft-lb
sec
936,000
275
10
100
(62.4)(240) 225
2
2
(62.4)(240) 125
2
936, 000 ft.
3403.64 sec 56.7 min
(d) In a location where water weighs 62.26 lb3 :
ft
a) W
b) t
c) W
(62.26)(240)(150)
2,241,360
275
2,241,360 ft-lb.
8150.40 sec
(62.26)(240) 125
2
2.264 hours
933,900 ft-lb; t
2 hr and 15.8 min
933,900
275
3396 sec
0.94 hours
56.6 min
In a location where water weighs 62.59 lb3 :
ft
a) W
b) t
c) W
(62.59)(240)(150)
2,253,240
275
2,253,240 ft-lb.
8193.60 sec
(62.59)(240) 125
2
2.276 hours
938,850 ft-lb; t
Copyright
2 hr and 16.56 min
938,850
275
3414 sec
0.95 hours
2014 Pearson Education, Inc.
56.9 min
472
Chapter 6 Applications of Definite Integrals
y 2
,
2
x2
15. The slab is a disk of area
work to pump the oil in this slab,
10 57
0 4
W
10 y 2
y 3 dy
thickness
y, and height below the top of the tank (10 y ). So the
y 2
.
2
W , is 57 (10 y )
10 y 3
3
57
4
The work to pump all the oil to top of the tank is
10
y4
4
11,875 ft lb 37,306 ft-lb
0
y 2
16. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is (14 y )( ) 2
the tank is half full and the volume of the original cone is V
3
500 57
4
0
So W
14 y 2
y 3 dy
14 y 3
3
57
4
y4
4
17. The typical slab between the planes at y and y
20 2
2
F
r 2h
1
3
52 (10)
ft 3 , and with half the volume the cone is filled to a height y, 250
6
250
6
volume
1
3
3
ft 3 , half the
250
3
y2
4
1
3
and since
y
y
3
500 ft.
500
60,042 ft-lb.
0
y has a volume of
(radius)2 (thickness)
V
100 y ft 3 . The force F required to lift the slab is equal to its weight:
y
51.2 V
51.2 100
y lb
F
5120
y lb The distance through which F must act is about
30
30
(30 y ) ft. The work it takes to lift all the kerosene is approximately W
W
5120 (30 y ) y ft-lb
0
0
which is a Riemann sum. The work to pump the tank dry is the limit of these sums:
30
W
0
5120 (30 y ) dy
5120
y2
2
30 y
30
900
2
5120
0
(5120)(450 ) 7,238,229.48 ft-lb
18. (a) Follow all the steps of Example 5 but make the substitution of 64.5 lb3 for 57 lb3 . Then,
ft
8 64.5
4
W
0
3
21.5
8
(10 y ) y 2 dy
10 y
3
64.5
4
3
4
y
4
8
64.5
4
0
10 83
3
84
4
ft
64.5
4
83 10
2
3
34,582.65 ft-lb
(b) Exactly as done in Example 5 but change the distance through which F acts to distance
8 57
0 4
Then W
64.5 83
3
(13 y ) y 2 dy
57
4
13 y 3
3
y4
4
8
0
57
4
13 83
3
84
4
57
4
83 13
2
3
(13 y ) ft.
57 83 7
34
2
(19 )(8 )(7)(2) 53,482.5 ft-lb
19. The typical slab between the planes at y and y
y
73
2
y
y has a volume of about
V
(radius) 2 (thickness)
y ft 3 . The force F ( y ) required to lift this slab is equal to its weight: F ( y )
2
y
73
V
73 y y lb. The distance through which F ( y ) must act to lift the slab to the top of the
reservoir is about (4 y ) ft, so the work done is approximately
W 73 y (4
y ) y ft-lb. The work done
n
lifting all the slabs from y
0 ft to y 4 ft is approximately W
73 yk 4
k 0
Copyright
2014 Pearson Education, Inc.
yk
y ft-lb. Taking the limit
Section 6.5 Work and Fluid Forces
of these Riemann sums as n
4
1 y3
3
0
2 y2
73
73
4
, we get W
64
3
32
2336
3
0
73 y (4 y ) dy
4
73
y 2 dy
4y
0
473
ft-lb 2446.25 ft-lb.
20. The typical slab between the planes at y and y
y has volume of about
V
(length)(width)(thickness)
2 25 y 2 (10) y ft 3 . The force F ( y ) required to lift this slab is equal to its weight:
F ( y)
53
53 2 25 y 2 (10) y 1060 25 y 2 y lb. The distance through which F ( y ) must act to
V
lift the slab to the level of 15 m above the top of the reservoir is about (20 y ) ft, so the work done is
approximately
1060 25 y 2 (20 y ) y ft-lb. The work done lifting all the slabs from y
W
n
1060 25 yk2 20 yk
5 ft is approximately W
y
5 ft to
y ft-lb. Taking the limit of these Riemann sums as
k 0
n
5
, we get W
5
1060
5
1060 25 y 2 (20 y )dy 1060
5
20 25 y 2 dy
5
5
5
(20 y ) 25 y 2 dy
y 25 y 2 dy . To evaluate the first integral, we use we can interpret
25 y 2 dy as the area of the semicircle whose radius is 5, thus
5
20
y
5
5
(5)2
1
2
5
u
5
1060
5
250 . To evaluate the second integral let u
5
0, thus
5
1 0
2 0
y 25 y 2 dy
5
20 25 y 2 dy
5
y 25 y 2 dy
u du
2
25 y 2
F ( y ) 9800
5
20 25 y 2 dy
25 y 2
0. Thus, 1060
1060(250
21. The typical slab between the planes at y and y
5
0)
du
5
5
20
5
2 y dy; y
5
25 y 2 dy
5
u
0,
20 25 y 2 dy
265000
832522 ft-lb
y has a volume of about
(radius) 2 (thickness)
V
y m3 . The force F ( y ) required to lift this slab is equal to its weight:
V
25 y 2
9800
2
y
25 y 2
9800
y N. The distance through which F ( y ) must
act to lift the slab to the level of 4 m above the top of the reservoir is about (4 y ) m, so the work done is
W
approximately
y
25 y 2 (4 y ) y N m. The work done lifting all the slabs from y
9800
0 m is approximately W
0
9800
5 m to
25 y 2 (4 y ) y N m. Taking the limit of these Riemann sums,
5
we get W
9800
0
5
100 y
9800
25 y 2
2
25 y 2 (4 y) dy
4 y3
3
y4
4
9800
2
y
100 y 2
5
0
9800
5
22. The typical slab between the planes at y and y
100 y 2
0
500
100 25 y 4 y 2
25 25
2
125
625
4
y has a volume of about
y ft 3 . The force is F ( y )
Copyright
4
3
y 3 dy
56 lb
ft 3
V
2014 Pearson Education, Inc.
15, 073, 099.75 J
V
56
(radius) 2 (thickness)
100 y 2
y lb. The
474
Chapter 6 Applications of Definite Integrals
distance through which F ( y) must act to lift the slab to the level of 2 ft above the top of the tank is about
(12 y ) ft, so the work done is
W
100 y 2 (12 y ) y lb ft. The work done lifting all the slabs from
56
10
0 ft to y 10 ft is approximately W
y
56
100 y 2 (12 y ) y lb ft. Taking the limit of these
0
10
Riemann sums, we get W
10
56
0
56
56
0
1200 100 y 12 y 2
10,000
2
12,000
100 y
y 3 dy
(12 y ) dy
56
10,000
4
4 1000
2
100 y 2
2
1200 y
10
56
0
12 y 3
3
5
2
(56 ) 12 5 4
100 y 2 (12 y ) dy
y4
4
10
0
(1000) 967,611 ft-lb. It would cost
(0.5)(967, 611) 483,805¢= $4838.05. Yes, you can afford to hire the firm.
m dv
dt
23. F
1
2
mv dv
by the chain rule
dx
m v2 ( x2 ) v 2 ( x1 )
24. weight
0.3125 lb
32 ft/sec2
1
2
26. weight
27. v1
W
1.6 oz
1
2
x2
0.1 lb
F ( x ) dx
28. weight
(132ft/sec)2
1
2
m
6.5 lb
16
6.5 oz
1
8
1
256
32
x1
1
2
slugs; W
dv
dx
v
1
256
0.3125 lb
132 ft/sec; m
0.1 lb
1
320
32 ft/ sec2
1 mv 2
1
2
m
x2
m
dx
m
x
1 v 2 ( x) 2
2
x1
slugs (160 ft / sec)2
0.3125
32
32 ft/ sec2
50 ft-lb
slugs;
85.1 ft-lb
153 mph
mv22
x1
mv dv
dx
dx
mv12 , as claimed.
weight
32
lb; mass
ft , v
0 sec
2
0 mph
x1
mv22
90 mi 1 hr
1 min 5280 ft
1 hr 60 min 60 sec 1 mi
25. 90 mph
W
2
16
2 oz
1
2
x2
W
slugs; W
ft ; 2 oz
224.4 sec
1 1
2 256
6.5
(16)(32)
1
320
0.125 lb
(224.4)2
slugs; W
1
2
1 1
2 256
1
2
m
(0) 2
6.5
(16)(32)
slugs (280ft/ sec)2 122.5ft-lb
0.125 lb
32 ft/ sec 2
1
256
slugs;
98.35 ft-lb
slugs (132ft/sec)2 110.6ft-lb
29. We imagine the milkshake divided into thin slabs by planes perpendicular to the y -axis at the points of a
partition of the interval [0, 7]. The typical slab between the planes at y and y
y has a volume of about
V
(radius) 2 (thickness)
weight: F ( y )
4
9
V
4
9
y 17.5 2
14
y 17.5 2
14
y in 3 . The force F ( y ) required to lift this slab is equal to its
y oz. The distance through which F ( y ) must act to lift this slab
to the level of 1 inch above the top is about (8 y ) in. The work done lifting the slab is about
W
4
9
( y 17.5)2
142
(8 y ) y in oz. The work done lifting all the slabs from y
Copyright
2014 Pearson Education, Inc.
0 to y
7 is approximately
Section 6.5 Work and Fluid Forces
7
4
9.142
W
0
475
( y 17.5)2 (8 y ) y in oz which is a Riemann sum. The work is the limit of these sums as the
norm of the partition goes to zero:
7 4
(y
0 9 142
W
y4
4
4
9 142
7
4
2 0
9 14
7
17.5)2 (8 y ) dy
9 y3
26.25
2
y 2 2450 y
2450 26.25 y 27 y 2
74
4
4
9 142
0
9 73
y 3 dy
72 2450 7
26.25
2
91.32 in-oz
30. We fill the pipe and the tank. To find the work required to fill the tank note that radius = 10 ft, then
V
100 y ft 3 . The force required will be F = 62.4 V = 62.4 100 y = 6240 y lb. The distance
through which F must act is y so the work done lifting the slab is about W1 6240 y y lb ft. The work it
takes to lift all the water into the tank is: W1
385
385
W1
360
385
with W1
360
6240 y dy
385
y2
2
6240
y
y lb ft. Taking the limit we end up
[3852 3602 ] 182,557,949 ft-lb
6240
2
360
6240
360
To find the work required to fill the pipe, do as above, but take the radius to be
y ft 3 and F
1
36
V
360
W2
W2
0
The total work is W
6,370,000
y. Also take different limits of summation and integration:
y dy
62.4
36
y2
2
0
W
3602
2
62.4
36
0
r
2
4
110,855 sec
dr 1000 MG
1
352,864 ft-lb
6.672 10
11
31 hr
35,780,000 dr
6,370,000 r
1
6,370,000
0 (23 10 29 )
d
1 ( 1)2
F( ) d
1 35,780,000
r 6,370,000
1000 MG
2
5.144 1010 J
1
35,780,000
23 10 29
1
0
1) 2
(
(23 10
1
29
)1
1
2
11.5 10
29
W1 W2 where W1 is the work done against the field of the first electron and W2 is the work done
be the x-coordinate of the third electron. Then r12
against the field of the second electron. Let
and r22
(
5 23 10
r22
29
3
23
12
29
d
d
5 23 10 29
3 ( 1) 2
5 23 10 29
3 ( 1)2
d
d
23 10
29
23 10 29
1
1
5
1 3
5
1 3
( 23 10
29
)
( 23 10 29 ) 16
1
4
1
4
10 29.
Therefore W
(
1)2
5 23 10
3
r12
W1
W2
360
be the x-coordinate of the second electron. Then r 2
32. (a) Let
(b) W
62.4
36
182,910,813
1650
W
1650
35,780,000 1000 MG
(1000) 5.975 10
ft. Then
W1 W2 182,557,949 352,864 182,910,813 ft-lb. The time it takes to fill the tank
and the pipe is Time
31. Work
1
6
in.
V
62.4
360 62.4
36
0
W2
4
2
W1 W2
23
4
10 29
Copyright
23
12
10 29
23
3
10 29
7.67 10 29 J
2014 Pearson Education, Inc.
1
2
23
4
23 10
12
10
29
29
, and
(3 2)
1) 2
476
Chapter 6 Applications of Definite Integrals
33. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate s right-hand
edge: y x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then x 5 y and
the total width is L ( y )
2(5 y ). The depth of the strip is ( y ). The force exerted by the water against
2x
2
one side of the plate is therefore F
2
124.8
y 2 dy 124.8
5y
5
5
117
(124.8) 105
2
3
y2
5
2
2
w( y ) L( y ) dy
1 y3
3
(124.8) 315 6 234
2
5
5
2
124.8
5
2x
0
F
3
( 124.8)
0
9
2
18
3
9
27
2
( 124.8)
0
62.4(10 y )(4) dy
249.6
5
3
0
(10 y ) dy
25 y 2
u
5
du
1 25 u1/2 du
2 0
1
3
4
0
6
(10 y )dy 187.2 10 y
2 y dy; y
25
0.
125 .
3
0
5
5
0
3
y2
2
u
25, y
Thus, 124.8
y3
3
0
3
5
u
a
strip
w depth
F ( y ) dy
b
F
a
6364.8 lb
strip
w depth
F ( y ) dy
4
187.2(40 8)
5990.4 lb
0
5
0
b
F
a
strip
w depth
F ( y ) dy
6 25 y 2 dy
5
0
y 25 y 2 dy
25 y 2 dy as the area of a quarter circle whose
6 14 (5)2
5
0
5
0
b
F
249.6 30 92
0
(6 y ) 25 y 2 dy 124.8
25 y 2 dy
0
y2
2
y 2 dy 124.8 6 y
y
2 25 y 2 , the depth of the strip is (6 y )
6 25 y 2 dy
u 3/2
y2
2
249.6 10 y
To evaluate the first integral, we use we can interpret
0
3
6
1684.8 lb
62.4 (6 y ) 2 25 y 2 dy 124.8
radius is 5, thus
0
3, the depth of the strip is (10 y )
62.4(10 y )(3) dy 187.2
36. The width of the strip is L( y )
0
y 3. Thus the total width is
x
4, the depth of the strip is (10 y )
(b) The width of the strip is L( y )
4
x 3
62.4 (2 y ) 2(3 y ) dy 124.8
35. (a) The width of the strip is L( y)
0
25 13 125
2( y 3). The depth of the strip is (2 y ). The force exerted by the water is
w(2 y ) L( y ) dy
3
5
2
4 13 8
1684.8 lb
34. An equation for the line of the plate s right-hand edge is y
L( y )
62.4 ( y ) 2(5 y ) dy
75
2
0, thus
5
0
5
6 25 y 2 dy
0
. To evaluate the second integral let
y 25 y 2 dy
y 25 y 2 dy
1 0
2 25
u du
124.8 752
125
3
9502.7 lb.
37. Using the coordinate system of Exercise 32, we find the equation for the line of the plate s right-hand edge to
be y
(a)
2x 4
F
0
4
x
y 4
2
w(1 y ) L ( y ) dy
( 62.4) ( 4)(4)
(b) F
and L( y )
0
4
(3)(16)
2
( 64.0) ( 4)(4)
2x
62.4 (1 y )( y 4) dy
64
3
(3)(16)
2
y 4. The depth of the strip is (1 y ).
62.4
( 62.4)( 16 24 64
)
3
64
3
Copyright
( 64.0)( 120 64)
3
0
4
4 3y
y 2 dy
( 62.4)( 120 64)
3
1194.7 lb
2014 Pearson Education, Inc.
62.4 4 y
1164.8 lb
3 y2
2
y3
3
0
4
Section 6.5 Work and Fluid Forces
38. Using the coordinate system given, we find an equation for the
line of the plate s right-hand edge to be y
and L( y)
2x
1
F
0
4 y
2
x
4 y. The depth of the strip is (1 y )
w (1 y )(4 y ) dy
y3
2x 4
1
62.4 3
5 y2
2
(62.4)(11)
6
114.4 lb
1
62.4
5
2
(62.4) 13
4y
0
y 2 5 y 4 dy
0
4
(62.4) 2 156 24
39. Using the coordinate system given in the accompanying figure,
we see that the total width is L ( y ) 63 and the depth of the strip
is (33.5 y )
33 64
0 123
64
123
F
33
0
w(33.5 y ) L( y ) dy
64
123
(33.5 y ) 63 dy
(63) 33.5 y
(64)(63)(33)(67 33)
(2)(123 )
y2
2
33
(63)
(33.5 y ) dy
2
(33.5)(33) 332
64 63
123
0
33
0
1309 lb
40. Using the coordinate system given in the accompanying figure,
1 y 2 so the total width
we see that the right-hand edge is x
is L( y ) 2 x 2 1 y 2 and the depth of the strip is ( y ). The
force exerted by the water is therefore
0
F
1
62.4
w ( y ) 2 1 y 2 dy
0
1
2
3
(62.4)
41. (a)
1 y 2 ( 2 y ) dy
(1 0)
3/2 0
62.4 23 1 y 2
41.6 lb
62.4 lb3 (8 ft) 25 ft 2
F
12480 lb
ft
(b) The width of the strip is L( y )
5
0
62.4(8 y )(5) dy
312
5
0
(c) The width of the strip is L ( y )
F
b
a
strip
w depth
F ( y ) dy
312 2 8 y
1
y2
2
5/ 2
0
5, the depth of the strip is (8 y )
(8 y ) dy
312 8 y
y2
2
5
0
F
312 40 25
2
b
a
strip
w depth
F ( y ) dy
8580 lb
5, the depth of the strip is (8 y ), the height of the strip is
5/ 2
0
62.4 (8 y )(5) 2 dy
312 2
5/ 2
0
312 2 40
25
4
Copyright
2014 Pearson Education, Inc.
2
9722.3
(8 y ) dy
2 dy
477
478
Chapter 6 Applications of Definite Integrals
3
4
42. The width of the strip is L( y )
b
F
a
93.6
3
3
2 3
strip
w depth
F ( y ) dy
12 y 3 3 y 2
2 3 y , the depth of the strip is (6 y ), the height of the strip is 2 dy
0
y3
3
y2 3
62.4(6 y ) 43 2 3
2 3
93.6
3
0
93.6 2 3
3 0
y 2 dy
3
72 36 12 3 8 3
2x
y and the total width is
2 y.
1
(a) The depth of the strip is (2 y ) so the force exerted by the liquid on the gate is F
1
0
y 2 dy
1571.04 lb
43. The coordinate system is given in the text. The right-hand edge is x
L( y )
12 3 6 y 2 y 3
1
1
0
0
50(2 y ) 2 y dy 100 (2 y ) y dy 100
4
3
100
100
15
2
5
(20 6)
(b) We need to solve 160
1
0
2 y1/2
0
w(2 y ) L( y ) dy
1
2 y 5/2
5
0
y 3/2 dy 100 43 y 3/2
93.33 lb
y ) 2 y dy for h. 160 100 23H
w( H
2
5
H
3 ft.
44. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for
x 52 y. The total width is L( y ) 2 x 45 y and the
the line of the end plate s right-hand edge is y 52 x
h
depth of the typical horizontal strip at level y is ( h y ). Then the force is F
where Fmax
6667 lb. Hence, Fmax
3
3
h
2
(62.4) 54
h
3
(62.4) 54
1
6
w
h
(h y ) 54 y dy
0
h3
(10.4) 45 h3
volume of water which the tank can hold is V
1
2
2h
5
(Base)
2 h2
5
V
(30) 12h 2
45. The pressure at level y is p( y)
1 b
b 0
pressure is p
2
1w y
b
2
b
w
b
0
p( y ) dy
b2
2
wb .
2
w y
1 bw
b 0
1
2
(62.4) 54
h
3 5
4
h
0
hy
Fmax
10.4
w(h
y ) L( y) dy
y 2 dy
(62.4) 54
0
3 5
4
6667
10.4
(Base)(Height) 30, where Height
12(9.288)2
Fmax ,
hy 2
2
y3
3
9.288 ft. The
h and
1035ft 3 .
the average
y dy
This is the pressure at
level b2 , which is the pressure at the middle of the
plate.
46. The force exerted by the fluid is F
w
ab 2
2
wb
2
(ab)
b
0
w(depth)(length) dy
b
0
w y a dy
(w a)
p Area, where p is the average value of the pressure.
Copyright
2014 Pearson Education, Inc.
b
0
y dy
y2
(w a) 2
b
0
h
0
Section 6.6 Moments and Centers of Mass
0
47. When the water reaches the top of the tank the force on the movable side is
(62.4)
0
2
1/2
y2
4
y2
(62.4) 23 4
( 2 y ) dy
3/2 0
2
(62.4) 23 43/2
2
479
(62.4) 2 4 y 2 ( y )dy
332.8 ft-lb. The force
x 3.33 ft. Therefore
compressing the spring is F 100 x so when the tank is full we have 332.8 100 x
the tank will overflow.
the movable end does not reach the required 5 ft to allow drainage
48. (a) Using the given coordinate system we see that the total
width L( y ) 3 and the depth of the strip is (3 y ).
3
Thus, F
0
(62.4)(3)
3
0
3
w(3 y )L( y ) dy
0
(62.4)(3 y ) 3 dy
(3 y ) dy (62.4)(3) 3 y
9
2
(62.4)(3) 9
(62.4)(3)
9
2
y2
2
3
0
842.4 lb
(b) Find a new water level Y such that FY
(0.75)(842.4 lb)
and Y is the new upper limit of integration. Thus, FY
(62.4)(3)
0
(Y
2 FY
(62.4)(3)
Y
6.6
Y
y ) dy (62.4)(3) Yy
1263.6
187.2
6.75
y2
2
Y
631.8 lb. The new depth of the strip is (Y
Y
w(Y
0
y )L( y ) dy
Y2
2
(62.4)(3) Y 2
0
2.598 ft. So,
Y
3 Y
62.4
Y
0
(Y
y)
y ) 3 dy
2
(62.4)(3) Y2 . Therefore,
3 2.598
0.402 ft
4.8 in
MOMENTS AND CENTERS OF MASS
1. Since the plate is symmetric about the y -axis and its density is
constant, the distribution of mass is symmetric about the y -axis
and the center of mass lies on the y -axis. This means that x
It remains to find y
Mx
M
0.
. We model the distribution of mass
with vertical strips. The typical strip has center of mass:
2
x, x 2 4 , length: 4 x 2 width: dx,
( x, y )
area: dA
4 x 2 dx, mass: dm
4 x 2 dx
dA
The moment of the strip about the x-axis is y dm
plate about the x-axis is M x
2
2
32 32
5
Therefore y
128
5
Mx
M
y dm
2
16 x 4 dx
22
. The mass of the plate is M
128
5
32
3
12
5
x2 4
2
4 x 2 dx
16 x
2
(4 x 2 ) dx
x5
5
2
2
2
2
4x
x3
3
The plate s center of mass is the point x , y
Copyright
16 x 4 dx. The moment of the
2014 Pearson Education, Inc.
5
16 2 25
2
2
2
0, 12
.
5
8 83
5
16 2 25
32
3
.
480
Chapter 6 Applications of Definite Integrals
2. Applying the symmetry argument analogous to the one
, we use the
in Exercise 1, we find x 0. To find y Mx
M
vertical strips technique. The typical strip has center of
2
x, 25 2 x , length: 25 x 2 , width: dx,
mass: ( x, y )
25 x 2 dx, mass: dm
area: dA
25 x 2 dx.
dA
The moment of the strip about the x-axis is
25 x 2
2
y dm
25 x 2 dx
2
25 x 2
2
dx.
The moment of the plate about the x -axis is M x
is M
5
dm
5
x5
5
625 x 50
x3
3
2
5
25 x 2 dx
5
55
5
2 2 625 5 50
53
3
x3
3
25 x
The plate s center of mass is the point ( x , y )
5
25 x 2
52
625 5 10
1
3
2
5
2
5
y dm
53
3
53
4
3
5
dx
2
5
625 50 x 2
x 4 dx
625 83 . The mass of the plate
53. Therefore y
Mx
M
54
53
8
3
4
3
10.
(0, 10).
3. Intersection points: x x 2
x 2 x x2 0
x(2 x) 0
x 0 or x 2. The typical vertical
strip has center of mass: ( x, y )
2
x, x2 , length: x x 2
x,
x x2
( x)
2
2 x x2 ,
( x)
2 x x 2 dx, mass: dm
width: dx, area: dA
dA
x2
2
2 x x 2 dx. The moment of the strip about the x-axis is y dm
it is x dm
2
2
0
2
0
4
5
(2 x x 2 ) dx. Thus, M x
x
2
x5
5 0
x4
2
2 x2
2
x3 dx
2 x x 2 dx
3
4
3
5
25
5
23
2 x3
3
2
2
x4
4 0
2
x3
3 0
x2
(x, y)
2
0 2
y dm
23 1 45
2
3
4
5
24
4
23
4 83
4
3
x2
2 x x 2 dx
; My
24
12
x dm
4
3
; M
. Therefore, x
2
0
x,
2 x2
x2 3
2
My
M
1, 53 is the center of mass.
2
x, x2 3 , length: 2 x 2
x2 3
3 1 x 2 , width: dx, area: dA 3 1 x 2 dx,
Copyright
x
dm
2 x2
3x2 3 0
4. Intersection points: x 2 3
3( x 1)( x 1) 0 x
1. or x 1 Applying the
symmetry argument analogous to the one in Exercise 1,
we find x 0 The typical vertical strip has center of mass:
( x, y )
2 x x 2 dx; about the y -axis
2014 Pearson Education, Inc.
4
3
2
2 0
2 x3
x 4 dx
2 x x 2 dx
2
0
3
4
2 x x 2 dx
1 and y
Mx
M
Section 6.6 Moments and Centers of Mass
mass: dm
x 2 3 1 x 2 dx
3
2
Mx
y dm
M
1 x 2 dx. The moment of the strip about the x-axis is
dA 3
y dm
dm
1
3
2
x4
1
1
3
x 4 3x 2
3
2
2 x 2 3 dx
1 x 2 dx
1
3
x5
5
3
2
1
x3
3
x
x 2 3 dx
2 x3
3
y 3 , width: dy, area: dA
length: y
mass: dm
dA
y2
y y3
2
2 y4
2
3
3
3 10 45
15
3
Mx
M
4 . Therefore, y
5
32
4
32
5
;
8
5
3
2y
5
y
3
1
y2
0
5
1
4
,
y3 dy,
y
y 3 dy
y
y3
y
2
2
dy
y 6 dy; the moment about the x-axis is y dm
y dm
1
2
2 15
y 3 dy. The moment of the strip about the
y
y -axis is x dm
2
2 x 2 3 dx;
3
2
1
y y3
2
5. The typical horizontal strip has center of mass: ( x, y )
Mx
1
3x
2 1 13
3
1
x4
3
2
0, 85 is the center of mass.
(x , y )
2
481
7
y
7
1
1
2 3
0
2
5
1
y5
5
1
3
0
35 42 15
2
35 7
1
7
My
M
. Therefore, x
4
y3
3
y 4 dy
4
105
4
2y
0
4 ;
105
16
105
2
15
1
5
y 3 dy
y y
; My
M
and y
1
2 0
x dm
1
dm
Mx
M
y2
0
2
15
y 4 dy. Thus,
y2
y2
2
y )3 dy
(y
8
15
4
2 y4
y 6 dy
y4
4
1
0
16 , 8
105 15
(x, y)
is the
center of mass.
y2
6. Intersection points: y
y ( y 2)
0
y
y2
y
0 or y
2 The typical horizontal
y2 y
strip has center of mass: ( x, y )
2
length: y
y2
y
area: dA
2y
y 2 dy, mass: dm
16
3
2
x
y 2y
16
4
16
12
40 32
5
My
M
4
5
4
5
y 2 dy
(4 3)
; M
3
4
2 y2
4
3
,y ,
and y
2
0
Mx
M
2y
y 2 dy.
y2 2 y
y 2 dy
dA
2
y 3 dy. Thus, M x
; My
dm
3
5
y2
2
,y
y 2 , width: dy,
2y
The moment about the y -axis is x dm
is y dm
y
2
x dm
0 2
y 2 dy
2y
4
3
Copyright
3
4
2 y3
2 y3
0
y 4 dy
3
y
3
(x , y )
y 4 dy; the moment about the x-axis
2
y dm
y2
1
2
2
2
2 y2
y4
2
4 83
0
3,1
5
2 y3
3
y 3 dy
y5
5
4
3
2
0
2
8 32
5
. Therefore,
is the center of mass.
2014 Pearson Education, Inc.
y4
4
2
0
482
Chapter 6 Applications of Definite Integrals
7. Applying the symmetry argument analogous to the one used in
Exercise 1, we find x 0. The typical vertical strip has center
cos x
of mass: ( x, y )
x, 2
, length: cos x, width: dx,
cos x dx, mass: dm
area: dA
dA
cos x dx. The moment
cos x
2
of the strip about the x-axis is y dm
2
cos 2 x dx
Mx
1 cos 2 x
2
2
/2
y dm
/2
/2
/2 4
cos x dx
dx
4
(1 cos 2 x) dx; thus,
(1 cos 2 x) dx
/2
/2
sin x
cos x dx
4
/2
sin 2 x
2
x
4
/2
Mx
M
2 . Therefore, y
2
42
0
6
4
2
(x, y)
8
; M
dm
0, 8 is the center of mass.
8. Applying the symmetry argument analogous to the one used in
Exercise 1, we find x 0. The typical vertical strip has center
of mass: ( x, y )
x,
sec2 x
2
, length: sec2 x , width: dx,
area: dA sec 2 x dx, mass: dm
sec2 x
2
moment about the x-axis is y dm
2
/4
2
/4
1
3
Therefore, y
2
9. M y
M
y
1
2
x
1
x
1
1 2x
21
dx
1 x
Mx
M
2
tan x
1
2 3
Mx
/4
sec4 x dx. M x
1
4
ln 2
2
/4
1 ( 1)
Mx
M
4
3
/4
y dm
2
/4
/4
sec2 x dx
sec 2 x dx. The
dA
2
/4
1
2
2
3
/4
sec4 x dx
; M
2
0,
/4
2
3
/4
/4
dm
(x, y )
tan 2 x 1 sec2 x dx
/4
(tan x )3
2
3
sec2 x dx
4
3
3
( sec2 x ) dx
/4
2
tan x
sec 2 x dx
is the center of mass.
dx 1,
1
x
1 2 1
2 1 x2
dx
ln x
2
1
ln 2
dx
X
2
1
2x 1
My
M
1,
4
1
ln 2
1.44 and
0.36
Copyright
2014 Pearson Education, Inc.
/4
/4
tan x
/4
/4
1 ( 1)
2 .
Section 6.6 Moments and Centers of Mass
483
10. (a) Since the plate is symmetric about the line x y and its
density is constant, the distribution of mass is symmetric
about this line. This means that x y The typical vertical
2
x, 9 2 x ,
strip has center of mass: ( x, y )
9 x 2 width: dx, area: dA
length:
mass: dm
9 x 2 dx. The moment about the
dA
9 x2
2
x-axis is y dm
Thus, M x
9 x 2 dx
3
9 x 2 dx
0 2
y dm
9 x 2 dx,
3
x3
3 0
9
4
9x
2
(Area of a quarter of a circle of radius 3)
4, 4
(x, y)
9 x 2 dx
2
2
(27 9)
9
9 ; M
. Therefore, y
4
dm
Mx
M
dA
(9 ) 9 4
dA
4
is the center of mass.
(b) Applying the symmetry argument analogous to the
one used in Exercise 1, we find that x 0. The
typical vertical strip has the same parameters as in
part (a). Thus, M x
2
3
y dm
9 x 2 dx
0 2
dA
3
9 x 2 dx
32
2(9 ) 18 ; M
dm
(Area of a semi-circle of radius 3)
same y as in part (a)
dA
9
2
9
2
. Therefore, y
0, 4 is the center of mass.
(x, y)
11. Since the plate is symmetric about the x-axis and its density is
constant, the distribution of mass is symmetric about this line.
This means that y 0. The typical vertical strip has center of
( x, 0), length:
mass: ( y, y )
width: dx, area: dA
2
1 x2
1
1 x2
1
1 x2
dx, mass: dm
,
dA
2
1 x2
dx.
The moment about the y-axis is
x dm
x
2
1 x2
My
12 x
0 1 x2
M
dm
x
My
M
dx
1
0
ln 2
/2
2x
1 x2
dx
dx
[ln(1 x 2 )]10
2
1 x2
2 ln 2
dx
ln 4
2 x
1 x2
dx. Thus,
ln 2.
2 [arctan x ]10
(x, y)
2 (arctan1)
ln 4 ,
Copyright
2
4
2
. Therefore,
0 is the center of mass.
2014 Pearson Education, Inc.
Mx
M
(18 ) 9 2
4
, the
484
Chapter 6 Applications of Definite Integrals
12. Since the plate is symmetric about the line x 1 and its density
is constant, the distribution of mass is symmetric about this line
and the center of mass lies on it. This means that x 1. The
typical vertical strip has center of mass:
( x, y )
2 x x2
x,
2 x2 4x
2
x, x 2 2 x ,
2
length: 2 x x 2
2 x2
3x 2 6 x
4x
3 2 x x2
width: dx, area: dA 3 2 x x 2 dx, mass: dm
dA
2 x x 2 dx. The moment about the x-axis is
3
x2
3
2
y dm
23
0 2
3
2
x4
8
5
8
5
2
5
16
13. M y
1
1
4
x5
5
3
2
dx [2 x1/2 ]16
1
x4
2
5
2 [ x3/2 ]16
1
3
dx
6
3
2
x4
2
4 x3
3
0
3
2
25
5
3
4 x3 4 x 2 dx. Thus, M x
24
4
3
23
2
x3
3 0
x2
24 52 1 23
3
2
4 83
3
y dm
4 . Therefore,
is the center of mass.
16
42; M x
My
M
x
dx
2 x x 2 dx
3
0
1,
2
2x
2
dm
(x, y)
x
1
x
x2
3
2
; M
16 1/2
x 1 dx
16 1
1
x
M
4 x3 4 x 2 dx
10
24 6 15
15
Mx
M
y
2 x 2 x x 2 dx
1
Mx
M
7 and y
1
2 x
1
x
1 16 1 dx
2 1 x
dx
1
2
16
1
ln x
ln 4,
ln 4
6
14. Applying the symmetry argument analogous to the one used in Exercise
1, we find that y 0. The typical vertical strip has center of mass:
1
( x, y )
x3
x,
2
x3
area: dA
My
2
a2
15. M x
My
2
2
x x2
2
1
dx
My
M
x2
2
2
x2
2
x2
2
1
x dm
1
a
1
x 1
2
x 2 dx
1
2
1
dx
2
2 x1
2
x2
2
2
1
width: dx ,
2 ( a 1)
;
a
1
a2
( a 2 1)
2a
a 1
2 1
x2
x2
2
x2
2
1
2
( 1)
1
1
M
a2
dm
(x, y )
1 x
3
2a , 0
a 1
x 2 3 dx
x
2
x2
a
1
2
x 1
dx
. Also, lim x
a
dx. Thus,
1
a2
1
2.
dx
2 12
1;
dx
x dx
2(2 1)
1
a
2
2
x 1
2
x
dx
2 ,
x3
dx. The moment about the y -axis is x dm
2 ( a 1)
a
2
2
2
x3
dA
dx 2
. Therefore, x
dx
1
x3
x
a2
1 x2
y dm
2 2
1 x2
( x, 0), length: 13
dx, mass: dm
x dm
( a 2 1)
1
x3
2
2 x2
2. So x
2
2 2 12
1
My
M
Copyright
3
2
4 1 3; M
dm
Mx
M
(x , y )
and y
1
2
2014 Pearson Education, Inc.
2
1
2
x2
3, 1
2 2
dx
2 2
1
x
2
x2
dx
is the center of mass.
Section 6.6 Moments and Centers of Mass
485
16. We use the vertical strip approach:
Mx
1 1
2 0
x2
My
y
x x2
x 4 12 x dx
1
x6
6 0
4
6 x4
M
2
1 x x
2
0
y dm
6 14
1
x dm
0
1
dm
0
Mx
M
0
6
4
1
6
x x x2
x x2
x3
x5 dx
1
1;
2
1
dx
dx 12
3, 1
5 2
1
2
1
6
dx
0
1
0
1
x3 12 x dx 12
0
x3 dx 12 x3
x 4 dx 12 x4
12 13
1
4
b
a
shell
radius
2
4
2 x3/2
3
1
shell
height
4
dx
2 x 4
x
(b) Since the plate is symmetric about the x-axis and its density
( x)
x
dx 16
2
3
16
32
3
1
224
3
1
12
12
1. So x
4
x
dx 16
8 23
1
x
1
5
My
M
3
5
12
20
and
approach to find x : M y
8(2 2 2) 16; M
4
x dm
dm
4
x
1
My
M
8. So x
1
4
4
x
x
16
8
4
x
4
x
2
dx
dx
(x , y)
4
8
1
x
1
8
x
1
x
dx
0. We use the vertical strip
x 8 1x dx
1
4
4 x
x
1
is a function of x alone, the
distribution of its mass is symmetric about the x -axis. This means that y
8 1 ( 2)
12 14
is the center of mass.
(8 1)
16
1
x5
5 0
4
x3
1
x4
4 0
3
x2
17. (a) We use the shell method: V
4 1/2
x2
dx
8
4
1
4
1
x 1/2 dx
x 3/2 dx
8
8 2 x1/2
2x
1/2 4
1
(2, 0) is the center of mass.
(c)
18. (a) We use the disk method: V
b
R ( x)
a
2
4
dx
4
x2
1
dx
4
4
1
x 2 dx
1 4
x 1
4
4
1
4
( 1)
[ 1 4] 3
(b) We model the distribution of mass with vertical strips: M x
2
4
1
x 3/2 dx 2
2 16
3
So x
2
3
My
M
28 ;
3
28
3
4
4
2
x 1
2
1 ( 2)
M
dm
42
1 x
7
3
and y
Mx
M
Copyright
2; M y
dx
2
4
2
1
2
4 x
1 x
4
x dm
dx
(x , y )
1
2
4
y dm
4
1
7, 1
3 2
x 2x
2
x
2
x
1 2
dx
x 1/2 dx
2
dx
4 1/2
x dx
1
2 2 x1/2
4
1
is the center of mass.
2014 Pearson Education, Inc.
4 2
1 x2
x dx
4
3/ 2
2 2 x3
1
2(4 2)
4.
4
1
3;
5
486
Chapter 6 Applications of Definite Integrals
(c)
19. The mass of a horizontal strip is dm
dA
L dy, where L is the width of the triangle at a distance of y above
h y
h
its base on the x -axis as shown in the figure in the text. Also, by similar triangles we have Lb
L bh ( h y ). Thus, M x
bh 2 12
Mx
M
y
1
3
bh 2
6
; M
2
bh
h
3
bh2
6
h
y dm
0
h
dm
y bh ( h
b
h
0
(h
b h
h 0
y ) dy
b h (h
h 0
y ) dy
hy
2
b hy
2
h
y 2 dy
y ) dy
b
h
hy
y2
2
h
0
y3
3
b
h
h
0
h2
b h3
h 2
h2
2
h3
3
bh .
2
So
the center of mass lies above the base of the triangle one-third of the way toward
the opposite vertex. Similarly the other two sides of the triangle can be placed on the x -axis and the same results
will occur. Therefore the centroid does lie at the intersection of the medians, as claimed.
20. From the symmetry about the y -axis it follows that x
0. It also follows
that the line through the points (0, 0) and (0, 3) is a median
1 (3
3
y
0) 1
(x , y )
(0, 1).
21. From the symmetry about the line x
y it follows that x
y . It also
follows that the line through the points (0, 0) and 12 , 12 is a median
y
x
2
3
1
2
0
1
3
(x, y)
22. From the symmetry about the line x
1, 1
3 3
.
y it follows that x
follows that the line through the point (0, 0) and
y
x
2 a
3 2
0
1a
3
(x, y )
a, a
3 3
a, a
2 2
y . It also
is a median
.
23. The point of intersection of the median from the vertex (0, b) to the
opposite side has coordinates 0, a2
x
a
2
0 23 a3
(x, y)
a,b
3 3
y
(b 0) 13
b
3
and
.
Copyright
2014 Pearson Education, Inc.
Section 6.6 Moments and Centers of Mass
a
2
24. From the symmetry about the line x
a.
2
it follows that x
487
It also
follows that the line through the points a2 , 0 and a2 , b is a median
1 (b
3
y
25.
x1/2
y
3/2
2 14
26.
x3
y
0
0
2
3
36 x3dx
du
u 1, x 1
M
sin 2
2
0
3 x 2 dx
2
1 3/2
4
0
2
3
27
8
2
13
6
1
8
1 9 x 4 dx;
u 10]
a 2k
2
0
ak sin d
10 1 1/2
u du
1 36
Mx
0
; My
ak
x3dx;
1 du
36
27. From Example 4 we have M x
a2k
2
x
1 3/2
4
(dx )2
dx
1 41x dx;
2
3
dx
3/2
9
4
( dy )2
1 9 x 4 dx;
x
[u 1 9 x 4
x
1
4
x
0
3 x 2 dx
dy
1 3
Mx
2
1 3/2
4
.
( dx )2
ds
x 1 41x dx
0
2
3
a,b
2 3
(x , y )
x 1/2 dx
1
2
dy
2
Mx
b
3
0)
a (a sin )(k sin ) d
a2k
2ak . Therefore, x
0
My
M
54
sin 2 d
0
a2k
a (a cos )(k sin ) d
0
cos
10
2 u 3/2
36 3
1
0
103/2 1
a2k
2 0
sin cos d
0 and y
Mx
M
(1 cos 2 ) d
a2k
2
sin 2
a2k
2
1
2ak
2a 2
a2k
0
a
4
0;
0, a4
the center of mass.
28. M x
y dm
a2
a2
/2
0
/2
0
a2
a
2
a2
x dm
/2
a2k
sin d
0 ( 1)
0
(a sin )
ad
/2
0
2
a k 12 0
0
/
a 2 sin
(a cos )
(cos )(1 k cos ) d
/2
0
2
a
a2
a2
a2
/2
cos
( 1) 0
ad
/2
sin d
d
a2k
2
a 2 k sin2
2
a k 0 12
a 2 cos
0
/2
1 k cos
(sin )(1 k cos ) d
sin cos d
2
a 2 k sin2
/2
0
0
a2
(sin )(1 k cos ) d
cos
My
0
1 k cos
a
/2
sin cos d
/2
a2 k
2
2
a2
d
(cos )(1 k cos )d
Copyright
2014 Pearson Education, Inc.
a2k
2
a 2 (2 k );
is
488
Chapter 6 Applications of Definite Integrals
/2
a2
0
a 2 sin
a2k
2
a (1 0)
29.
a(
2k ). So x
x 6,
g ( x)
x2
x 6
0
2
3
125
0 and y
x 2 , f ( x)
2 12
6
125
8
3
( x 6)2
9 54 108 243
5
3
6 x 13 x3
x2
x2
1
M
0
2
0
1
1 x3
3
0
1
1
x
17/12 0
12
17
x2
8
3
11
1
17/12 0 2
y
6
17
4 71
1
1 x4
4
0
1
3
22
1
5
2 14
12
17
x 2 ( x 1)
0
698
595
a2k
4
ak
a 2
0;
(1 k cos ) d
k
a
2
k
1
3
2ak
0, 2 a 2ka
is the center of mass.
k
6 1 x3
125 3
3x 2
3
1 x4
4
2
x 2 12 x 36 x 4 dx
4
1,4
2
3 1 x3
125 3
3
6 x 2 36 x 15 x5
is the center of mass.
2 x 2 ( x 1)
x 2 dx
17 ;
12
0
12 1
17 0
1 15
2
a
2
24 72 32
5
2 x3
0
2 x 2 ( x 1) dx
1 x5
5
2
a2
1
1
2 x 2 ( x 1) dx
2 x 14 x 4
x
x 1;
/2
0)
a (2 k )
2k
6 x x3 dx
3 3
125 2
f ( x) 2, g ( x) x 2 ( x 1), f ( x) g ( x)
x3
a
a2k
4
1;
2
dx
3
125
d
x2
x 6
6 3
125 2
2
a2
0
2
a (
a 2 (2 k )
a ( 2k )
1 cos 2
2
sin 2
2
0)
0
/2
1
12 4
x2
a2k
d
(1 k cos ) d
k
2
Mx
M
2;
x( x 6) x 2 dx
9 27 81
4
0
g ( x)
125
6
3 1
1
125/6 2 2
y
My
M
8
3
cos
a2k
2
/2
a
/2
1 x2
2
18 9
6
125
d
k sin
3, x
/2
2
a 2 (0 1) a2k (
(0 0)
( x 6) x 2 dx
3
1
125/6 2
x
0
a
x
a2
d
a 2 sin
1 k cos
0
0
f ( x)
3
0
/2
k sin
9
2
30.
a
a
M
sin 2
2
2
ad
0
/2 1 cos 2
2
0
a2 k
2
0
2
M
a2k
cos d
dx
2 x x4
1
4
0
6 1
17 0
33 , 698
85 595
x3 dx
33 ;
85
4 x6
2 x5
x 4 dx
6
17
4 x 17 x7
is the center of mass.
Copyright
2014 Pearson Education, Inc.
1 x6
3
1
1 x5
5
0
2
Section 6.6 Moments and Centers of Mass
31.
x 2 , g ( x) x 2 ( x 1), f ( x) g ( x)
f ( x)
x2
x 2 ( x 1)
2
M
x2
0
1
4/3 0
0
16
3
2
1 x5
5
0
x2
4
3
4
2
0, x
2 x2
0
2;
1
x3 dx
4;
3
0
3 2
4 0
x 2 ( x 1) dx
x x
1 21
4/3 0 2
x
2
x 2 ( x 1) dx
3 1 x4
4 2
y
x3 2 x 2
2
1 x4
4
0
2
2
2 x3
3
x
489
8 32
5
0
x 2 ( x 1)
2
0, x
0, x
2 x3
x 4 dx
6;
5
3 2
8 0
dx
2 x5
x 6 dx
2
1 x7
7
0
3 1 x6
8 3
3 64
8 3
128
7
8
7
0
6,8
5 7
is the
center of mass.
32.
2 sin x, g ( x)
f ( x)
2
M
2
1
4
x
1
4
1
4
2
0
4
1
8
[u
2 x dx
1
4
2
1
4
2
1
8
0
0
2
0
2
(0 2 ) 0
2
du
2dx, x
2
1
8
4 x 4 cos x 0
center of mass.
1) (0 1)
4 ;
2 x x sin x dx
x sin x dx
4 4 sin x dx 81
0
2x
(4
1 2
4 0
4 4sin x sin 2 x dx
0
2
2 x cos x 0
x 2 sin x 0 dx
0
1;
2
2 sin x dx
0
2 ;
2
1
4
x2
1;
y
2
1
8
0, x
1
16
x0
2
1
32
0
sin x x cos x 0
2 1
2
1
4
0
(2 sin x) 2 (0)2 dx
2
1 cos 2 x
2
u
2
1
4
sin 2 x dx
4 4 sin x dx 81
0
0
0
2
2
1
8
dx
u
2
4 ]
4
1
8
sin u 0
2
1
16
4 x 4 cos x 0
2
1
8
2
dx 161
0
4 x 4 cos x 0
cos 2 x dx
0
2
1
16
4) 81 (0 4) 161 (2 ) 0 0
(8
4
1
32
x0
0
9
8
cos u du
2
2
1, 9
8
is the
33. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we
(dx )2
have that the length of a particular segment is ds
and M
ds. If
is constant, then x
My
M
x ds
ds
(dy )2 . This implies that M x
x ds
length
Mx
M
and y
y ds
strip has center of mass: ( x, y )
mass: dm
2 pa
2
2 pa
dA
a2
x,
a
2
2
a 4x p dx. Thus, M x
x4
16 p 2
dx
5
x
a2 x
2
80 p 2
Copyright
, length: a
y dm
2 pa
2 pa
x2 ,
4p
width: dx, area : dA
2 pa 1
2 pa 2
x2
4p
a
5
a
2 pa
x
2 2 a2 x
80 p 2 0
2014 Pearson Education, Inc.
x2
4p
x ds
y ds
.
length
ds
34. Applying the symmetry argument analogous to the one used in Exercise 1, we find that x
x2
4p
y ds, M y
a
x2
4p
0. The typical vertical
dx,
dx
2a 2 pa
25 p 2 a 2 pa
80 p 2
490
Chapter 6 Applications of Definite Integrals
2a 2
pa 1 16
80
pa
3
pa 808016
2
ax 12x p
0
2 pa
3
ax 12x p
8a
2a 2
2 pa
2 pa
3
8a 2
Mx
M
. So y
2a 2
pa
64
pa 80
2
3
5
8a
2a pa
3
5
pa
8a 2
pa
23 pa pa
12 p
(2 )( y )( L)
(2 )(2) 4 8
32 2
(where
36. The midpoint of the hypotenuse of the triangle is 32 , 3
the line y
equation of the median
3,3
2
is
3 5
2
2
x 32
x 2 3 x 94
4 x 2 12 x 9
dx
pa 1212 4
4a
(2 )( y )( A)
(2 )(2)(8)
32
and the surface
2 x is an
the x-coordinate of the centroid
(2 x 3) 2
5
2
5
4
(2 )(4)(9)
37. The centroid is located at (2, 0)
4
pa 1 12
4a
x2
4p
8 is the length of a side).
y
5 x 2 15 x 9
1 x 2 3 x 2 ( x 2)( x 1) 0
y 2. By the Theorem of Pappus, the volume is V
2 (5 x ) 12 (3)(6)
a
2 pa
2 x contains the centroid. The point
units from the origin
solves the equation
2 pa
dm
a, as claimed
35. The centroid of the square is located at (2, 2). The volume is V
area is S
; M
5
x 1 since the centroid must lie inside the triangle
(distance traveled by the centroid)(area of the region)
72
V
(2 )( x )( A)
(2 )(2)( )
4 2
38. We create the cone by revolving the triangle with
vertices (0, 0), (h, r ) and (h, 0) about the x-axis (see the accompanying
figure).Thus, the cone has height h and
base radius r. By Theorem of Pappus, the lateral surface area swept out by the
h2
r
2
2
hypotenuse L is given by S 2 yL
r2
r r2
h 2 . To
calculate the volume we need the position of the centroid of the triangle.
From the diagram we see that the centroid lies on the
line y
r
2h
4h2 r 2
4h 2
39. S
2 yL
40. S
2
41. V
2 yA
L
4 h2 r 2
2h
x2
y
triangle
( x h) 2
x. The x-coordinate of the centroid solves the equation
r
2h
x
4 a2
2
a
4
3
ab 2
x
r.
3
2 r 2 4h 2
r2
4
9
0
x
2h
3
By the Theorem of Pappus, V
(2 y )( a)
2a
( a)
(2 y )
y
2a ,
2 a2 (
ab
2
y
Copyright
or 43h
2
x
1 hr
2
r
3
and by symmetry x
2h ,
3
0
2)
4b
3
and by symmetry x
0
2014 Pearson Education, Inc.
r
2h
x
r
2
2
1
3
h2
r2
4
since the centroid must lie inside the
1
3
r 2 h.
Section 6.6 Moments and Centers of Mass
42. V
2
A
43. V
2
A
V
a3 (3
3
a2
2
a 34a
2
4)
(2 ) (area of the region) (distance from the centroid to the line y
distance from 0, 34a
to y
4 a 3a
6
2
4 a 3a
6
,
4 a 3a
6
4a
3
x 34 a . The intersection of y
. Thus, the distance from the centroid to the line y
4a
6
2
3a
6
2 (4 a 3a )
6
V
intersection of the two perpendicular lines occurs when x a
2a a
2
a (2 )
.
2
2a
x
. Thus the distance from the centroid to the line y
x
x a is
Therefore, by the Theorem of Pappus the surface area is S
2
1 ab, V
2
1
3
45. If we revolve the region about the y -axis: r
of Pappus:
a 2b
1
3
r
b, h
a
1
3
b2a
2 y
1 ab,
2
A
1 ab
2
2 x
1 ab
2
1
3
V
y
b
3
x
a;
3
a, h
2 a 3 (4 3 )
6
A
2a a
2
2a
2
a
a (2 )
2
2
0
x
x
2a a
2
2
2a
( a)
a b, and
2
2a .
a
2 a 2 (2
The
2a 2
2
).
x . By the Theorem
If we revolve the region about the x-axis:
b 2 a, and
a,b
3 3
b
x 34a is
x a and y
x a is
a2
2
2 (4 a 3a )
6
(2 )
x a has slope
x a and passing through the centroid 0, 2 a has equation y
44. The line perpendicular to y
y
x a ). We must find the
x a. The line containing the centroid and perpendicular to y
1 and contains the point 0, 34a . This line is y
the point
491
y . By the Theorem of Pappus:
is the center
of mass.
46. Let O(0, 0), P(a, c), and Q(a, b) be the vertices of the given triangle. If we revolve the region about the
x-axis: Let R be the point R (a, 0). The volume is given by the volume of the outer cone, radius RP c,
minus the volume of the inner cone, radius
RQ
1
3
b, thus V
given by the area of triangle OPR minus area of triangle OQR, A
the Theorem of Pappus:
1
3
a c 2 b2
2 y
1 a (c
2
b)
c 2 a 13 b 2 a
1 ac
2
c b;
3
y
1 ab
2
1
3
a c 2 b 2 , the area is
1 a (c
2
b), and
y . By
If we revolve the region about the
y -axis: Let S and T be the points S (0, c) and T (0, b), respectively. Then the volume is the volume of the
cylinder with radius OR a and height RP c, minus the sum of the volumes of the cone with radius
SP a and height OS c and the portion of the cylinder with height OT b and radius TQ a
with a cone of height OT b and radius TQ a removed. Thus
V
a2c
1
3
a 2c
same as before, A
2
3
a 2 (a b)
2 x
a 2b 13 a 2b
1 ac
2
1 a (c
2
1 ab
2
b)
2
3
1 a (c
2
x
a 2 c 23 a 2 b
b), and
2a ( a b )
3(c b )
Copyright
2
3
a 2 (a b). The area of the triangle is the
x . By the Theorem of Pappus:
2a (a b) c b
, 2
3(c b)
is the center of mass.
2014 Pearson Education, Inc.
492
Chapter 6 Applications of Definite Integrals
CHAPTER 6
1.
A( x)
4
b
V
a
x2
2
4
2.
1 (side) 2
2
3
4
4x 4x x
b
a
32 3
4
A( x)
3 4
4 0
4
x3
3 0
8 x5/2
5
4
7
1
5
2 x
0, b
x
4
4 x 4 x3/2
3
4
4
x 2 dx
8 32
5
32
/4
a
a
cos 52x
2
A( x) dx
A( x)
4
4
4 0
5 /4
x cos22 x
/4
6
6
0
2
0
6
x
36 24 6 x 36 x 4 6 x3/2
(diameter)2
(35 40 14)
2
2
x
6 6 6 18 62
4 x x5/2
(1 sin 2 x);
A( x) dx
cos 2
2
4
(edge)2
216 16
8
35
b
V
(1 sin 2 x) dx
5
4
b
64
3
2
3
5
4
,b
A( x)
2
8 3
(15 24 10) 8153
15
(diameter)2 4 (2sin x 2 cos x) 2
4
2
2
8
5
1
5 /4
5.
3
4
3
x 4 dx
4 sin x 2sin x cos x cos x
4
4.
1
4 2
x2 ; a
A( x )dx
2x2
3
4
sin
0, b 1
x 2 x5/2
x
5 0
9
280
(35 40 14)
A( x)
a
1
4 0
5 1
A( x)dx
2
x2
x
4
x4 ; a
4 x 7/2
7
V
3.
(diameter)2
x 2 x x2
4
4 70
PRACTICE EXERCISES
4
x4
16
dx
8
5
2 x
4
6 6 62
x2
4
2x2
2
4
2 x 7/2
7
63
3
4
36 24 6 x 36 x 4 6 x3/2
x 2 dx
1728
5
216 576 648
4 x x5/2
4
x5
5 16 0
x4
16
4
2 x3/2
3
36 x 24 6
; a
0, b
32 32
8
7
2
5
32
72
35
Copyright
18 x 2 4 6
72 360
4
2014 Pearson Education, Inc.
V
32
4
x2 ; a
b
a
1
1728
5
2 x5/2
5
1800 1728
5
A( x) dx
8
7
0, b
2
5
6
V
6
x3
3 0
72
5
Chapter 6 Practice Exercises
6.
1 (edge) 2 sin
2
3
A( x)
3
4
2
4 x
b
V
a
3
4
4 3x; a
1
A( x) dx
0
2 x
493
2
2 x
0, b 1
1
2 3x 2
4 3x dx
2 3
0
7. (a) disk method:
b
V
R ( x)
a
1
1
2
9 x8 dx
2
1
dx
3x 4 dx
1
x
9 1
2
1
(b) shell method:
b
V
a
shell
radius
2
shell
height
1
dx
0
2 x 3x 4 dx
1
3 x5 dx
2
2
0
3
1
x6
6 0
Note: The lower limit of integration is 0 rather than 1.
(c) shell method:
b
V
a
shell
radius
2
shell
height
dx
1
2
1
(1 x) 3 x 4 dx
2
3 x3
5
b
2
r ( x)
1
x6
2
1
2
3
5
3
5
1
9 9 1 x4
1
2
12
5
1
2
(d) washer method:
3, r ( x ) 3 3 x 4
R( x)
1
9
1
1
1 2 x4
3 1 x4
x8
V
1
9
dx
1
R( x)
a
2x4
x8 dx
2
2 x5
5
9
dx
1
1
x9
9
2
5
18
1
1
9
2
2 13
5
dx
26
5
8. (a) washer method:
R( x)
4
x3
1
2
, r ( x)
16
5 32
16
5
1
2
b
V
a
1
4
R ( x)
2
1
10
1
2
r ( x)
16
5
2
1
4
2
dx
20
4
x3
1
2
1 2
2
16
5
dx
57
20
( 2 10 64 5)
(b) shell method:
2
V
2
1
x 43
x
1
2
dx
2
4x
2
x2
4 1
2
(2 x)
4
x3
1
4
2
1
4
1
4
2
5
4
5
2
(c) shell method:
2
V
2
b shell
a radius
4
x2
4
x
shell
height
x
2
x2
4 1
dx
2
2
2
1
( 1 2 2 1)
Copyright
1
2
dx
2
4 4 1
2 8
1 x3
1
4
4
x2
3
2
2014 Pearson Education, Inc.
1
x
2
dx
x 5
2
x
4 1
494
Chapter 6 Applications of Definite Integrals
(d) washer method:
b
V
2
R( x )
a
7 2
2
2
1
49
4
16
49
4
16
49
4
16
r ( x)
2
1
2
4
x3
4
x 6 dx
2
x 5
5 1
x x 2
1
4
2
dx
dx
3
1 2x
2
1
5
49
4
16
( x 1) dx
x2
2
x
2
r ( y)
1
5 32
1 1
1
4
1
160
1
5
49
25
2
5
1
2
16
(40 1 32)
103
20
9. (a) disk method:
2
5
V
5
x 1 dx
1
1
5
1
24
2
1
4
(b) washer method:
R( y )
y2 1
5, r ( y )
2
2
2
d
V
R( y )
c
2
25 y 4 2 y 2 1 dy
32
5
24 2
2
3
8
32
dy
24 y 4 2 y 2 dy
2
3
2
2
5
32
15
1
3
(45 6 5)
2
2
24 y
25
y5
5
1088
15
(c) disk method:
R( y)
y2 1
5
d
V
2
2
2
R( y ) dy
c
2
4 y2
16 8 y 2
64
3
2
y 4 dy
2
32
32
5
64
64
15
(15 10 3)
512
15
2
4 y2
8 y3
3
16 y
2
3
1
dy
y5
5
2
2
1
5
10. (a) shell method:
d
V
c
4
0
2
2
shell
radius
2 y y
y3
3
y4
16
y2
4
shell
height
dy
2
2
64
3
4
0
dy
4
0
64
4
y3
4
y2
2
12
Copyright
dy
64
32
3
2014 Pearson Education, Inc.
y2 1
2 y3
3
2
2
2
dy
8
49
4
71
10
Chapter 6 Practice Exercises
495
(b) shell method:
b
V
a
4
5
2
shell
radius
2
shell
height
64
3
32
4
dx
2 x 2 x
0
4
2
x dx
2 x3/2
0
x 2 dx
4
x3
3 0
4 x5/2
5
2
128
15
(c) shell method:
b
V
shell
radius
2
a
16 x3/2
3
2
shell
height
2x2
4
dx
2 (4 x ) 2 x
0
4
x3
3 0
4 x5/2
5
16
3
2
4
5
8 32
4
2
x dx
8 x1/2 4 x 2 x3/2
0
64
3
32
4
3
64
4
5
1
x 2 dx
2
3
64
64
5
4
5
1
(d) shell method:
d
V
c
4
2
shell
radius
2
0
shell
height
y3
4
4 y 2 y2
4
dy
0
2 y2
2
dy
y2
4
2 (4 y ) y
2
2
32
0
4
y4
16
2 y3
3
4
dy
2
3
0
y2
4y
y3
4
y2
64 16
32
dy
2
8
3
32
3
1
11. disk method:
R( x)
tan x, a
0, b
/3
V
3
0
/3
tan 2 x dx
sec 2 x 1 dx
0
3 3
/3
tan x x
0
3
12. disk method:
V
0
(2 sin x)2 dx
4 x 4cos x
x
2
4 4sin x sin 2 x dx
0
sin 2 x
4
0
4
4
2
0
1 cos 2 x
2
4 4sin x
0
9
2
(0 4 0 0)
dx
8
(9
2
16)
13. (a) disk method:
2
V
0
16
15
x2
2x
2
(6 15 10)
2
dx
0
x4
x5
5
4 x3 4 x 2 dx
2
4 x3
3
0
x4
32
5
16
32
3
16
15
(b) washer method:
2
V
12
0
x2
2x 1
2
2
dx
dx
0
2
0
( x 1) 4 dx
5
x 1
5
2
2
8
5
2
5
2
0
(c) shell method:
b
V
2
2
2
3
shell
radius
2
a
0
shell
height
4 x 2 x2
dx
2 x2
2
2
0
x3 dx
x2
(2 x )
2
2
0
2x
x3 4 x 2
2
dx
4 x dx
2
2
0
(2 x ) 2 x x 2 dx
x4
4
4 x3
3
2x 2
2
0
2
4
8
3
(36 32)
(d) washer method:
2
V
0
2
0
x5
5
2
x2
2x
2
dx
4 4 x 2 8x x 4
x4
4x2
4x
2 2
2
0
0
2 dx
4 x3 4 x 2 dx 8
2
0
8
32
5
Copyright
4 4 x2
2
0
16 16 8
x4
8
2x
x2
2x
2
dx 8
4 x3 8 x 4 dx 8
5
(32 40) 8
2014 Pearson Education, Inc.
72
5
40
5
32
5
32
3
8
496
Chapter 6 Applications of Definite Integrals
14. disk method:
/4
2
V
0
4 tan 2 x dx
/4
8
sec2 x 1 dx
0
8
tan x x
/4
2 (4
0
)
15. The material removed from the sphere consists of a cylinder
and two caps. From the diagram, the height of the cylinder
is 2h, where h 2
Vcy1
(2h)
2
3
2
3
22 , i.e. h 1. Thus
6 ft 3 . To get the volume of a cap,
use the disk method and x 2
2
4 y 2 dy
1
2
22 : Vcap
2
y3
3
4y
5 ft 3 . Therefore,
3
28 ft 3 .
3
y2
8
3
8
1
b
11/2
12
11/2
264
3
17.
1
2
18. x
19.
y
x3/ 2
3
4
L
1
2
3
4
dy
dx
1 1
4 x
4 x3
363
x
8
2
dx
dy
2
3
y 1/3
9 y 2/3 4 y
L
1 40 u1/2 du
18 13
ln x
8
1 ( y )2
2
1
x 1/2
11/2
11/2
dx
4
363
11 3
2
2 x
L
41
1 2
x 1/2
2
4 x2
121
12 1
11/2
11
2
24
11/2
1/3
1
4
x 1/2
14
3
10
3
1
1
2
2
2
dx
dy
4y
2x
1
8x
1
2x
1
8x
1 ( y )2 dx
2/3
2
1
27
2
dx
8
L
9 y 2/3 4
40
1 2 u 3/2
18 3
13
1 1
4 x
x1/2
9
dy; [u
y
dy 2
dx
1 x1/2
2
4
2 x dx
1 8
3 1
Length
1
2
2
3
y 2/3
x2
dx 12
2
and the x -axis around the x-axis. To find the
11/2
dx
11/2
4
363
132
1
4
1 1
4 x
4 x2
121
12 1
112
4
132
276 in 3
88
x1/2
y
4 x2
121
1
4 x2
121
12 1
R( x)
a
10
3
6
16. We rotate the region enclosed by the curve y
volume we use the disk method: V
1
3
4
Vcy1 2Vcap
Vremoved
x 2 dy
1
1
du
1
2x
1
8x
Copyright
6y
dx
dy
1/3
403/2 133/2
dx
(8 x )
x2
2
dy
x1/2 dx
8
1
dy; y 1
1
2
2 x dx
4
9 y 2/3
dy
u 13, y
8
1
4
2 x3/2
3
1
2 x1/2
8 9 y 2/3 4
1
3 y1/3
40]
u
7.634
(16 x 2 1) 2
256 x 4 32 x 2 1
64 x 2
2
1
1
1
2
2
1 ln x
8
1
16 x 2 1
8x
4
2x
1 ln 2
8
2014 Pearson Education, Inc.
1
8x
1
1 ln1
8
3
1 ln 2
8
dy
1
dx
1
3
Chapter 6 Practice Exercises
1 y3
12
20. x
2
8
12
1
2
b
21. S
a
2
1 x4
2
S
1
d
24. S
2
2 (4 y
4 3
13
12
dx; dx
2x 2
2x 1
dx
3/2 1
9
2
dx
dy
2
dx
dy
1)3/2
6
x4
3
1
S
0
2
x3
3
2
1
2
3
1
y4
1
2x 1
dx
1
2
(4 2 y )
4y y
4y y
2
4
dy
2 y
2
1
1
b
2 y
(125 27)
b
F ( x) dx
a 1
40
F2 ( x) dx
a
is W
W1 W2
0
2
2 (8
3
1)
1 x 4 dx
1
6 0
1 x 4 4 x3 dx
3
6
dx
dy
1
4y
4y 1
4y
4 y y2 4 4 y y2
4 y y2
4
4 y y2
4
dx
dx
dy
1
2
1
2
2
1
6
S
2
2
y
4y 1
4y
40
0
100 dx
0.8(40 x ) dx
4000 640
dy
6
2
4 y 1 dy
49
3
(98)
100 x
40
0
40
x2
2 0
0.8 40 x
F1 ( x ) 100 N . The
4000 J; the rope alone: the force required to lift the
rope is equal to the weight of the rope paid out at elevation x
W2
28
2 2
0
25. The equipment alone: the force required to lift the equipment is equal to its weight
work done is W1
dy
2
1
y 1
2x 1 1
1)3/2
2 (x
3
y4
1 y3
12
dy
0
1
16
1
1
1
y2
S
2 2
2
L
2 2 1
dx
dy; dy
2
dy 2
dx
1
y4
1
2x 1
x 1 dx
0
1
2
2 1 2
y
1 4
dy
dy 2
dx
x2
4
4 y y2
6
2
1
y2
3
2 2
y4
1
16
1
2x 1
dx
dy; dy
y2
4y
y2
dy
dy
dy 2
dx; dx
dx
2 x 1
c
1
2
dy 2
dx
2 x 1
c
1
1
4
7
12
0
d
23. S
2
dy
2
dx
dy
1
y2
1
2 y 1
2
6 3
y2
1
4
1
y4
1
12
2x 1
0
a
1
2
2 y 1
3
b
22. S
y4
1
16
1
dx
dy
1
y
497
F2 ( x)
0.8 402
0.8(40 x ). The work done is
402
2
(0.8)(1600)
2
640 J; the total work
4640 J
26. The force required to lift the water is equal to the water s weight, which varies steadily from 8 800 lb to
8 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight
is F ( x) 8 800
b
W
a
F ( x ) dx
6400 x
2 4750 x
2 4750
4750
0
4750
x2
2 9500 0
x
9500
(6400) 1
6400 1
x
9500
6400 4750
lb. The work done is
dx
47502
4 4750
3
4
(6400)(4750)
22,800,000 ft-lb
27. Using a proportionality constant of 1, the work in lifting the weight of w lb from r a to a is
r
r a
wt dt
2
w t2
r
r a
w
2
r 2 ( r a )2
Copyright
w (2ar
2
a 2 ).
2014 Pearson Education, Inc.
498
Chapter 6 Applications of Definite Integrals
28. Force constant: F
300
250
F
k
x
1.2
0
200
kx
k (0.8)
250 N/m; the 300 N force stretches the spring
k
1.2
1.2 m; the work required to stretch the spring that far is then W
125 x 2
250 x dx
1.2
125(1.2)2
0
0
F ( x ) dx
1.2
0
250 x dx
180 J
29. We imagine the water divided into thin slabs by planes
perpendicular to the y -axis at the points of a partition
of the interval [0,8]. The typical slab between the planes at
y and y
y has a volume of about
(radius)2 (thickness)
V
5
4
y
2
25
16
y
y 2 y ft3 .
The force F ( y ) required to lift this slab is equal to its
weight: F ( y ) 62.4 V
(62.4)(25)
16
y 2 y lb. The distance through which F ( y ) must act to lift this slab to the level 6 ft above the
top is about (6 8 y ) ft, so the work done lifting the slab is about
work done lifting all the slabs from y
8
W
0
(62.4)(25)
16
0 to y
25
16
y 2 (14 y ) y ft lb. The
8 to the level 6 ft above the top is approximately
y 2 (14 y ) y ft lb so the work to pump the water is the limit of these Riemann sums as
the norm of the partition goes to zero: W
(62.4)
(62.4)(25)
16
W
8
y4
4
14 y 3
3
(62.4)
0
25
16
8 (62.4)(25)
(16)
14
3
84
4
83
(62.4)(25)
16
y 2 (14 y ) dy
0
8
0
14 y 2
y 3 dy
418,208.81 ft-lb
30. The same as in Exercise 29, but change the distance through which F ( y ) must act to (8 y ) rather than
(6 8 y ). Also change the upper limit of integration from 8 to 5. The integral is:
W
5 (62.4)(25)
16
0
(62.4)
25
16
8
3
y 2 (8 y ) dy
3
5
4
5
4
(62.4)
25
16
5
0
8 y2
y3 dy (62.4) 25
16
8 y3
3
y4
4
5
0
54,241.56 ft-lb
31. The tank s cross section looks like the figure in Exercise 29 with right edge given by x
horizontal slab has volume
slab is its weight: F ( y )
60
work to pump the liquid is W
to empty the tank is
y 2
2
(radius)2 (thickness)
V
y
4
5
10
y
y
.
2
A typical
y 2 y. The force required to lift this
y 2 y. The distance through which F ( y ) must act is (2 10 y ) ft, so the
60
22,500 ft-lb
275 ft-lb/sec
10
0
(12 y )
y2
4
dy 15
12 y 3
3
y4
4
10
22,500 ft-lb; the time needed
0
257sec
32. A typical horizontal slab has volume about
to lift this slab is its weight F ( y )
V
(20)(2 x ) y
(57)(20) 2 16 y 2
(20) 2 16 y 2
y. The distance through which F ( y ) must act is
(6 4 y ) ft, so the work to pump the olive oil from the half-full tank is
Copyright
y and the force required
2014 Pearson Education, Inc.
Chapter 6 Practice Exercises
0
57
W
4
(10 y )(20) 2 16 y 2 dy
2880
0
4
0
10 16 y 2 dy 1140
22,800 (area of a quarter circle having radius 4)
3/2 0
16 y 2
2 (1140)
3
16 y 2
4
1/2
( 2 y ) dy
(22,800)(4 ) 48,640
4
335,153.25 ft-lb
33. Intersection points: 3 x 2 2 x 2
3x 2 3 0
3( x 1)( x 1) 0 x
1 or x 1. Symmetry
suggests that x 0. The typical vertical strip has center of
mass: ( x, y )
x,
length: 3 x 2
2 x2
3 x2
2
x, x 2 3 ,
2
2 x2
3 1 x 2 , width: dx,
area: dA 3 1 x 2 dx, and mass: dm
x 2 3 1 x 2 dx
3
2
y dm
x5
5
3
2
2 x3
3
3
x x3
3
1
3x
6
1
1
1
1
3
1
1
5
4
34. Symmetry suggests that x
3
y
Mx
M
( 3 10 45)
32
54
x 2 dx
dA
moment about the x-axis is y dm
Mx
3
15
2
3
Mx
8.
5
3
2
y dm
32
5
; M
1
x4
1
dm 3
1
1
2 x 2 3 dx
1 x 2 dx
Therefore, the centroid is ( x , y )
0, 85 .
2
x, x2 , length: x 2 , width: dx,
x 2 dx, mass: dm
x 4 dx
2
2 x 2 3 dx
the moment about the x-axis is
0. The typical vertical strip
has center of mass: ( x, y)
area: dA
x4
3
2
3
3 1 x 2 dx
dA
2
y dm
2
2
x
2
x dx
x 4 dx
2
the
2
x5
10
2
2
35. The typical vertical strip has: center of mass: ( x, y )
x,
x2
4
4
2
area: dA
4
x2
4
4
x2
4
4
0
16
4
2
4x
64
12
y dm
x3
4
dA
the moment about the x-axis is
x2
4
Thus, M x
, width: dx,
dx, mass: dm
dx
4
y dm
x2
4
, length: 4
x2
4
4
2 0
16
2x2
dx
32
3
dx
x
My
M
2
16
x4
16
dx
4
x4
16 0
16 3
32
x4
16
dx; moment about: x dm
2
16 x
4
x5
5 16 0
2
(32 16) 16 ; M
3
2
Copyright
and y
Mx
M
64
64
5
128
5
4
dm
128 3
5 32
4
0
12 .
5
4
x2
4
; My
x2
4
4x
x dx
x dm
dx
Centroid is ( x , y )
2014 Pearson Education, Inc.
x3
4
4x
4
x3
12 0
3 , 12
2 5
.
dx.
499
500
Chapter 6 Applications of Definite Integrals
36. A typical horizontal strip has: center of mass:
y2 2 y
,y
2
( x, y )
area: dA
y 2 , width: dy,
, length: 2 y
y 2 dy , mass: dm
2y
dA
y 2 dy; the moment about the x-axis
2y
is y dm
y 2 dy
y 2y
2 y2
y 3 dy;
y2 2 y
the moment about the y -axis is x dm
2
0
2
2 0
2 y2
y 3 dy
4 y2
y 4 dy
8
3
4
4
3
y5
5
4 y3
2 3
My
M
x
2
y4
4
2 y3
3
0
32 3
15 4
8
16
4
48
2 3
32
5
32
15
Mx
M
4 3
34
2
3
0
2
8
5
y 2 dy
2y
2
and y
16
3
16
4
; M
4 y2
2
16
12
4
3
2
dm
0
y 4 dy
Mx
; My
x dm
y dm
y 2 dy
2y
y2
1. Therefore, the centroid is ( x , y )
y3
3
2
0
8 ,1
5
.
37. A typical horizontal strip has: center of mass:
y2 2 y
,y
2
( x, y )
area: dA
y 2 dy
x-axis is y
2 y2
2 y2
0
My
4
2 y3
y3
y4 dy
21
0 2
x dm
4
3
y2
2
y3
3
y 4 dy
(1 y ) 2 y
y3
8
6
4
5
y4
4
dA
the moment about the
y 2 dy
y (1 y ) 2 y
dm
y2 2 y
2
x dm
2
y 2 dy, mass: dm
2y
(1 y ) 2 y
y 2 , width: dy,
, length: 2 y
1
2
4
5
8
3
0
y 2 dy
4 y3
4 2
4
y3
2 y3
3
4 y2
2
2 y2
y4
4
16
4
y 4 (1 y ) dy
2
16
3
0
y 5 dy
M
8
3
9 , 11
5 10
the center of mass is ( x , y )
4 y2
y5
5
y4
24 ;
5
y 4 dy; the moment about the y -axis is
My
M
dA
3
x3/ 2
about the y -axis is x dm
dx
x
2
0
16
y5
5
y4
(1 y ) 2 y
3
8
24
5
9
5
4 y2
1
3
4 y3
1
4
y6
6
2
y 2 dy
0
and y
Mx
M
x,
3
2 x3/ 2
, length:
3
x3/ 2
the moment about the x-axis is y dm
3
x3/ 2
1 4 23
2 3
0
y 5 dy
16 (20
60
2
5
2
y4
2y
44
15
24
y2
3
8
Mx
4 (11)
15
15 24)
25
5
y dm
44 ;
15
26
6
y 3 dy
44
40
11 .
10
Therefore,
.
38. A typical vertical strip has: center of mass: ( x, y )
mass: dm
32
5
1 4 y3
2 3
dm
x
16
4
1
2
dx
3
x1/ 2
Copyright
, width: dx, area: dA
3
2 x3/ 2
dx.
2014 Pearson Education, Inc.
3
x3/ 2
dx
9
2 x3
3
x3/ 2
dx,
dx; the moment
Chapter 6 Additional and Advanced Exercises
91 9
1 2 x3
(a) M x
9 3
1 x3/ 2
M
x 9
dx
1 2 x3
9
(b) M x
12
b
39. F
a
W
x
strip
depth
b
a
75
b
a
25
9
W
7
3
4
3
F
y 2 y 2 dy
75
250
3 216
strip
depth
75
9 216
L( y ) dy
4
2 y 5/2
5
0
849 h 2
2
849 h 2 .
2
h
2
s 2 h 12 9.707
CHAPTER 6
1. V
b
a
f ( x)
2. V
a
0
x
9
2 x3/2
dx
12 ;
1
20
9
Mx
M
3 and y
9
2 x1/2
3
dx
3/ 2
5
9
4
9
52; M
1
1
x
3
x
3/ 2
dx
6 x1/2
9
1
1
3
2
(62.4)(2 y )(2 y ) dy
0
2
249.6
2y
0
y3
3
y 2 dy 249.6 y 2
2
0
F
5/6
0
10
3
5
6
75
5/6
2 y3
3
0
y2
7
6
y
y (2 y 4) dy
4
0
2
5
(62.4) 6 8
y
2
(9 y ) 2
62.4
5
32
y, L( y ) 1
Now solve
h
F
0
849 h 2
2
5/6 5
3
0
50
18
(75)
(25 216 175 9 250 3)
62.4
75
7
6
(75)(3075)
9 216
62.4
dy
h
W
strip
depth
9 y1/2 3 y 3/2 dy
0
(62.4)(176)
5
h
849
125
216
2
3
2196.48 lb
L( y ) dy , h
849(h y ) 1 dy
40000 to get h
25
36
4 y dy
118.63 lb.
4
(48 5 64)
0
y 10
2 y2
3
0
the height of the mercury
(h y ) dy
849 h y
y2
2
h
0
9.707 ft. The volume of the mercury is
9.707 ft 3 .
ADDITIONAL AND ADVANCED EXERCISES
f ( x)
2
dx
b2
ab
dx
a2
a
1
f (t ) dt
x
a
2
x2
f (t ) dt
ax for all x
f ( x)
a
2
2x a
2x a
2
f ( x)
x
3. s ( x) Cx
f ( x)
h
3
12
4
9 2 3
x
1
x3/ 2
42. Place the origin at the bottom of the tank. Then F
column, strip depth
x
332.8 lb
L( y ) dy
175
216
62.4 6 y 3/2
2
F
1
My
M
x
4; M y
Mx
M
and y
(249.6)
strip
depth
5/6 10
3
0
(75)
41. F
W
13
3
1 9
x 1
9
2
9
; My
4
1
L( y ) dy
8
3
(249.6) 4
40. F
My
M
20
9
9
x 1/2
6
dx
9
x 2
2 1
9
2
dx
501
0
x
0
2
x
a
2
f (t ) dt
Cx
1
C 2 1 dt k . Then f (0)
x2
f ( x)
a
a
x for all x
2
C
0 k
f ( x)
a
x
0
C 2 1 dt a
where C 1.
Copyright
2x 1
f ( x)
2x 1
C 2 1 for C 1
f ( x)
f ( x)
2
2014 Pearson Education, Inc.
f ( x)
x C 2 1 a,
502
Chapter 6 Applications of Definite Integrals
4. (a) The graph of f ( x) sin x traces out a path from (0, 0) to ( , sin ) whose length is
L
1 cos 2
0
d . The line segment from (0, 0) to ( , sin ) has length
2
(
0)2 (sin
0) 2
sin 2 . Since the shortest distance between two points is the length of
the straight line segment joining them, we have immediately that
1 cos2
0
2
d
(b) In general, if y
1
0
if 0
2
.
f ( x ) is continuously differentiable and f (0)
2
f (t )
sin 2
2
dt
f 2 ( ) for
0, then
0.
5. We can find the centroid and then use Pappus Theorem to calculate the volume. f ( x)
f ( x)
1
2
g ( x)
1
3
1;
6
0
1 11
1/6 0 2
y
x2
x
x2
1
x2
2
x2
1, 2
2 5
is the distance from
on y
1
3
dx
0, x 1;
x
x x 2 dx
1
x
1/6 0
x
0
x
x2
0
1
6
0
x 4 dx
x2
3
x3 dx
2
5
1 x
1
2
9
10
1
2
Thus,
x
to the axis of rotation, y
2
9.
10
y
9
20
2
5
2
0
6
x x 2 dx
1
1 x4
4
0
1 x3
3
1
1 x5
5
0
1 x3
3
3
1
3
1
5
Thus V
1
3
2
3
0
1
4
0
1;
2
The centroid is
1, 2
2 5
.
x. To calculate this distance we must find the point
The point of intersection of the lines x
1 .
10 2
1
1 x3
3
0
1 x2
2
6
x2 ,
x that passes through 12 , 52 . The equation of this line
x that also lies on the line perpendicular to y
is y
1
1; M
x, g ( x)
1
10 2
2
1
6
30 2
y
9
10
and y
x is
9 , 9
20 20
.
.
6. Since the slice is made at an angle of 45 , the volume of the wedge is half the volume of the cylinder of radius
1
2
7.
2 x
y
1
x
ds
1 2 (1)
2
1
2
and height 1. Thus, V
1 dx
A
3
0
8
2 x
1
x
.
1 dx
(1 x)3/2
4
3
3
0
28
3
8. This surface is a triangle having a base of 2 a and a height of 2 ak . Therefore the surface area is
1 (2
2
9. F
x
W
a)(2 ak )
ma
2
d2
dt 2
t2
0 when t
F dx
12 mh 12mh
18m
0
2 2
a k.
a
t2
m
v
dx
dt
t3
3m
C1
0
x
t4 .
12 m
Then x
(12 mh)1/ 4
0
2h
3
F (t )
2 3mh
dx dt
dt
4h
3
C; v
0 when t
h
(12 mh)1/ 4 2 t 3
t 3m dt
0
t
0
C
0
t2
3m
x
t4
12 m
C1 ;
(12 mh)1/4 . The work done is
(12 mh)1/ 4
1 t6
3m 6 0
3mh
Copyright
dx
dt
2014 Pearson Education, Inc.
1
18 m
(12mh)6/4
(12mh)3/ 2
18m
Chapter 6 Additional and Advanced Exercises
2 lb 12 in
1 ln 1 ft
10. Converting to pounds and feet, 2 lb/in
1/2
12 x 2
and v1
0 at t
s
16
ds
dt
0)
v0 2
32
1 mv 2 ,
1
2
1 v2
320 0
1
2
0 ft/sec, we have 3
s
1 mv 2
0
2
3 ft lb. Since W
0
24 lb/ft. Thus, F
where W
v02
24 x
v02
64
v
v0 320
3 640
64
lb
0
24 x dx
1
32 ft/sec2
0
v0
32
t
30 ft.
we use the vertical strips technique. The typical strip has center of mass: ( x, y )
1 xn
y dm
1 x n dx, mass: dm 1 dA
2
2
dx
n
1 1 x
2
1
Mx
( n 1)(2 n 1) 2(2 n 1) ( n 1)
( n 1)(2n 1)
2
1
0
1 x n dx
2
2
dx
11
02
2 n 2 3n 1 4 n 2 n 1
( n 1)(2 n 1)
1
xn 1
n 1 0
2 x
slugs,
and the height is
11. From the symmetry of y 1 x n , n even, about the y -axis for 1 x 1, we have x
width: dx, area: dA
1
320
16t 2 v0 t (since
3 640. For the projectile height, s
32t v0 . At the top of the ball s path, v
v
1
10
3 ft lb, m
1/2
W
503
is the location of the centroid. As n
x 2n dx
2n2
.
( n 1)(2 n 1)
1
2
,y
,
n
x, 1 2x , length: 1 x n ,
1 x n dx. The moment of the strip about the x-axis is
1 2 xn
2n .
n 1
1
n 1
21
Mx
M
0. To find y
x
1
Also, M
Therefore, y
2 xn 1
n 1
1
Mx
M
dA
1
x2n 1
2n 1 0
1
n
1
1 x
( n 1)
2n 2
( n 1)(2n 1) 2 n
2
n 1
1
1
2n 1
dx
n
2n 1
so the limiting position of the centroid is 0,
1
2
0,
n
2n 1
.
12. Align the telephone pole along the x-axis as shown
in the accompanying figure. The slope of the top
length of pole is
5.5
8 40
1 9
8
14.5
8
9
8
1
8
9
8
40
11 . Thus, y
8 80
11 x is an equation
80
1 (14.5
40
11 x
8 80
9)
of the line
representing the top of the pole. Then
b
My
a
b
M
a
x
2
11 x
x 81 9 80
dx
0
40
y 2 dx
40
y 2 dx
0
1
8
9
11 x
80
2
dx
1
64
40
1
64
40
0
11 x
x 9 80
0
9
11
80
2
dx;
2
x dx. Thus, x
calculator to compute the integrals). By symmetry about the x-axis, y
from the top of the pole.
My
M
129,700
5623.3
23.06 (using a
0 so the center of mass is about 23 ft
13. (a) Consider a single vertical strip with center of mass ( x, y ). If the plate lies to the right of the line, then the
moment of this strip about the line x b is ( x b) dm ( x b) dA
the plate s first moment about
x
b is the integral ( x b ) dA
x dA
b dA
M y b A.
(b) If the plate lies to the left of the line, the moment of a vertical strip about the line x
(b x)
dA
the plate s first moment about x
Copyright
b is (b x) dA
2014 Pearson Education, Inc.
b dA
b is (b x ) dm
x dA
b A M y.
504
Chapter 6 Applications of Definite Integrals
14. (a) By symmetry of the plate about the x-axis, y
0. A typical vertical strip has center of mass: ( x , y )
( x, 0), length: 4 ax , width: dx, area: 4 ax dx, mass: dm
kx 4 ax dx, for some
dA
proportionality constant k. The moment of the strip about the y -axis is M y
a 5/2
x dx
0
4k a
a 3/2
4k a
0
x
4k a
dx
5a ,0
7
(x , y )
a
2 x 7/2
7
0
a
2 x5/2
5
0
4k a
4k a1/2
2 a 7/2
7
8k a 4
.
7
Also, M
4k a1/2
2 a 5/2
5
8k a 3
.
5
Thus, x
y2
4a
width: dy, area: a
0
y4
4a
My
M
8a 4 4a 2
2a
dm
2a
y2
4a
4 a4
3
0
2a
3
1
2 a3
16a 4 y
y 4a 2
b2
b2
b2
length:
Mx
b2
b2
a2
a2
a1
0 2
x2
a2 x
4k x ax dx
8k a 4
5
7
8k a3
5a
7
y
2
0
2a 2
y
0
1
4a
y2
32a 6
0
32a
3
6
8a 4
3
0
0
2a
1
2
16a 2 3
4a 2 y
16a 4
4
1
2a
1
32 a 2
dy
y6
6
8a 4 y 2
1 0
4a 2 a
2a 2 4 a 2
2a
y5
20a
4a2 y 2
4a
4a 2
1
32a 2
dy
32a5
20 a
2a
x 2 , width: dx, area: dA
b2
x2
x2
b2 x
2
b2
a2
dx
x2
b
2 a
b
x3
3 a
Copyright
2
32 a5
20 a
y 4 dy
2a
1
32 a 2
32a6
8a 4 y 2
y6
6
2a
0
4 a4 ;
3
1 2a
4a 0
y3 dy
8a 4
3
y 16a 4
8a 4 4 a 4
4a 2 y
y 3 dy
2a 3 . Therefore,
2
2
2
2
x, b x 2 a x ,
b2
x2
a2
x2
dx, mass: dm
dA
2
2
x, b 2 x ,
x 2 dx. On [a, b] a typical vertical strip has center of mass: ( x, y )
a2
a
y2
4a
a
a y3
3
2a
y 2 dy
y2
,
4a
, y , length: a
0 is the center of mass.
x 2 , width: dx, area: dA
y dm
a
2 0
2
x2
x2
4k x 2 ax dx
dy. Thus,
15. (a) On [0, a ] a typical vertical strip has center of mass: ( x, y)
length:
0
My
M
y2 4a 2
8a
,y
0
y5
20a
1
16a 2
2a
Mx
M
and y
dy
6
1 2a
4a 2a
y4
4
y2
4a
y 5 dy
64 a
6
2a 2 y 2
y2
4a
1 2a
8a 2a
dy
8a 4 4a 2
dy
1
4a
2a
y2
4a
2
a y3
3
dy
a
y a
y2 a
2a
2a
1
32a 2 0
1
32 a 2
4a 2 y 2
4a
y
y4
4
2a 2 y 2
My
M
64 a
6
0
y4
4a
y a
y5 dy
6
dA
dy
ay 2
0
16a 4 y
2a
1
4a
2a
dy
y 2 4a 2
8a
2a
0
1
32 a 2
y2
4a
2a
x dm
1
32 a 2
dy, mass: dm
y y a
2a
ay 2
2a
x
2a
y dm
a
dm
0
is the center of mass.
(b) A typical horizontal strip has center of mass: ( x , y )
Mx
a
x dm
x 2 dx, mass: dm
b2
b2
b2
x 2 dx
a2 a
x2
a
2 0
2
dA
b2
x 2 dx. Thus,
a2
x 2 dx
b1
a 2
b2
x2
b2
a 2 dx
b
2 a
b2
x 2 dx
b3
b3
3
2014 Pearson Education, Inc.
3
b2 a a3
b2
x 2 dx
0;
Chapter 6 Additional and Advanced Exercises
ab 2
2
My
a3
2 b3
2 3
a
x dm
a
0
x b2
b2
x
1/2
x2
2 b2 x2
2
0
a
dx
0
a
3/ 2
a3
3
ab 2
x2
a2
x a2
x2
2 a2 x2
3
2
b3
3
b
3
a
1/2
(b)
4
a3
lim
b
a
2 b2 x 2
3
b
4
2
2 3/2
3
4
3
b2 a 2
a 2 ab b2
a b
centroid as b
b a ).
b
dx
0
4
3
b3 a 3
b2 a 2
x b2
x 2 dx
x2
1/2
dx
b
a
a
4
3
b2
x
3/ 2
;
3
2 3/2
3
We calculate the mass geometrically: M
3
b
a
0
2 3/2
b3 a 3
b3 a 3
3
x 2 dx
a
3/ 2
0
2
a3
3
A
0
b2
4
b
2
a
2 3/2
a2
4
4
(b a ) a 2 ab b 2
4 a 2 ab b 2
(b a )(b a )
3 (a b)
a2 a2 a2
a a
3a 2
2a
4
3
2a
b3
3
(x, y)
b2
a3
3
3
a 2 . Thus, x
; likewise y
2a , 2a
b3 a 3
Mx
M
M x;
My
M
4 a 2 ab b2
3 (a b)
a. This is the centroid of a circle of radius a (and we note the two circles coincide when
a , 24
3 a
. The shaded portion is
144 36 108. Write ( x y ) for the centroid of the
remaining region. The centroid of the whole square
is obviously (6, 6). Think of the square as a sheet of
uniform density, so that the centroid of the square
is the average of the centroids of the two regions,
weighted by area: 6
6
36
and y
24
a
108( y )
144
8( a 1)
.
a
36
a
3
108( x )
144
and
which we solve to get x
Set x
8
a
9
7 in. (Given). It follows that a
9, whence y
64
9
7 19 in. The distances of the
centroid ( x , y ) from the other sides are easily computed. (Note that if we set y
x
.
is the limiting position of the
16. Since the area of the triangle is 36, the diagram may
be labeled as shown at the right. The centroid of
the triangle is
505
7 19 . )
Copyright
2014 Pearson Education, Inc.
7 in. above, we will find
506
Chapter 6 Applications of Definite Integrals
17. The submerged triangular plate is depicted in the
figure at the right. The hypotenuse of the triangle
has slope 1 y ( 2)
( x 0) x
( y 2)
is an equation of the hypotenuse. Using a typical
horizontal strip, the fluid pressure is
strip
depth
(62.4)
F
2
6
(62.4)( y )
2
62.4
y2
6
8
3
(62.4)
(62.4)
208
3
strip
length
dy
( y 2) dy
2 y dy
62.4
216
3
4
y3
3
6
36
(62.4)(112)
3
32
2
y2
2329.6 lb
18. Consider a rectangular plate of length and width
w. The length is parallel with the surface of the fluid
of weight density . The force on one side of the
0
plate is F
w
y2
2
( y )( ) dy
0
w
w2
2
.
The average force on one side of the plate is
Fav
0
w
w
( y ) dy
w
y2
2
0
w
w.
2
Therefore the force
w2
2
w
2
(the average pressure up and down ) (the area of the plate).
Copyright
2014 Pearson Education, Inc.
( w)
CHAPTER 7 INTEGRALS AND TRANSCENDENTAL FUNCTIONS
7.1
1.
3.
THE LOGARITHM DEFINED AS AN INTEGRAL
21
dx
3 x
2y
y 2 25
2
3
ln x
ln y 2
dy
25
5. Let u = 6 + 3 tan t
7.
dx
2 x 2x
dx
2 x1
du
u
x
ln u C
8. Let u = sec x + tan x
sec x dx
ln(sec x tan x )
9.
11.
ln 3 x
e dx
ln 2
4 (ln x )3
1 2x
ln 3
13.
ln 9 x /2
e dx
ln 4
14. Let u = ln(cos x)
ln 9
1 (
cos x
u du
r1/2
1 r 1/2 dr
2
e
r
r
dr
du
1/ 2
er
r 1/2 dr
C
ln 1
sin x)dx
u2
2
2 eu du
2eu
C
Copyright
sec x dx
du ;
u
2 ln(sec x tan x) C
1)
dx
8e( x
1)
C
(ln 4) 4
8
ln(ln x) x ln1 x dx
2(eln 3 eln 2 )
tan x dx
1/2
C
C
8e( x
ln(ln x )
dx
x ln x
r
2 du
x
(ln1)4
8
(ln(cos x ))2
2
C
C
ln 2 sec y
(sec x)(tan x sec x) dx
(ln 4) 4
8
1 dx;
x ln x
C
ln 6 3 tan t
C
ln 52
dx;
10.
1
C
ln 2 ln 5
1
ln 4t 2 5
ln u
2(ln u )1/2 C
4
(ln x ) 4
8
dx
tan x ln(cos x)dx
15. Let u
x
2 e(ln 9)/2 e(ln 4)/2
ln 4
du
1
2 x
du
du
u
dy
3 2 1
1 1 dx
ln x x
du
2e x /2
ln 1
eln 3 eln 2
ln 2
12. Let u = ln(ln x)
x
(ln u ) 1/2 u1 du
1 4 (ln x)3 1
2 1
x
dx
sec y tan y
2 sec y
ln u
0
ln 3 x 2
8r dr
4r 2 5
du
u
dt
(sec x tan x sec 2 x) dx
du
du
u ln u
ex
3sec 2 t
6 3 tan t
3sec2 t dt ;
; let u 1
x
3 dx
1 3x 2
4.
du = sec y tan y dy;
dx
2 x 1
0
2.
C
du
6. Let u = 2 + sec y
ln 32
ln 2 ln 3
du
2(3 2)
u du
u2
2
C
(ln(ln x )) 2
2
C
2
tan x dx;
C
dr ;
2e r
1/ 2
C
2e
r
C
2014 Pearson Education, Inc.
507
508
Chapter 7 Integrals and Transcendental Functions
r1/2
16. Let u
r
e
1/ 2
e r
r 1/2 dr
dr
r
t2
17. Let u
ln x
x ln 2 x 1
1
x
19. Let u
1/ x 2
x
e x
21. Let u
sec t
esec(
t)
2
du
t)
23. Let u
ev
ln( /2)
ln( /6)
ln
du
2
26.
e
e
27.
1
0
2
dx
x
x
ev dv
2 du
1
dx
d
e
x
x
1
C
ln x dx;
x
eu du
dx
u
1e x
2
C
eu
2
e1/ x
C
2
1 e 1/ x
2
C
du
sec t tan t dt;
C
esec(
/6
eu du
2
dx
cos u du
1
er dr ;
t)
C
C
C
ln u
d
cos u du
sin u 1
er
1 er
1 du
u
C
ln
1
2
0
6
2 sin
ln
2
u
C
ln 2
sin
ln u
C
u
1
2
ln
1
2
1
1
2
Copyright
ln
1
2
ln
1
2
sin(1)
e x dx;
1
2(ln1 ln 2)
;
1
2
1
;
ln(1 er ) C
du
2
21
6
eln
sin( ) sin(1)
e x dx
du
,v
t)
ln(e x 1) C
1
1
2
u
/2
/6
u = 1, x
dr
ecsc(
C
ln 6
2sin u
e x 1
dx; let u
eu
2ev dv; v
2
1 du
u
1 1
0 2
/2
2
du
e
e
2 xe x dx; x = 0
2 xe x cos e x
1
1 ex
1 eu
2
eu du
t )dt
du
25. Let u 1 er
eu du
t ) cot(
2
0
2
e t
C
du = csc( + t) cot( + t)dt;
2ev cos ev dv
ex
24. Let u
csc(
eu
x 3 dx;
1 du
2
1
C
eu du
dt
1 du
2
e1/ x
x2
dx;
sec t tan t dt
sec( t ) tan( t )dt
csc(
1
x2
du
1
2
r
2e
C
ln 2 x 1 C
C
dx
x 3dx
22. Let u = csc( + t)
e
u
t2
2te
2 ln x dx
x
1 dx
x
2 x 3dx
du
dx
3
du
1
x2
du
x 2
20. Let u
e
1
2 u
dx
r1/ 2
2e
2t dt ;
du
2 ln x
du
r 1/2 dr ;
2 du
2 eu du
2t dt
du
ln 2 x 1
18. Let u
1 r 1/2 dr
2
du
1
2 ln 2
2014 Pearson Education, Inc.
0.84147
Section 7.1 The Logarithm Defined as an Integral
28.
0
2
5
1
2 5
d
x2
29. Let u
2
0
du
1
x1/2
30. Let u
42 x
x
1
4 x1/ 2
1
7
0
32. Let u = tan t
x=2
u
4 2x
2
x
sec t dt
ln u
24
35.
37.
3
dx
1 du
u dx
1
2
x=1
2 1
x
u = 1, t
log10 x
dx
x
ln x
ln10
1
x
4 log 2 x
dx
x
4 ln x
ln 2
dx
u
u du
1
3
1
2 1
1
ln10
1 u2
2
C
u du
x 2 x (1 ln x)dx;
1 du
2
2u (ln x 1)
ln x
1
ln 2
Copyright
du
1u
2
16)
65,520
2
32, 760
(ln x )2
2 ln10
1 dx;
x
2 ln 4
0
ln x e
ln 2 1
e (ln 2) 1
x
dx
1
36.
1 dx
x
1
ln 2
2
3ln 3
2ln 2 1
ln 2
20 )
du
ln 4
0
0
u = ln 2;
ln x
u
1
3
du
dx
1 (65,536
2
(2ln 2
1
ln 2
dx;
0;
u 1;
4
u = 0, x = 2
1
x
1
1
x
dx;
1
ln10
4 ln x
ln 2
1
1
dx
1
x
u
2
65,536;
65.536
3
4
ln 2
0
u 16
3
u = 2;
6
ln 7
2 ln x (2 x) 1x
48
u
2;
(23 22 )
1
ln 2
1
ln 3
1
3
ln
ln 2
2u
ln 2 0
ln 2 u
2 du
0
u = 1, x = 4
1
u
u
24
ln 5
1
ln 2
0
ln x
ln10
38.
1 dx;
x
du
2 1 x 2 dx
0
du
1 65.536 du
2 16
(1 ln x )dx
u = 0, t
1
3
2
(70 7)
1
ln 7
24
ln1 ln 5
(1 25)
1
5
21 )
2
2( u 1)
ln 2 1
0
7u
ln 7 1
2 x ln x
16, x = 4
34. Let u = ln x
2 2ln x
1 x
1 1 u
0 3
(22
; x=1
2 du
sec2 t dt ; t = 0
2
ln
du = sin t dt; t = 0
0 u
7 du
1
du
x2 x
33. Let u
dx
x
2 u
1
du = sin t dt
sin t dt
/4 1 tan t
3
0
2
1
1
5
u = 1, x
1
2 ln 2
2 du
x 1/2 dx
2
31. Let u = cos t
x 1/2 dx
ln
x dx; x = 1
2
1 2u
2 ln 2 1
2u du
1
2
2
1 du
2
1
5
ln
2
1
5
1
1
5
ln
2 x dx
du
dx
/2 cos t
d
2 1
1 2
2
x 2( x ) dx
0
1
5
x 1
1
ln 2
eln 2 1ln 2
ln 2
2 1
ln 2
C
u
0, x
1 (ln 4) 2
2
4
u
(ln 4)2
2 ln 2
2014 Pearson Education, Inc.
ln 4
(ln 4) 2
ln 4
ln 4
1
ln 2
509
510
39.
Chapter 7 Integrals and Transcendental Functions
4 ln 2 log 2 x
dx
x
1
4 ln 2
x
1
ln x
ln 2
4
1 (ln x ) 2
2
1
4 ln x
dx
1 x
dx
1 [(ln 4) 2
2
(ln1) 2 ]
1 (ln 4) 2
2
1 (2ln 2) 2
2
2(ln 2)2
40.
e 2 ln10(log10 x)
dx
x
1
41.
2 log 2 ( x 2)
dx
x 2
0
4(ln 2)
2
1
ln 2
42.
44.
45.
47.
2
2
dx
x (log8 x )2
dy
dt
et sin(et
e
1
x
1
x
dx
t
2
du
t
e dt
y
cos(e
ln 2
2) C
y
1
x
(ln( x 1))2
2
dx;
u
(ln10) ln u
2
(ln x )
x
et sin(et
y
2
ln 2
1
ln x
(ln10)
(ln 8)2
dx
(ln 4)2
2
(ln 2)2
2
10
(ln100)2
20
10
ln10
1/10
(ln1) 2
2
(ln x )
1
2
ln10
0
3
(ln10) 2
2
(ln 2)2
2
2
ln 2
2
ln x
C
9
C
(ln 8)2
ln x
C
C
(ln1)2
2
ln10
ln 2
1 dx
x
du
(ln10) ln ln x
1
(ln1)2
2
2)dt ;
sin u du
0
cos(et
cos u C
cos(2
2) C ;
C = cos 0 = 1; thus, y 1 cos(et
2) + C = 0
2)
e t sec2 ( e t )dt ;
e t
du
e t dt
y (ln 4)
2
1
tan( e ln 4 ) C
thus, y
3
1
0
(ln( x 1))2
2
2
ln10
1) x1 1 dx
(ln 8)2
ln x 2
ln 8
e t sec 2 ( e t )
let u
1
1
ln 2
(ln(10 x )) 2
20
10
ln10
dx
1) x1 1 dx
(ln10) u1 du
dx
2)
y(ln 2) = 0
dy
dt
2 3 ln( x
ln 2 2
dx
x
2
(ln( x 2))2
2
1
ln 2
(ln1)2
3 ln 2
2
9
2
ln( x
ln10 0
ln10
ln x
1
ln x
2)] x 1 2 dx
(ln e)2
2 ln10
3 2 log 2 ( x 1)
dx
x 1
2
dx
x log10 x
2
dx [(ln x ) 2 ]1e
10 10 [ln(10 x )] 1
ln10 1/10
10 x
9 2 log10 ( x 1)
dx
x 1
0
let u
48.
(ln 2)
2
4(ln10)
20
(ln10)
46.
1 2 [ln( x
ln 2 0
10 log10 (10 x )
dx
x
1/10
10
ln10
43.
e (ln10)(2 ln x ) 1
x
(ln10)
1
1
du
e t dt
2
1
tan
1
y
1
4
sec 2 u du
C
2
1
1
(1) C
tan( e t )
Copyright
2014 Pearson Education, Inc.
1
tan u C
2
C
tan( e t ) C ;
3;
Section 7.1 The Logarithm Defined as an Integral
49.
d2y
dy
dx
2e x
dx 2
dy
dx
2e x
2
d2y
1 e 2t
2
t
dy
dt
t
1 e 2t
2
1 e2
2
1
y
t = 1 and y = 1
51.
52.
dy
dx
1 1x at (1, 3)
d2y
0
ln sec 0
53. V
2
2
1/2
54. V
9x
0
x3 9
y
x2
8
L
56.
4
y 2
4
x
L
12
4
dy
1 e 2t
4
1 t2
2
0
2
dx
C
2; thus
1 e2
2
1 C1
3
27
0
1
2
C1
ln sec x
2
dx
27
ln x
1 e2
2
C
1; thus
1 t C1;
C ; y = 3 at x = 1
y
x) 1
0 1 12 e 2 C
0
1 e2
2
2 x 1 2(e x
1 e2
4
C=2
dy
dx
1 t2
2
y
y
tan x 1
1 e 2t
4
x ln x
y
1 e2
2
1 t
1
2
1 e2
4
2
(tan x 1) dx
ln sec x
x C1 and
x
2
1/2
ln 2 ln 12
2
ln( x3 9)
3
2 (2 ln 2)
27 (ln 36 ln 9)
0
ln 24
ln16
27 (ln 4 ln 9 ln 9)
54 ln 2
1 ( y )2
ln x
8
2e x
y
C ; t = 1 and dt
2 1
dx
1/2 x
2
x
3
1
x ln x
C1
x 12 dx
2e0 C
0
tan x C and 1 = tan 0 + C
0 C1
27 ln 4
55.
y
dy
dx
sec2 x
dx 2
C1
1 e2
4
1
2
1
0
2 x C1 ;
1 2e0 C1
1 e 2t
dt 2
dy
dt
C ; x = 0 and dx
2e x
y
x = 0 and y = 1
50.
dy
2e x
511
y
1
dx
dy
dx
dy
2
x
4
2
1
x
8 x2 4
dx
4 4x
1 ( y ) 2 dx
2 ln 4
1
dy
y
8
2
y
8 x
4 4
1
dx
dy
12 y 2 16
dy
8y
4
2
x2 4
4x
1
1
x
2
x2 4
4x
x2
8
dx
y
8
1
12 y
4 8
2
y
dy
2
y
2
ln x
2
y2
16
1
8
4
(8 ln 8) (2 ln 4)
y 2 16
8y
2
y 2 16
8
6 ln 2
2
12
2 ln y
(9 2 ln12) (1 2 ln 4)
4
8 2 ln 3 8 ln 9
57. (a)
L( x)
f (0)
f (0) x , and f(x) = ln(1 + x)
(b) Let f(x) = ln(x + 1). Since f ( x)
1
( x 1)2
f ( x) x 0
1
1 x x 0
1
L(x) = ln 1 + 1 x
0 on [0, 0.1], the graph of f is concave down on this interval
and the largest error in the linear approximation will occur when x = 0.1. This error is
0.1 ln(1.1) 0.00469 to five decimal places.
Copyright
L(x) = x
2014 Pearson Education, Inc.
512
Chapter 7 Integrals and Transcendental Functions
(c)
The approximation y = x for ln(1 + x) is best for smaller
positive values of x; in particular for 0 x 0.1 in the
graph. As x increases, so does the error x ln(1 + x). From
the graph an upper bound for the error is 0.5 ln(1 + 0.5)
0.095; i.e., E ( x ) 0.095 for 0 x 0.5. Note from the
graph that 0.1 ln(1 + 0.1) 0.00469 estimates the error in
replacing ln(1 + x) by x over 0 x 0.1. This is consistent
with the estimate given in part (b) above.
58. (a)
f ( x)
ex
e x ; L( x)
f ( x)
(b) f(0) = 1 and L(0) = 1
ex
(c) Since y
f (0)
f (0)( x 0)
error = 0; f (0.2)
e
0.2
L( x) 1 x
1.22140 and L(0.2) = 1.2
error
0.02140
0, the tangent line approximation
e x . Thus L(x) = x + 1
always lies below the curve y
never overestimates e x .
59. Note that y = ln x and e y
ln a y
0
e dy
60. (a)
y
a
1
ln x dx are under the curve between 1 and a;
area to the left of the curve between 0 and ln a. The sum of these areas is equal to the area of the
a
rectangle
x are the same curve;
1
ex
ln a y
ln x dx
y
e dy
0
ex
a ln a.
0 for all x
ln b x
(b) area of the trapezoid ABCD <
ln b x
e dx
ln a
the graph of y
ln a
e dx
e x is always concave upward
1 ( AB
2
area of the trapezoid AEFD
CD)(ln b ln a)
eln a eln b
2
(ln b ln a ). Now 12 ( AB CD ) is the height of the midpoint M
since the curve containing the points B and C is linear
e(ln a ln b )/2 (ln b ln a )
(c)
ln b x
e dx
ln a
ex
ln b
ln a
eln b
e(ln a ln b )/2 (ln b ln a )
eln a /2 eln b /2
ln b x
ln a
eln a
b a
b a
ln b ln a
e dx
eln a eln b
2
(ln b ln a )
b a, so part (b) implies that
eln a eln b
2
a b
2
Copyright
(ln b ln a)
eln a eln b
e(ln a ln b)/2
b a
ln b ln a
a b
2
2014 Pearson Education, Inc.
b a
ln b ln a
ab
a b
2
b a
ln b ln a
a b
2
e(ln a ln b )/2
Section 7.1 The Logarithm Defined as an Integral
513
61. y = ln kx y = ln x + ln k; thus the graph of
y = ln kx is the graph of y = ln x shifted vertically by ln k, k >
0.
62. To turn the arches upside down we would use the formula
y
ln sin x ln 1 .
sin x
63. (a)
(b)
cos x .
a sin x
y
Since sin x and cos x are
less than or equal to 1, we have for a >
1, a 11 y a1 1 for all x. Thus,
a
lim y
0 for all x
the graph of y
looks more and more horizontal as a
+ .
64. (a) The graph of y
x ln x appears to be concave
upward for all x > 0.
(b)
y
x ln x
Thus, y
y
1
2 x
1
x
0 if 0 < x < 16 and y
1
4 x3/ 2
y
1
x2
1
x2
x
4
1
0
x
4
x 16
0 if x > 16 so a point of inflection exists at x = 16. The graph of
y
x ln x closely resembles a straight line for x
inflection visually from the graph.
Copyright
10 and it is impossible to discuss the point of
2014 Pearson Education, Inc.
514
Chapter 7 Integrals and Transcendental Functions
65. From zooming in on the graph at the right, we
estimate the third root to be x
0.76666
66. The functions f ( x) x ln 2 and g ( x) 2ln x appear
to have identical graphs for x > 0. This is no
accident, because x ln 2
eln 2 ln x
(eln 2 )ln x
2ln x.
67. (a) The point of tangency is (p, ln p) and mtangent
the equation of the tangent line is y
(p, ln p)
(b)
ln p
1
p
p 1
p
1
p
1
p
dy
1.
x
since dx
The tangent line passes through (0, 0)
x. The tangent line also passes through
1
e
e, and the tangent line equation is y
x.
d2y
1 for x 0
y = ln x is concave downward over its domain. Therefore, y = ln x lies below the
x2
dx 2
graph of y 1e x for all x > 0, x e, and ln x ex for x > 0, x e.
(c) Multiplying by e, e ln x < x or ln x e
(d) Exponentiating both sides of ln x
(e) Let x =
to see that
e
x.
e
x, we have eln x
e
e . Therefore, e
68. Using Newton s Method: f(x) = ln(x)
1
f ( x)
e x , or x e
e x for all positive x
e.
is bigger.
1
x
xn 1
xn
ln( xn ) 1
1
xn
xn 1
xn [2 ln( xn )].
Then x1 2, x2 2.61370564, x3 2.71624393, and x5 2.71828183. Many other methods may be used.
For example, graph y = ln x 1 and determine the zero of y.
ln 8
ln 3
69. (a) log 3 8
1.89279
(b) log 7 0.5
ln 0.5
ln 7
0.35621
(e) ln x
(log10 x)(ln10)
2.3ln10
5.29595
(f) ln x
ln 7
2.80735
ln 0.5
(log 2 x)(ln 2) 1.4 ln 2
(g) ln x
(log 2 x)(ln 2)
1.5ln 2
1.03972
(h) ln x
(log10 x)(ln10)
(c) log 20 17
70. (a)
ln10
ln 2
ln17
ln 20
log10 x
0.94575
ln10 ln x
ln 2 ln10
(d) log 0.5 7
ln x
ln 2
log 2 x
Copyright
(b)
ln a
ln b
log 4 x
ln a ln x
ln b ln a
2014 Pearson Education, Inc.
0.97041
0.7 ln10
ln x
ln b
logb x
Section 7.2 Exponential Change and Separable Differential Equations
7.2
EXPONENTIAL CHANGE AND SEPARABLE DIFFERENTIAL EQUATIONS
1. (a)
y
e x
(b)
y
e x
e 3x 2
(c)
y
e x
Ce 3 x 2
2. (a)
y
1
x
(b)
y
(c)
y
3.
y
5.
y
2y
3 Ce 3 x 2
2
e x
3y
1 t 4 dt
2
2y
3y
dt
2
x
1
1 x
4
1
1
2
1 t 4 dt
ex
x
1
x
x et
1 t
x2 y
x
4 x3
1 x
3
4
1
1
y
2
( x 2)e x
7.
y
cos x
x
y
y
y
x
ln x
dy
9. 2 xy dx
y
1
2 32 y 3/2
10.
dy
dx
x2 y
3 Ce 3 x 2
2
e 3x 2
e x
3 e x
Ce 3 x 2
ex
xy e x
x t
x 1x et dt
dt e x
1 t 4 dt 4
2 x3
1 x4
y
e x tan 1 2e x
y
2
1 4e2 x
y
6.
8.
e x
2
3 e x
e x
y2
1
e x
1
y
y
2
e x
y
2
1 4e 2 x
ln x x
(ln x )
1
x
2
y
2 x1/2 y1/2 dy
2 x1/2 C1
dy
x 2 y1/2 dx
1
ln x
dx
2 y 3/2
3
sin x
x
2 x3
1 x4
y
2e x
x 2
y 1
e x tan 1 2e x
e ( ln 2) tan 1 2e ln 2
; y ( ln 2)
2
2 xe x ( x 2)
x sin x cos x
y
x2
cos( 2)
sin x; y 2
( 2)
1 x4
1
1 x
y 1
1 2e
xy
3 e 3x 2
2
y2
( x 3)
1
( x C)
y
e x tan 1 2e x
y
e x
2
ex
2 x3
1 x4
y
3 e 3x 2
2
1
1 x et
x2 1 t
3e x
y2
1
( x C )2
y
xy
x
1
1 x4 1
y
2
1
x
e x
2
e x
y
1
( x 3) 2
y
dt
3y
e x
y
y
1
x C
2y
1
x2
y
x 3
1 x et
x 1 t
e x
y
1
x2 y
4.
515
y
2
e x
1 cos x
x
x
2 tan 1 1 2 4
2
(2 2)e 2
2 xy; y (2)
sin x
x
y
2
1 4e 2 x
y
x
xy
sin x
2
0
y
0
x2 y
1
(ln x )2
2 y1/2 dy
x1/2
y 1/2 dy
Copyright
x2
ln x
x 1/2 dx
C , where C
x 2 dx
x2
(ln x )2
x2 y
2 y1/2 dy
xy
y 2 ; y (e )
e
ln e
e.
x 1/2 dx
1C
2 1
y 1/2 dy
x 2 dx
2014 Pearson Education, Inc.
2 y1/2
x3
3
C
2 y1/2
1 x3
3
C
516
Chapter 7 Integrals and Transcendental Functions
11.
dy
dx
ex y
12.
dy
dx
3x 2 e y
13.
dy
dx
y cos
e x e y dx
dy
dy
3 x 2 e y dx
y
dy
14.
dy
2 xy dx
2
15.
dx
1
y3 2
dy
ey
x dx
2 tan u
1
2 xy
dy
x1 2
dy
3
2
y
dx
e ye x
x
dy
dx
x
right-hand side, substitute u
e y dy
2 eu du
dy
e y dy
17.
18.
19.
2x 1 y2
| y|
1
dy
dx
e2 x
ex
2y
dy
y 2 dx
1 ln
3
20.
dy
dx
dy
dx
y
e2 x
ex
dy
2e x
y
y
dy
e ye x
x
dx
x
du
1
2 x
e y
2
e
2 du
2e x
3
x
3 x
dy
ey
x3
2 y1 2 dy
e y dy
dx
x
1
x
C
x
e
x
x 1 2 dx
3C
2 1
dx. In the integral on the
dx, and we have
esin x
C1
e y dy
esin x cos x dx
C , where C
dy
2 x dx
1 y2
C
dx. In the integral on the left-
C , where C
C , where C
e y
C1
dy
x 1 2 dx
e y esin x cos x dx
dy
esin x
y
x3 C
ex
dy, and we have
y
e y dy
e y
C1
sec2 y
2 y1 2 dy
dx
dx
ey
3 x 2 dx
ey
C
C
3C
2 1
e y sin x cos x
y3 2
dx
C where C
3x 2 y3 6 x 2
y 2 dy
1 y2
C1
sin 1 y
2 x dx
dy
e2 x e y
e xe y
ex
e2 y
dx
e2 y dy
dx
e x dx
e2 y dy
x 2 C since
1 x2
2
e x dx
2C1
3x 2 y 3 2 dx
y2
y
3
2
dy
3 x 2 dx
y2
y
3
2
dy
x3 C
xy 3 x 2 y 6
ln| y 3|
3 x
ex
sin x 2 C
y
y
1
x
ey
dx
1
y
2 du
2 y3 2
2eu
dy
x 2 tan y
2 x 1 y 2 dx
dy
y
dy
e x dx
e y dy
sec2 y
2 ydy
esin x cos x dx
dy
dx
e
e y
e y sin x
16. (sec x) dx
1
2 y
du
e y dy
3 x 2 dx
y dx
x C
C1
1
2
e x dx
e y dy
y cos 2
hand side, substitute u
sec 2 u du
e y dy
( y 3)( x 2)
1 dy
y 3
( x 2)dx
1 dy
y 3
( x 2)dx
2x C
Copyright
2014 Pearson Education, Inc.
3 x 2 dx
e2 y
2
ex
C1
Section 7.2 Exponential Change and Separable Differential Equations
21.
1 dy
x dx
ye x
2
2 ye x
1
y y 2
22.
dy
dx
ex y
ey
1 ey
23. (a)
(c)
24. (a)
e y 1
y
y0 e
(20,000) k
dp
dh
kp
p
(c) 900 1013e
25.
dy
dt
26.
A
0.6 y
A0 e kt
0.2
C
y
1
1
y 2 y
dy
2
ex
y
e x 1 dx
ln 1 e y
ln 0.99
1000
y0 (0.82)
4 ln
dy
x C
ln (0.9)
p0 ekh where p0
2
1
e
k
( 0.00001)t
2
xe x dx
dy
1 ex
2
2
ex
y0 e1000k
2
L( x)
y
ex
y0 e 0.6t ; y0
L0
2
ln (0.9)
0.00001
t
y
L0 e 18k
Therefore, y
30.
y
y 100e 0.6t
y 100e 0.6
0.121
0.9777 km
54.88 grams when t 1 hr
A 1000e(ln (0.8) 10)t , where A represents the amount of sugar
e kt
e(ln 4)t
y
e24 ln 4
at y
ln 2
18
k
ln10
2 and t
424
0.0385
0.0385 x
e5 k
4e3k
e2k
10, 000 8 y0
y0
10,000
8
1250
7500
ek
0.75
ln 0.1 (ln 0.75)t
t
ln 0.1
ln 0.75
0.5 we have 2
585.35 kg
L( x)
x
L0 e 0.0385 x ; when the intensity is
59.8 ft
e0.5k
ln 2
t
40 ln (0.1)
0.5k
k
92.1 sec
ln 2
0.5
ln 4.
2.81474978 1014 at the end of 24 hrs
y0 e3k ; also y (5)
10, 000
4 y0 e3k
(b) 1 10, 000e(ln 0.75)t
18k
L0 e 0.0385 x
y
31. (a) 10, 000e k (1)
C
x C
ln (90) ln(1013)
20
k
ln(1013) ln(900)
0.121
ln 12
1
y0 e kt and y (3) 10, 000
y0 e5k
2
e x 1 dx
dy
V0 e t 40 when the voltage is 10% of its original value
0.1V0
y0 e kt and y0
ex
2
10,536 years
h
ln (0.8)
10
k
L
29.
y
82%
900
ln 1013
100
one-tenth of the surface value, 100
28. V (t ) V0 e t 40
1
4 ln
0.00001
1013; 90 1013e 20k
0.121h
800 1000e10 k
L0 e kx
C
1
e
that remains after time t. Thus after another 14 hrs, A 1000e(ln (0.8) 10)24
27.
2
xe x dx
2.389 millibars
( 0.121) h
y
y
ln 1 e y
y0 e
p 1013e 6.05
(b)
2 ln
e y 1 ex 1
0.99 y0
e( 0.00001)t
1
y 2 y
y 2 y
e x 1 dx
dy
(b) 0.9
2
2
y0 ekt
y
ex
xe x dx
dy
ex
2
517
40, 000
ln 2. Thus, y
y0 e5k . Therefore
y0 e(ln 2)t
10, 000
y0 e3ln 2
y0 eln 8
4
k
k
ln 0.75 and y 10, 000e(ln 0.75)t . Now 1000 10, 000e(ln 0.75)t
8.00 years (to the nearest hundredth of a year)
ln 0.0001 (ln 0.75)t
t
ln 0.0001
ln 0.75
32.02 years (to the nearest hundredth
of a year)
Copyright
2014 Pearson Education, Inc.
518
Chapter 7 Integrals and Transcendental Functions
32. Let z
dz
dt
r ky. Then
k
dy
dt
k ( r ky )
kz. The equation dz / dt
1
r ce kt .
k
1
(a) Since y (0) y0 , we have y0
( r c ) and thus c
k
1
r
r
y
r [ r ky0 ]e kt
y0
e kt
.
k
k
k
kz has solution z
ce kt , so
ce kt and y
r ky
(b) Since k
0, lim
r
e kt
k
y0
t
r
k
r ky0 . So
r
.
k
y
y
r /k
y
y0
t
33. Let
y (t ) be the population at time t, so t (0) 1147 and we are interested in t (20). If the population
continues to decline at 39% per year, the population in 20 years would be 1147 (0.61) 20
species would be extinct.
34. (a) We will ignore leap years. There are (60)(60)(24)(365)
0.06
1, so the
31,536,000 seconds in a year. Thus, assuming
314,419,198ekt , with t in years, and
exponential growth, P
31,536,000
314,419,199
ln
0.0083583.
12
314,419,198
(You don t really need to compute that logarithm: it will be very nearly equal to 1 over the denominator
of the fraction.)
314,419,199
314,419,198e12k /31,536,000
k
(b) In seven years, P 314,419,198e (0.0083583)(7) 333,664,000 . (We certainly can t estimate this
population to better than six significant digits.)
35. 0.9 P0
36. (a)
P0 ek
k
0.2 P0
P0 e
dp
dx
1
100
p (100)
(b)
e
xp( x)
(ln 0.9)t
1 dx
100
r ( x)
ln p
ln (0.2)
1
100
C1e( 0.01)(100)
20.09
54.61e( 0.01)(10)
(ln 0.9)t
t
ln 0.2
ln 0.9
x C
p
e( 0.01x C )
C1
20.09e
54.61
$49.41, and p(90)
54.61e( 0.01)(90)
r ( x)
0
eC e 0.01x
p( x)
$22.20
p( x) xp ( x );
(54.61 .5461x)e 0.01x . Thus,
54.61 .5461x
x 100. Since
r 0 for any x 100 and r 0 for x 100,
then r ( x) must be a maximum at x 100.
Copyright
0.2 P0
15.28 yr
.5461e 0.01x
p ( x)
r ( x)
0.2
dp
p
p
20.09
p(10)
(c) r ( x)
ln 0.9; when the well s output falls to one-fifth of its present value P
(ln 0.9)t
2014 Pearson Education, Inc.
C1e 0.01x ;
54.61e 0.01x (in dollars)
Section 7.2 Exponential Change and Separable Differential Equations
37.
A0 e kt and A0
A
A 10ekt , 5 10ek (24360)
10
then 0.2(10) 10e 0.000028454t
38.
A0 e 0.00499t
0.05 A0
39.
A0 e139k
A0 e kt and 12 A0
A
y0 e kt
y
y0 e ( k )(3 k )
40. (a)
A
A0 e kt
(b)
1
k
3.816 years
(c) (0.05) A
ln
7
4
e10 k
65
43. T Ts
33 Ts
46 Ts
y0
20
0.00499; then
(0.05)( y0 )
after three mean lifetimes less than 5% remains
0.262
ln 2
2.645
ln 20
t
20 C, T
t
2.645ln 20
ln 2
t
60 C
60 20
27.5 min is the total time
T0 Ts e kt , T0
(b) T Ts
30
ln(0.5)
139
11.431 years
70e 10 k
4
7
e 10 k
0.05596
10
65
56563 years
k
ln 2
2.645
k
90 C, Ts
(a) 35 20 70e 0.05596t
reach 35 C
42. T
y0
ln 2 t
2.645
A exp
A 10e 0.000028454t ,
0.000028454
600 days
e3
e 2.645k
1
2
e139k
1
2
ln 0.05
0.00499
y0 e 3
T0 Ts e kt , T0
41. T Ts
k
t
ln 0.2
0.000028454
t
ln (0.5)
24360
k
519
65 e kt
T0
15 C
65
65 e 10k and 50
35
ln 2
10
and 30
To Ts e kt
39 Ts
60
e 20k
k
35 15 105e 0.05596t
90 C, Ts
T0 65 e 10 k and 15
2
it will take 27.5 10 17.5 minutes longer to
T0
65
T0 65 e 20 k simultaneously
T0 65
30 e10
10 k
e
ln 2
10
T0 65
t
13.26 min
T0 65 e 20 k . Solving
T0 65 e 10 k
T0
65
2 T0 65 e 20k
30 eln 2
5
2
e 10 k
1518 79Ts Ts2
33 Ts
46 Ts
46 Ts e 20 k
46 Ts e 10 k and 33 Ts
39 Ts 2
46 Ts
1521 78Ts Ts2
33 Ts
Ts
3
46 Ts
Ts
39 Ts
39 Ts
46 Ts
e 10k and
2
3C
44. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room
temperature the silver will be 120 min from now, and t0 the time the silver will be 10°C above room
temperature. We then have the following time-temperature table:
time in min. 0
temperature Ts
T Ts
70
T0 Ts e kt
(a) T Ts
20 (Now) 35
Ts 60 Ts
60 Ts
T0 Ts e 0.00771t
140
t0
x Ts y Ts 10
Ts
Ts
70 Ts
x
Copyright
Ts
Ts e 20k
70 Ts
70e 20k
k
Ts e (0.00771)(35)
x
60
2014 Pearson Education, Inc.
ln 76
0.00771
70e 0.26985
53.44 C
1
20
520
Chapter 7 Integrals and Transcendental Functions
T0 Ts e 0.00771t
(b) T Ts
y
70e 1.0794
(c) T Ts
y
Ts
70 Ts
Ts e (0.00771)(140)
23.79 C
T0 Ts e
ln 17
Ts
0.00771t
0.00771t0
Ts 10
Ts
1
0.00771
t0
70 Ts
ln 17
252.39
Ts e (0.00771)t0
252.39 20
10
70e 0.00771t0
232 minutes from now the
silver will be 10°C above room temperature
1c
c0 e k (5700)
k
2 0
ln(0.445)
6659 years
0.0001216
45. From Example 4, the half-life of carbon-14 is 5700 yr
c0 e 0.0001216t
c
(0.445)c0
46. From Exercise 45, k
(a) c
c0 e
c0 e 0.0001216t
t
(0.17)c0
c0 e 0.0001216t
t
14,571.44 years
(b) (0.18)c0
c0 e 0.0001216t
t 14,101.41 years
12,101 BC
(c) (0.16)c0
0.0001216t
t 15,069.98 years
13, 070 BC
47. From Exercise 45, k
y
ln(0.995)
0.0001216
49. e (ln 2/5730)t
50. (a)
0.0001216 for carbon- 14
y0 e 0.0001216(5000)
48. From Exercise 45, k
t
0.0001216
0.0001216 for carbon-14.
0.0001216t
c0 e
ln 2
5700
0.5444 y0
y
y0
y
12,571 BC
y0 e 0.0001216t . When t
0.5444
0.0001216 for carbon-14. Thus, c
5000
approximately 54.44% remains
c0 e 0.0001216t
(0.995)c0
c0 e 0.0001216t
41 years old
0.15
e (ln 2/5730)(500)
ln 2
t
5730
ln(0.15)
t
5730ln(0.15)
15,683 years
ln 2
0.94131, or about 94%.
(b) We ll assume that the error could be 1% of the original amount. If the percentage of carbon-14 remaining
5730ln(0.93131)
were 0.93131, the Ice Maiden s actual age would be
588 years.
ln 2
Copyright
2014 Pearson Education, Inc.
Section 7.3 Hyperbolic Functions
7.3
HYPERBOLIC FUNCTIONS
3
4
1. sinh x
coth x
2. sinh x
cosh x
5,
3
1
tanh x
4
3
1
3 2
4
1
cosh x
4,
5
and csch x
sech x
1 sinh 2 x
cosh x
3,
5
1 sinh 2 x
sech x
1
cosh x
3. cosh x
17 , x
15
0
sinh x
coth x
1
tanh x
17 ,
8
sech x
4. cosh x
13 , x
5
0
sinh x
coth x
1
tanh x
13 ,
12
sech x
ln x
2 e 2e
6. sinh (2 ln x)
e2 ln x e
2
7. cosh 5 x sinh 5 x
e5 x e
2
9. (sinh x cosh x)4
ex e
2
15 ,
17
1
cosh x
5 ,
13
elnx
1
eln x
2
2
eln x
5x
eln x
2
e5 x e
2
x
ex e
2
sinh( x x)
(b) cosh 2 x
ex e
2
1
4
y
2e x
2e x
6sinh 3x
1
4
dy
dx
4e 0
17 2
15
1
x
1 (4)
4
6 cosh 3x
1
3
x
289
225
4
3
5
3
144
25
12 ,
5
1
tanh x
4 , coth x
5
1
sinh x
5
12
3,
5
5,
4
tanh x
sinh x
cosh x
8. cosh 3 x sinh 3x
e3 x e
2
12
5
13
5
sinh x
cosh x
8
15
17
15
12 ,
13
1
x
x2
1
x4 1
2 x2
x2
4
ex
1
4
3x
e3 x e
2
3x
e4 x
ln cosh 2 x sinh 2 x
ex
e x
ln1 0
2sinh x cosh x
cosh 2 x sinh 2 x
ex
e x
ex
1
2 cosh 3x
Copyright
8 ,
15
tanh x
1
and csch x
2
64
225
1
169
25
x
4
ex e
2
3
4
5
4
sinh x
cosh x
tanh x
sinh x
cosh x
tanh x
15
8
cosh x cosh x sinh x sin x
2
5,
4
4
3
1
sinh x
sinh x cosh x cosh x sinh x
cosh( x x)
12. cosh 2 x sinh 2 x
5,
3
e5 x
10. ln(cosh x sinh x) ln(cosh x sinh x )
11. (a) sinh 2 x
1
sinh x
25
9
2
5x
x
25
16
and csch x
cosh 2 x 1
1
cosh x
9
1 16
3
4
cosh 2 x 1
ln x
2ln x
1 16
9
1
sinh x
and csch x
5. 2cosh (ln x)
13.
521
2014 Pearson Education, Inc.
e x
ex
e x
e 3x
8 ,
17
522
Chapter 7 Integrals and Transcendental Functions
dy
dx
14.
y
1 sinh
2
15.
y
2 t tanh t
16.
y
t 2 tanh 1t
17. y
ln(sinh z )
dy
dz
cosh z
sinh z
coth z
18. y
ln(cosh z )
dy
dz
sinh z
cosh z
tanh z
19.
(sech )(1 ln sech )
20.
21.
y
y
y
2x 1
y
sech tanh
sech
ln sech
tanh
csch coth
1 ln csch
csch
ln csch
coth
x 2 1 sech ln x
24.
y
4 x 2 1 csch ln 2x
25.
y
y
(1
4 x2 1
) tanh 1
(1
)
tanh t
t
sech 2 1t 2t tanh 1t
tanh
1 ln sech
1 ln sech
1 ln csch
csch coth
1 1 ln csch
tanh v sech 2 v
tanh v
csch 2 v
x2 1
1/ 2
x
1 x
2
Copyright
dy
dx
2
x x
1
4 x2 1
ln 2 x
1/ 2 2
1
(sech tanh ) 1
csch coth
ln x
1
2
1
sech
sech 2 t
coth v
coth v csch 2 v
x2 1
2x
x2 1
coth 3 v
1/2
cosh 1 2 x 1
dy
d
2t tanh t 1
sech
2 coth v
2
eln 2 x e
dy
dx
tanh t1/2 t 1/2
tanh 3 v
1
2
2
eln x e
t2
2 tanh v sech 2 v
1
2
cosh v
sinh v
x2 1
sinh 1 x1/2
cosh 1 2 x 1
coth
(coth v) coth 2 v
y
27.
dy
dv
ln sinh v 12 coth 2 v
2t1/2
csch coth
csch
csch
(tanh v) tanh 2 v
23.
y
(1 ln sech )
sinh v
cosh v
1 t 1/2
2
t 2
sech tanh
sech
csch
dy
dv
ln coth v 12 tanh 2 v
sinh 1 x
sech 2 t 1
dy
d
1 ln csch
cosh(2 x 1)
sech 2 t1/2
dy
d
sech tanh
csch
dy
dt
dy
dt
t 2 tanh t 1
(coth v) 1 csch 2 v
26.
cosh(2 x 1) (2)
2t1/2 tanh t1/2
(tanh v) 1 sech 2 v
22.
1
2
2
2 x (2 x )
1
2 x 1 x
(2)
1
2
4 x2 1
4x
4 x2 1
1
2 x (1 x )
( x 1)
1
1/ 2
2
2( x 1)1/ 2
( 1) tanh 1
1
dy
dx
2x
1
1
1
x 1 4x 3
tanh 1
2014 Pearson Education, Inc.
1
4 x2 7 x 3
2
4x
dy
dx
4
Section 7.3 Hyperbolic Functions
28.
( 2
y
2
2
1)
dy
d
2) tanh 1 (
1)
2 ) tanh 1 (
2
2
(2
29.
y
(1 t ) coth 1 t
30.
y
1 t 2 coth 1 t
31.
y
cos 1 x x sech 1 x
32.
y
ln x
dy
dx
33.
34.
y
y
1 x2
dy
dt
csch 1 2
1 (
dy
dt
dy
dx
1 x2
ln x
1 x2
1
1
x 1 x
1
2
2
ln
dy
d
1
2
1/2
1 x2
2
35.
y
sinh 1 (tan x)
dy
dx
36.
y
cosh 1 (sec x)
dy
dx
1 2
1/2
1
sec 2 x
1 (tan x )
sec 2 x
2
(sec x )(tan x )
coth 1 t
1 2t coth 1 t
2 x sech 1 x
1
2
1
1
1 x2
1 x2
sech 1 x
sech 1 x
1
x
x
1
x
1 x2
sech 1 x
x
1 x2
sech 1 x
ln 2
2
1
sec 2 x
|sec x|
2
sec x
(sec x )(tan x)
2
2
sec x 1
tan x
cosh x
1 sinh 2 x
(b) If y
dy
sin 1 (tanh x ) C , then dx
sech 2 x
sech 1 x 12 1 x 2
1
2 t
1
2
2
1 22
dy
tan 1 (sinh x) C , then dx
x2
2
( 1) coth 1 t1/2
2
ln 2
2
37. (a) If y
38. If y
1/ 2
1 t1/ 2
ln(1) ln(2)
1
2
(ln 2)2
dy
d
t
sech 1 x
1
2
1
1
2
(1) sech 1 x
x 1 x2
2
1
2
(1 t )
1
x
1)
1) 1
2t coth 1 t
1
1 t2
2) tanh 1 (
(2
1)2
2) tanh 1 (
(2
1 t2
1/2
csch 1 12
1
2
(1 t ) coth 1 t1/2
1 x 2 sech 1 x
1
x
2
523
2
1 tanh x
dy
C , then dx
|sec x||sec x|
|sec x|
(sec x )(tan x )
|tan x|
| sec x |
sec x, 0
x
2
cosh x
cosh 2 x
sech x, which verifies the formula
sech 2 x
sech x
sech x, which verifies the formula
x sech 1 x
x2
2
2x
1
x 1 x
2
4 1 x
2
x sech 1 x, which verifies the
formula
39. If y
x2 1
2
coth 1 x
x
2
dy
C , then dx
x coth 1 x
Copyright
x2 1
2
1
1 x2
1
2
x coth 1 x, which verifies the formula
2014 Pearson Education, Inc.
524
Chapter 7 Integrals and Transcendental Functions
40. If y
x tanh 1 x
1
2
dy
ln 1 x 2
tanh 1 x x
C , then dx
tanh 1 x, which verifies the
2x
1
2 1 x2
1
1 x2
formula
41.
42.
1 sinh
2
cosh u
2
sinh 2 x dx
sinh 5x dx
u du, where u
cosh 2 x
2
C
43.
2 dx
C
5 sinh u du, where u
5cosh u C
2 x and du
x
5
1 dx
5
and du
5cosh 5x C
6 cosh 2x ln 3 dx 12 cosh u du , where u
x
2
ln 3 and du
12 sinh u C 12sinh 2x ln 3
44.
45.
46.
47.
4 cosh (3 x ln 2) dx
tanh 7x dx
coth
3
4 cosh
3
4 sinh u
3
sinh u
7 ln e x /7
C
e x /7
u du, where u
3 cosh
sinh u
d
3 ln sinh u
C1
3 ln e / 3
e
sech 2 x 12 dx
/ 3
dt
csch ( ln t ) coth (ln t )
t
1 dx
7
x /7
7 ln e 2e
csch 2u du , where u
7 ln e x /7
/ 3
3 ln e / 3
/ 3
e
2
e / 3
dx
C
(5 x) and du
coth u C
2 sech t
dx
coth (5 x ) C
t
t1/2 and du
dt
2 t
C
csch u coth u du, where u
csch u C
3 ln e
x 12 and du
2 sech u tanh u du , where u
dt
C1
3
C1
3
x /7
d
and du
3 ln 2 C1
tanh x 12
3 dx
ln 2) C
C1
3 ln sinh
sech 2u du , where u
csch 2 (5 x) dx
sech t tanh t
t
3
C
3x ln 2 and du
and du
7 ln cosh 7x
2( sech u ) C
50.
x
7
7 ln | cosh u | C1
( coth u ) C
49.
4 sinh(3 x
3
C
7 cosh u du, where u
tanh u C
48.
u du, where u
1 dx
2
ln t and du
dt
t
csch(ln t ) C
Copyright
2014 Pearson Education, Inc.
C1
C
e x /7
7 ln 2 C1
Section 7.3 Hyperbolic Functions
51.
ln 4
ln 2
x
ln 4 cosh x
ln 2 sinh x
coth x dx
ln 2
u
eln 2 e
2
sinh(ln 2)
15/8
ln 2
x
0
ln2
53.
2 ln 2
4e
0
1
2
17/8
ln 2
ln 4
e
sinh
/4
ln 17
8
e
e
2
2e
1
8
0
e
4e
0
0 e2
cosh(tan ) sec2
2 ln 2 81
1
d
1
/2
0
2 cosh(ln t )
t
ln 2
dt
1
0
58.
4 8cosh x
x
1
dx 16
2
1
16 sinh u
59.
0
ln 2
1
2
3
8
cosh 2 2x dx
0
ln 2
2
1
dx
3
8
x
1
t
e
eln 2 e
2
ln 2
2
0
1
2
e2 e
2
e
ln 2
2
eln 2
1
e
1
2
1
1
2
2
1/2
e1
e e
2
1
2
1
e e
1
u
0, x
2
e
1, x
1
e
u 1,
4
e e
1
u 1
2
2
u
ln 2
3
4
1
, x 1
8 e2 e
sinh x x
ln 2
0
2
dx
2 x
dx
u
4
e e
0, x
2
2
1)dx
1
u
0
x
d , x
1
2
ln 2
Copyright
ln 2
2
cos d , x
2 e 2e
ln 2
x1/2 , du
(0 0)
sec2
dt , x 1
1 0
(cosh x
2 ln 2
1
2
ln 4
3
32
2
e1 e
2
sin , du
16(sinh 2 sinh1) 16
(sinh 0 0) (sinh( ln 2) ln 2)
1 ln 2
2
ln t , du
sinh(ln 2) sinh(0)
cosh x 1
2
d
2(cosh1 cosh 0)
cosh u du where u
ln 2
ln 2 2ln 2
tan , du
0
ln 2
0
17
8
2
2 ln 2 14 1 ln 4 34
1
2
2 sinh u du where u
1
0
3
32
1 e 2
0
1
4
4
ln 4
e2
2
sinh(1) sinh( 1)
1
cosh u du where u
sinh u
ln 2
1
2sinh(sin ) cos d
2 cosh u
57.
1
15
8
2
2sinh (2 x) dx,
eln 4 e
2
1 d
ln 4
cosh u du where u
sinh u
56.
2
cosh 2 x, du
e2
ln 4
d
1
4
4
ln 4
ln 17
8
1
2
ln 2
ln 2
eln 4 e
2
sinh(ln 4)
cosh (ln 4)
ln1
1
32
e
2
u
du where u
d
cosh x dx;
ln 25
cosh (2 ln 2)
1
2
ln 4
2
ln 2
2 ln 2 e 2
/4
2 ln 4
d
2 ln 2
55.
u
ln | u | 1
d
ln 2
2
ln 2
ln 2
ln 4
ln 15
.4
8 3
1 17/8 1
2 1
u
0
cosh 0 1, x
2e cosh
ln4
e
54.
u
ln 34
sinh x, du
3,x
4
2
ln 2 sinh 2 x
dx
cosh 2 x
tanh 2 x dx
0
where u
1
2
2
ln 2
ln 15
8
ln | u | 3/4
52.
15/8 1
du
3/4 u
dx
525
2014 Pearson Education, Inc.
u 1, x
2
e e
4
u
2
1
0
ln 2
1
2
2
2
ln 2
1
2
1
1
4
ln 2
526
60.
Chapter 7 Integrals and Transcendental Functions
ln10
0
2
1
63. tanh
1
65. sech
1 3
5
5
12
5
12
ln
1
ln
2 3
4 x
1
1/3
2
3
6 dx
0
25
144
2
1 9 x2
2
1
5/4 1 x 2
(b) coth
70. (a)
1/2
0
dx
1 ln (9/4)
2
(1/4)
ln 3
66. csch
1
15
4
tanh
1
2
1
12/13
dx
x 1 16 x
4/5
2
sech
ln
72. (a)
(b)
2
1 12
13
5 3
4
dx
1 x 4 x2
1
2
csch
11
2
sech
ln
14
5
1/2
1
ln
25
9
3
1
ln 3
1 ln
2
9
ln 3
4/3
1/ 3
ln
3 2
3
3 dx, a 1
1
2sinh 1 1
0
2
1
15
4
2 coth
9/4
1/4
1
2
11
2
ln
1
3
tanh
1
0
tanh
11
2
1 ln 3
2
du
u a2 u2
, u
1
1
2
1 x 2
2 1
ln 2
1 (12/13) 2
(12/13)
ln 2
1
2
4 dx, a 1
1 12
13
sech
4/5
ln
1
2
4 x, du
12/13
ln 2 ln 32
csch 11
3x, du
tanh
0
sinh
1
3
5
3
9.9 2ln10
3 2
coth
5/4
13 5
12
csch
3 sinh 0
2 ln 1
2
sech 1u
(b)
ln
ln 3 ln
x
1
sinh
2 sinh 1 1 sinh
1 ln 1 1/2
2
1 1/2
11
2
3/13
1/5
1 5
4
0
x
2ln10
64. coth
12 1
1
1
10
2 ln10 10
ln10
0
ln 3
3
1 dx
, where u
0 a2 u2
1 1
coth
2 coth
1
1 x2
(b) tanh
71. (a)
1
dx
ln10
e
2 sinh x x
ln
3 1
u
(cosh x 1)dx
1 5
3
3
(b) 2sinh 1 1 2 ln 1
69. (a)
e
0
ln10
2
3
ln
1x 2 3
2 0
2sinh
ln10
2
62. cosh
1
sinh
ln
dx
(sinh 0 0)
1 (9/25)
(3/5)
dx
0
cosh x 1
2
1 ln 1 (1/2)
2
1 (1/2)
1
2
(b) sinh
68. (a)
4
0
sinh(ln 10) ln 10
61. sinh
67. (a)
ln10
4sinh 2 2x dx
sech
ln
2
3
1 (4/5) 2
(4/5)
1
Copyright
ln
13
169 144
12
ln 43
csch 11 csch
5/4
(1/2)
14
5
ln 1
2
11
2
1
2
csch
1 ln 2
2
1
11
2
5
2
2014 Pearson Education, Inc.
csch 11
ln
5
25 16
4
Section 7.3 Hyperbolic Functions
cos x
73. (a)
0
2
1 sin x
dx
0
1
0
1 u2
1
sinh
(b) sinh
1
0 sinh
1
e
dx
1 x 1 (ln x )2
74. (a)
0
1 du
0 a2 u2
1
0
u
0
ln 0
1
sinh
(b) sinh 1 1 sinh
du where u
0
1
sinh
0 1
1
0
0
1
0
1 dx, a
x
ln x, du
12 1
ln 1
cos x dx;
0 1
sinh 1 1 sinh
0
1
0 sinh
ln 0
, where u
u
sin x, du
1
sinh 1 1
0
02 1
ln 0
527
ln 1
2
f ( x) f ( x )
f ( x) f ( x )
f ( x) f ( x )
f ( x) f ( x)
2 f ( x)
and O( x)
. Then E ( x ) O ( x )
f ( x).
2
2
2
2
2
f x f ( x)
f ( x) f ( x )
f ( x ) f ( ( x ))
Also, E x
E ( x) E ( x) is even, and O ( x)
2
2
2
f ( x) f ( x )
O ( x) O ( x ) is odd. Consequently, f ( x) can be written as a sum of an even and an odd
2
f ( x) f ( x )
f ( x) f ( x)
f ( x) f ( x )
function. f ( x)
because
0 if f is even, and f ( x)
because
2
2
2
f ( x) f ( x)
2 f ( x)
2 f ( x)
0 if f is odd. Thus, if f is even f ( x)
0 and if f is odd, f ( x) 0
2
2
2
75. Let E (x )
76.
y
ey
ey
1
sinh
x
dv
dt
mg sech 2
lim v
t
160
0.005
t
gk
m
2x
t
sinh
ey
1
2 xe y
1
ey
x
y
ln x
t
gk
m
e2 y 1
x2 1
e2 y
2 xe y 1 0
Since e y
0, we cannot choose
0.
mg
k
sech 2
mg 1 tanh 2
gk
m
gk
m
gk
m
g sech 2
t . Thus
mg kv 2 . Also, since tanh x
t
0 when x
0, v
0.
lim
t
s (t )
gk
m
tanh
y
x2 1
x
mg
k
when t
78. (a)
ey
x2 1
m dv
dt
(c)
ey e
2
x
x 2 1 because x
77. (a) v
(b)
sinh y
4 x2 4
2
2x
x
x
mg
k
160,000
5
tanh
400
5
a cos kt b sin kt
kg
t
m
mg
lim
k t
80 5
ds
dt
tanh
kg
t
m
mg
k
mg
k
(1)
178.89 ft/sec
ak sin kt bk cos kt
d 2s
dt 2
ak 2 cos kt bk 2 sin kt
k 2 (a cos kt b sin kt )
k 2 s (t )
acceleration is proportional to s. The negative constant k 2
implies that the acceleration is directed toward the origin.
(b) s (t )
a cosh kt b sinh kt
ds
dt
2
ak sinh kt bk cosh kt
d 2s
dt 2
ak 2 cosh kt bk 2 sinh kt
k 2 ( a cosh kt b sinh kt ) k s (t ) acceleration is proportional to s. The positive constant k2 implies
that the acceleration is directed away from the origin.
Copyright
2014 Pearson Education, Inc.
0
528
Chapter 7 Integrals and Transcendental Functions
2
79. V
ln 3
sech 2 x dx
80. V
2
81.
1 cosh 2 x
2
y
0
1 e2 x e
2
2
82. (a)
(b)
(c)
(d)
2x
y
ln 5
x
lim tanh x
x
x
x
lim sinh x
x
lim
ln 3
2
ln 5
3 1/ 3
3 1/ 3
x
ex
lim
x
ex
x
x
x
x
x
ex
x
(f )
(g)
(h)
(i)
83. (a)
lim sech x
x
x
0
lim coth x
x
0
lim csch x
y
H
w
lim
ex e
ex e
x
lim coth x
x
2
ex e
x
lim coth x
x
lim
x
x
x
e
ex
2
x
ex
2
e
x
x
x
x
lim e x e
x
0 e
x
lim
w x
cosh H
e
lim
2
ex e
tan
0 e
x
lim
x
x
lim
dy
dx
ex
H
w
(b) The tension at P is given by T cos
H cosh Hw x
84. s
1 sinh ax
a
1 sinh 2 ax
a
sinh ax
1
1
a
as
a2s2 1
ax
1
ex
1
ex
0 e
x
x
x
2
1
ex
w
H
x
lim
e2 x 1
e2 x 1
1
0 1
0 1
1
0
1
ex
1
ex
lim
x
1
1
1
e2 x
1
e2 x
1 0
1 0
1
1
2x
lim e2 x 1
ex
ex
. ex
e
x
0 e
1
lim
2e x
e2 x 1
sinh Hw x
T
H sec
0
0 1
0
sinh Hw x
H 1 tan 2
H 1
w x
sinh H
wy
sinh 1 as
s2
ex
0
1 0
2x
lim e2 x 1
H
w x
w Hw cosh H
1
x
ex
ex
1
x
ex
1
e2 x
ex
ex
1
ex
ex
ex
1 0
1 0
0
x
1
ex
1
ex
1
x
1
e2 x
1
e2 x
2
lim
ex
1
ex
x
0 e
x
x
ex
x
1
ex
1
lim
1
2ex
lim
1
ex
1
ex
ex
x
e
1
ex
ex
lim
x
x
1 sinh 2 x ln 5
2
0
0
2
ex
1
ex
2
ex
x
x
0 e
cosh 2 x dx
1
ex
lim
lim
2
lim
x
lim e x e
x
0
ex
1
ex
1
ex
1
ex
1
ex
1
ex
lim
1
ex
ex
x
ex
x
ex e
2
lim
ex
lim
lim
ex
1
ex
1
ex
1
(e)
ln 5
1 (sinh 2 x ) 2 dx
0
6
5
ex e
ex e
ex e
2
lim
2
L
5 15
lim
1 dx
tanh x 0
ex e
ex e
lim
x
x
1
4
0
lim sinh x
2
0
sinh 2 x
lim tanh x
x
2
cosh 2 x sinh 2 x dx
0
x
1 sinh 1 as;
a
y
1 cosh ax
a
1
a2
Copyright
2014 Pearson Education, Inc.
1
a
cosh 2 ax
2
Section 7.4 Relative Rates of Growth
1
a
85. To find the length of the curve: y
cosh ax
y
sinh ax
b
1 sinh ax
1 sinh ab. The area under the curve is A
a
a
0
1 1 sinh ab which is the area of the rectangle of height
a a
b1
0 a
1
a
below.
x 2 1 from A to T
1 cosh
2
(b) A(u )
(c)
7.4
1 cosh 2 u
2
A (u ) 12
cosh u
u sinh u
x 2 1 dx
1
1 sinh 2 u
2
A(u ) u2
sinh 2 u
1
2
A(u )
sinh ax
b
0
1
a2
0
cosh ax dx
sinh ab
area of the triangle OTP minus the area
cosh u
x 2 1 dx.
1
cosh 2 u sinh 2 u
cosh 2 u sinh 2 u
1
2
1
a2
b
L
and length L as claimed, and which is illustrated
cosh u sinh u
1
2
A (u )
1 (sinh ax)2 dx
0
cosh ax dx
86. (a) Let the point located at (cosh u , 0) be called T. Then A(u )
under the curve y
b
L
529
1
2
C , and from part (a) we have A(0)
cosh 2 u 1 sinh u
1
2
(1)
0
0
C
A(u )
u
2
u
2A
RELATIVE RATES OF GROWTH
1. (a) slower, lim
x
(b) slower, lim
x
x 3
ex
x3 sin 2 x
ex
x
(d) faster, lim
x
(e) slower, lim
x
(f ) slower, lim
x
(g) same, lim
x
(h) slower, lim
x
2. (a) slower, lim
x
(b) slower, lim
x
x
x1/ 2
ex
x
4x
ex
x
x
ex/ 2
ex
x
lim
ex
2
lim
ex
x
log10 x
e
x
x ln x x
ex
x
1
2
x
1
ex / 2
1/ 2
lim
ex
x
4
e
since
1
2 xe x
6 4sin 2 x
ex
lim
x
2
ex
0
lim
x
0 by the Sandwich
10
ex
0
1
0 since
3
2e
1
0
1
2
ln x
(ln10)e x
lim
x
lim
x
lim
x
for all reals, and lim
x
1
2
10 x 4 30 x 1
ex
x
10
ex
3 x
2e
lim
e
6 x 2 cos 2 x
ex
lim
ex
lim
x
4
e
lim
3 x
2
x
3 x 2 2sin x cos x
x
6 4sin 2 x
ex
lim
ex
0
lim
2
ex
Theorem because
(c) slower, lim
1
ex
lim
x
lim
x
40 x3 30
ex
x (ln x 1)
ex
Copyright
lim
x
120 x 2
ex
ln x 1 x
ex
1
(ln10) xe x
lim
(ln10)e x
x
lim
x
1
x
1
x
lim
x
lim
x
240 x
ex
0
240
ex
lim
x
ln x 1 1
ex
2014 Pearson Education, Inc.
lim
x
0
ln x
ex
lim
x
1
x
x
e
lim
x
1
xe x
0
530
Chapter 7 Integrals and Transcendental Functions
1 x4
ex
(c) slower, lim
x
x
x
e x
ex
x
1
e2 x
lim
x
xe x
ex
(f ) faster, lim
5 x
2e
lim
e
x
(e) slower, lim
x
5 x
2
x
(d) slower, lim
1 x4
e2 x
lim
4 x3
2 e2 x
lim
x
5
2e
0 since
lim
x
x
e 1
ex
0
(h) same, lim
x
3. (a) same, lim
x
e1
ex
x
e
ex
x 4 x3
x2
(f ) slower, lim
2x
x2
x
(g) slower, lim
x
(h) same, lim
x
4. (a) same, lim
x
x
(f ) slower, lim
(g) faster, lim
x
(h) same, lim
x
ecos x
ex
0
2
2
1
lim 1
x
2
2
lim
1
x
lim
1 lim 1
ln10 x
x2
0
1 1
1
0
1
x
1
x
(ln 2)2 2 x
2
lim
x
x lim 1
ex x
ex
0
lim 8 8
lim 1
x
1
x3/ 2
1
lim 10 10
x
x 2e
x2
x
0
ln x 2
ln10
lim
x2
1
10
2
1
ex
lim
x
log10 x 2
1 lim 2 ln x
ln10 x
x2
x2
x
1
2 lim x
ln10 x
2x
lim ( x 1)
x
x
x
(1.1) x
x
24
162 x
e1
ex
x
x3 . x 2
x2
(e) faster, lim
x
lim
x
x
lim
x
(ln 2)2 x
2x
x
10 x 2
x2
(d) slower, lim
ln x
x
x
x2
x
1
e
x 4 x3
x4
lim
x
x2
(c) slower, lim
1
e
x
2 x 3
2x
lim
xe
x2
lim
x
lim
x
8 x2
x2
x
(b) same, lim
x
2
3
x 1)
lim
x
x ln x
x
ecos x
ex
e1
lim x3 1
2
(e) slower, lim
ecos x
x
( x 3) 2
x
1
2x 4
2x
lim
(c) same, lim
x
e( x
x
x5 x 2
x2
e 1
x
lim
x2 4 x
x2
x
e 1
ex
lim
0
0
, so by the Sandwich Theorem we conclude that lim
x
(b) slower, lim
(d) same, lim
x
lim x
x 1
x
24 x
8e 2 x
x
lim
x
lim
x
1
(g) slower, since for all reals we have 1 cos x 1
lim
12 x 2
4e 2 x
2
x 2 100 x
x2
lim
x
lim
x
1
10 x x 2
(ln1.1)(1.1) x
2x
lim 1
x
0
100
x
Copyright
lim
x
(ln1.1)2 (1.1) x
2
1
2014 Pearson Education, Inc.
and also
0
Section 7.4 Relative Rates of Growth
5. (a) same, lim
x
log3 x
ln x
ln 2 x
ln x
(c) same, lim
ln x
ln x
(d) faster, lim
x
ln x
x
(e) faster, lim
x
ln x
x
(f ) same, lim
5ln x
ln x
x
x
x
x
ex
ln x
x
6. (a) same, lim
x
(b) same, lim
x
lim
x1/ 2
ln x
lim
1
ln x
1
2
lim
x
1
2
lim
1
2
1/ 2
x
lim
1
x
x
x
x
2 x
x
2
lim
x
lim x
1
x
x
lim 5 5
x
1
x ln x
lim
x
lim xe x
1
x
x
ln x 2
ln 2
lim
lim
x
ln10 x
ln10
x
1 lim 2 ln x
ln 2 x
ln x
1 lim ln10 x
ln10 x
ln x
ln x
1
x (ln x )
lim
ln x
1 lim ln x 2
ln 2 x
ln x
ln x
x
1
x
0
ex
lim
x
log10 10 x
ln x
x
1
ln 3
1
ln x
x
log 2 x 2
ln x
(c) slower, lim
1
2
lim
ln x
x
(h) faster, lim
x
1
x
(g) slower, lim
x
2
2x
1
x
lim
1
ln 3
lim
ln x
x
(b) same, lim
x
ln x
ln 3
lim
1 lim
ln 2 x
10
10 x
1
x
1 lim
ln10 x
2
1 lim
ln10 x
2
ln 2
1
1
ln10
0
1
x2
(d) slower, lim
lim
ln x
x
x
x 2 ln x
ln x
(e) faster, lim
x
lim
x
(f ) slower, lim
e x
ln x
(g) slower, lim
ln(ln x )
ln x
x
x
(h) same, lim
x
ex
ex /2
7. lim
x
lim
x
ln x
e
e
ln(2 x 5)
ln x
lim e x /2
x
xx
(ln x ) x
x
x /2 x
lim
lim
x
x
x
1
x 2 ln x
x
ln x
1
e x ln x
lim
1/ x
ln x
1
x
lim
2
2x 5
1
x
x
x
0
2
lim
x
x
ln x
2
lim
x
1
2
1
x
lim x
x
2
0
lim
x
1
ln x
lim
x
2x
2x 5
0
lim
x
2
2
lim 1 1
x
e x grows faster then e x /2 ; since for x
ee we have ln x
e and
lim
ln x x
e
(ln x) x grows faster then e x ; since x
lim
x x
ln x
x x grows faster then (ln x ) x . Therefore, slowest to fastest are:
x
x
x
ln x for all x
, e , (ln x) , x x so the order is d, a, c, b
Copyright
2014 Pearson Education, Inc.
0 and
531
532
8.
Chapter 7 Integrals and Transcendental Functions
lim
(ln 2) x
x2
x
x
2
x 2 ; lim x x
x
2x
(ln 2)2 x
x
lim
2
2
ln(ln 2) (ln 2) x
2x
lim
x
ln(ln 2) (ln 2) x
2
lim
x
2
(ln 2)2 2 x
lim
x
0
ln(ln 2)
2
2
lim (ln 2) x
x
x
x
x
x
x
x 5
(c) true; x
x 5
(d) true; x
2x
x
2x
ln x
ln 2 x
1
x
(b) true;
x
1
x
1
2 if x 1 (or sufficiently large)
lim 1 1
x
x 5
x
1
5
x
6 if x 1 (or sufficiently large)
1 if x 1 (or sufficiently large)
1
x
ex
1
2 cos x
2
x
ex
x ln x
x2
x
ln(ln x )
ln x
(h) false; lim
x
3
1
1
x
lim 1
1
x
x
(f ) true; lim
2 if x 1 (or sufficiently large)
x2
(d) true; 2 cos x
x
1
x
1
x
x
and
x
ln x
ln x
x
ex
ln x
x
lim
3
2
1
if x is sufficiently large
0 as x
lim
x
1
x
ln x
ln( x 2 1)
L 1
g ( x)
f ( x)
O( f ).
g
2 if x is sufficiently large
0
1
1 if x is sufficiently large
sufficiently large
1
x
ex
1
lim
x
1
x
2x
x2 1
lim
x
x2 1
2 x2
11. If f ( x) and g ( x) grow at the same rate, then lim
1
L
, x 2 , 2 x and e so the order is c, b, a, d
1
x2
(c) false; lim
(g) true;
x
x
1
x
2
2x
lim
x
x 3
1
x
e
2x
0
0
1
( x 5) 2
x
1
x 3
1
x
10. (a) true;
(e) true;
ln x
x
x2 5
x
(h) true;
x
1 if x 1 (or sufficiently large)
1
ex
lim
1
x
x
2
e
1 if x 1 (or sufficiently large)
x
x ln x
x
(g) false; lim
lim
1
x
x 5
e
e2 x
x
(f ) true;
1
1
x
(e) true; lim
e
x
1
x
(b) false; lim
(ln 2) x grows slower than
x 2 grows slower than 2 x ; lim 2 xx
grows slower than e . Therefore, the slowest to the fastest is: ln 2
9. (a) false; lim
0
x
f ( x)
g ( x)
L 1
x
f ( x)
g ( x)
lim
x
f ( x)
g ( x)
1
2
L
1
2 x2
0
1
2
lim
x
1
L
L 1 if x is sufficiently large
12. When the degree of f is less than the degree of g since in that case lim
x
Copyright
g ( x)
f ( x)
f ( x)
g ( x)
2014 Pearson Education, Inc.
0.
0. Then
f
f ( x)
g ( x)
L
1 if x is
O ( g ). Similarly,
Section 7.4 Relative Rates of Growth
13. When the degree of f is less than or equal to the degree of g since lim
than the degree of g, and lim
x
x
f ( x)
g ( x)
a
b
f ( x)
g ( x)
533
0 when the degree of f is smaller
(the ratio of the leading coefficients) when the degrees are the same.
14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the
same degree grow at the same rate.
15.
16.
17.
lim
ln( x 1)
ln x
lim
ln( x a )
ln x
lim
10 x 1
x
x
x
x
1
x 1
1
x
lim
x
x
lim
x
10 x 1
x
x
x
x a
lim
x
1
1
lim
x 1
x
1
x a
1
x
lim
x
lim
lim
x
ln( x 999)
ln x
1 and lim
x
1
1
lim
x
x4 x
x2
lim
x
x 1
x
10 and lim
x
x4 x
x4
lim
x
conclude that x 4
19.
lim
x
xn
ex
lim
x
an x n
20. If p( x)
lim
p( x)
ex
x
nx n
ex
1
lim
x
x
e
p( x)
x
lim
x
x1/ n
ln x
lim
x
17,000,000
(b) ln e
n!
ex
x 1
x
lim
x
1 1. Since the growth rate is transitive, we
x 4 x3
x4
x ).
1. Since the growth rate is transitive, we
o e x for any non-negative integer n
a1 lim xx
e
x
a0 lim 1x where each limit is zero (from Exercise
e
x
x
0
e grows faster than any polynomial.
x (1 n )/ n
n
1
a1x a0 , then
an 1 lim xex
x
ex
lim
x
xn
0
n 1
an lim xx
x
x 999
x have the same growth rate (that of x 2).
an 1 x n 1
19). Therefore, lim
21. (a)
x
3
x and x 4
n
x 4 x3
x2
1 and lim
lim
x
1. Therefore, the relative rates are the same.
conclude that 10 x 1 and x 1 have the same growth rate (that of
18.
1
x 999
1
x
1 lim
n x
1
n
17, 000, 000
x1/ n
17 106
e
o x1/ n for any positive integer n
ln x
1/106
e17
24,154,952.75
(c) x 3.430631121 1015
(d) In the interval 3.41 1015 ,3.45 1015 we have
ln x 10ln(ln x ). The graphs cross at about
3.4306311 1015.
22.
lim
x
lim
ln x
an x n an 1 x n
1
x
a1x a0
lim an
x
an 1
x
ln x
xn
lim
x
a1
xn 1
a0
xn
1/ x
nx n 1
an
lim
x
than any non-constant polynomial (n 1)
Copyright
2014 Pearson Education, Inc.
1
an nx n
0
ln x grows slower
534
Chapter 7 Integrals and Transcendental Functions
23. (a)
lim
n
n log 2 n
1
log 2 n
lim
n (log 2 n) 2
n
0
(b)
n log 2 n
grows slower then n(log 2 n)2 ;
lim
n
n log 2 n
n
ln n
ln 2
1/ 2
lim
3/2
n
n
2 lim 1
ln 2 n
n1/2
3/2
0
1
n
1 lim
ln 2 n
1
2
1/ 2
n
n log 2 n grows slower
than n . Therefore, n log 2 n grows at the
slowest rate the algorithm that takes
O(n log 2 n) steps is the most efficient in the
long run.
24. (a)
lim
n
log 2 n
n
log 2 n
1 lim
ln 2 x
log 2 n
2
lim
n
2
ln n 2
ln 2
n
lim
n
(ln n )2
n(ln 2)
log 2 n
grows slower then n; lim
1
n
2
n
2 lim 1
ln 2 n
n1/ 2
1/ 2
ln 2
n
1
n
2
2
n log2 n
n
1
2
2(ln n )
lim
2
0
lim
2
ln 2
log 2 n
n
n
2
lim lnnn
n
lim
n
ln n
ln 2
1/ 2
n
1
2
lim 1n
(ln 2)2 n
0
1 lim ln n
ln 2 n
n1/ 2
(b)
grows slower than n log 2 n.
Therefore log 2 n
2
grows at the slowest rate
the algorithm that takes O log 2 n
2
steps
is the most efficient in the long run.
25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because
219
524, 288 1, 000, 000 1, 048,576
220.
26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because
218 262,144 450, 000 524, 288 219.
Copyright
2014 Pearson Education, Inc.
Chapter 7 Practice Exercises
CHAPTER 7
1.
PRACTICE EXERCISES
e x sin e x dx
et cos 3et
1
3
2 dt
cos e x
0
sin
tan 3x dx
0 cos
x
3
x
3
dx
3
1/2
1/4
1/6
2cot x dx
1/2 1
du ,
u
3et
2 and du
2
C
ln |1|
2 1/ 2 1
1/2 u
du , where u
1/ 2
/6
cos t
/2 1 sin t
5.
1/2 1
du,
u
dt
2
ln | u | 1/2
2
2
1/2
e x sec e x dx
ln 12
sec u du , where u
ln sec u tan u
7.
ln( x 5)
dx
x 5
cos 1 ln v
v
ln sec e x
u du, where u
u2
2
8.
C
dv
ln( x 5)
2
C
10.
73
dx
1 x
32 1
5x
1
dx
3
71
dx
1 x
2
ln 2
sin x, du
1
6
tan e x
2
u
2 ln 2
7
ln x
C
1 dx
x 5
1 dv
v
C
3(ln 7 ln1) 3ln 7
32
1
1
5
ln 32 ln1
Copyright
1 ln 32
5
ln
u 1, x
5
32
ln 2
2014 Pearson Education, Inc.
1
2
2 1 ln 2
2
2, t
ln 4
C
sin 1 ln v
3 ln | x | 1
1
5
u
e x dx
cos u du, where u 1 ln v and du
1 32 1 dx
5 1 x
1
4
ln1 12 ln 2 ln1 ln 2
cos t dt ; t
ln( x 5) and du
0
cos x dx;
1,x
2
u
ln1 ln 2 ln 2
e x and du
dx; x
ln 8
2
sin u C
9.
ln 12
where u 1 sin t , du
ln | u | 2
6.
ln 1
1 sin x
3
3
ln 23
3ln 12
x
2
x
3et dt
cos 3x , du
where u
1
1/4 cos x
dx
1/6 sin x
2
3et
3 ln 12
3 ln | u | 1
4.
1 sin
3
u C
e x dx
C
cos u du , where u
1 sin
3
3.
e x and du
sin u du , where u
cos u C
2.
535
6
u
ln 2
1
2
u
1
2
536
11.
Chapter 7
e2
e
1
x ln x
Transcendental Functions
e2
dx
e
2
2 u1/2
12.
4
2
2
(ln x) 1/2 1x dx
4
(1 ln t )(t ln t )dt
2
1
1
u 1/2 du, where u
2 1
4 ln 4
2ln 2
u du , where u
13. 3 y
2y 1
14. 4 y
ln 3 y
3y 2
(ln12) y
4 ln 4
u2
(4 ln 4) 2 (2 ln 2) 2
1
2
2 ln 2
ln 2 y 1
y (ln 3)
ln 4 y
ln 3 y 2
2 ln 3
y
( y 1) ln 2
y ln 4
x2
9
ln 3 y
ln(3ln x)
17. ln( y 1)
x ln y
eln( y 1)
e( x ln y )
18. ln(10 ln y )
ln 5 x
eln(10ln y )
eln 5 x
16. 3 y
19 . (a)
x2
3ln x
log x
lim log 2 x
3
x
(b) lim
x
(c)
lim
x
(d) lim
x
x
1
x
x
x
100
x
xe
x
tan 1 x
lim
x
ln e 2 y
ln x
ln 2
ln x
ln 3
2
lim 2x
x
x
1
2
2
u
(f )
20. (a)
1
lim csc1 x
2
ln x9
x
lim sinhx x
x
e
lim 3
x
x
(b) lim
x
(c)
x
y ln 3 ln(3ln x )
ln 3
lim ln
2
x
lim 22 xx
1
lim
x
x
sin
x
lim
1
x
x
1
ex e
x
x
x
x
lim 23
2
4
4 ln 4
(2 ln 2)2
2
ln 32 y
ln 2
u
1 ln t dt ;
(16 1)
ln 2
y
30(ln 2)2
ln 2
ln 32
2 ln 3 (ln 3 ln 4) y
2
ln x9
2 y (ln e)
10 ln y
ln 3
ln 2
5x
ln y
2
ln x9
ye x
y
eln y
x
2
1
e x /2
same rate
lim 1 1
same rate
x
2e
x
0
1
lim
faster
1
x
2
x 1
x
2x
lim 1 e2
x
1
lim
2
x
1
2
1
1
same rate
1
x2
same rate
slower
2
x
ln 2 x
2 ln x
lim ln2(ln
lim 2lnln2x 12
x)
ln x 2
x
x
2
10 x3 2 x 2
lim 30 x x 4 x lim 60 xx 4
ex
e
e
x
x
Copyright
1
2
same rate
lim 60x
x
e
0
ln 3x
ln | x | ln 3
ln 3 ln(ln x )
ln 3
ye x
y 1
1 ln x 2
2
9
y
ln(3ln x )
ln 3
y
e x eln y
xe x
ex
lim 100
lim 100
x
x
x
faster
lim
u
ln t 1 dt
(8ln 2)2 (2 ln 2)2
(ln 3 ln 2) y
( y 2) ln 3
2 ln 2, t
x 2
(e)
e2
u 1, x
ln 9
ln12
e2 y
15. 9e 2 y
e
(t ) 1t
t ln t , du
t
1
2
x
2 2 2
(t ln t )(1 ln t )dt
2
1 dx;
x
ln x, du
slower
2014 Pearson Education, Inc.
y 1 ex
y
e x /2
1
y
1
1 ex
Chapter 7 Practice Exercises
(d)
1 1
x
1
x
tan
lim
x
1
x
x
1
tan
lim
x
x 2
1 x 2
2
1
lim
lim
x
x
x
1
1
1
1
same rate
x2
x 2
(e)
(f )
lim
lim
x4
2
lim
lim
x
e
2
x 1
1
2x
3
2
ex e
x
x
e
x
1
x2
1
1
x2
1
(d)
x
2
ex e x
x
lim
x
e
x
1
x
x
2 1
lim
x
lim
x
faster
1
x2
2
1 e 2x
2
same rate
1
x2
2 for x sufficiently large
true
1
x2
(b)
lim
sech x
1
sin
x
1
x2
x
1
(c)
sin
x
21. (a)
1 1
x
x4
x 2 1 M for any positive integer M whenever x
1
x4
lim x xln x
lim 1 1
x
ln(ln x )
ln x
lim
x
1
(e) tan1 x
2
1
2
(f ) coshx x
e
1
x
lim
1
the same growth rate
x
1/ x
ln x
lim ln1x
1
x
x
for all x
true
2x
1 (1
2
1 e
0
x
1) 1 if x
false
false
grows slower
0
M
true
true
1
x4
22. (a)
1
x2
1
x2 1
1
x4
1 if x
0
true
1
(b)
lim
x
1
x2
x
lim ln
x 1
x
(d)
ln 2 x
ln x
(e)
sec 1 x
1
e
df
dx
ex 1
1
x2 1
lim
x
1
x4
(c)
(f ) sinhx x
23.
x4
0
true
1
lim 1x
x
ln 2
ln x
true
1 1 1 2 if x
cos
1 1
x
2
1
1
2
0
1
1 e 2x
df 1
dx
2
1
2
2
true
if x 1
if x
0
true
true
1
x f (ln 2)
df
dx x ln 2
Copyright
df 1
dx
x f (ln 2)
1
ex 1
1
2 1
x ln 2
2014 Pearson Education, Inc.
1
3
537
538
24.
Chapter 7
y
y 1 1x
f ( x)
f f 1 ( x)
25.
26.
y
y
9e
1
4
27.
K
1
x 1
df
dx
dy
dx
ln x
x /3
1
x
1
1
1
x2
f ( x)
Transcendental Functions
y 1
x
1 ( x 1)
x;
1
dy
dx
e3 e3 1
3e
df 1
dx
1 ;
x 1
1
( x 1) 2 f ( x )
f ( x)
f 1 f ( x)
1
1
1
x
1
1
1
dy dx
dx dt
x /3
;
dy
dt
dx
dt
1
x
( dy / dt )
( dy / dx )
x
dx
dt
dy
dt e2
1
x
29.
1
4
9 y
3e
x /3
; x
9
y
9e
ln 5 ln x ln 3 ln x
ln 5 ln 3 ln 53
Yes, because there is only one intersection
log 4 x
log 2 x
ln x
ln 4
ln x
ln 2
ln x ln 2
ln 4 ln x
1
x
x and
1 m/sec
e
28 . (a) No, there are two intersections: one at x = 2 and the
other at x = 4
(b)
1
1
x2 ;
2
5ft/sec
ln(5 x) ln(3x)
1
x
1
f ( x)
f ( x)
1 ; dy
x
dt
f 1 ( x)
1
y 1
ln 2
ln 4
ln 2
2 ln 2
1
2
Copyright
2014 Pearson Education, Inc.
3
dx
dt x 9
1
4
9
3
e3
9
e3
Chapter 7 Practice Exercises
30.
ln 2 ,
ln x
(a) f ( x)
539
ln x
ln 2
g ( x)
(b) f is negative when g is negative, positive when g is
positive, and undefined when
g = 0; the values of f decrease as those of g increase.
dy
y cos 2
31. dx
3 y ( x 1)2
y 1
32. y
33.
dy
y
y cos
( y 1)
dy
y
dy
e x y 2
36.
( x 2)
e
dy
dx
y ln y
1 x2
1
e tan
37. x dy
y
38. y
ln
2 dx
dy
y3
3
ex
C
y 1
e x
y
1/ 2
2e
ln 2
dx
x
ln 4. So 2 ln
eln(4 x)
1
3
y
y
y3
3 ex
dy
2
2ln
y3
3
ee
y
tan
ln 2
C
2
cos( x )
y
2,so e
C1
2
y
e x
1
y
tan 1 ( x ) C
2
e
. We have y (0)
( ) ln 2
ln 2
y
ee
C
C
2e
2
ln(4 x )
2 x 1
ln
y 1
e2
ee
and
tan 1 (0) C
2 ln
1 1
1 ln(4 x)
2
ln1 C
ln(4 x)1/2
2
C. We have y (0) 1
3 ex
e2
1 x
ln x C . We have y (1) 1
ln x ln 4
ex
Copyright
C
C. We have y (0)
y 1
y 1
e x
1
cos( x )
2 tan x C1
2
0 C
y 1 2 x
e 2 x 1 dx
ex
y2
2
tan 1 ( x) C
ln(ln y )
dy
0
ln 22
ex
e2 x 1
e
sin y 2
tan x C
2
e ( x 2)
( x 2)
tan 1 (0) C
2
y dx
2 ln 2
e
(0) C
ln
dx
1 x2
sin y 2
ey
2
tan 1 x 2C
y
( x 1)3 C
sin x dx
cos2 ( x )
y dy
y
x C
y ln y
sec2 x dx
e ( x 2) dx
2
dy
y ln y
2 tan y
3( x 1)2 dx
0
e y dy
2e
y
sec y 2
34. y cos 2 ( x) dy sin x dx
35. dx
ey
dx
ydy
sec y 2 sec 2 x
yy
2
e x
1
1/3
2014 Pearson Education, Inc.
(1)3
3
e0 e 0 C
C
1.
3
So
540
Chapter 7
Transcendental Functions
A0 ekt we have
39. Since the half life is 5730 years and A(t )
ln(0.5)
. With
5730
ln(0.5)
ln(0.1) 5730 t
k
40. T Ts
(5730)ln(0.1)
ln(0.5)
t
kt
1.
t
A(t )
(a)
(b)
(c)
2. (a)
0
(220 40)e
e x dx
e
lim A(t )
t
t
lim
0
(220 40)e
4 ln(9/7)t
t
x t
ln 6
4 ln 97
ln(0.5)
t
5730
5730k
0.1 e
k /4
, time in hours,
k
4 ln
7
9
1.78 hr 107 min, the total time
4ln
0
lim 1 e
t
V (t )
A( t )
1 e
2t
1 e
t
2
lim
t
the time it took to cool
V (t )
A(t )
t
0
lim log a 2
a 1
lim log a 2
a 1
lim log a 2
a
lim
0
1 e
1 e
t
t
ln 2
lim ln
a
0
ln 2
lim ln
a
1 e
t
dx
2
e
2x t
0
2
1 e
2t
lim
2
1 e
0
t
1 e
t
t
lim
0
2
1 e
t
;
ln 2
lim ln
a
;
a 1
ln 2
lim ln
a
2x
0;
a 1
a
e
2
a
x
4. In the interval
2
t
0
1
2t
0
2 the function
sin x
sin x 0 (sin x)
is not defined for all values in
that interval or its translation by 2 .
5. (a) g ( x) h( x) 0 g ( x)
h( x); also g ( x) h( x) 0 g ( x) h( x)
g ( x) h( x); therefore h( x) h( x) h( x) 0 g ( x) 0
(b)
f ( x) f ( x )
2
f ( x) f ( x )
2
(c) Part b
ln(0.5)
t
5730
9
7
92 min
1 e t ,V (t )
t
lim log a 2
a
A0e
ln(0.5)
ADDITIONAL AND ADVANCED EXERCISES
t
lim
e5730k
1
2
19,035 years (rounded to the nearest year).
180 40
from 180 F to 70 F was 107 15
CHAPTER 7
A0e5730k
10% of the original carbon-14 remaining we have 0.1A0
To Ts e
70 40
A0
2
f E ( x ) fO ( x )
f E ( x ) fO ( x )
2
f E ( x ) fO ( x ) f E ( x ) fO ( x )
2
f E ( x ) f O ( x ) f E ( x ) fO ( x )
2
f E ( x);
f E ( x ) f O ( x ) f E ( x ) fO ( x )
2
fO ( x)
such a decomposition is unique.
Copyright
0
2014 Pearson Education, Inc.
g ( x ) h( x )
0
Chapter 7 Additional and Advanced Exercises
6. (a)
g (0) g (0)
1 g (0) g (0)
g (0 0)
g (0) g 2 (0) 1
(b) g ( x)
7. M
g (0)
g ( x h) g ( x)
h
0
h
0
dy
dx
(c)
0
lim
lim
h
1 g 2 (0) g (0)
g (h)
h
1 g ( x)
1 g ( x ) g ( h)
dy
1 y2
1 y2
0
tan 1 g ( x)
1 2
0 1 x2
dx
1
2 tan 1 x
g ( x)
tan 1 y
x
2
0
0
g ( x)
g ( x ) g ( h) g ( x) g 2 ( x) g (h)
h 1 g ( x ) g ( h)
0
lim
h
h
1 1 g 2 ( x)
dx
C
g ( x) g (h)
1 g ( x) g (h)
0
h
g 3 (0) g (0)
2 g (0)
0
lim
2
g (0) g 3 (0)
2 g (0)
541
1 g 2 ( x) 1
x C
tan 1 g ( x)
g ( x)
2
x C ; g (0)
tan 1 0
0
0 C
tan x
1 2x
0 1 x2
and M y
1
ln 1 x 2
dx
0
ln 2
My
M
x
ln 2
ln 4 ;
y
0 by
2
symmetry
4
8. (a) V
1/4 2 x
4
x 1
1/4
2 x
4 1
1
1/4 2 2 x
4 1
(b) M y
Mx
M
y
9.
2
1
dx
1/4 2 x
Mx
1 ln 2
M
2
A0 ert ; A(t )
A(t )
4 1
4
dx 4 ln | x | 1/4 4 ln 4 ln 14
ln16 4
4 1/4 x
4
4
4 1/2
8
64 1 63 ;
1 x 3/2
1
dx 12
x dx
3
3 24
24
24
1/4
1/4
1 dx 1 4 1 dx
1 ln | x | 4
1 ln16 1 ln 2;
8 1/4 x
8
2
2 x
1/4 8
4 1 1/2
M
1/2 4
x
dx
x
2 12 23 ; therefore, x My
1/4 2
1/4
dx
2
3
350
450 x
2 A0
( 1 (
d
dx
d
dx
1
1
0
350 2
450 x
1
2 ))
A0 ert
2 A0
e rt
2
rt
x
350 and tan
tan 1 450
2
x
1 350
1 200
tan
tan
450 x
x
350
(450 x )2
350
(450 x )2 122500
3 x 2 3600 x 1020000
first derivative test, ddx
600 100 70
63
24
ln 2
2
3
21
12
7
4
and
ln 2
3
ln 2
t
ln 2
r
10. In order to maximize the amount of sunlight, we need to maximize the angle
line segments to their vertex. The angle between the two lines is given by
have tan 1
ln 24
x 100
200 2
x
200
x 2 40000
0
200
x2
1
1
x
9
3500
0
200
x
350
(450 x )2 122500
2
.7
r
t
Copyright
70
( r %)
formed by extending the two red
1 (
2 ) . From trig we
tan 1 200
x
200
x 2 40000
200 (450 x) 2 122500
350( x 2
40000)
600 100 70. Since x 0, consider only x
9
0 and ddx
0 local max when
5000
236.67 ft.
70
100 r
x 400
2014 Pearson Education, Inc.
600 100 70. Using the
542
Chapter 7
Transcendental Functions
Copyright
2014 Pearson Education, Inc.
CHAPTER 8
8.1
TECHNIQUES OF INTEGRATION
USING BASIC INTEGRATION FORMULAS
1
1.
16 x
8x2 2
0
dx
u 8 x 2 2 du 16 x dx
u 2 when x 0, u 10 when x 1
1
0
16 x
8x
2
2
10
dx
2
1
du
u
ln u
ln10 ln 2
x2
2.
x2 1
10
2
ln 5
dx
Use long division to write the integrand as 1
x2
x
2
1
dx
1
1 dx
x
1
x tan
2
1
1
x
2
1
.
dx
x C
2
3.
sec x tan x dx
Expand the integrand: sec x tan x
2
sec2 x 2sec x tan x tan 2 x
sec2 x 2sec x tan x (sec2 x 1)
2sec2 x 2sec x tan x 1
sec x tan x
2
dx
2 sec 2 x dx 2 sec x tan x dx
2 tan x 2sec x
1 dx
x C
We have used Formulas 8 and 10 from Table 8.1.
/3
4.
/4
u
1
2
cos x tan x
dx
tan x du sec2 x dx
u 1 when x
/3
/4
1
2
cos x tan x
4, u
dx
1
dx
cos2 x
3 when x
3
1
1
du
u
ln 3 ln1
/3
ln u
3
1
1
ln 3
2
Copyright
2014 Pearson Education, Inc.
543
544
Chapter 8 Techniques of Integration
1 x
5.
dx
1 x2
Write as the sum of two integrals:
1 x
1 x
1
dx
2
1 x
2
x
dx
1 x2
dx
For the first integral use Formula 18 in Table 8.1 with a 1.
For the second:
u 1 x2
du
x
1 x
2 x dx
1
2
dx
2
1
1/2
u
1 x2
u
1 x
So
1 x
1
6.
x
u
x
2
dx
1
x
sin 1 x
1
dx
2 x
1
2
du
u
e cot z
sin 2 z
u
sin 2 z
C
2ln
x 1
C
dz
cot z
e cot z
C
dx
2ln u
7.
1 x2
dx
x 1 du
x
du
csc2 z dz
du
1
sin 2 z
dz
e u du
dz
e u
C
e cot z
C
3
2ln z
dz
16 z
8.
u
ln z 3
3ln z
du
3
dz
z
Using Formula 5 in Table 8.1,
3
2ln z
dz
16 z
1
2u du
48
2u
C
48ln 2
3
2ln z
C
48ln 2
Copyright
2014 Pearson Education, Inc.
Section 8.1 Using Basic Integration Formulas
1
9.
e
z
z
e
dz
ez
Multiply the integrand by
ez
1
dz
e z
ez
ez
u
du
ez
e
2z
1
e2 z 1
ez
dx
e z du
1
dx
u
2
1
du
tan 1 e z
tan 1 u C
2
10.
8
x
1
u
u
2
.
2x 2
dx
x 1 du dx
0 when x 1, u 1 when x
2
x
1
1
8
2
2x 2
dx 8
0
2
1
u
2
du
1
1
8 tan 1 u
8
0
0
11.
4
11
u
u
(2 x 1)2
2 x 1 du
1 when x
0
C
4
2
dx
2dx
1, u 1 when x
1
2
1 1 (2 x 1)
0
4
dx
2
1
2
11 u
1
2 tan 1 u
0
du
2
1
4
4
3
4x2 7
dx
1 2x 3
12.
Use long division to write the integrand as 2 x 3
3
4x2 7
dx
1 2x 3
3
1
2 x dx
3
1
3 dx
3
1
3
2 x dx
x2
1
3
1
3 dx
3
3x 1
For the last integral,
u 2 x 3 du 2 dx
u 1 when x
1, u 9 when x
3
2
1 2x
8 12
3
2
.
2x 3
dx
4
3
Copyright
2014 Pearson Education, Inc.
545
546
Chapter 8 Techniques of Integration
3
9
2
dx
x
2
3
1
ln u
1
du
1 u
9
1
ln 9 ln1 2 ln 3
3
4x2 7
dx
1 2x 3
So
13.
4 2ln 3
1
dt
1 sec t
Multiply the integrand by
1 sec t
.
1 sec t
1
1 sec t
1 sec t
cos t
cot 2 t
2
1 sec t 1 sec t
tan t
sin 2 t
1
cos t
1 dt
csc2 t dt
dt
dt
1 sec t
sin 2 t
t cot t csc t C
cos t
1 csc2 t
sin 2 t
Here we have used Formula 9 in Table 8.1 for the second integral, and the substitution u
1
for the third integral, which gives it the form
14.
u
2
du
1
u
sin t , du
1
.
sin t
csc t sin 3t dt
Write sin 3t as sin(2t t ) and expand.
cos 2t sin t (2sin t cos t ) cos t
sin t
csc t sin 3t
cos 2t 2cos 2 t
csc t sin 3t dt
2cos 2t 1
2cos 2t dt
1 dt
sin 2t t C
/4
15.
1 sin
cos2
0
d
Split into two integrals.
/4
1 sin
cos
0
2
/4
d
1
cos
0
/4
0
2
/4
d
0
/4
sec2 d
0
tan
/4
sin
cos 2
d
d
2
The second integral is evaluated with the substitution u
cos
cos
2
d
1
u
2
du
1
u
(1
cos 2
2) (0 1)
sin
sec 0
sin
du
sin d , which gives
1
.
cos
Copyright
2014 Pearson Education, Inc.
cos t dt
Section 8.1 Using Basic Integration Formulas
1
16.
2
2
d
1
Write the integrand as
1)2
1 (
1
2
2
1
d
1)2
1 (
1
1 u
2
. With u
1, du
d ,
d
du sin
1
u C
sin 1(
1) C
We have used Formula 18 in Table 8.1 with a 1.
ln y
17.
y 4 ln 2 y
dy
ln y
1
.
y 1 4 ln 2 y
8 ln y
dy
y
Write the integrand as
u 1 4 ln 2 y
ln y
2
y 4ln y
du
ln y
1
dy
y 1 4ln 2 y
1 1
1
ln u C
du
8 u
8
dy
1
ln(1 4 ln 2 y ) C
8
Note that the argument of the logarithm is positive, so we don t need absolute value bars.
2 y
dy
2 y
18.
u
y
du
1
2 y
dy
Using Formula 5 in Table 8.1,
2 y
dy
2 y
2u du
1 u
2
ln 2
19.
1
sec
tan
C
sec
tan
u 1 sin
C
d
Multiply the integrand by
1
1
2 y
ln 2
cos
d
cos
du
cos
.
cos
cos
d
1 sin
cos d
Copyright
2014 Pearson Education, Inc.
547
548
Chapter 8 Techniques of Integration
cos
d
1 sin
1
du ln u C
u
ln (1 sin ) C
We can discard the absolute value because 1 sin
1
20.
dt
t 3 t2
Use Formula 5 in Table 7.10, with a
1
t 3 t
1
t
csch 1
3
3
dt
2
4t 3 t 2 16t
21.
t2
4
3.
C
dt
4
Use long division to write the integrand as 4t 1
4t 3 t 2 16t
t
2
is never negative.
4
dt
4t dt
1
1 dt 4
t
2
t
t 2 tan 1
2
2t 2
t
2
4
.
dt
4
C
To evaluate the third integral we used Formula 19 in Table 8.1 with a
2.
x 2 x 1
dx
2x x 1
22.
Split into two integrals.
x 2 x 1
dx
2x x 1
1
dx
2 x 1
x 1 ln x
For the first integral we used u
23.
/2
0
0
/2
u 1 cos
2 when
0
1
dx,
2 x 1
sin
1 cos
du
/2
1 cos2
1 cos
1 cos d
0
/2
x 1, du
du
u C
1 cos
.
1 cos
(Note that when 0
u
C
1 cos d
Multiply the integrand by
/2
1
dx
x
/ 2, sin
0
sin
1 cos
0 so sin 2
sin . )
d
d .
sin d
0, u 1 when
1
d
2
/2
1
du
u
2
1
1
du
u
Copyright
2 u
2
1
2 2 2
2014 Pearson Education, Inc.
Section 8.1 Using Basic Integration Formulas
(sec t cot t )2 dt
24.
Expand the integrand:
2
sec t cot t
sec2 t 2sec t cot t cot 2 t
sec2 t 2sec t cot t csc2 t 1
(sec t cot t )2 dt
sec2 t dt 2 csc t dt
tan t 2ln csc t cot t
csc2 t dt
1 dt
cot t t C
We have used Formulas 8, 9 and 15 from Table 8.1.
1
25.
e
dy
2y
1
ey
Multiply the integrand by
1
e
1
e
ey
e
y
e
ey
dy
2y
y
e
2y
sec
1
e y dy
du
1
u u2 1
1
ey
dy ; u
1
dy
2y
.
ey
u
du
C
sec
1 y
e
C
We have used Formula 20 in Table 8.1.
6
dy
y 1 y
26.
u
y
du
1
dy
2 y
6
1
dy 12
du
y 1 y
1 u2
12 tan
2
27.
x 1 4 ln 2 x
u
2 ln x
du
2
x 1 4 ln 2 x
1
y
C
dx
2
dx
x
1
dx
1 u2
sin
1
du
u C
sin
1
2 ln x
Copyright
C
2014 Pearson Education, Inc.
549
550
Chapter 8 Techniques of Integration
1
28.
dx
( x 2) x 2
u
x 2
4x 3
du
dx
1
( x 2) x
1
dx
2
u u2 1
4x 3
sec 1 u
du
sec 1 x 2
C
C
We have used Formula 20 in Table 8.1 with a 1.
29.
(csc x sec x )(sin x cos x ) dx
Expand the integrand and separate into two integrals.
(csc x sec x )(sin x cos x ) 1 cot x tan x 1 cot x tan x
(csc x sec x )(sin x cos x ) dx
cot x dx
tan x dx
ln sin x
ln sec x
C
ln sin x
ln cos x
We have used Formulas 12 and 13 from Table 8.1.
30.
x
2
3sinh
ln 5 dx
x
ln 5
2
u
1
dx
2
du
x
ln 5 dx
2
3sinh
6 sinh u du
6cosh u C
3
2 x3
31.
2
x2 1
6cosh
x
ln 5
2
dx
Use long division to write the integrand as 2 x
3
3
2 x3
2
x
2
1
dx
2x
2
2x
x
2
2
3
2 x dx
2
2x
x
2
1
dx
x2
3
dx
1
2
x 2 , du
For the second integral we use u
3
C
3
2
2x
x2 1
.
3
2 x dx
2
2x
x
2
1
dx
2 x dx.
ln x 2 1
3
2
(9 2) (ln8 ln1)
7 ln 8 9.079
Copyright
2014 Pearson Education, Inc.
C
Section 8.1 Using Basic Integration Formulas
1
32.
1
0
33.
1
1 x 2 sin x dx is the integral of an odd function over an interval symmetric to 0, so its value is 0.
1 y
dy
1 y
1 y
Multiply the integrand by
1 y
dy
1 y
1 y
1 y
sin
1
2
and split the indefinite integral into a sum.
1 y
1
dy
1 y
1 y2
y
dy
2
y
1 y2
dy
C
The first integral is Formula 18 in Section 8.1, and for the second we use the substitution
u 1 y 2 , du
0
1
2 y dy . So
1 y
dy
1 y
sin
1
y
ez
ez
0
1
(0 1)
34.
1 y2
0
2
1
2
dz
z
Write the integrand as e z ee and use the substitution u
z
e z e dz
z
e z ee dz
eu
C
( x 1) x 2
u
x 1,
du
e z dz.
eu du
ee
7
35.
e z , du
z
C
dx
2 x 48
dx;
x 2 2 x 48 u 2 72
We use Formula 20 in Table 8.1.
7
( x 1) x
2
1
7sec
7
7
dx
2 x 48
1
u
7
u u
C
sec
1
2
72
x 1
7
Copyright
du
C
2014 Pearson Education, Inc.
551
552
Chapter 8 Techniques of Integration
1
36.
dx
(2 x 1) 4 x 4 x 2
u
2 x 1,
du
4 x 4 x2
2dx;
u 2 12
We use Formula 20 in Table 8.1.
1
(2 x 1) 4 x 4 x
1
2
dx
2
1
37.
12
u
C
u u
1
sec
2
1
du
2
1
sec
2
1
2x 1
2 3 7 2 7
d
2 5
Use long division to write the integrand as 2
2 3 7 2 7
d
2 5
2
d
1 3
3
d
1 2
2
1
1
cos
1
1
cos
1 cos
csc2
2
5
2
5
ln 2
2
5
.
d
C
5
2
5, du
2d .
d
Multiply the integrand by
cos
5
5
1d
In the last integral we have used the substitution u
38.
C
1
1
cos
cos
1
.
1
cos
1
1 cos
cos2
1
sin 2
csc cot
csc2
csc2 d
d
cot
csc
csc cot
csc cot d
C
We have used Formulas 9 and 11 from Table 8.1.
39.
1
1 ex
dx
Use one step of long division to write the integrand as 1
1
dx
1 ex
1 dx
ex
1 ex
dx
x ln 1 e x
ex
1 ex
.
C
For the second integral we have used the substitution u 1 e x , du
Copyright
e x dx. Note that 1 e x is always positive.
2014 Pearson Education, Inc.
Section 8.1 Using Basic Integration Formulas
x
40.
dx
1 x3
x 3/2 ,
u
x
1 x
3
3 1/2
x dx
2
du
2
3
dx
1
1 u2
du
2
tan 1 u C
3
/4
41. The area is
/4
2
tan 1 x 3/2 C
3
2cos x sec x dx
/4
/4
2sin x ln sec x tan x
2 ln
2 1
2 1
2 1
2 2 ln
/4
42. The volume using the washer method is
2 ln
/4
2 1
2 2 ln 3 2 2
1.066
4cos 2 x sec2 x dx.
Split into two integrals; for the first write 4cos2 x as 2 1 cos 2x and for the second use Formula 8 in
Table 8.1.
/4
/4
/4
4cos 2 x sec 2 x dx
/4
/4
/4
43. For y
/3
1
0
/3
0
ln (cos x ) , dy / dx
tan x
sec x dx
2
44. For y
/4
0
/4
0
0
3
ln (sec x ) , dy / dx
1
tan x
sec x dx
2
dx
2 1
2
tan x
1
1 ( 1)
2
/3
ln 2
3
tan x. The arc length is given by
sec x dx since sec x is positive on the interval of integration.
ln sec x tan x
ln
1
/4
/4
sec2 x dx
/4
/4
/4
0
ln 1 0
/4
0
/4
sec x dx since sec x is positive on the interval of integration.
ln sec x tan x
ln 2
sec 2 x dx
tan x. The arc length is given by
/3
dx
/4
2 1 cos 2 x dx
2 x sin 2 x
2
/4
4 cos2 x dx
/4
0
ln 1 0
ln
Copyright
2 1
2014 Pearson Education, Inc.
553
554
Chapter 8 Techniques of Integration
45. Since secant is an even function and the domain is symmetric to 0, x
0.
For the y-coordinate:
y
1
2
/4
/4
/4
/4
sec2 x dx
1 tan x
2
/4
/4
/4
/4
ln sec x tan x
sec x dx
1
ln
2 1
ln
2 1
1
1
2 1
2 1
ln
0.567
ln 3 2 2
46. Since both cosecant and the domain are symmetric around
y
1 5 /6 2
csc x dx
2 /6
5 /6
/6
ln 2
3
3
3
ln 2
3
1 3 x 3 e x dx
47.
1
48.
3
3
3
3
xe x
3
C
Multiply the integrand by
1
dx
1 sin 2 x
2
1 2 tan x
2u,
1
1 2u
2
sec 2 x
sec 2 x
.
sec 2 x
sec2 x tan 2 x
sec2 x
dx
1 2 tan 2 x
dx
du sec2 x dx
tan x,
sec2 x
v
0.658
ln 7 4 3
dx
1 sin 2 x
u
5 /6
/6
ln csc x cot x
3
2
ln
2
/ 2.
1 cot x 5 /6
2
/6
csc x dx
1
2
/ 2, x
dv
du
dx
1
1 2u 2
du
2 du
1
1
dv
2 1 v2
1
tan 1 v C
2
1
tan 1 2 tan x
2
C
Copyright
2014 Pearson Education, Inc.
Section 8.2 Integration by Parts
x 7 x 4 1 dx
49.
x 4 1, du
u
4 x 3dx;
u 1
du
4
1
(u 1) u du
4
1 3/2
1 1/2
u du
u du
4
4
1 5/2 1 3/2
u
u
C
10
6
3/2
1 3/2
1 4
3u 5 C
3x 4
u
x 1
30
30
x 7 x 4 1 dx
2/3
( x 2 1)( x 1)
50.
x 7dx
2
C
dx
x 1
, du
x 1
The easiest substitution to use is probably u
2
(1 x )2
dx.
The integral can be written as
1
x 1
x 1
2/3
3 1/3
u
C
2
8.2
dx
( x 1)2
3 x 1
2 x 1
1
u 2/3 du
2
1/3
C
INTEGRATION BY PARTS
1. u
x, du
x sin 2x dx
2. u
, du
cos
sin 2x dx, v
dx; dv
2 x cos 2x
d ; dv
d
2 cos 2x dx
cos
sin
2 cos 2x ;
d ,v
1
sin
1
2 x cos 2x
sin
;
d
sin
4sin 2x
1
2
cos
C
C
cos t
3.
2
( )
2t
( )
cos t
2
( )
sin t
t
0
sin t
t 2 cos t dt
t 2 sin t 2t cos t 2sin t C
Copyright
2014 Pearson Education, Inc.
555
556
Chapter 8 Techniques of Integration
4.
sin x
2
( )
cos x
2x
( )
sin x
2
( )
x
cos x
x 2 sin x dx
0
5. u
dx ; dv
x
ln x, du
2
1
6. u
x2
2
x ln x dx
e 3
x ln x dx
1
7. u
x, du
x e x dx
8. u
x4
4
xe x
x e3 x dx
e
e x 4 dx
1 4 x
1
e x dx
x e3 x
3
2ln 2 34
ln 4 43
;
e4
4
3e4 1
16
e
x4
16 1
ex ;
xe x
e3 x dx, v
e3 x dx
1
3
2
x2
4 1
2 ln 2
x4
4
x3 dx, v
e x dx, v
dx; dv
;
2 x 2 dx
1 2 x
1
ln x
dx; dv
x, du
2
ln x
dx ; dv
x
ln x, du
x2
2
x dx, v
x2 cos x 2 x sin x 2cos x C
ex
C
1 e3 x ;
3
x e3 x 1 e3 x
3
9
C
e x
9.
x2
( )
2x
( )
2
( )
e x
e x
e x
x 2 e x dx
0
x 2e x
2x e x
2e x
x2
2 x 1 e2 x
C
e 2x
10.
x2
2x 1
( )
2x 2
( )
2
( )
1 e 2x
2
1 e2 x
4
1 e2 x
8
x2
0
2 x 1 e 2 x dx
1
2
1 x2
2
11. u
tan 1 y, du
tan 1 y dy
dy
1 y2
; dv
y tan 1 y
dy, v
y;
y dy
y tan 1 y
1 y2
Copyright
1
2
3
2
x
5
4
ln 1 y 2
e2x
C
1 (2 x
4
2)e 2 x
1 e2 x
4
C
C
y tan 1 y ln 1 y 2
2014 Pearson Education, Inc.
C
Section 8.2 Integration by Parts
dy
sin 1 y, du
12. u
sin 1 y dy
13. u
x, du
1 y2
dx; dv
1 y2
C
tan x;
tan x dx
2 x, dy
y;
y sin 1 y
1 y2
sec 2 x dx, v
x tan x
4 x sec 2 2 x dx; [ y
dy, v
y dy
y sin 1 y
x sec 2 x dx
14.
; dv
x tan x ln |cos x | C
y sec2 y dy
2dx ]
y tan y
tan y dy
y tan y ln | sec y | C
2 x tan 2 x ln |sec 2 x | C
ex
15.
x3
( )
ex
3x 2
( )
ex
6x
( )
ex
6
( )
ex
x3e x dx
0
x 3e x
3x 2 e x
6 xe x
6e x
C
x3 3 x 2
6x 6 ex
24 pe p
24e p
C
e p
16.
p4
( )
4 p3
( )
12 p 2
( )
24 p
( )
24
( )
e p
e p
e p
e p
e p
p 4 e p dp
0
p4e p
p4
4 p3e p 12 p 2 e p
4 p 3 12 p 2 24 p 24 e p
C
C
ex
17.
x2 5x
( )
ex
2x 5
( )
ex
2
( )
ex
0
x 2 5 x e x dx
x2
7x 7 ex
Copyright
x2 5x e x
(2 x 5)e x
2e x
C
C
2014 Pearson Education, Inc.
x2e x
7 xe x
7e x
C
557
558
Chapter 8 Techniques of Integration
er
18.
( )
er
2r 1
( )
er
2
( )
er
r2
r 1
r2
0
r 1 e r dr
r2
r 1 er
(2r 1) 2 er
r 1
(2r 1)er
r2
C
2e r
C
r 2 er
C
ex
19.
x5
( )
ex
5x4
( )
ex
20 x3
( )
ex
60 x 2
( )
ex
120 x
( )
ex
120
( )
ex
x5e x dx
0
x5e x
x5 5 x 4
5 x4 e x
20 x3e x
60 x 2 e x 120 xe x 120e x
20 x3 60 x 2 120 x 120 e x
C
C
e4t
20.
t2
( )
2t
( )
2
( )
1 e 4t
4
1 e 4t
16
1 e 4t
64
t 2 e 4t
4
t 2 e4t dt
0
t2
4
21.
r2
I
e sin d ; [u
[u
cos , du
e sin
1
32
sin , du
sin d ; dv
e cos
t
8
I C
2 t e 4t
16
e 4t
2 e 4t
64
2I
t 2 e 4t
4
t e 4t
8
1 e 4t
32
C
C
cos d ; dv
e d ,v
C
e d ,v
e ]
e sin
I
e ]
e sin
e cos
C
I
e sin
e cos
I
1
2
e cos d ;
e sin d
e sin
e cos
C , where C
is another arbitrary constant
22.
e y cos y dy;[u
I
I
[u
e y cos y
sin y, du
cos y, du
sin y dy; dv
e y ( sin y ) dy
cos y dy; dv
e y dy, v
Copyright
e y dy, v
e y cos y
e y]
I
e y]
e y sin y dy;
e y cos y
e y sin y
2014 Pearson Education, Inc.
e y cos y dy
C
2
Section 8.2 Integration by Parts
e y cos y e y sin y I C
C
2
C
23.
1 e2 x
2
1 e2 x
2
I
I
cos 3 x
3 1 e2 x
2 2
1e y
2
1e y
2
2I
sin 3x C
2 x, du
sin y cos y
1 e2 x
2
1e y
4
I
cos 3 x
I ; [u
3 e 2 x sin
4
3 x 94 I
e
e y dy , v
cos y dy; dv
2x
I
C
sin 2 x cos 2 x
4
xe x dx
2
3
26. u
x, du
1
xe x
2
3
dx; dv
27. u
x, du
3
0
3
3
x ln x x 2
x (1 x)3
ln 12
2
18
x x2
(2 x 1) dx
x 1
ln x
sin (ln x) dx; du
1 dx
x
u
dx
3
3
(2 x 1) dx
u
1
2
xe x
1
xe x dx ; [u
2
3
ex
C , where
2 1
3 0
(1 x)3 dx
3
3
0
0 23
sin 2 x
cos2 x
tan 2 x dx
0
e x dx, v
dx; dv
3s 9e 3s 9
2
3
C
x, du
e 3s 9
C
2
5
1 cos2 x
cos 2 x
dx
(tan x x) dx
(1 x)5 2
3
3
1
0
4
15
dx
cos 2 x
dx
ln |cos x |
3
dx
tan x x;
x2
2 0
3
2
ln 2 18
; dv
dx, v
x ln x x 2
x; ln x x 2 dx
2
1
x 1
dx
x ln x x 2
x ln x x 2
(sin u ) eu du. From Exercise 21,
C
Copyright
2x 1
x ( x 1)
2014 Pearson Education, Inc.
x dx
2 x ln | x 1| C
(sin u ) eu du
e du
x cos (ln x) x sin (ln x)
e x ];
(1 x )3 ;
2
3
0
x tan x x
ln x x 2 , du
28. u
2
3
tan 2 x dx, v
dx; dv
x tan 2 x dx
3
e x dx
1 x dx, v
2
3
x 1 x dx
e x 23 x dx
e y]
e y]
sin y cos y
C
C
4C
13
C , where C
e y dy, v
1e y
2
sin y cos y
1 e2 x ]
2
dx; v
sin y, du
sin y ; dv
e y ( sin y ) dy
C
e
3 sin 3x 2 cos 3x
cos y, du
e y cos y
1 e2 x ]
2
2x
3cos 3 x, dv
e y sin y dy
e y cos y dy [u
1
2
e2 x
13
I
1
2
2dx]
3s 9 x 2
e 3s 9 ds;
ds 23 x dx
0
C , where
C
2
C
29.
sin y
sin 3 x, du
sin 3 x 32 e 2 x cos 3 x dx
3 e2 x
4
cos 3 x
e y sin y
1
2
I
25.
e y sin y e y cos y
1
2
I
e2 x dx; v
3 sin 3x dx, dv
e2 x sin 3x dx; [u
3
2
e 2 x sin 2 x dx; [ y
I
cos 3x; du
cos 3 x
1 e2 x
2
13 I
4
24.
C
is another arbitrary constant
e2 x cos 3 x dx; [u
I
e y sin y cos y
2I
559
eu
sin u cos u
2
C
560
Chapter 8 Techniques of Integration
u
ln z
z (ln z ) dz ; du
1 dz
z
u
2
30.
dz
e
u2
( )
2u
( )
2
( )
eu u 2 eu du
e du
2u
1 e 2u
2
1 e 2u
4
1 e 2u
8
z2
4
x sec x 2 dx Let u
1 ln |sec x 2
2
cos x
x
32.
e 2u
2(ln z ) 2
x 2 , du
u e 2u
2
1 e 2u
4
2 ln z 1
1 du
2
2 x dx
C
e2 u
4
2u 2
2u 1
C
C
x dx
x sec x 2 dx
1
x
cos x
x
1
2
1 ln |sec u
2
sec u du
dx Let u
u
x(ln x) dx; du
dx
1
2 x
x , du
dx
2du
dx
dx
2 cos u du
2 sin u C
ln x
1
x
u
eu u 2 eu du
dx
e2u u 2 du ;
e du
e 2u
u2
( )
2u
( )
2
( )
1 e 2u
2
1 e 2u
4
1 e 2u
8
x2
4
35. u
1
x (ln x )2
dx Let u
ln x, du
ln x
x2
36.
u2
2
u 2 e2u du
0
34.
dx
(ln x )3
x
tan u | C
tan x 2 | C
2
33.
u2
2
u 2 e2u du
0
31.
e 2u u 2 du ;
1
x
ln x, du
1
x2
dx; dv
ln x
x
dx Let u
1
x
ln x, du
1
x
1 e 2u
4
2 ln x 1
1
x (ln x )2
dx
dx
e2 u
4
2u 2
x2
2
ln x
2
du
1
u
C
1 u4
4
C
C
C
1
u2
2u 1
x2
2
C
ln x
x2
4
1
ln x
C
C
1;
x
dx, v
ln x
x
u e 2u
2
2
2 ln x
dx
1
x2
e 2u
1
x
dx
C
(ln x)3
x
Copyright
dx
u 3 du
1
4
2014 Pearson Education, Inc.
(ln x) 4 C
2 sin x C
Section 8.2 Integration by Parts
4
x3e x dx Let u
37.
38. u
x3 , du
x3e x x 2 dx
1 x2
3
x 2 sin x 3 dx Let u
40.
1 cos x 3
3
41. u
sin 3 x, du
x3 , du
1 sin
2
1 sin
2
sin 3 x cos 2 x dx
sin 2 x, du
1 x3 e x
3
x2 1
1
3
32
x2 1
1 du
3
cos 2 x dx, v
32
cos 2 x, du
1 sin
4
cos 2 x
1 sin
2
3
1 x2
3
x 2 dx
1 sin
4
1 sin
2
1 ex
3
3
2 sin 3 x sin
5
x2 1
32
2
15
x 2 sin x3 dx
x ln x dx
1
3
3
2
cos 2 x
3 cos 3 x
4
sin 3 x cos 2 x dx
cos 2 x
2 x 53 cos 3 x cos 2 x C
cos 4 x dx, v
1 sin
4
4 x;
2 x sin 4 x 12 cos 2 x sin 4 x dx
sin 4 x dx, v
1 cos
4
1 cos
4
Let u
2 3/2
x ln x
3
2 3/2
x ln x
3
ln x, du
2
x dx
3
4 3/2
x
C
9
1
dx, dv
x
4x ;
2 x cos 4 x 12 sin 2 x cos 4 x dx
x dx, v
2 3/2
x
3ln x 2
9
Copyright
2 3/2
x
3
C
2014 Pearson Education, Inc.
52
sin u du
2 x;
1 sin 2 x sin 4 x 1 cos 2 x cos 4 x 1 sin 2 x cos 4 x dx
4
8
4
3 sin 2 x cos 4 x dx 1 sin 2 x sin 4 x 1 cos 2 x cos 4 x
4
4
8
1
1
sin 2 x cos 4 x dx 3 sin 2 x sin 4 x 6 cos 2 x cos 4 x C
x ln x dx
x2 1
sin 3 x cos 2 x dx
2 x sin 4 x 12
1 ex
4
4
C
C
2 x;
1 cos
2
1 cos 3 x
2
3x sin 2 x
2sin 2 x dx; dv
sin 2 x cos 4 x dx
3
2
9
4
C
;
2 x dx
sin 2 x dx, v
3x sin 2 x
2 cos 2 x dx; dv
sin 2 x cos 4 x dx
43.
1 eu
4
3x sin 2 x 23 cos 3 x sin 2 x dx
3 cos 3 x
4
3 x sin 2 x
sin 3 x cos 2 x dx
u
e x 3 x 2 dx
3 x 2 dx
eu du
1
4
;
3
1
3
1
3
3sin 3x dx; dv
sin 3 x cos 2 x dx
42. u
32
3cos 3 x dx; dv
cos 3x, du
1 sin
2
5
4
x2 1
3
x3e x dx
C
sin 3 x cos 2 x dx
u
3
x 2 1 x dx, v
2 x dx; dv
x3 x 2 1 dx
1 x 3e x
3
1 ex
3
4
x3 dx
1 du
4
x 2 e x dx, v
3
x5e x dx
x 2 , du
4 x3 dx
3
3 x 2 dx; dv
3
39. u
x 4 , du
C
1 cos u
3
C
561
562
Chapter 8 Techniques of Integration
e x
x
44.
45.
dx Let u
y
x
cos x dx; dy
1
2 y, du
1
2 x
2 eu du
dx
2eu
C
2e x
C
2 y cos y dy;
sin y;
2sin y dy
2 y sin y 2 cos y C
2 x sin x
2 cos x C
y e y 2 y dy
dx
2 y 2 e y dy ;
2 y dy
dx
y
2 y2
( )
4y
( )
ey
4
( )
ey
ey
2 y 2 e y dy
0
47.
e x
x
dx
x
x e x dx; dy
e
cos y dy, v
2 y sin y
y
46.
1
x
2du
2 y dy
2dy ; dv
2 y cos y dy
dx
cos y 2 y dy
dx
2 x
dx
u
1
2 x
x , du
2 y2 e y
4y ey
4e y
C
2x e x
4 xe x
4e x
C
sin 2
( )
1 cos
2
2
( )
1 sin 2
4
2
( )
2
1 cos
8
2
2
2 2
0
0
2
sin 2 d
2
8
( 1)
cos 2
2
4
1
4
0
( 1)
1 cos
4
sin 2
2
0 0
1
4
1
2
2
8
2
0
1
2
2
8
4
cos 2 x
48.
3
( )
3x2
( )
1 cos 2 x
4
6x
( )
1 sin 2 x
8
6
( )
1 cos 2 x
16
x
0
1 sin 2 x
2
2 3
0
x cos 2 x dx
3
x3
2
2
sin 2 x 34x cos 2 x 34x sin 2 x 83 cos 2 x
0
2
0 316 ( 1) 38 0 83 ( 1)
16
Copyright
0 0 0 83 1
2014 Pearson Education, Inc.
3 2
16
2
3
4
34
16
2
Section 8.2 Integration by Parts
sec 1 t , du
49. u
2
2
3
t2 1
1
2
t2
2
sec 1 t
2
2
1
2
0
5
9
3
2 x dx
1 x4
12
1
0
x tan 1 x dx
x tan 1 x dx
2
2
2
2
3
1
2
; dv
t2
3 2
4
3
3
dt
t t
3
4
12
Let u
1
2
1 2
x tan 1 x
2
1
2
1
2
3
3
3
t dt
3 2 t2 1
2
3
3
5
9
5
1
2 2
2 x dx
x
0
1 x
4
1
2
6
1 1
2 0
6 3 12
12
1
1
tan 1 x, du
1 2
x tan 1 x
2
1
2
2
3 6
2 3
x2
1 x2
1 x
dx, dv
2
x dx , v
x2
2
dx, dv
2
x 2 dx, v
dx
1
1
dx
1 x2
1 2
1
1
tan 1 x C
x tan 1 x
x
2
2
2
1 2
x
x 1 tan 1 x
C
2
2
52.
x
x 2 tan 1
dx
2
Let u
x
dx
2
x3
x
tan 1
3
2
1
3
x3
x
tan 1
3
3
1
3
x3
x
tan 1
3
2
1 2
x
3
2
x tan
1
1/ 2
x
tan 1 , du
2
In the remaining integral, let w 1
1 ( x / 2)
1 3
x
2
dx
x2
1
4
2x
2x
dx
x2
1
4
1
3
2x
1
x2
, dw
4
Copyright
3 3
9
x2 ;
2 x dx, v
0
2
2
5
9
1
1 2
x 2 sin 1 x 2
2 x sin 1 x 2 dx
1 x4
t2 ;
2
t dt , v
1
sin 1 x 2 , du
50. u
; dv
2
t t
t sec 1 t dt
5
9
51.
dt
x2
4
dx
x
.
2
2014 Pearson Education, Inc.
x3
3
2
1 x4
1/2
4 x3 dx
563
564
Chapter 8 Techniques of Integration
1
3
2x
1
3
dx
x2
1
4
4
dw
w
4
x2
ln 1
3
4
4
ln w
3
Thus the original integral is equal to
x3
x
tan 1
3
2
53. (a) u
1 2
x
3
x, du
S1
dx; dv
2
3
S3
54. (a) u
5
2
3
2
S3
7
2
2
5
( 1)n
ln 2 x
e
0
(2 ln 2) e x
2 ln 2 2
dx 2
ln 2
0
2 ln 2
ln 2
0
ex
ln 2
0
3
5
( n 1)
sin x n
3
2
2
2
5
2
3
2
2
3
2
sin x dx
sin x dx
7
2
5
2
5
2
(2 n 1)
2
x sin x (2 n 1) 2
( 1)n 1
cos x (2 n 1) 2
2
5
2
(2 n 1)
2
(2 n 1)
2
1
2
ln 2
ln 2 x
0
0
2n
e dx
2 ln 2 2
2 (1 ln 2)
Copyright
3
2
2
2
cos x 3 2
7
2
( 1) n
(2 n 1)
5
3
2
sin x dx
cos x
2
xe x dx
xe x
2
2
(2n 1)
2
2
7
(2 n 1)
2
sin x 2
( n 1)
x sin x 5 2
x cos x dx
3
5
x cos x n
0
sin x
ln 2 x e x dx
2
2 ln 2
2
2
(2n 1)
2
( 1) n
ln 2
(2 n 1)
(2 n 1)
3
x sin x 3 2
x cos x dx
3
sin x;
x sin x
x cos x dx
cos x dx
( 1)n 1
cos x dx, v
sin x 0
cos x dx
2
n ( 1)n 1
x cos x dx
5
( 1)n
(d) Sn
0
2
2
(b) S2
55. V
dx; dv
3
S1
(c)
( n 1) ( 1) n
x, du
3
3
n
( 1) n 1
2
2
x cos x 2
( n 1)
( 1) n 1
x sin x dx
(d) Sn 1
cos x dx
0
x cos x
x sin x dx
2
cos x;
x cos x 0
x sin x dx
0
C
sin x dx, v
x sin x dx
0
(b) S2
(c)
x2
4
ln 1
3
4
2014 Pearson Education, Inc.
2
4
7
2
cos x 5 2
sin x dx
2n
2n
6
Section 8.2 Integration by Parts
1
56. (a) V
2 xe x dx
0
1
(b) V
u 1 x, du
1
e x
[0 1( 1)]
1
e x
2
57. (a) V
0
2
V
2
2
58. (a) V
1
4
e
2
2
2
2
2
e x;
e x dx
2
(
0
sin x dx
2)
x, du
2
2
2
0
0
0 1
2
x cos x dx; u
x sin x
2
2
x sin x 0
2
2
e
1 1e 1
2
0
cos x 0
0
1
0
0
2 x cos x dx
2
2
(b) V
1
e
e x dx, v
dx; dv
(1 x)
2
2
1
e
2
0
e x dx
2 (1 x)e x dx;
0
V
1
0
0
1
e x
1
e
2
1
xe x
2
sin x dx
dx; dv
0 2
cos x dx, v
cos x 0
2
sin x;
2 (0 1)
2
2 x( x sin x) dx;
0
sin x
x2
( )
cos x
2x
( )
sin x
2
( )
cos x
0
V
(b) V
2 (
0
2
0
x 2 sin x dx
x) x sin x dx 2 2
x 2 cos x 2 x sin x 2 cos x
2
x sin x dx 2
0
0
x 2 sin x dx 2 2
8
59. (a)
A
e
1
ln x dx
(e ln e 1 ln 1)
(b) V
e
1
e
x1
(ln x)2 dx
e(ln e)2 1(ln1)2
e
e
e
x ln x 1
1
dx
e (e 1) 1
x(ln x )2
e
e
1
1
e
2 x ln x 1
(2e ln e 2(1) ln 1)
e
2x 1
Copyright
2 ln x dx
e
1
2 dx
e
2e (2e 2)
(e 2)
2014 Pearson Education, Inc.
0
2
x cos x sin x 0
2
4
2 3 8
565
566
Chapter 8 Techniques of Integration
e
(c) V
2 ( x 2) ln x dx
1
1 e2
2
2
e
(d) M
1
1 e 1 (ln
1 1 2
1
2
60. (a)
e
0
1 e2
2
x)2 dx
x (ln x)2
e
e
1
1
1
(b) V
0
2
2
0
x2
2
1
x tan 1 x
2
0
1
2
2
2
2
e t
2
e t sin t
x n sin x
1
2
1
4
x
2
0
e
e 1
1 2
1
1 e2
4
e1
1 2
1
x 2 dx
tan 1 x
1 (e
2
2)
(x, y)
e
e
2 x ln x 1
e2 1 , e 2
4
2
1
8
1
2
0
1 1 x2
2 0 1 x2
dx
2
1 tan 1 1 (0 0)
1 tan 1 1
2
2
8
2
sin t cos t
e t
2
0
1 e 2
2
sin t cos t
e t
2
0
nx n 1 dx; dv
cos x dx, v
sin x]
nx n 1 sin x dx
Copyright
2014 Pearson Education, Inc.
1
2
1
0 12 1
1
1 x2
1 4
(
0
2)
2
2 dx
is the centroid.
dx
1
e2 9
2
x dx
e(ln e) 2 1 (ln 1)2
2e 2e 2)
9
4
2e
e2 1 ;
e t cos t dt
x n , du
e
2e
0
x n cos x dx; [u
I
2
sin t cos t
2
0
63. I
1 (e
2
1 x 2 ln
2
1 (1) 2
4
2 ln x dx
1 e2
2
2
1
4e t sin t cos t dt
e t sin t dt
0
2
2
0
1
1
2
av ( y )
e
2 x ln x
0
1 ln 2
2
4
2e t cos t dt
1 e2
4
1 x
0 1 x2
0
1
2
(see Exercise 22)
62. av( y )
ln 1 x 2
1
2
2 ln 1)
1
2
e
2x 1
1
x tan 1 x
2 x tan 1 x dx
8
1
2
61. av( y )
1
2
2x
1 x2
2
2
1 e x ln x dx
1 1
x
e
1 x2
4
1
tan 1 x dx
1 (ln
2
1 x2
4
1 (1) 2 ln1
2
tan 1 1 0
4
1
( x 2) ln x dx
2 ln 1
(2e ln e 2(1) ln 1)
1
A
1
2
2e ln e
e
ln x dx 1 (from part (a));
1 e 2 ln e
2
y
2
dx
Section 8.2 Integration by Parts
x n sin x dx; [u
64. I
x n cos x
I
65.
x n , du
x(ln x )n
cos x]
e ax dx, v
1 e ax ]
a
0
n (ln x )n
x
(ln x) n , du
n
(ln x )n 1 dx, dv
x
1 m 1
(ln x )n 1 and
x
m 1
uv
sin x dx, v
1
dx; dv 1 dx, v
x]
n(ln x) n 1 dx
(ln x )n , du
u
nx n 1 dx; dv
x n 1 e ax dx, a
n
a
(ln x ) n dx; [u
I
I
67.
x n eax
a
nx n 1 dx; dv
nx n 1 cos x dx
x n e ax dx; [u
I
I
66.
x n , du
567
/2
68. First to show that
0
n
x m (ln x )n 1 dx
m 1
v du
/2
cosn x dx
xm 1
m 1
x m dx, v
0
sin n x dx note that cos x over the interval [0, / 2] is the reflection of
sin x over the same interval around the line x
/ 4.
Each iteration of the reduction formula in Example 5 for the definite integral produces an expression like
/2
cos n 1 x sin x
n 1 /2 n 2
cos
x dx
n 0
n
0
The evaluation on the left will be 0 as long as n
n
accumulate in front of the
n 1
integral on the right. When the initial n is even, the last iteration will have n 2 and the remaining integral
n 1 n 3
1 /2
13
( n 1)
1 dx
. When the initial n is odd the last iteration will have
will be
0
n n 2
n
2
2
2 4
n 1 n 3
2 /2
2 4
( n 1)
cos x dx 1
.
n 3 and the remaining integral will be
n n 2
n
3 0
3
69.
b
a
( x a ) f ( x) dx;
( x a)
0
70.
x
b
b
b
a
x
x 1 x2
x a, du
b
b
x
a
a
b
f (t ) dt
f (t ) dt dx
1 x 2 dx; u
x 1 x2
u
f (t ) dt dx
b
b
a
x
1 x 2 , du
x2
1 x2
dx
1 x 2 dx
dx; dv
2, and factors of the form
x
f ( x) dx, v
(b a )
b
b
b
b
f (t ) dt
x
f (t ) dt (a a )
a
b
f (t ) dt
f (t ) dt
f (t ) dt dx
x
1 x2
1 x2
dx, v
1 x 2 1 dx
x 1 x2
1
dx; dv
1 x2
x
x 1 x2
1 x2
1 x2
dx
dx
Copyright
2014 Pearson Education, Inc.
1
1 x2
dx
b
b
a
x
f (t ) dt dx
568
Chapter 8 Techniques of Integration
1 x 2 dx
x 1 x2
1 x 2 dx
x
2
1
1 x
1 x2
1
2
2
1 x 2 dx
dx
1
1 x2
1 x 2 dx
2
sin 1 x dx
x sin 1 x
72.
tan 1 x dx
x tan 1 x
tan y dy
x tan 1 x ln | cos y | C
73.
sec 1 x dx
x sec 1 x
sec y dy
x sec 1 x ln | sec y tan y | C
x sec 1 x ln sec sec 1 x
74.
log 2 x dx
x log 2 x
x sin 1 x cos y C
tan sec 1 x
x sin 1 x cos sin 1 x
x log 2 x ln 2 C
x2 1
x sinh 1 x
x sinh 1 x cosh y C
sinh y dy
check: d x sinh 1 x cosh sinh 1 x
(b)
sinh 1 x dx
x sinh 1 x
check: d x sinh 1 x
78. (a)
tanh 1 x dx
1
x
1 x2
x tanh 1 x
1 x
12
2
C
sinh 1 x
dx
x sinh 1 x 12
sinh 1 x
C
(b)
tanh 1 x dx
x
1 x2
x
1 x2
dx
x tanh 1 x
check: d x tanh 1 x
1 ln
2
x sinh 1 x cosh sinh 1 x
sinh sinh 1 x
x
1 x2
1 x2
x
1 x
x
2
1 x2 .
1 x
2
12
dx
2 x dx
1
1 x2
dx
x sinh 1 x
sinh 1 x dx
C;
check: d x tanh 1 x ln cosh tanh 1 x
tanh 1 x
C
x tanh 1 x ln | cosh y | C
tanh y dy
x tanh 1 x ln cosh tanh 1 x
C
1 x2 .
76. Yes, tan 1 x is the angle whose tangent is x which implies sec tan 1 x
sinh 1 x dx
C
x log 2 x lnx2 C
75. Yes, cos 1 x is the angle whose cosine is x which implies sin cos 1 x
77. (a)
dx
x tan 1 x ln cos tan 1 x
x sec 1 x ln x
C
2y
2 y dy
1
1 x2
dx C
71.
sin y dy
x 1 x2
C
tanh 1 x
x
1 x2
sinh tanh 1x
1
2
cosh tanh 1 x 1 x
dx
tanh 1 x dx
x
1 x2
1 x2
dx
tanh 1 x 12
2x
1 x2
C
tanh 1 x
x
1 x2
Copyright
dx
x tanh 1 x
x
1 x2
dx
2014 Pearson Education, Inc.
1 ln
2
1 x2
tanh 1 x dx
C
C;
sinh 1 x dx
1 x2
12
C
Section 8.3 Trigonometric Integrals
8.3
1.
2.
TRIGONOMETRIC INTEGRALS
cos 2 x dx
0
3 sin 3x dx
1
2
cos 2 x 2dx
9
0
1 sin
2
2x C
9
cos 3x
sin 3x 13 dx
3.
cos3 x sin x dx
cos3 x ( sin x) dx
4.
sin 4 2 x cos 2 x dx
1
2
5.
sin 3 x dx
6.
cos3 4 x dx
sin 5 x dx
sin x dx
8.
0
10.
cos3 x dx
6
0
/6
6
0
1
4
cos 0
1
2
9
9
2
1
x C
1 sin 5 2 x
10
1 cos2 x sin x dx
cos2 4 x cos 4 x dx
sin 2 x
2
C
cos2 x sin x dx
sin x dx
1 sin 2 4 x cos 4 x 4 dx
1 cos 2 x
sin x dx
2 cos 2 x sin x dx
4
3
cos3 2x
6
0
2
1
4
cos x 13 cos3 x C
cos 4 x 4dx 14 sin 2 4 x cos 4 x 4 dx
0
2 cos5 x
5
2
sin 2x dx
0
cos x
2 34
(0)
2
6
cos 3 x 3dx
0
2 cos3
3
x 15 cos5 x C
2 cos2 2x sin 2x dx
0
1 sin 2 x cos x dx
cos 2 3x
1 2 cos 2 x cos 4 x sin x dx
sin x dx
cos4 x sin x dx
cos 2 x cos x dx
3cos5 3x dx
2
5
0
cos 4 2x sin 2x dx
16
15
cos x dx
1 sin 2 3 x
2
sin 2 x cos x dx
sin x 13 sin 3 x C
cos 3 x 3dx
1 2 sin 2 3 x sin 4 3 x cos 3x 3dx
0
cos 3 x 3dx 2
1 23
11.
sin 2 x sin x dx
cos 3
1 cos 4
4
sin 4 2 x cos 2 x 2dx
sin 5 2x dx (using Exercise 7)
2cos 2x
9.
9
0
1 sin 3 4 x C
4 x 12
1 sin
4
7.
569
1
5
(0)
sin 3 x cos3 x dx
1 sin 4
4
6
0
sin 2 3 x cos 3 x 3dx
6
3
sin 4 3 x cos 3x 3dx
sin 3x 2 sin3 3 x
sin 3 x 1 sin 2 x cos x dx
sin 3 x cos x dx
0
sin 5 3 x
5
0
6
8
15
sin 3 x cos2 x cos xdx
x 16 sin 6 x C
Copyright
2014 Pearson Education, Inc.
sin 5 x cos x dx
570
12.
Chapter 8 Techniques of Integration
cos3 2 x sin 5 2 x dx
1 sin 2 2 x sin 5 2 x cos 2 x 2dx
1
2
1 sin 6
12
13.
14.
1 cos 2 x
2
1 sin
4
x
2
0
2
1
dx
2 0
15.
2
0
0
0
2
1
cos
4 0
(0 0)
2
0
sin y dy 3
3
7 sin t 3 sin3 t
8sin 4 x dx
2
2
18.
1
2
2 x 2dx
0
dx 2
x
1
2
dx
0
8
0
cos5 y
8cos 4 2 x dx
x 14 sin 2 x
5
3 sin5 t
2
cos 2 x dx
2
1
dx
2 0
1
2 2
0
8
0
2
1
2
dx 14 cos 2 x 2dx
2
1
cos
2 0
1 sin 2
4
2
4 x 2 dx 2 cos 4 x dx
0
3
2 x dx
1 (0)
2
1 sin 2(0)
4
sin y dy
2
cos4 y sin y dy
1 1 53
(0)
0
1
7
cos6 y sin y dy
16
35
cos t dt 3 sin 2 t cos t dt 3 sin 4 t cos t dt
7 sin t 7sin 3 t
2
21 sin 5 t
5
sin 6 t cos t dt
sin 7 t C
1 2 cos 2 x cos2 2 x dx
0
1 cos 4 x
2
dx
2 x 2sin 2 x 0
0
dx
0
cos 4 x dx
3
1 cos 4 x 2
dx
2
16 sin 2 x cos 2 x dx 16
2
0
C
dx
0
1 cos 2 y
2
7
sin 7 t
7
1 cos 2 x 2
2
0
4x
2
0
cos7 y
7
3 5
cos 2 x 2dx 2
1 sin
4
1 cos 2 x dx
cos2 y sin y dy 3
3 dx 4 cos 4 x dx
19.
2
1
2 0
dx
7 cos 7 t dt (using Exercise 15)
0
1
2
1 cos 2 x dx
sin 6 y sin y dy
2
cos3 y
17.
sin 5 2 x cos 2 x 2dx 12 sin 7 2 x cos 2 x 2dx
1
2
4
cos y 3 3
16.
1
2
dx
2 1 cos 2 x
2
0
sin 7 y dy
2
cos 2 x cos 2 2 x sin 5 2 x 2dx
2x C
sin 2 x dx
4
1
2
1 sin 8 2 x C
2 x 16
cos 2 x dx
1
2
cos3 2 x sin 5 2 x 2dx
1
2
2 1 2 cos 4 x cos2 4 x dx
cos 8 x dx
1 cos 2 x
2
3x
1 cos 2 x
2
1
sin 4 x 81 sin 8 x C
dx
4 x 2 x 12 sin 4 x C
2 x 2sin x cos x (2cos 2 x 1) C
1 cos 8
2 dx 4 cos 4 x dx2 2
4 1 cos 2 2 x dx
2 x 12 sin 4 x C
4 dx 4
dx
2 x sin 2 x cos 2 x C
2 x 4sin x cos3 x 2sin x cos x C
Copyright
1 cos 4 x
2
2014 Pearson Education, Inc.
x
dx
Section 8.3 Trigonometric Integrals
20.
0
8 sin 4 y cos 2 y dy
1 sin
2
y
1
2 0
1
2
21.
22.
2
0
25.
26.
2
0
dx
2
1 cos 2
0
2 sin 2 x
3 1 cos x
/2
3
2
6
0
0
2
0
cos4 2
C
0
|sin | d
1 cos x
1 cos x
1 sin 3 2
2
3
2
sin d
2cos x
cos t dt
cos
3
3
cos
2 y dy
2
3
3
0
0
4
2
0
sin t 0
2
2
sin t
2 2
1 0 0 1 2
2
1 1 2
0
1 cos 2 x
2
2
1 sin 5 2
2
5
2 2
2 sin 2 x 1 cos x
dx
cos x)3 2
2 (1
3
0
cos 2 d
0
cos t dt
0
cos2 2 y dy
0
C
2 sin x dx
0
2
|cos t | dt
cos 2 y dy
2
2
2cos 2x
sin 2x dx
2 | sin x | dx
0
sin 2 2 y cos 2 y dy
sin 4 2 cos 2 d
2
dy
0
2 sin 2 x 1 cos x
dx
3
1 cos 2
sin 2 x
32
2
3
dx
32
1 cos 3
2
3
2
3
cos4 x
6 1 sin x
6 1 sin x
1
0
dx
6
0
5
cos 4 x 1 sin x
cos x
6
5
2
0
sin x 1 cos x dx
0
sin 2 2 1 sin 2 2
2 sin 2 x
3 1 cos x
dx
2(1 sin x )1 2
5
0
0
1 sin x dx
0
5
2
0
d
3
1 sin 2 y
2
3
sin 2x dx
0
cos 2 y dy
0
2y
dy
1 sin 2 2 y cos 2 y dy
0
1 cos4 2
2
4
sin 2 2 cos 2 d
1 sin 2 t dt
0
/3
29.
1 sin
2
8
1 cos 2 x dx
0
27.
28.
y 81 sin 4 y
1 cos x
2
dy
cos 4 y dy
sin 2 2 cos3 2 d
2
24.
1
2 0
dy
1 cos 4 y
2
0
0
8 cos3 2 sin 2 d
0
23.
2y
1 cos 2 y 2 1 cos 2 y
2
2
0
8
6
1 sin x
1 sin x
dx
cos x 1 sin x dx
5
1 sin x
1 sin x
6
5
6 1 sin 2 x
0
1 sin x
dx
2 1 sin 6
cos 4 x
6 1 sin x
2 1 sin 0
dx
5
6
6
dx
2 12
cos 4 x 1 sin x
cos3 x 1 sin x dx
6
571
2
1 sin x
5
6
0
2 1
dx
cos2 x
1 sin x
2
6
0
cos4 x 1 sin x
5
cos x
1 sin x
2
6
cos x 1 sin 2 x
cos x sin 2 x 1 sin x dx;
Copyright
dx
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cos2 x
dx
1 sin x dx
dx
2 3
3 2
32
572
Chapter 8 Techniques of Integration
Let u 1 sin x
2 (1
3
sin x)3 2
2 (1
3
sin )3 2
2 3
3 2
2
3
30.
7 12
2
2
32.
1
5
2
7
2
cos 2 x
1 sin 2 x
1 cos 2 t
0
0
7 12
1 sin 2 x
1
2
dx
32
2
0
cos t
cos3 t
3
sin 2 t
0
32
cos t
72
1 sin 2 x
0
sin t dt
cos3 t
3
0
tan x sec2 x dx
34.
sec x tan 2 x dx
sec x tan x tan x dx; u
sec x tan x
35.
sec3 x tan x dx
36.
sec3 x tan 3 x dx
sec 2 x tan 2 x dx
6
u 1 sin
2u 3 2
32
2 3
7 2
72
7 12
dx
2
2
cos
sin 3 t dt
1 sec
2
18
35
1
2
sin
dx
1 1
2
0
1
2
2(1)
sin 3 t dt
0
cos 2 t sin t dt
1 13 1 13
8
3
sec 2 x dx, dv
sec x tan x dx, v
x C
tan x, du
sec2 x sec x dx
sec x dx
sec x tan x
sec x tan x ln | sec x tan x |
1 sec3
3
tan 2 x sec x dx
tan 2 x sec x dx
x C
sec 2 x tan 2 x sec x tan x dx
sec2 x sec x tan x dx
tan 2 x sec2 x dx
1 tan 3
3
Copyright
sec2 x sec2 x 1 sec x tan x dx
1 sec5
5
sec x;
tan 2 x 1 sec x dx
x tan x 12 ln | sec x tan x | C
sec 2 x sec x tan x dx
1
u du
cos2 2 x
1 sin 2 x
1 sin 2 2
sin d
0
52
4 3
5 2
sec x tan x ln | sec x tan x |
sec4 x sec x tan x dx
37.
0
sec x tan x ln | sec x tan x |
2 tan 2 x sec x dx
tan 2 x sec x dx
1 tan 2
2
sec x tan x
tan 2 x sec x dx
sec x tan 2 x dx
2
0
1 13 1 13
sec2 x tan x dx
sec3 x dx
2
5
7 12 1 sin 2 2 x
2
1 sin 2 x
sin 3 t dt
33.
sec x tan x
u5 2
1 sin 2 712
2
dt
1
32
x
1
2 u3 2
3
32
2 3
3 2
dx
1 sin 2 x
3,
2
sin x)3 2
4 u5 2
5
52
4 3
5 2
u 1 sin 56
1 cos2 t sin t dt
cos 2 t sin t dt
0
2 u7 2
7
2 | sin | d
0
dt
2 (1
3
7 12
1 sin 2 x
1 cos 2 t sin t dt
sin t dt
32
2 3
7 2
5
6
cos x dx, x
(u 1)2 u du
2
3
4
5
1 cos 2 d
0
du
1 sin 56
2
3
32
32
6
1 sin 2 x dx
7 12
31.
u 1 sin x
x 13 sec3 x C
x C
2014 Pearson Education, Inc.
2 1
2
2
Section 8.3 Trigonometric Integrals
38.
sec 4 x tan 2 x dx
sec2 x tan 2 x sec 2 x dx
tan 4 x sec2 x dx
39.
0
3
0
3
2 sec3 x dx; u
2 sec3 x dx
4 3 2
2
0
3
0
3
sec3 x dx 2
2 sec3 x dx
4 3
e x sec3 e x dx; u
2 e x sec3 e x dx
sec 4
tan
42.
0
1 tan
3
0
3
0
3
sec2 x dx, v
0
2
3
sec x tan 2 x dx
2 sec3 x dx
0
2
3
0
2 sec3 x dx
3
2 3 ln 2
sec e x tan e x e x dx, dv
sec e x
sec e x tan e x
sec3 e x e x dx
C
sec2 e x
ln sec e x
sec e x tan e x
1
2
0
3
sec x sec2 x 1 dx
d
sec 2
1 tan
3
sec 2
C
C
tan 2 sec 2
2 tan
3
1 tan 2 (3 x) sec2 (3 x) 3dx
tan e x ;
sec e x e x dx
tan e x
d
3
1 e x dx
tan e x
ln sec e x
4 3 2ln |1 0 | 2 ln 2
sec2 e x e x dx, v
sec e x tan e x
sec2
3
3
sec e x tan 2 e x e x dx
sec2
tan x;
21 0 2 2
sec e x tan e x
1
2
3sec 4 (3 x) dx
x 13 tan 3 x C
sec x dx;
sec e x , du
1 tan 2
d
1 tan 5
5
sec x tan x dx, dv
sec e x tan e x
e x sec3 e x dx
tan 2 x 1 tan 2 x sec2 x dx
2 ln sec x tan x
3
e x sec3 e x dx
41.
sec x, du
2sec x tan x
4 3 2 ln 2
40.
tan 2 x sec 2 x dx
573
d
tan
1 tan 3
3
C
C
sec2 (3 x)3dx
tan 2 (3 x)sec 2 (3 x) 3dx
2
2
tan(3 x) 13 tan 3 (3 x ) C
43.
2
4
csc4
(0)
44.
2
d
1
sec6 x dx
4
1
3
1 cot 2
csc2
d
4 tan 3 x dx
d
4
cot 2 csc2
d
cot
4
3
sec4 x sec2 x dx
2
tan 2 x 1 sec2 x dx
tan 4 x sec 2 x dx 2 tan 2 x sec 2 x dx
45.
4
csc 2
4 sec2 x 1 tan x dx
2 tan 2 x 4 ln |sec x |
C
sec2 x dx
tan 4 x 2 tan 2 x 1 sec2 x dx
1 tan 5
5
x
2 tan 3
3
4 sec2 x tan x dx 4 tan x dx
2 tan 2 x 2 ln sec2 x
Copyright
C
x tan x C
2
4 tan2 x
2 tan 2 x 2 ln 1 tan 2 x
2014 Pearson Education, Inc.
4 ln |sec x | C
C
cot 3
3
2
4
574
Chapter 8 Techniques of Integration
4
46.
4
6 tan 4 x dx
4
6
4
sec2 x 1 tan 2 x dx
4
4
sec2 x tan 2 x dx 6
2 1 ( 1)
47.
4
6
4
6 tan x
tan 5 x dx
4
6x
4
4
sec2 x 1 tan x dx
48.
cot 6 2 x dx
cot 4 2 x csc2 2 x dx
cot 2 2 x csc 2 2 xdx
csc 2 2 x 1 dx
cot 4 2 x csc2 2 x dx
cot 2 2 x csc2 2 xdx
3
3
csc2 x 1 cot x dx
6
ln 2
ln 2
3
1
2
52.
sin 2 x cos 3 x dx
1
2
55.
csc2 2 x dx
3
6
dx
csc2 x cot x dx
C
cot 4 2 x dx
cot 2 2 x csc2 2 x 1 dx
1 cot 5 2 x
10
3
6
1 cot 3 2 x
6
cot 2 x
2
cot x dx
8 csc2 t cot 2 t dt 8 cot 2 t dt
1 cos
2
(sin x sin 5 x) dx
1
2
sin( x) sin 5 x dx
1
2
sin 3 x sin 3 x dx
2
x sec2 x ln |sec x | C
1 cot
2
2x x C
ln |csc x |
8 cot 3 t
3
8 csc2 t 1 dt
8 cot t 8t C
sin 3 x cos 2 x dx
2
dx
ln 3
8 csc2 t 1 cot 2 t dt
51.
2
4
3
4
8
cot 4 2 x csc2 2 x dx
cot 2 2 xdx
8cot 4 t dt
0
cot 4 2 x csc2 2 x 1 dx
cot 2 2 x csc 2 2 x dx
8 cot 3 t
3
54.
1 sec 4
4
tan x dx
cot 4 2 x csc2 2 x dx
cot 3 x dx
4
sec 2 x dx 6
ln |sec x | C 14 tan 4 x 12 tan 2 x ln|sec x |
cot 4 2 x cot 2 2 x dx
tan 2 x dx
sec 4 x 2sec2 x 1 tan x dx
cot 4 2 x csc2 2 x dx
6
53.
3
cot 2 2 x cot 2 2 x dx
1 1
2 3
50.
tan 2 x 1
4
3
2
cot 4 2 x csc2 2 x dx
3
49.
2
4
4
6
4
tan x dx
sec3 x sec x tan x dx 2 sec x sec x tan x dx
tan 2 x 1
4
3
2
2
tan 4 x tan x dx
4
sec2 x tan 2 x dx 6
4 6 1 ( 1)
sec 4 x tan x dx 2 sec 2 x tan x dx
1
4
4
3
6 tan3 x
sec 2 x 1 dx
4
4
6
1 cos 5 x C
x 10
sin x sin 5 x dx
1
2
cos 0 cos 6 x dx
dx 12
1 cos
2
cos 6 x dx
1 cos 5 x C
x 10
1
2
1 sin 6 x
x 12
0
sin x cos x dx
cos 3 x cos 4 x dx
2
1
2 0
1
2
sin 0 sin 2 x dx
cos( x) cos 7 x dx
Copyright
2
1
sin
2 0
1
2
2 x dx
1
4
cos x cos 7 x dx
cos 2 x 0
1 sin
2
2014 Pearson Education, Inc.
2
1(
4
1 1)
1 sin 7 x C
x 14
1
2
3
6
Section 8.3 Trigonometric Integrals
2
56.
57.
2
sin 2 cos 3 d
1
2
1
2
58.
1
2
1
4
cos 3 d
cos 3 d
cos 2 2 sin
4 cos 4
2
1 cos 2
2
1 cos(2
2
3)
2
sin
sin
cos3 sin 2 d
cos3
60.
sin 3
sin 2 cos 2 sin
cos 2 d
2 cos 4
2 cos5
5
61.
sin
cos
1 1
2 2
1 cos
4
62.
sin
1
2
1
4
63.
3cos 2
cos3
2sin
sin(2 3)
2sin
sin(2 3)
1 cos 5
20
sin 2
sec3 x
tan x
dx
1 sin
20
4 cos2
2 cos4 sin
5
) cos 5
C
1 sin d
4 cos3
3
cos
2 cos5
5
d
C
C
1 cos 2
2 cos 2
1 sin d
sin d
3 cos 2
sin
2 cos 4
d
d
sin
d
cos
1
4
d
1
2
cos 3 d
sin(
sin 2 cos 3 d
) sin 5
1
4
d
sin
sin 5
d
cos(
) cos 3
C
1
2
sin 2 sin 3 d
sin 3 cos
cos(
C
1
2
cos 3 d
1
4
4 cos5
5
d
d
d
1 sin d
cos
sin
0
2
cos 2 cos 3 d
1 sin
4
3
2
6 x 81 sin 8 x
cos 3 d
4cos 4
d
cos
1
2
d
1 sin
6
d
59.
1
2
cos 3 d
cos 5 d
1
4 cos2
d
1
2
cos(2 3)
1
4
2 cos 2
1 1 sin
2 6
cos 6 x cos 8 x dx
cos 3 d
cos d
d
sin
2
1
2
cos 7 x cos x dx
575
1
2
d
sin 4
cos(1 2)
1
4
sin 6 d
tan 2 x 1 sec x
dx
tan x
tan 2 x sec x
tan x
dx
1
2
sin 3 d
1 1 sin(3 1)
2 2
1 cos 2
1 cos
8
16
sin 3 cos 3 d
d
sec2 x sec x
tan x
cos(1 2)
sin(3 1)
1 cos
24
4
sec x
tan x
dx
1
4
d
6
dx
sin 3 d
2 sin 3 cos 3 d
C
tan x sec x dx
csc x dx
sec x ln |csc x cot x | C
64.
sin 3 x
cos 4 x
dx
sin 2 x sin x
cos 4 x
sec 2 x sec x tan x dx
65.
tan 2 x
csc x
dx
sin 2 x
cos 2 x
sec x tan x dx
66.
cot x
cos 2 x
dx
1 cos 2 x sin x
dx
cos4 x
sin x dx
cos x
1
sin x cos 2 x
dx
1 cos2 x
cos 2 x
cos4 x
1 sec3
3
sec x tan x dx
sin x dx
sin x
dx
sin x dx
cos 2 x sin x
dx
cos4 x
dx
sec3 x tan x dx
sec x tan x dx
x sec x C
1 sin
cos 2 x
x dx
2
cos x
cos2 x
sin x dx
sec x cos x C
2
2 sin x cos x
Copyright
dx
2
sin 2 x
dx
csc 2 x 2dx
2014 Pearson Education, Inc.
ln |csc 2 x cot 2 x | C
576
Chapter 8 Techniques of Integration
x sin 2 x dx
67.
1 x2
4
x
1 1
2 2
x sin 2 x
x cos3 x dx
68.
1 cos 2 x
2
dx
1
2
1 sin
2
2 x dx
x sin x
x sin 2 x cos x dx
1 x sin 3
3
x
1
3
1 x sin 3
3
x
1 cos
3
y
1 x sin 3
3
4
71. V
0
2
72.
(
A
sin 2 x dx
0)
sin 2 x
2
73. M
1
2
2
2
4
2 1
2 1
1 cos 2 x
2
x, du
x, du
dx, dv
1
2
4
3
1
4
x3
1
2 2
2
2
0
2
1 x sin 3
3
ln(0 1)
4
1 cos
3
x
1 cos3
9
x
1
2
cos x
2
0
0
x2
x
x
C
2 0
2 1
| sec x | dx
cos 2 x dx
2 1
;
2 1
ln
1
1 ( 1)
2 1
2 1
2 ln
dx
2 0
ln
1
4
0
2 1
2 1
4
tan x
4
1 tan 2 x dx
0
ln
ln
4
dx
4
tan 2 x;
2 cos 2 x dx
0
2
2
2
1 sin 3
3
cos2 x sin x dx
1
3
ln
x
2
(x, y)
2 1
2 1
0
1
2 1
2 1
0, ln
sin 2 x
4
3
sin 2 x
4
4
1 x2
2
2
2
1
2
x2
0
0
3
cos 2 x dx
2
2
4
3
2
2
2
(1 0)
1
x cos x dx
2
2
2
0
4
4
)2 sin (2 )
1 (2
2
0
2
2
sin 2 x
2
sin x
4
2
cos 2 x dx
x sin x
4
3
2
2
0
1
2
0
2
sin xdx
cos 2
2 x cos x cos2 x dx
cos 0
1
4
Copyright
2
2
0
1
6
2
4
3
x 2 dx
3
8
4
3
0
1
2
0
1 (0)2
2
x2 dx
2
2
0
1
2
; y
2
2
2
( 1 1)
0
1
2
2
2
1
2
x cos xdx
2014 Pearson Education, Inc.
1
4
2
2 2
;
dx, dv
0
2
0
x cos x
2
2
cos 2 x dx
2
2
2
x, du
2 1
2
4
x cos xdx
0
0
3
(0 1)
2 sin 2
2
2
sin (0)
u
6
sin x
sin 2 x cos x dx, v
dx, dv
sin x dx
cos x dx, v
2
x x cos x dx
0
1
3
x
x sin 2 x cos x dx;
x cos x dx
2
(0 0)
0
2 1
2 ln
x cos x dx
0
x
0
2x
x C
tan x;( y )2
1
1 cos 4 x dx
0
2
2
4
1 cos3
9
x
ln
dx
u
x sin x cos x
ln sec x tan x
4 sec2 x
4 2
2 1
2 1
ln
4
0
1 sin
2
cos 2 x dx, v
x sin 2 x 18 cos 2 x C
x dx;
1 x sin 3
3
dx, dv
x;
sec x tan x
sec x
sec x dx
1
y
2 cos
3
x
ln |sec x tan x |
4
1 sin 3
3
x
x sin x cos x dx
ln(sec x); y
70. M
1 cos3
9
2
x
x, du
x sin x cos x; u
sin x dx
1 x sin 3
3
1
4
u
x 1 sin 2 x cos x dx
1 cos2 x sin x dx
x cos x dx
69.
1 x2
4
x cos 2 x cos x dx
x cos x dx
x sin x
x dx 12 x cos 2 x dx
0
cos x dx, v
sin xdx
2
cos 2 x dx
dx
sin x
Section 8.4 Trigonometric Substitutions
1
12
2
3
1
16
3
3
8
3
2 0
0
3
1
2
3
3
dx
3
3
0
8
2
2
1
cos 2 x dx
0
2
4
3
The centroid is
sec 2 x dx
ln sec 3
4
1
0
, 812
3
8
2
2
0
dx
.
sin 2 x 2sin x sec x sec2 x dx
0
3
2
3
8 2 3
12
1
4
0
2 ln 2
0
2 cos 2 x 1
dx
2
0
2
4
2 tan x dx
2 ln 2
8
2
0
x
2
cos 2 x dx 2
sin 2 x
4
6
2
0
3
dx
2
1.
1
sin x sec x
3
x
2
0
sin 2 x dx
2 0
2
2
2
0
x sin x cos x
2
sin 2 x
2
0
0
1
0
3
74. V
8.4
2
x3
2
577
3 1 cos 2 x
2
0
ln sec 0
3
tan
0
2 3
4
3
ln |sec x | 0
tan x
3
0
tan 0
3
sin 2
dx 2
sin 2(0)
3
2 ln 2
3
21 3 48 ln 2
24
TRIGONOMETRIC SUBSTITUTIONS
x
3 tan ,
2
2
because cos
dx
9 x2
3 dx
2.
du
4.
5.
6.
7. t
dx
2 4 x2
u,3 dx
32
dx
0
9 x2
2 dx
1 4x
2
5sin ,
25 t 2 dt
sin
du
1 u2
sec t dt
1x 2
2 2
tan
ln | sec
du ]
1 1
2 2
tan
1x 2
2 0
11
2
1x 32
3 0
sin
; t
2 x, dt
2 dx
2
2
, dt
5 cos
cos
C
;u
tan | C
ln
tan t ,
t
1 tan 1 (
2
1 1
2 2
sin
1
1
2
0
2
tan 1 1
0
0
6
dt
1 t
25
2
sin
d
1
|cos |
3
1
3|sec |
9 x2
Copyright
t
5
x
3
C
dt ,
cos2 t
, du
ln 9 x 2
cos
3
C
1
2
4
1
tan
0
4
1
2
1
2
C
1 u 2 | sec t | sec t ;
ln u 2 1 u
4
x
ln 1 9 x 2
0
4
3x
16
6
sin
2
1
t
1/ 2
0
sin
1 1
2
sin
1
0
0
4
4
5 cos ;
25 cos 2
1 t
5
1
2
9 x2
3
2
1
2
1)
5 cos d , 25 t 2
5 cos
9 sec2
ln | sec t tan t | C
1 tan 1 1
2
sin
9 1 tan 2
;
2
d
cos
1 2 dx
2 0 4 x2
dx
0 8 2 x2
25
2
3cos
1
2
2
0
d
2
, 9 x2
cos2
2
dt
cos2 t (sec t )
2
12 2
cos
; [3 x
1 9 x2
1 u2
3.
0 when
3
3d
, dx
d
25 t 2
5
25
C
1 cos 2
2
d
25 sin 1 t
2
5
2014 Pearson Education, Inc.
25
2
sin 2
4
t 25 t 2
2
C
C
C
;
578
Chapter 8 Techniques of Integration
1
3
8. t
sin ,
2
2
1 9t 2 dt
9.
7
2
x
1
3
7
2
10.
3
5
x
2
sec , 0
25 x
y
12.
y
sec
x
x
sec
sec
1
2 dx
2 tan
9 x2
9 x2
tan
1
10
sec
, dx
sec
tan
d
d , y2
d
tan
, dx
sec
C
du
1
2
sec
du
tan | C
1
2
9 sec 2
9
7 tan ;
C
3 tan ;
25 x 2 9
3
5x
3
ln
4 x 2 49
7
ln 27x
C
7 tan ;
7 sec2
d , x2 1
sin
1 d
7 tan
C
1
2
1
x
5
y
C
sin 2
sec
d
1 y
5
Copyright
1 cos 2
y 2 25
10
2 y2
tan ;
d
C
1
sec
sin
x
x2 1
x2
cos
C
x dx;
9 x2
1
10
C
1 cos 2
2
2
1
5
d
x2 1
x
C
d
x2 1
5 tan ;
tan ;
d , x2 1
sec
u C
y 2 25
y
1 y
5
2 cos2
1 du
2
25
tan 2 cos 2
1
5
tan
x
y2
d ,
d
d
2 x dx
1
u
49
49
sin 1 (3t ) 3t 1 9t 2
1
6
C
49 sec2
tan | C
7 tan 2 d
tan
d
sec
sec3 tan
cos 2
ln |sec
tan
5 sec
5 sec
tan
2
x3 x 2 1
x dx
, dy
C
2
sec , 0
tan
d
cos
C
cos
2
sin
d , 25 x 2 9
125 sec3
sin
2
7 sec
1
6
d
1 ln |sec
2
d
tan
7 sec tan
7 sec
5 tan
sec , 0
2
, dy
sec
sec
cos ;
d , 4 x 2 49
tan
sec
d
1 y
7
2
dy
dx
15. u
2
5 sec , 0
13. x
tan
7 tan
y 2 49
7
1
10
3
5
3 tan
dy
y3
x
sec
9
y 2 25
14.
3
5
7 sec , 0
7
1
2
, dx
2
5
y 2 49
y
d
cos2
1
3
d
sec
7 tan
5 dx
11.
sec tan
49
2
7
2
d , 1 9t 2
cos
cos
, dx
2
dx
4x
cos
sec , 0
1
3
, dt
C
2014 Pearson Education, Inc.
C
C
d
C
Section 8.4 Trigonometric Substitutions
16. x
2 tan ,
x 2 dx
4 x2
2
4 tan 2
2sec2
2
C
2 tan ,
x
4
x2
x
4
x
x
19. w
2
sec
1
4 w
dw
cot
x 1
dx
1 x
21.
x
2
9 sin
9 w2
w
x2
4 C
d , x2 1
1
sin
x 1
2 cot
2
; [t
sec
3
C
8
C
2 4 w2
w
C
C
3 cos ;
d
csc2
1 d
C
1 x
.
1 x
1 x 1
so that cos
2
0 and 1 x 2
cos .
sin
1
cos d
cos
sin
sin 1 x
1 d
cos
1 x2
C
C
Copyright
2 d
cos , dt
2 cos ;
1 sin 2
sin 2
cot 2 d
3
x2 1
x
d , 9 w2
dx where
d
sec ;
d , 4 w2
sin 1 w3
sin
sec
C
Multiply the integrand by
cos d ,
1 x
x2 8
3 cos
2
1
cos 4
8
d
sin 2
3cos 3cos d
sin , dx
dx
, dw
cos 2
C
2 cos
2
2 cos
1 x2
2
, dw
d
x 1
dx
1 x
x 1
1
3
2 sec 2 d
1 d
;
1
3t 3
sin 2
2
C
1
t
8
sec 2
2
cos
8
cos4
4sec2 ;
2
4
sin 3 d
cos d
2
2
, x2
sec2
d
4 sin
20. w 3 sin ,
9 w2
w2
dt
, dx
8 2 cos
2
2d
4 C
2
8 dw
w
2
C
8
tan 2 sec
2 sin ,
2
1
t4
1 x
2
cos2
d
4 x2
2
dx
2
1
t2
8
32
tan ,
2
, dx
cos
d , 4 x2
2 tan 2 d
cos2
t
18.
2
8 tan 3
2
8 t 4 1 dt
1
3
d
x 2 tan
2
x3 dx
2
2 sec 2
, dx
4 sec2
2 tan
17. x
2
2014 Pearson Education, Inc.
x2 4
2
sin d ]
x2 4
83
32
C
579
580
Chapter 8 Techniques of Integration
x2
22. u
4
du
x x 2 4 dx
23.
x
1
2
sin , 0
, dx
3
24.
1 x
x
2sin , 0
1
dx
0 4 x2
25.
x
6 2 cos
32
x
x
1
sec , 0
x
x
2
32
2
dx
x
1
2
x
12
2
dx
cos
4x
30. t
2
1
3
1
9t
2
1
2
2
tan ,
1
cos
cot 4
cos
sec
d
2
1
3
sec2
sec
, dx
4
2
6
2
sec2
4
d
, dt
d , 1 x2
cot 2
1
2
4 cos 2
1
3
csc 2
sec 2
sec 2
2 cos 2
6
32
x
52
3
x
C
cos3 ;
d
cot 5
5
3 x2 1
1
5
C
32
C
1 x2
x
5
C
cos ;
cot 3
3
d
2
sin
Copyright
sin
0
1
4 3
1
3
C
3
1 x2
x
C
sec4 ;
cos
C
2 tan x 1 2 x
4
4x2 1
d , 9t 2 1 sec 2 ;
d
3
4 tan
tan 5 ;
32
2
1 d
C
x2 1
d , 4x2 1
d
sec2
tan 3 ;
C
12
0
3
12
0
1
3 sin 3
csc 2
3
4
8 cos3 ;
tan
d
d , 1 x2
cos
d
cos
2
8
6 dt
2
, dx
1
4
1
sin
sin 4
sin 4
tan ,
8dx
d
, dx
2
x4
29.
tan
d
32
d , x2 1
tan
sin 6
sin ,
1 x2
sec
cos3 cos d
x6
28.
d
tan 5
2
3 1 cos 2
cos2
0
d , x2 1
tan
sin 2
sec 2 sec
sin ,
1 x2
4
C
cos3 ;
d , 4 x2
cos
d
, dx
2
52
1
sec
tan 3
x 2 dx
27.
tan
32
6 d
1
4 0 cos2
d
, dx
2
32
d
2 cos
8 cos3
0
sec
dx
26.
, dx
6
sec , 0
2
cos
cos3
0
32
x2 4
1
3
C
d , 1 x2
cos
3 4 sin 2
2 32
x dx;
1 u3 2
3
u du
3 2 4 x 2 dx
0
1 du
2
2 x dx
cos
C
tan 1 3t
3t
9t 2 1
2014 Pearson Education, Inc.
C
C
4 3
4
3
Section 8.4 Trigonometric Substitutions
x2 1
31. u
du
x3 dx
x2 1
x
x2 1
x
25 4 x 2
32. u
x
25 4 x 2
33. v
du
1
8
dx
sin ,
1 v2
34. r
1 du
u
2
sin 2
52
dr
tan , t
et dt
38.
14
2 dt
1 12 t 4t t
1
2 du
1
31 u
2
y
e tan , 0
e
dy
1 y 1 ln y
39.
x
sec , 0
dx
x x2 1
sec2
cos d , 1 r 2
32
; u
1
3
cot 6
tan
1
2
5/2
cos5 ;
d
cot 7
7
tan 1 (4 3)
1
tan (1 3)
1
t
d
, dy e tan
4
4 e tan sec 2
0
e tan sec
, dx
2
sec tan d
sec tan
sec
d
4
6
2 4
sec2
0
tan
C
1
tan
1
6
1 x2
2
1 ln
2
x2 1 C
C
1 v2
sec 2
tan
d
7
1 r2
r
1
7
C
C
d , e 2t
ln sec
9
9 tan 2
9
3 sec ;
tan 1 (4 3)
tan 1 (1 3)
tan
sec 2
tan
sec
(4 3)
(3 4)
;u
cos
d , 1 e 2t
d
4
ln y
d
d , x2 1
, du
2
1 tan 2
1
(4 3)
(3 4)
sec2
ln sec
sec2
tan
1
0
4
tan ;
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ln 1
sec2 ;
4
5
3
5
d , 1 u2
sec ;
sec 1 x C
Copyright
1
tan
tan
sin
tan , 6
1 tan 2
6
d , 1
4
d
sec
tan
1
2 du
1 3 1 u2
dt
2
sec2
C
3
v
1
3
C
tan 1 34 , dt
sec3
(3 4)
1 ln | u |
2
10
sec 2
tan
(4 3) tan
2 t , du
4 2 sec2
6
tan 3
3
tan 1 34 , dt
ln(tan ), tan 1 34
1
1 x2
2
du
cos5 ;
d
csc2
ln 9 ln 1
tan
1
u
C
52
ln(3 tan ), tan 1 13
10
3
ln
ln (3 4) 1 e 2t
37.
tan 2
tan 1 (4 3) 3 tan sec2 d
tan 1 (1 3) tan 3 sec
4
3
ln (4 3)
25 4 x 2
1 ln
8
d , 1 v2
d
1
2
x dx;
C
sin 8
ln 4 et dt
0
e2t 9
36. Let et
d
cos5 cos
3 tan , t
ln 53
1 du
8
cos
; dr
2
r8
35. Let et
, dv
1 x2
2
x dx
x2 1
x dx
cos5
2
x dx;
1 ln | u |
8
cos
52
sin ,
1 r2
dx
8 x dx
2
v 2 dv
1 du
2
2 x dx
2
1
5
sec 2 ;
581
582
40.
Chapter 8 Techniques of Integration
x
x
2
x
sec , dx
x dx
42.
x
2
sec
1
sin , dx
cos
2
cos d
cos
43. Let x 2
tan , 0
dx
1 x
x
1 x
44. Let ln x
45. Let u
2
1
2
sec
1
dx
2
4 u 2 du
2 2 cos
4 sin 1 2x
x3 2
46. Let u
x
1 x3
47. Let u
u
x
x
u2 3
dx
1
4
u2
cos
2 u 2 1 u 2 du
d
C
1 u
sin , du
1
4
4 x x2
dx
d ,
ln
4 x
x
sin
1 ln x
ln x
2
4 u2
u2
dx
d
8
4 sin 1 u2
C
C
cos
ln | csc
1
ln x
2u du
2
2
cot | cos
C
C
4 u 2 du;
2 cos
8 cos 2
d
d
2
x2
sec
1 cos 2
2
d
4 d
4 u2
2
4 u2
C
4 cos 2 d
4 sin 1 2x
x 4 x C
2 u 1 3 du
3
2
u1 3
2
13
1 u 2 3u
du
, 1 u2
cos
1 sin 4
16
Copyright
C
1
1 u2
2 sin 1 u
3
du
u 1 u 2 2u du
C
2 sin 1
3
x3 2
C
2 u 2 1 u 2 du ;
cos
2 sin 2
d
2
3
du
x 1 xdx
2
cos
1
4
d
ln x
1 tan 2
1 x4
1 ln
2
tan | C
d , 1
1
, 4 u2
d ; 1 x4
cos
csc
2u du
2 sin 2
cos 4 d
, 1x dx
C
cos
2u 13
3
23 3
2
2 cos
4sin
u2 3
dx
cos ;
1 sec 2
2
x dx
d
2
4
x
x2 1 C
1 sin 2
1 ln |sec
2
2u du
2 cos d , 0
C
tan ;
C
d
2
ln x
2sin , du
2 sin 2
tan
d
1 sin 2
sin
d
u
4
1
, 1 x2
0 or 0
u2
x
d
sec2
, 2 x dx
2
1 ln x
ln x
x
2
sec2
sin 1 x C
d
cos 2
sin
dx
ln ln1 x
2
2
2
sec2
C
sin ,
1 ln x
x ln x
d
d ,
sec2
sec
1
2
dx
4
d , x2 1
tan
sec sec tan
tan
sec2 ;
tan 1 x C
C
sec 2
1
x
d , 1 x2
sec 2 d
dx
41.
sec 2
tan , dx
1
4
cos 2
1 sin
8
d
2 cos 2
1
2
sin 2 2 d
C
1
4
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1 sin
4
1 1 cos 4
2
2
cos
d
2 cos2
1
C
Section 8.4 Trigonometric Substitutions
1
4
cos3
1 sin
2
1 sin 1
4
1
2
x
48. Let w
w2
sec
w2 1 dw
2
2 tan sec
tan
d ,0
tan | 2 tan 2 sec d
tan
dy
x2
x dx
2sec , 0
x2 4
2
2
sec 1 2x
dy
x 2 9 dx
3 sec tan
3 tan
ln 3
ln 3 C
dy
4 dx
3 tan 1 x
2
2
y
x2 1
2 dy
dx
sec2
d
1 0 C
C
x2 4
x
2 sec
d , x2
tan
2 and y
dx
;y
0
3 dx
x2 4
3
8
x 2 1, dy
sec3
4 dx
;y
x
x2 9
sec d
C
cos
d
1
y
C
w2 1 dw
2
2
tan 2
sec
tan
2 sec2
sec
d ,v
sec
d
sec
w w2 1 ln w
w2 1
sec
d
d
C
x 2| C
C; x
d
tan
tan | C
2 tan 2 d
x2 9
3, dy
ln |sec
d , dv
2 sec
2 sec tan
d
dx
1, dy
y
x2
y
, dx
2
2 sec tan
2 sec
2 tan
y
x2
4; dy
d
sec2
d
2 ln |sec
x
52.
2 sec3
tan
w2 1
2 w dw
w
dx
tan , du
2sec tan
sec
1 u2
1u
4
tan
1 sec
x 1 x 2 ln | x 1
51.
, w2 1
2 tan 2
2 tan 2 sec d
50.
2
tan
2 sec
49.
x 2
x 1
dx
tan d ; u
2 sec
32
1 u2
1u
2
x 1 x C
2 w dw
sec
d
1 sin 1 u
4
C
1
4
x 1
2 tan
tan
cos
32
x 1 x
x 1
w sec , dx
1 sin
4
583
;y
3
dx
x2 1
dx
x2 4
32
;x
sin
C
x
1
x2 1
4
2 tan
2 sec2
0
; x
0
1 d
0 C
tan | C
2 tan
C
3sec , 0
ln |sec
ln 3x
y
dx;
2
0
, dx
y
x2 4
2
2
3sec
x2 9
3
ln 3x
C
sec 1 2x
d , x2 9
tan
C; x
5 and y
3 tan
ln 3
x2 9
3
3
2
tan 1 2x C ; x
tan , dx
tan
Copyright
cos
sec2
C
3
2
tan 1 1 C
2 and y
0
0
d , x2 1
32
sec3 ;
x
C; x
tan
sec
C
x2 1
2014 Pearson Education, Inc.
C
0 and y 1
3
8
584
Chapter 8 Techniques of Integration
3 9 x2
3
53. A
2 3 cos
A
54.
dx; x
0
3 cos
3
0
y2
x2
a2
b
x
4b
2
a
2
0
A
d
2
4b
0
x
0
sin
1
6 3 12
12
12
6 3 12
x
sin ,
12
6 3 12
1
12
6 3 12
12
6
6 3 12
72
2
2
1 x 2 dx
0
a
2
0
0
cos 2
ab sin 2 0
dx, dv
sin 1 12 0
a sin
0, x
d
4ab
2
2ab 2
dx, v
2 1 cos 2
2
0
0
a
a sin
2
d
ab sin
sin 0
ab
x
12
1 x2
6 3 12
12
0
12
12
x sin 1 x
6 3 12 0
, dx
1 1 2 x2
2 0
1 x2
6 1 cos 2
1
2 0
2
4
1 sin
8
2
2
sin 1 12
1 2
2
0
u
sin 1 x, du
cos , x
0
sin
cos
6
0
3 3
6 3 12
4
6
1
d
4 0
1 2 2 x sin
1
0
2
1 x
x
dx
2 sin
Copyright
12
2x 0
0, x
1
2
1 x2
2
x dx, v
sin
6
1
sin 2
2 0
6
1
cos
4 0
x
2
dx, dv
sin 1 x, du
12
0
0
1
1 x
u
2 1 x 2 sin 1 x
sin 1 12
dx, dv
6
d
2 d
;
2
sin 1 x , du
u
1
1 x2
12
6 3 12 48
d
12
6 3 12 48
d
2
sin 1 x dx
2 1
6 sin 2
1
2 0
cos
0
dx
dx
d , 1 x2
cos
0
6
6 3 12
1
1
2
2 12
x sin 1 x
1
2
2
0
1 x2
dx
1 1 2 sin 1 1
2 2
2
121
0 2
4ab
3
4
0
cos , x
a
3 cos ;
6 3 12
;
12
2
6
6 3 12
d , 1 x2
2ab
12
1 x 2 sin 1 x
2
0
12
6 3 12 48
1
2
d
a
9 9 sin 2
2
cos
4b
a
x sin 1 x dx
12
6 3 12 48
y
1 x
2
sin
2
b 1 x 2 dx
a cos
x
0
2
3
2
sin 1 x, du
x dx
0
0
d , 9 x2
3 cos
d
a
4
cos 2 d
12
0
12
cos2
a cos
cos
sin 1 x dx u
12
x sin 1 x
(b) M
0
, dx
2
0
2
2ab
12
0
2
3
a
a
2
d
, dx
2
2
b 1 x2 ; A
y
2
1 x 2 dx
2ab
55. (a)
1
a sin ,
0
3 sin , 0
1 2 2 1 x2
0
1 x2
dx, v
1
1 x
dx, dv
2x
1 x
2
dx, v
dx
6 2
6 3 12 72
2014 Pearson Education, Inc.
2
x
3
6
1
2
12
12 3 72
6 3 12
2 1 x2
Section 8.5 Integration of Rational Functions by Partial Fractions
56. V
1
x tan 1 x
0
1 x2
2
8
2
1
tan 1 x
0
1 1 x2
2 0 1 x2
0
1 1 dx
2 0
1
dx
1 1 1
2 0 1 x2
1 x2
3
1 x2
1
3
1
2
8
x 2 , du
x2
1
3
1 u
x 2 1 x 2 x dx
3/2
1
5
(c) Trig substitution: x
x3 1 x 2 dx
5/2
1 x2
sin ,
1
2
1 1
2 0
tan 1 1 0
2
2
32
du
2 x dx
900 x 2
x
f ( x)
30 cos
30sin
8.5
1.
0
x
cot
0
30cos
1 cos 3
3
32
32
1 x2
2
15
1 u 3/2
3
A B
A
x 3
5
2 A 3B 13
B
x 2
cos2
sin
C
5 x 13
B
(10 13)
52
C
1 u 5/2
5
C
cos
1 cos5
5
1 cos 2
1 x2
1
3
C
cos 2 sin d
32
1
5
1 x2
52
C
900 x 2 . The slope of the tangent line is also
d
2
30
30 ln 30
x
30cos d , 900 x 2
, dx
1 sin 2
sin
d
900 x 2
x
900 302
C
30 csc d
900 x 2
C
30 cos
30 sin d
C;
30 ln 30
x
f ( x)
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
5 x 13
( x 3)( x 2)
2)
4
.
900 302
30
30 ln 30
30
(
x dx
cos d , 1 x 2
, dx
30sin , 0
30
1 x2
dx
tan 1 1 0 0
u u 3/2 du
1
2
u du
1 x2
1
1 x2
1
f ( x ) is given by f ( x). Consider the triangle whose hypotenuse is
900 x 2
x
30 cos d
30 ln csc
f (30)
dx
1 x2
3
1 du
2
the 30 ft rope, the length of the base is x and height h
(b)
1
3
1
2
1 x2
2
1 1
2 0
8
1
2
8
2 x dx
x dx, v
dx
sin 2 cos 2 sin d
cos4 sin d
, thus f ( x)
1
1 x2
1
x 1 x 2 dx, v
1 x2
dx, dv
C
58. (a) The slope of the line tangent to y
900 x 2
x
1
1 x2
1
1 tan 1 x
2
0
x
1 u
sin 3 cos cos d
cos 2 sin d
tan 1 x, du
u
2 x dx, dv
32
1 x2
(b) Substitution: u 1 x 2
x3 1 x 2 dx
1
2
dx
dx
57. (a) Integration by parts: u
x3 1 x 2 dx
x tan 1 x dx
585
A( x 2) B ( x 3)
B
Copyright
3
A
( A B) x (2 A 3B )
x 13
2; thus, ( x 53)(
x 2)
2
x 3
2014 Pearson Education, Inc.
3
x 2
900 x 2
x
900 x 2
586
2.
3.
4.
Chapter 8 Techniques of Integration
5x 7
x2 3x 2
5x 7
( x 2)( x 1)
A B
5
A 2B
7
B
x 4
( x 1)2
A
x 1
B
( x 1)2
x 4
( x 1)2
1
x 1
3
( x 1) 2
2x 2
x2 2x 1
A
A
x 2
2x 2
( x 1)2
2 and B
B
x 1
A 3; thus,
2
x 4
A
x 1
5x 7
5x 7
x 2 3x 2
A( x 1) B
B
( x 1) 2
2x 2
2x 2
x2 2 x 1
4; thus,
A( x 1) B( x 2)
2
x 1
3
x 2
( A B ) x ( A 2 B)
2
x 1
A 1
Ax ( A B )
A( x 1) B
A B
A 1 and B
4
A
Ax ( A B )
3; thus,
2
A B
2
4
( x 1) 2
A C
5.
z 1
z 2 ( z 1)
A
z
B
6.
1
z
z3 z 2 6 z
B
z2
C
z 1
A
2
7.
t2 8
t 2 5t 6
1
2; thus,
1
( z 3)( z 2)
1
5
B
5t 2
t 2 5t 6
A
t2 8
t 5t 6
1 t173
8.
t4 9
t 9t 2
4
9t 2 9
t 9t 2
1
4
9t 2 9
A C
z 1 ( A C) z
( A B) z B
A
z 3
1;
5
B
z 2
2 A 3B
1
1
z2
1
5
9t 2 9
t t2 9
2
At t 2 9
9
9A
0
9B
9
A
A B 1
5
B
2
A
t 3
(10 2) 12
B
t 2
B
5t 2
(Ct
D )t 2
9t 2 9
t t2 9
A
t
2
B
t2
0
1
A
B
1 A(1
1 x 1 x
1 x2
dx
1 dx
1 dx
1
2 1 x 2 1 x
2
1 x2
C
0; B 1
D
Ct D
t2 9
( A C )t 3 ( B D)t 2 9 At 9 B
x) B(1 x); x 1
ln 1 x
ln 1 x
Copyright
4
10; thus, 4t 92
t
A
1;
2
x
9t
1
1
1
t2
B
1;
2
C
2014 Pearson Education, Inc.
A(t 2) B(t 3)
A 17; thus,
12
10
t2 9
0
2 A 3B 1
z 2
5t 2
(t 3)(t 2)
(after long division);
B t2 9
A B
1
5
z 3
5t 2
t 2 5t 6
( A B) z (2 A 3B)
12
t 2
1
0
2
z 1
A( z 2) B ( z 3)
z
z3 z 2 6 z
thus,
A B
2
z
0
B D
9.
z 1
z 2 ( z 1)
(after long division);
( A B )t ( 2 A 3 B )
2
Az ( z 1) B ( z 1) Cz
2
B 1
C
1
z2 z 6
5B 1
z 1
2
Section 8.5 Integration of Rational Functions by Partial Fractions
10.
11.
1
A
B
1 A( x
x x 2
x2 2 x
dx
1 dx 1 dx
1
2 x
2 x 2
2
x2 2 x
x 4
x2 5x 6
A
x 6
B
x 1
x 4 dx
x2 5x 6
12.
2x 1
x 2 7 x 12
2 x 1 dx
x 2 7 x 12
13.
y
8 y dy
4 y2 2 y 3
1 ln 5
2
14.
y 4
1
12y
2
y
B
x 3
2x 1
4
1
t 3 t 2 2t
t 1
16.
A
t
1 dy
12 y
x 3
2 x3 8 x
A
x
x 2
C
x3
x2 2 x 1
5;
7
2 ln
7
5 ln
7
x 6
x 1 C
1 ln
7
A( x 3) B ( x 4); x
3
B
9ln x 4
C
ln
3 ln
4
1;
2
C
7 ln x 37
A( y 1) B ( y 3); y
3
1
dy
12y 1
1
1 ln
4
y 3
1
8
6
7
1
7; x
( x 4)9
2;
7
C
4
9
1
A
9;
C
( x 3)
1;
4
3 ln 5
4
4
2
7
A
( x 6)2 ( x 1)5
1
4
B
y 1
x
5 ;
16
1 (x
2
x 3 dx
2 x3 8 x
y
3
A
3;
4
3 ln1
4
1 ln 9
4
1 ln 5
4
A
4; y
3ln y 1
1
12
1
dt
t
3;
4 ln 12 3ln 32
(4 ln1 3ln 2)
ln 27
4
dt
t 2
1
6
3)
3
8
3
1
B
1 dt
3 t 1
1 ln | t |
2
0
1 ln | t
6
1;t
2
A
2|
1 ln | t
3
A( x 2)( x 2) Bx ( x 2) Cx( x 2); x
dx
x
1
16
dx
x 2
5
16
dx
x 2
3 ln
8
x
1 ln
16
x 2
0
B
1;
6
3 ;x
8
2
2
1| C
A
5 ln
16
x 2
1;
B 16
C
C
3x 2
( x 1)2
( x 2)
0
A(t 2)(t 1) Bt (t 1) Ct (t 2); t
1
2
C
x 2
B
x 2
4ln y
1 16
ln 27
8 8
dt
t 3 t 2 2t
5
1 ln ( x 2) ( x 2)
16
x6
17.
B
ln x 2
A( y 1) By; y
C
t 1
B
t 2
1;
3
C
A( x 1) B ( x 6); x 1
1 8 dy
4 4 y 1
y 4
1 ln 27
ln 18 ln 16
8
15.
A
ln 15
2
B
y 1
dy
B
dx
x 1
y
3 8 dy
4 4 y 3
A
y
y 4
5
7
B
y 1
1 ln 3
2
y2 y
2
9 xdx4 7 xdx3
A
y 3
y2 2 y 3
x 4
0
1;
2
x
ln x
dx
x 6
2
7
A
x 4
2) Bx; x
587
A
3, A B
2
x2
2
2 x 3ln x 1
A
(after long division); 3 x 22
( x 1)
3, B
1
1
x 1 0
1;
1
2
1
x3dx
0 x 2x 1
2
2 3ln 2 12
Copyright
1
0
A
x 1
B
( x 1)2
( x 2) dx 3
(1)
1 dx
0x 1
3x 2
1
dx
0 ( x 1) 2
3ln 2 2
2014 Pearson Education, Inc.
A( x 1) B
Ax ( A B)
588
18.
19.
Chapter 8 Techniques of Integration
x3
x 2x 1
A
3,
A B
x2
2
2 x 3ln x 1
1
( x 2 1)2
A
x 1
x
C
1
1;
4
dx
x2 1
x2
( x 1) x 2 2 x 1
x 1
1
4
21.
dx
x 1
22.
1
( x 1) x 2 1
Bx C
x2 1
A
x 1
A
t
A
1
x2
(
2ln 2)
8
Bt C
t2 1
3t 2
t2 1
1 ln
2
3
Ay B
Cy D
y2 1
y2 1
0, B 1; A C
3
1
1;
A C
1 ln 2
2
A B
x
2 x2 1
1
2( x 1)
1
1
y2 1
0; constant
4
4
2
y2
2 y 1 ( Ay B ) y 2 1
2
C
2; B D 1
1 ln1
2
A B C
1
D
1;
2
C
3 dt
1 t
1
2( x 1)
C
1 1 dx
2 0x 1
1 ln1
4
3 ( t 1)
1
t2 1
A B
1 1 ( x 1)
2 0 x2 1
dx
1 tan 1 0
2
4; coefficient of t 2
A
tan 1 3
0;
1
dx
0 ( x 1) x 2 1
1;
2
0
D ( x 1) 2 ;
A 1 2 ; coefficient of x 2
1 tan 1 1
2
2
A B
A B
dt
4ln1 12 ln 2 tan 1 1
12
Cy D
y2 2 y 1
2
1
2
dy
Ay 3
By 2
1 dy
y2 1
C
Copyright
2 3ln 2
C
1
y
tan 1 y
dx
1 ( x 1) 2
ln ( x 1)( x 1)3
C
C
ln 9
D
0
Ax ( A B)
x 2 dx
( x 1) x 2 2 x 1
3;
4
B
1 ln 2
4
1 ln 4
2
0 dx
1x 1
B ( x 1)( x 1) C ( x 1); x
3 3t 2 t 4
dt
1
t3 1
4 ln 3
A( x 1) B
B( x 1)( x 1) 2 C ( x 1) 2
( Bt C )t ; t
2 ln 3 12 ln 2 12
4
1 ln
4
3x 2
2 3ln 2 ( 12)
1;
4
x 1
x 1
B
ln x 1
A t2 1
C
( x 2)dx 3
( Bx C )( x 1); x
1
1 tan 1 x
2
0
tan 1 t
1 ln 2
2
3
4
x 1
B
( x 1)2
A B
A B 1
A C
C
x3
A( x 1)2
A x2 1
t 4
1
A( x 1)( x 1)2
A B
constant
1; coefficient of t
2
1
2
1 ln
4
1
1;
2
B
1
2 4
2 ln 3 ln 2
y2 1
0 0 3ln1 ( 11)
D
( x 1) 2
dx
1
2 ( x 1)2
1| 14 ln x 2 1
y2 2 y 1
0
C
( x 1) 2
1 ln | x
2
4 ln t
23.
1
B
x 1
0
B
x3 dx
1 x2 2x 1
D
A B
3t 2 t 4
t3 t
0
1
x 1
x 1
3 dx
4 x 1
1 ln 2
4
0
3, B 1;
coefficient of x 2
1;
4
A
A
C
( x 1) 2
A
x 1
A
x 1
( x 1)
1 ; coefficient of
4
D 1 A B 12 ; thus, A 14
dx
dx
dx
1 dx
1
1
x 1 4 x 1 4 ( x 1) 2 4 ( x 1)2
1
4
2
(after long division); 3 x 22
2
B
x 1
A B C
20.
3x 2
( x 1)2
( x 2)
2
2014 Pearson Education, Inc.
( A C ) y ( B D)
y
2
y
2
1
2
dy
3
Section 8.5 Integration of Rational Functions by Partial Fractions
24.
8 x2 8 x 2
4x
A
25.
2
Ax B
4 x2 1
2
1
Cx D
4x
0, B
2; A C
tan 1 2 x
1
4 x2 1
2s 2
s 2 1 ( s 1)3
1
8
2
8x2 8x 2
C
8; B D
( Ax B ) 4 x 2 1
2
D
C
s 1
D
( s 1)2
4 x2 1
dx
4 x2 1
2
2
( 3 A B ) s3 (3 A 3B) s 2
3 A B 2C
D
A 3B 2C
( A 3B ) s B C s 4
2 s3
2s 2 2 s 1
( A 3B 2C
s s
2
0 summing all equations
D
2
B C D E
2
9
ds
s2 1
ds
Bs C
s2 9
A
s
2
ds
( s 1)2
2E
s
2
2
9
27.
0
x2 x 2
x3 1
E
0; 18 A 9 B D
x2
Bx C
x2 x 1
A
x 1
A B 1, A B C
3A
2
2
3
1 dx
x 1
1
3
2
3
1 dx
x 1
1
3
2 ln
3
x 1
2
3
A
1
2
9
2
u 2 34
1 ln
6
u
x 2
B 1 A
2
3
4
du
2
x 12
dx u
2
3
1
3
2
D s3
s2
E s2 1
s 1
D)s ( B C D E )
4
E
2;
( Bs C ) s s 2 9
x 1
s
81
s s2 9
2
ds
s
ds
1
3
u
u2
x
tan 1
Copyright
1
2
3
4
du
1
2
3 2
x
du
3
1
2 u2
C
x
4;
3
3
4
2 ln
3
18
s ds
( s 2 9)2
( A B) x2
( Bx C )( x 1)
1 A B
u
A
0;
( Ds E ) s
81A 81 or A 1; A B 1
4
adding eq(2) and eq(3)
C
2
Es
E ) s 81A
A x2
2A 2
( s 1) 1 tan 1 s C
Ds 2
18;
x 12
3
3
2
9
D
2
1 dx
x 1
3
4
(9C
0
1, A C
x 4
x
A s2
Bs 4 Cs 3 9 Bs 2 9Cs
( A B) s 4 Cs 3 (18 A 9 B D) s 2
E
( s 1) 2
ds
( s 1)3
2
s 4 81
Ds E
A s 4 18s 2 81
9C
4 x2 1
0
summing eqs (2) and (3)
2 B 2 0 B 1; summing eqs (3) and (4)
C 0 from eq (1); then 1 0 D 2 2 from eq (5)
D
1;
s 4 81
x dx
8
0
3 A 3B 2C D E
26.
( A C ) x ( B D );
D s 2 1 ( s 1) E s 2 1
D ) s3 (3 A 3B 2C D E ) s 2
( 3 A B 2C
A C
2s 2
s 2 1 ( s 1)3
4 Bx 2
E
( s 1)3
( As B)( s 1)3 C s 2 1 ( s 1) 2
( A C )s4
8 x 2 8 x 2 dx
0;
4 Ax3
Cx D
C
As B
s2 1
2s 2
As 4
2
589
B
ln | s |
0; C
9
s2 9
C
( A B C )x ( A C)
2 A B 1, add this equation to eq (1)
2
x 2 dx
x3 1
23
x 1
(1 3) x 4 3
x2 x 1
dx
dx
du
x 1
0;
1 ln
6
x2
2014 Pearson Education, Inc.
x 1
3 tan 1 2 x 1
3
C
590
28.
Chapter 8 Techniques of Integration
1
x4 x
A
x
Cx D
x2 x 1
B
x 1
( A B C ) x3 ( B C
2B C
1 dx
x4 x
0
C
1
x
13
x 1
A( x 1) x 2
1
D) x2
( 2 3) x 1 3
x2
A
x 1
x4 1
A
(C x D ) x( x 1)
A 1, B D
1 B 2B
1 dx
x
x 1
0
0
D
B,
C
2
3
1
3
B
B C
D
0
1;
3
D
1 2 x 1 dx
3 x2 x 1
1 1 dx
3 x 1
x 1| C
x2
Cx D
x2 1
B
x 1
0
dx
x2 x 1
2
ln | x | 13 ln | x 1| 13 ln | x
29.
( B D) x
2 B, A B C
B x x2
x 1
A( x 1) x 2 1
B ( x 1) x 2 1
(Cx D)( x 1)( x 1)
( A B C ) x3 ( A B D ) x 2 ( A B C ) x A B D
A B C 0, A B D 1,
adding eq (1) to eq (3) gives 2 A 2 B 0, adding eq (2) to eq (4) gives
A B C 0, A B D 0
B
1,
4
0;
x2
2 A 2 B 1, adding these two equations gives 4 B 1
A B D
1
4
30.
0
1 dx
x 1
x2 x
x 3x2 4
and using A B C
1 1 dx
4 x 1
A
x 2
4
1,
2
D
1
2
1 dx
x2 1
x2
Cx D
x2 1
B
x 2
1 ln
4
A B 4C
1, 2 A 2 B 4 D
3
5
1
3
10
31.
3
2
3
10
B
1 ;
10
1 dx
x 2
1
10
1 dx
x 2
1
5
2
5
2
8
2
2
2
A
4
2
2
D
2
1
5
2B D
4
(2
2) d
2
ln
32.
4
4
2
2
2
2
3
(A
2
2
2
4
2; 2
D
1
3
2
3
2
5
2
8
2
2
(2
d
2
2
2
tan 1 (
1
A
2
4 3
2 2 3
B)
4
B
1
2 2 1
3 10
x 2
2
1)
C
2
4d
2
2
1
1 (A
2
1
2
E
C 3
1
2
2
(A
2; 2 A B
2
d
2
1 ln
10
x 2
2
2
2
2
2
x 2
(Cx D )( x 2)( x 2)
0, 2 A 2 B D 1,
D 2 C
Copyright
1,
5
C
1,
5
subtracting
and substituting
(C
D
D)
E
2
1
x2 1
1 tan 1 x
5
5
B 1; 2 A 2 B C
2
2
d
d
1) 2 1
(
1 ln
10
2
2
1
2
3
2
x C
B)
C
1
1 tan 1
2
dx
x2 1
dx
x2 1
F
2
B)
1
2
2
using
12
1 ln x 1
x 1
4
A B C
( 1 5) x 1 5
4
A
2
ln
2
2
2
D
2
B ( x 2) x 2 1
1 10
x 2
3 ln
10
1 dx
x2 1
(2 B D)
2) d
2
C
1,
4
A
adding this equation to the previous equation gives
2 3 5 2 8
2
2
A 3 (2 A B) 2 (2 A 2 B C )
3
2,
5
x 2 x dx
x4 3x2 4
C
2
1 tan 1 x
2
1
14
x 1
substituting for C in eq (1) gives A B
A B
x dx
x2 1
B
2
1,
5
D
4
5
A
x 1
14
x 1
0
subtracting eq (1) from eq (3) gives 5C 1
eq (2) from eq (4) gives 5D
2A
1 ln
4
x 1
x
dx
4
( A B 4C ) x 2 A 2 B 4 D
0
for D in eq (4) gives 2 A 2 B
C
A( x 2) x 2 1
x
( A B C ) x3 (2 A 2 B D ) x 2
0
using 2 A 2 B
E
F
2014 Pearson Education, Inc.
F
2
2
C
C
D
8
C
2;
Section 8.5 Integration of Rational Functions by Partial Fractions
A 5
B 4
2 A 3 2B 2
A 5 B 4
2A C
4
4
3
4
33.
3
1
2 x3 2 x 2 1
x2 x
3
x4
x
1
35.
2 x dx
1;
2
9 x3 3 x 1
x3 x 2
1 ; 1
x ( x 1) x( x 1)
dx
x
dx
x 1
1 ln
2
y3 y
y
4
1
y y2 1
y
B
2y
0,
4
39.
dy
2 x2
1
y y2 1
A
y
By C
1; C
0;
0 or C
2 ( y 1) dy
y 4 y2 1
y
3
y
Ay 2
1 ln
2
y, et dt
B, A C
dy
y 1
y2 1
dy ]
y
y2 1
dy
y
B
x 1
1
0
3y 2
Copyright
2
1
A
C
1; x 1
B 1;
C
A( x 1) B ( x 1);
C
A
x
C
B
x2
C
x 1
0
B
7; x
1
x
1; A C
9
A y2 1
dy
y dy
2
y 2 1 ( y 1)
A
A
2;
7 ln x 1 C
12 x 4
A(2 x 1) B
y
2
1
2
dy
dy
y2 1
dy
y 1
ln | y | 12 ln 1 y 2
C
y2 1
A 1, B
( y 1) 2
dy
y 2
A
By C
By 2 Cy By C
A B
y2
2
y dy
dy
y
2 C1
( A B ) y 2 Cy
( By C ) y
A
y 1
tan 1 y C , where C
2
ln xx 1
2
1
4
0;
dx
4 ( x 1) dx 6 2dx
2
x 1
(2 x 1)2
1
2
2
;
y3 y 2 y 1 y3 y 2 y 1
B C
1
1
4 x 3ln 2 x 1 (2 x 1) 1 C , where C
y2 1
( By C )( y 1)
2 y ln y 1
et dt
;[et
e 3et 2
2t
;
C1
2
2
x2
A
x 1
B
(2 x 1)2
16 x3 dx
4 x2 4x 1
2;
F
A( x 1) Bx; x
9 x 2 ln x
A
2x 1
1
2x 1
B
2y 2
y3 y 2 y 1
y2
0
A y2 1
A B
1
1 ln x 1
x 1
2
x
7 xdx1
A B
y2 y 1
2
dx
x2
E
x 2 1 dx 12 xdx1 12 xdx1
x3
3
12 x 4 ; 12 x 4
4 x 2 4 x 1 (2 x 1)2
2 y4
3
B
x 1
x ( x 1)
9 dx 2 dx
x
A 1; A B
38.
tan 1
3
ln x 1 C
x 4 dx
x2 1
x 1 C
D
(B D F )
A 0; B 1;
3 E 1; B D F 1 F
1
1
;
( x 1)( x 1) ( x 1)( x 1)
(4 x 4)
6;
y4 y2 1
1
A
x
x 2 ln x
1;
2
B
2( x 1) 2 3ln 2 x 1
37.
2
Ax( x 1) B ( x 1) Cx 2 ; x 1
9 x 3 x 1 dx
x3 x 2
A
1
d
2
x ( x 1)
3
16 x3
4 x2 4 x 1
D 2 C
2
2
9 9 x2 3 x 1 (after long division); 9 x2 3 x 1
9 x 2 3x 1
36.
2
2x
x 1
x 12 ln x 1
d
4
1
x2 1
1
x2 1
A
1 x3
3
2
1
x2 x
x2 1
x2 1
d
1d
2x
2 x3 2 x 2 1
x2 x
34.
C 3
B
(2 A C ) 3 (2 B D ) 2 ( A C E )
C
4; 2 B D 2 D 0; A C E
2
2
2
A
591
( A B) y 2
1, C
( B C) y ( A C)
1;
ln y 1
1 ln
2
y2 1
C1 1
y 1
ln y 2
C
t
ln et 1
2014 Pearson Education, Inc.
e
2
C
tan 1 y C1
592
40.
Chapter 8 Techniques of Integration
e 4t 2 e 2 t e t
e 2t 1
y2
2
41.
42.
43.
cos y dy
2
1 ln cos
3
cos
1
2
4x
1 ( x 2)
1 ( x 1)
2
2
tan 1 3 x
1
x3 2 x
2
u2 1
dx
A
u 1
13
x
6
u2 1
6 du
1
1
A
u 1
2
u2 1
2 du
9x2 1
2
y2
2
dy
1
y
y2 1
dy
dy
y2 1
C
1 ln t 2
5
t 3
1
t 2
1
t 3
dt
dy
1
3
dy
y 2
1
3
1
2
2 dx
tan 1 (2 x) 2
1 ln sin y 2
5
sin y 3
C
dy cos 1 2 y 2
ln y 1
y 1 3
C
1 ln
3
cos
C
C
1
6u
dx
1
u 1
6u 5 du ]
3
u 1
du
dx
( A B)u
A(u 1) B (u 1)
Let x 1 u 2
2
2 du
u2 1
1
3
dx
1
2 x
du
1 du
u 1
3
u 1
3 xdx2 6
4x
1
1
dx
x 1
dx
( x 2) 2
3 dx
tan 1 (3 x) 2
9x
dx
( x 1) 2
du
2du
A B
1
x
2 du ;
u2 1
dx
A B
0,
A B
ln | u 1| ln | u 1| C
1
u 1 u3
2
( A B )u
6u 5 du
6u 2 du
u2 1
A B
A B
6u 3 u1 1 du 3 u1 1 du
B 1
x 1
x 1
ln
6
2
6
u2 1
0, A B
6
A
1;
C
du
6 du;
u2 1
6 du
B
3
A
3;
6u 3ln | u 1| 3ln | u 1| C
C
1
B
u 1
x
du
u6
dx
C
1
u 1
6 du
u2 1
2 x 1 ln
y
1
5
x
( x 2)2
x
( x 1) 2
dx
A(u 1) B (u 1)
6
A
u 1
y 1
y
tan 1 et
y2 y 2
dx 3
2
B
u 1
x 1
dx;
x
1
dy
C
tan 1 (3 x)
dx
Let u
dx; [Let x
x
6
x 2
1
dx;
x ( x 1)
16
6 x1 6 3ln x1 6 1
47.
dy ]
4 x2 1
1
x 1
1
u 1
x
e 2t 1
dt
t2 t 6
tan 1 (2 x )
dx
ln x 1
B
u 1
2 du
u2 1
46.
2
( x 1)2 tan 1 (3 x ) 9 x3 x
9x
1 ln
2
y
2
C
2
tan 1 2 x
3ln | x 2|
2
y3 2 y 1
et dt
dt ]
y, sin d
( x 2)2 tan 1 (2 x ) 12 x3 3 x
2
1 e 2t
2
t , cos y dy
; [cos
et , dy
y
tan 1 y C
; [sin y
sin d
cos 2 cos
1
6
45.
y2 1
1 ln
2
sin 2 y sin y 6
1
4
44.
e3t 2et 1 et dt ;
e 2t 1
dt
u 2u du
u2 1
2u du
A(u 1) B (u 1)
1
u 1
2u
x 1 1
x 1 1
dx
1
u 1
du
( A B)u
2u
1 du
u 1
2u 2 du
u2 1
A B
1 du
u 1
A B
2
du
0, A B
2 du
2
B 1
2u ln | u 1| ln | u 1| C
C
Copyright
2
u2 1
2014 Pearson Education, Inc.
2 du;
u2 1
A
1;
Section 8.5 Integration of Rational Functions by Partial Fractions
1
x x 9
48.
2
1
x x4 1
1
1
4
13
u 3
1 du
u (u 1)
10
x
1
16
C
1 ln | u |
80
1
20u
t 2 3t 2 dx
dt
3t 4
1 1 du
3 u 3
du
1
u 1
x4
x5 4
A
1
4
du
A 1
1 du
u
1 ln | u
80
1 ex
2
x
3
2t dx
dt
x 1
3
4
ln 13 C
x2 1
tan 1 0
dx
x2 1
0
2.5 2
55. V
0.5
2
1
0
tan 1 x
2.5 9
0.5 3 x x 2
y dx
xy dx
C
2
1
1 du; 1
u (u 1)
u (u 1)
1 ln | u |
4
1 ln | u
4
1
5
x 9 3
x 9 3
C
B
u 1
1 ln | x5 |
80
14
u2
1 16
u 4
ln tt 12
A C
1 ln x 4
4
x4 1
1 1
20 u 2
3 and x
C
B
u2
C
u 4
0, 4 B 1
1
4
B
1
1 du
du 80
u 4
1 ln x5 4
80
x5
4| C
Ce x ; t
C ; tt 12
A
u
0, 4 A B
1 1 du
80 u
du
1 ln | x5
80
1
20 x5
1| C
1
du; 2 1
u 2 (u 4)
u (u 4)
(4 A B )u 4 B
dt
t 1
1 ln
3
A
u
1;
3
B
1
20 x5
1
2
0
C
C
ln | t 2| ln | t 1| ln 2
2 3
2 x 2; 12 xdx1
ln 2
54. (t 1) dx
dt
C
2 3; x
3
4
4
t2
ln 2 tt 12
1
4
B
u 3
1
3
A
3| 13 ln | u 3| C
5 x 4 dx
1 16
u
dt
t 2
2
A
u 3
1;
du
1
5
4| C
dt
t 2 3t 2
1; x
B
( A C )u 2
1
du
u 2 (u 4)
1 ln | u
3
1 1 du
4 u 1
x5
dx; Let u
0, 3 A 3B
4 x3 dx
du
2 du; 2
u2 9
u2 9
2u du
A B
1 1 du
3 u 3
x4
dx; Let u
1 ; 1
16 5
4t 2 1 dx
dt
x
56. V
( A B )u 3 A 3B
( A B )u
1
u
1
u2 9 u
2u du
Au (u 4) B (u 4) Cu 2
t 2
t 1
53.
1
4
dx
6
A
52.
4
A(u 1) Bu
1
51.
13
u 3
x3
x x4 1
dx
1
x x5 4
50.
dx
A(u 3) B (u 3)
2 du
u2 9
49.
u2
dx; Let x 9
593
dt
3t 4 4t 2 1
C
C
dt
t 2 2t
x
1 ln | x
2
ln 2 ln 3
ln 6
tan 1 x
dt
t 1
ln | t 1|
dx
2x
dx
0 ( x 1)(2 x )
x
2.5
3
0.5
4
dt
t 2 13
3
1
0
Copyright
1|
3 tan 1
1
2
dt
t
3 tan 1
dt
t2 1
3
1 dt
2 t 2
ln | x 1| ln t t 2
ln | x 1| ln 6 t t 2
tan ln(t 1) , t
1
x 3
1 1
3 x 1
1
x
dx
2 1
3 2 x
3 tan 1 t C; t
1 and
3 tan 1 t
3t
ln | t 1| C; t
3t
0 and x
x 1
0
6t
t 2
x
tan 1 0
C; t
6t
t 2
1 and
1, t
0
ln |1| C
1
2.5
3 ln x x 3
0.5
dx
4
3
2014 Pearson Education, Inc.
3 ln 25
ln | x 1| 2 ln |2 x |
1
0
4
3
(ln 2)
594
57.
Chapter 8 Techniques of Integration
3
A
0
3
3
tan 1 x dx
1 ln
2
1
A 2
58.
5 dx
3 x
x
x 4 x 2 13 x 9
1 5
A 3 x3 2 x 2 3 x
dx
1
A
dx
dt
kx( N
1 ,
250
N
1000, t
499 x
ln 1000
x
(b) x
dx
dt
1
2
N
k (a x)(b x)
(a) a
dx
( a x )2
b:
1
a x
(b) a
x
1.
500
b:
ln ba
b a
1
0
e 4t
dx
( a x )(b x )
akt 1
a
a
dx
( a x )(b x )
k dt
dx
N x
1
N
k dt
2
3
tan 1 x 3 3
dx
x x 4
1
4
ln
x 4
x 4
1000e4t
e 4t
kt C ; t
x
a
dx
1
b a a x
ln ba xx
0 and x
0
a
a 2 kt
akt 1 akt 1
dx
1
k dt
b a b x
(b a) kt ln ba
b x
a x
1
a
C
1
a x
C
kt C ;
t
250
1 ln 1
1000
499
499
t
1 ln
2
x 4 2
x 4 2
(We used FORMULA 13(b) with a 1, b
Copyright
kt
x
1 ln 499
4
1
a
1 ln b x
b a
a x
kt C ; t
b e (b a ) kt
a
x
C
4
4
3.90
499 e 4t x 1000e4t
INTEGRAL TABLES AND COMPUTER ALGEBRA SYSTEMS
dx
x x 3
ln 125
;
9
ln N x x
x
1 ln
1000
1000 x
C
5
3
11ln 2 3ln 6)
e4t (1000 x)
499 x
(We used FORMULA 13(a) with a 1, b
2.
1
N
1 (8
A
2 ln x 1
k dt
a
akt 1
x
dx
x
1.10
ln x 3
500 499 500e4t
1
a x
C
3ln x
1 ln 2
1000
998
2
1000e4t
499 e 4t
k dt
1
N
k dt
0 and x
500
5 dx
3 x 1
dx
3
2
1 2
A 3
6
5
5
5
4 x 3 3 xdx3 2 xdx1
3
3
499 x
1000 x
4t
2
3 x2
1
2 0 1 x2
0
3
2
5 dx
3 x 3
dx
x( N x)
x)
dx
3
tan 1 x
1
A 2
0
3
k
8.6
3
5 4 x 2 13 x 9
dx
3 x3 2 x 2 3 x
x
1 x2
ln 2;
1 x2
2
1
A
x tan 1 x
1
2
0
3
0
3
0
3
A
59. (a)
60.
3
x2 1
3
1
x tan 1 x dx
A 0
x
3
x tan 1 x
3)
C
4)
2014 Pearson Education, Inc.
0 and
ab 1 e( b
a ) kt
a be( b
a ) kt
1000e4t
499 e4t
1.55days
Section 8.6 Integral Tables and Computer Algebra Systems
3.
x dx
x 2
( x 2) dx
x 2
x 2
2
1
3
x 2
2
1
2
3
1
dx
x 2
2
x 2 dx 2
x dx
(2 x 3)3 2
2( x 2)
3
x 2
1
1
2
1 (2 x 3) dx
2 (2 x 3)3 2
1
2x 3
1
(2 x
2 2x 3
dx
3
dx
2 (2 x 3)3 2
3
2
3
2x 3
( x 3)
2x 3
3 3) C
1
2
x 2 x 3 dx
1
2
2
2
2x 3
3
2
5
2, b
x(7 x 5)3 2 dx
1
7
2
7
1
7
7x 5
9 4x
x2
5
7
9 4x
x
dx
7x 5
2
7
8.
2 ln
3
9
9 4x
9 4x 3
9 4x 3
4x 9
( 9) x
dx
x2 4 x 9
3, n
3 and a
2
9
2
9
7, b
4 tan 1
27
4, b
dx
4
18 x 4 x 9
4x 9
9
5, n
2x 3
1
2(7 x 5)
7
5 and a
2, b
C
3, n
3
2
7, b
7x 5
C
5, n
3)
9)
9)
C
4, b
9)
C
Copyright
5
dx 57
(7 x 5)5 2 14 x 4
49
7
C
4, b
C
3, n 1 )
9)
4, b
1
2 x 3 dx
(2 x 3)3 2 ( x 1)
5
2, b
1
7
3)
3
2
dx
C
tan 1 4 x9 9
C
( 1)
C
9
1
2x 3
2
2
C
(We used FORMULA 13(a) with a
4x 9
9x
(7 x 5)5 2
49
C
(We used FORMULA 15 with a
4x 9
9x
1
2
2, b
(We used FORMULA 13(b) with a
9 4x
x
3
2
1 and a
2x 3
5
dx
x 9 4x
2 1 ln 9 4 x
9
3, n
(2 x 3)3 2
2
5
( 4)
2
1)
3
1
1
2 x 3 dx
5
(We used FORMULA 14 with a
9 4x
x
2x 3
2
2
(7 x 5)(7 x 5)3 2 dx 75 (7 x 5)3 2 dx
7
7
dx
2x 3
C
3
(We used FORMULA 11 with a
7.
3
2
3
(We used FORMULA 11 with a
6.
3
2
2, n
C
2x 3
2
2
C
dx
2x 3
1
2
dx
(2 x 3) 2 x 3 dx
5
4
2, n 1 and a 1, b
1
2
(We used FORMULA 11 with a
5.
dx
1
(We used FORMULA 11 with a 1, b
4.
1
x 2
2014 Pearson Education, Inc.
7x 5
C
3
dx
595
596
9.
Chapter 8 Techniques of Integration
x 4 x x 2 dx
( x 2)(2 x 3 2) 2 2 x x2
6
x 2 2 x x 2 dx
23
2
( x 2)(2 x 6) 4 x x 2
6
( x 2)( x 3) 4 x x 2
4sin 1 x 2 2 C
3
(We used FORMULA 51 with a 2 )
10.
x x2
x
2 12 x x 2
dx
x
1
2
(We used FORMULA 52 with a
11.
dx
x 7 x
dx
2
x
7
2
1
7
x2
12.
dx
x 7 x
dx
2
x
7
2
x
1
7
2
4 x2
x
dx
22 x 2
x
22
dx
x2 4
x
dx
x 2 22
x
x2
x2
dx
22
(We used FORMULA 42 with a
15.
e2t cos 3t dt
e2t
22 32
e 3t sin 4t dt
7
e 3t
( 3)2 42
x2
C
1
7
2
ln 7 x7 x
C
x2
C
1
7
2
ln 7 x7 x
C
2 ln 2
22 x 2
x
x2
2
cos 1 x
1 1 sin 1 x
2 2
2sec 1 2x
2)
2, b
e 2t
13
3, b
x2
C
4 x2
x
x 1 x2
4 2sec 1 2x
C
3)
e 3t
25
( 3sin 4t 4 cos 4t ) C
4)
C
x2
2
x2
2
x 2 dx
cos 1 x 12
cos 1 x
1 sin 1 x
4
1 x2
1
4
x 1 x2
(We used FORMULA 33 with a 1 )
18.
1 1
tan 1 (1x ) 111 x 2dx 2
1 (1) x
(We used FORMULA 101 with a 1, n 1 )
x tan 1 x dx
x1 tan 1 (1x ) dx
C
(2 cos 3t 3sin 3t ) C
x1 1
1 1
1 1
2 2
2ln 2
2)
( 3sin 4t 4 cos 4t ) C
x1 cos 1 x dx
4 x2
C
1 1
cos 1 x 111 x dx2
1 x
(We used FORMULA 100 with a 1, n 1 )
x cos 1 x dx
1) C
7)
(We used FORMULA 107 with a
17.
2
(2 cos 3t 3sin 3t ) C
(We used FORMULA 108 with a
16.
2
x
(We used FORMULA 31 with a
14.
1 sin 1 (2 x
2
7)
(We used FORMULA 34 with a
13.
7
7
ln
C
)
x
(We used FORMULA 26 with a
x x2
C
2
7
ln
C
4sin 1 x 2 2
1
1 sin 1 x 2
1
2
2 12 x x 2
dx
sin 1 x 2 2
x1 1
1 1
Copyright
x2
2
2
tan 1 x 12 x dx2
1 x
2014 Pearson Education, Inc.
C
Section 8.6 Integral Tables and Computer Algebra Systems
19.
x2
2
tan 1 x 12
x2
2
tan 1 x 12 dx
1
1
2 1 x2
x2
2
dx
tan 1 x 12 x
dx
x dx
x dx
tan 1 x dx
x2
x2
2
1 x2
1 ln
2
1 x2
dx
x 1 x2
x dx
dx
x
1 x
cos 5 x
10
sin 3x cos 2 x dx
ln | x | 12 ln 1 x 2
2
cos x
2
cos 5 x
10
sin 2 x cos 3x dx
cos x
2
8sin 4t sin 2t dx
8 sin 7t
7
2
8 sin 9t
9
2
sin 3t sin 6t dt
3sin 6t
sin 2t
cos 3 cos 4 d
6 sin 7
7
12
6sin 12
1 sin 13
13
2
cos 2 cos 7 d
(We used FORMULA 62(c)
27.
x3 x 1 dx
x2 1
2
x dx
x2 1
dx
x2 1
2
tan 1 x
x 1 tan 1 x
( 1)
tan 1 x dx
x2
C
1
x
x2
6
1 ln
6
x 1
1 x2
dx
1 x2
3, b
2)
2, b
3)
C
sin
8
7t
2
7
4, b
1
2
)
1,b
3
1
6
)
1
4
)
9t
2
sin
C
9
C
1,b
3
1 sin 15
15
2
with a 12 , b
1 2 x dx
2 x2 1
sin
C
13
2
13
sin
15
2
15
C
7)
dx
x2 1
2
1 ln
2
x2 1
x
2 1 x2
(For the second integral we used FORMULA 17 with a 1 )
Copyright
2014 Pearson Education, Inc.
C
tan 1 x ln | x | 12 ln 1 x 2
C
(We used FORMULA 62(c) with a
26.
C
C
(We used FORMULA 62(b) with a
25.
x 2 1 tan 1 x x
C
used FORMULA 62(b) with a
24.
x3
3
x 2 tan 1 x dx
C
( 2 1)
tan 1 x ( 21 1) x 2 dx
1 x
(We used FORMULA 101 with a 1, n
2)
(We used FORMULA 62(a) with a
23.
1
2
C
x ( 2 1)
( 2 1)
x 2 tan 1 x dx
(We used FORMULA 62(a) with a
22.
1 tan 1 x
2
2 1
3
3
tan 1 x 21 1 x 2 dx x3 tan 1 x 13 x 2 dx
1 x
1 x
(We used FORMULA 101 with a 1, n 2 )
x 1dx
1 x2
21.
dx (after long division)
x2 1
2 1
x 2 tan 1 x dx
x3
1 x2
20.
1
1 x2
1
597
1 tan 1 x
2
C
C
598
28.
Chapter 8 Techniques of Integration
x2 6 x
x
2
1
3
3
2
6 x dx
dx
x2 3
dx
tan 1 x
x
3
x2 3
3
2
3dx
2
3
x
2
dx
2
3
x
2
x
3
2
2
3
3
1
3
2
2
x
2
3
3
2 x dx
3
2
x
1
2 3
29.
3
dx
3
2
x
tan 1 x
2
3
2 2
C
3
the first integral we used FORMULA 16 with a
with a
2
3; for the third integral we used FORMULA 17
3)
tan 1 x
x
2 x2 3
3
x2 3
3
u
x
x
u2
sin 1 x dx;
dx
C
1 1
2 u1 1 sin 1 u 111
2 u1 sin 1 u du
2u du
u1
1
1 u
2
u 2 sin 1 u
du
u 2 du
1 u2
(We used FORMULA 99 with a 1, n 1 )
u 2 sin 1 u
1 sin 1 u
2
1 u2
1u
2
u2
C
1
2
sin 1 u
1u
2
1 u2
C
(We used FORMULA 33 with a 1 )
x 12 sin 1 x
30.
cos
1
x
x
dx;
x x2
1
2
u
x
x
u2
dx
C
cos 1 u
u
2u du
2 cos 1 u du
2u du
2 u cos 1 u 11 1 u 2
C
(We used FORMULA 97 with a 1 )
x cos 1 x
2
31.
x
1 x
dx;
1 x
u
x
x
2
dx
u
C
u 2u
1 u
2u du
du
2
u2
2
1 u
2
2 12 sin 1 u
1u
2
2
u
2
du
1 u2
C
sin 1 u u 1 u 2
C
(We used FORMULA 33 with a 1 )
sin 1 x
32.
2 x
x
dx;
x 1 x C
u
x
x
2
dx
u
sin 1 x
2 u2
u
2u du
(We used FORMULA 29 with a
u 2 u2
2 sin 1 u
2
C
x x2
2u du
2
C
2
2
u du
2
2
2)
2 x x2
Copyright
2sin 1 2x
C
2014 Pearson Education, Inc.
2
u
2
2
2
2
sin 1 u
2
C
Section 8.6 Integral Tables and Computer Algebra Systems
33.
1 sin 2 t (cos t ) dt
;
sin t
(cot t ) 1 sin 2 t dt
u
sin t
du
1 u 2 du
u
cos t dt
2
ln 1 1u u
1 u2
599
C
(We used FORMULA 31 with a 1 )
2
t
ln 1 1sinsin
t
1 sin 2 t
34.
C
dt
cos t dt
(tan t ) 4 sin 2 t
(sin t ) 4 sin 2 t
; [u
sin t , du
(We used FORMULA 34 with a
4 sin 2 t
sin t
1 ln 2
2
35.
y 3 (ln y )
2
;
2)
eu du
eu
y
e
u
dy
e du
t
tan 1 y dy;
C
ln y
du
3 u2
u
(We used FORMULA 20 with a
36.
u 4 u2
C
u
dy
4 u2
u
1 ln 2
2
du
cos t dt ]
3 u2
ln u
3 u2
C
3 (ln y ) 2
ln ln y
C
3)
y
t2
y
dy
2
2 t2 tan 1 t
2 t tan 1 t dt
2t dt
t2
1
2 1 t2
dt
t 2 tan 1 t
t2
1 t2
dt
(We used FORMULA 101 with n 1, a 1 )
t 2 1 dt
t2 1
t 2 tan 1 t
37.
1
x2 2 x 5
1
dx
t 2 tan 1 t t tan 1 t C
dt
1 t2
( x 1) 2 4
dx; [t
x 1, dt
(We used FORMULA 20 with a
( x 1) 2
ln ( x 1)
38.
x2
4
C
x2
dx
4x 5
1 ln
2
1 ln
2
7 ln
2
t
( x 2)
t2 1
( x 2)
( x 2)
2
dx;
1
t t2 1
2
( x 2)2 1
x2
4x 5
t2 4
dt
ln t
tan 1 y
t2
4
y
C
C
2)
x2
ln ( x 1)
t
1
dx]
x 2
4 t2 1
2x 5
(t 2)2
dt dx
(We used FORMULA 25 with a 1 )
x
2
y tan 1 y
t
1
t 2 4t 2 dt
dt
t
2
1
t2
t2 1
dt
4t
t2 1
4
dt
t2 1
(We used FORMULA 20 with a 1 )
4 ln t
( x 2) ( x 2)2 1
2
( x 6) x 2 4 x 5
2
Copyright
2
C
t2 1
C
4 ( x 2) 2 1 4 ln ( x 2)
C
2014 Pearson Education, Inc.
( x 2) 2 1
C
dt
600
39.
Chapter 8 Techniques of Integration
5 4 x x 2 dx
9 ( x 2) 2 dx; [t
(We used FORMULA 29 with a
x 2
2
40.
9 ( x 2)
2
x 2 2 x x 2 dx
x 2, dt
5 4x x2
x 2
2
C
2t 1 t 2 dt
x 1, dt
sin 1 1t
1
8
41.
1t
8
1 t 2 12
2t 2
2
3
5 sin 1 ( x
8
32
1) 32 2 x x 2
5 1
5
2 sin 2
15
sin 2 x cos 2 x
10
42.
8cos 4 2 t dt
8
cos 2 t sin 2 t
4 1
4
sin(2 2 t )
42
cos 2 t sin 2 t
43.
3sin 4 t
4
3t
C
2
sin 2 cos 2
10
44.
2
5
sin t cos
t
cos3 2 t sin 2 t
3 1
3 2
2, m
sec 2 t tan t
3
sin t cos
2 tan t
3
4
5
sin 2 2 x cos 2 x
32
1 sin 1 ( x
2
1) C
3 1
3
sin 2 x dx
3)
4
sin 2 x cos 2 x
10
2 sin 2 2 x cos 2 x
15
4 cos 2 x
15
C
3
2
C
2 1
5 2
sin t cos 3 t
2 4
2 1
2 4
sin 3 2 cos 2 2
10
sin 2 2 (cos 2 )2d
sin 3 2
15
cos 4 t dt
4)
2, m
4
t
3t C
sec t dt
sec2 t tan t
4 1
sin t cos 3 t
4 2
4 1
sec2 t dt
4)
2 sec 2 t tan t
3
2 tan t
3
C
2 tan t
3
An easy way to find the integral using substitution:
2sin 2 t cos 4 t dt
1 ( x 1)2
2)
3, n
3
(We used FORMULA 92 with a 1, n
sin t cos 3 t
x 1
2
sin 2 2 cos 2 d
2
t dt
32
C
3cos 2 t sin 2 t
2
2 sin 2 t cos 4 t dt
cos
1 ( x 1)2
C
4)
sin 2 cos 2
10
4
2
3
sin 1 1t
cos 2 2 t dt
(We used FORMULA 68 with a 1, n
3
1 t2
2, n
cos 2 x C
12
2
t
2
sin 4 2 x cos 2 x
10
sin 2 2 cos 2 d
2sin 2 t sec4 t dt
1 t 2 dt
2t 1
2 )
(We used FORMULA 69 with a
3
t2
C
sin 3 2 cos 2 2
2(2 3)
sin 2 2 cos3 2 d
1
2
2 ,n
(We used FORMULA 59 with a
3
C
(t 1) 2 1 t 2 dt
4x 5
5 and a
2, n
8
2 x cos 2 x 15
cos3 2 t sin 2 t
42
6 2t
2( x 1)2
sin 3 2 x dx
(We used FORMULA 61 with a
3
1 t
2 32
2 x x2 2 x2
x 1
8
(We used FORMULA 60 with a
4
C
(We used FORMULA 29 with a 1 )
1) 18 ( x 1) 1 ( x 1) 2 12
sin 4 2 x cos 2 x
52
sin 1 3t
1 t 2 dt
1 sin 1 ( x
8
sin 5 2 x dx
9 sin 1 x 2
2
3
dx]
(We used FORMULA 30 with a 1 )
4
32
2
9 t2
t
2
3)
9 sin 1 x 2
2
3
x 2 1 ( x 1)2 dx; [t
t 2 1 t 2 dt
9 t 2 dt
dx]
2 tan 2 t sec2 t dt 2
Copyright
tan t
2
sec2 t dt
2 tan 3 t
3
C
2014 Pearson Education, Inc.
sec2 t 1
C
2 tan 3 t
3
C
C
Section 8.6 Integral Tables and Computer Algebra Systems
45.
2
4 tan2 22 x
tan 2 x dx tan 2 2 x 4 tan 2 x dx
(We used FORMULA 86 with n 3, a 2 )
4 tan 3 2 x dx
tan 2 2 x
46.
4 ln
2
sec 2 x
8
cot 3 t
3
8cot 4 t dt
C
tan 2 2 x 2 ln sec 2 x
cot 2 t dt
(We used FORMULA 87 with a 1, n
1 cot 3 t
3
8
cot t t
C
4)
C
(We used FORMULA 85 with a 1 )
47.
2sec3 x dx
x tan x 3 2 sec x dx
2 sec (3
1)
3 1
(We used FORMULA 92 with n 3, a
)
1
1
sec x tan x
ln |sec x tan x | C
(We used FORMULA 88 with a
48.
)
2
3 x tan 3 x 4 2 sec 2 3 x dx
3 sec 3(4
1)
4 1
(We used FORMULA 92 with n 4, a 3 )
3sec 4 3 x dx
sec2 3 x tan 3 x
3
2 tan 3 x
3
C
(We used FORMULA 90 with a
49.
csc3 x cot x
5 1
csc5 x dx
5 2
5 1
3)
(We used FORMULA 93 with n
1 csc3
4
3 csc x cot x
8
x cot x
csc3 x cot x
4
csc3 x dx
5, a 1 and n
3 ln |csc x
8
cot x |
csc x cot x
3 1
3
4
3 2
3 1
csc x dx
3, a 1 )
C
(We used FORMULA 89 with a 1 )
50.
16 x3 (ln x)2 dx 16
x 4 (ln x ) 2
4
2
4
x3 ln x dx
(We used FORMULA 110 with a 1, n
16
51.
4
x (ln x )
4
2
4
x (ln x )
8
et sec3 et 1 dt ; x
4
x
32
C
et 1, dx
1 ln |sec x
2
tan x | C
3, m
4 x 4 (ln x ) 2
et dt
(We used FORMULA 92 with a 1, n
sec x tan x
2
16
1
2
x 4 (ln x )2
4
4
1 x (ln x )
2
4
2 and a 1, n
2 x 4 ln x
sec3 x dx
4
x
2
x3 dx
3, m 1 )
C
sec x tan x
3 1
3 2
3 1
sec x dx
3)
sec et 1 tan et 1
Copyright
1
4
ln sec et 1
2014 Pearson Education, Inc.
tan et 1
C
601
602
Chapter 8 Techniques of Integration
t
csc3
52.
t2
d ;
d
2 csc3 t dt
2t dt
(We used FORMULA 93 with a 1, n
csc
53.
1
0
cot
ln csc
2 x 2 1 dx; x
cot
dy
0
52
1 y2
;[y
55.
4
2 ln
0
sin x, dy
3
2 tan x
3
0
dr ; [ r
3
0
1 3
dt
t
2
1
; [t
7 2
4
3
0
sec 2
6
5 1
5
4 cos 2
15
0
3
2
1
2
4
15
3
2
2 y 1
y
5
2
57. S
0
2
2
x2
0
2 2
2
0
4
0
C
sec3 t dt
2
2
2 1
8
15
1
2
1
2
3 tan 3
(sec
sec
0
3
3 3
3
0
3
tan ) d
3
3
0
x2
x
2
3
4 2
sec 2
4 1 0
tan 4 d
3
6 sec2 d
0
sec
cos 4 sin
5
6
7
0
6
0
4
5
cos5 d
cos2 sin
3
6
0
6
0
9
160
5 and a 1, n
1
10
4
15
3)
3 9 48 32 4
480
203
480
dx
2
3
4 and FORMULA 84 with a 1 )
d ]
8 sin
15
sec 2 x tan x
4 1
0
sec4 x dx
2 3
tan
(We used FORMULA 61 with a 1, n
4
2
4)
cos3 d
sin
0
cos x
3
tan 3
3
3
5
sec tan d ]
tan , dt
0
cos 4 sin
5
sec t sec2 t dt
3 cos x dx
2
3
3
tan 2 d
6
cos 4 sin
5
cot t |
2 1
cos x dx]
sec , dr
3
0
1 ln |csc t
2
3, a 1 )
(We used FORMULA 86 with a 1, n
0
0
32
2
2 r 1
r
1
tan 3
4 1
56.
4
2
(We used FORMULA 92 with a 1, n
sec2 x tan x
3
csc t cot t
2
2
4
3 2
sec t dt
3 1 0
sec t tan t ln sec t tan t
3 2
csc t dt
C
(We used FORMULA 92 with, n
54.
3 2
3 1
3)
sec2 t dt
tan t , dx
4
sec t tan t
3 1
0
2
csc t cot t
3 1
2
dx
x 2 1 dx
Copyright
2014 Pearson Education, Inc.
3 1
3
6
0
cos d
x dx
Section 8.6 Integral Tables and Computer Algebra Systems
x x2 1
2
2 2
1 ln
2
2
x2 1
x
0
(We used FORMULA 21 with a 1 )
2
58.
6 ln
3 2
L
0
2
3
1 (2 x ) 2 dx
2
2
3
3 2
1
4
0
2 ln
x 2 dx
1
2
used FORMULA 2 with a
1 4 x2
x
2
3
(2)
4
59.
1 ln
4
A
3 dx
0 x 1
x
1 3 x dx
A 0 x 1
3
2
2 x 1
1 3
A 0
3
( x 1)3 2
1 2
2 3
1
1
4
1
2
ln x
3
2
1
2
3
4
1 ln
4
1 4 43
1 ln
4
y
1
4
3
1 ln 4
4
ln( x 1) 0
1 ln 2
2
1
1
2
4
u
4 8
2
x 2 1 4 x 2 dx; [u
2 x, du
1)
3
18 x 27 ln |2 x 3| 0
18 3 27 ln 9 ( 27 ln 3)
2 dx]
u 2 1 u 2 du
1 2u 2
1 u2
1 ln
8
u
1 u2
2
2
(We used FORMULA 22 with a 1 )
2 (1
4 8
9
4 2
2 4) 1 4
5 18 ln 2
2
5
5
1 ln 1
4
2
ln 2
0
2
1 4 43
4;
3
3
3
x 236
dx 18 22 xx 33 dx 54 2 dx
x 3
0
0 x 3
54 27 2 ln 3 27 ln 3 54 27 ln 3
61. S
0
1 3 dx
A 0 x 1
3
60. M y
3 2
x2
3 2
(We used FORMULA 11 with a 1, b 1, n 1 and a 1, b 1, n
1 3 dx
2A 0 x 1
1
4
2;
0
x 1 dx
1
0
3
x2
)
0
3
2
1 ln 2
4
3
2 2x 14
3 2
x 12 1 4 x 2
1 ln
4
2
1 ln
8
2
1 4
2 (1
8
2 4) 1 4
1 ln
8
2
1 4
7.62
Copyright
2014 Pearson Education, Inc.
603
604
Chapter 8 Techniques of Integration
62. (a) The volume of the filled part equals the length
of the tank times the area of the shaded region
shown in the accompanying figure. Consider
a layer of gasoline of thickness dy located at
height y where r y
r d . The width of
this layer is 2 r 2
r d
A
2
V
L A
r d
(b) 2 L
r2
r
r
y 2 . Therefore,
y 2 dy and
r d
2L
r2
r
r2
y 2 dy
y 2 dy
y r2 y2
2
2L
r2
2
(We used FORMULA 29 with a
2L
(d r )
2
r2
2
2rd d 2
r d
y
sin 1 r
r
r)
r2
2
sin 1 d r r
2L
2
d r
2
r2
2
d2
2rd
sin 1 d r r
2
63. The integrand f ( x)
x x 2 is nonnegative, so the integral is maximized by integrating over the function s
entire domain, which runs from x 0 to x 1
1
1
2
x x dx
0
0
2
1
2
x x dx
(We used FORMULA 48 with a
2
2
x x2
2
0
x x
2
1 2
2
2
sin
1
2
1 x
1
2
1 sin 1 (2 x
8
x 2 x x 2 dx
1)
0
1
8 2
1
8
( x 1)(2 x 3) 2 x x 2
6
2
x 2 x x 2 over the largest domain on which g is
1 sin 1 ( x
2
2
1)
0
1
2
2
2
CAS EXPLORATIONS
65.
Example CAS commands:
Maple:
q1: Int( x*ln(x), x );
# (a)
q1 value( q1 );
q2 : Int( x^2*ln(x), x );
q2
0
8
(We used FORMULA 51 with a 1 )
1
2 2
1
1)
2
64. The integrand is maximized by integrating g ( x)
nonnegative, namely [0, 2]
2
1
2
1
1
2
x
1
2
x
2
# (b)
value( q2 );
q3 : Int( x^3*ln(x), x );
# (c)
Copyright
2014 Pearson Education, Inc.
Section 8.6 Integral Tables and Computer Algebra Systems
q3
value( q3 );
q4 : Int( x^4*ln(x), x );
q4
# (d)
value( q4 );
q5 : Int( x^n*ln(x), x );
q6
# (e)
value( q5 );
q7 : simplify(q6) assuming n::integer;
q5
66.
collect( factor(q7), ln(x) );
Example CAS commands:
Maple:
q1: Int( ln(x)/x, x );
# (a)
q1 value( q1 );
q2 : Int( ln(x)/x^2, x );
q2
# (b)
value( q2 );
q3 : Int( ln(x)/x^3, x );
q3
# (c)
value( q3 );
q4 : Int( ln(x)/x^4, x );
q4
# (d)
value( q4 );
q5 : Int( ln(x)/x^n, x );
# (e)
q6 : value( q5 );
q7 : simplify(q6) assuming n::integer;
q5
67.
collect( factor(q7), ln(x) );
Example CAS commands:
Maple:
q : Int( sin(x)^n/sin(x)^n cos(x)^n), x 0..Pi/2 );
q
# (a)
value( q );
q1: eval( q, n 1 ):
# (b)
q1 value( q1 );
for N in [1, 2,3,5, 7] do
q1: eval( q, n N );
print( q1 evalf(q1) );
end do:
qq1: PDEtools[dchange]( x Pi/2-u, q, [u] );
# (c)
qq2 : subs( u x, qq1 );
qq3 : q q
q qq2;
qq4 : combine( qq3 );
Copyright
2014 Pearson Education, Inc.
605
606
Chapter 8 Techniques of Integration
qq5 : value( qq4 );
simplify( qq5/2 );
65-67. Example CAS commands:
Mathematica: (functions may vary)
In Mathematica, the natural log is denoted by Log rather than Ln, Log base 10 is Log[x, 10]
Mathematica does not include an arbitrary constant when computing an indefinite integral,
Clear[x, f, n]
f[x_]: Log[x] / x n
Integrate[f[x], x]
For exercise 67, Mathematica cannot evaluate the integral with arbitrary n. It does evaluate the integral (value
is /4 in each case) for small values of n, but for large values of n, it identifies this integral as Indeterminate
x n 1 ln x
n 1
x n ln x dx
65. (e)
x n dx, n
1
n 1
1
(We used FORMULA 110 with a 1, m 1 )
x n 1 ln x
n 1
xn 1
( n 1) 2
x n ln x dx
66. (e)
xn 1
n 1
C
n 1
x
ln x
n 1
ln x n1 1
C
x n dx, n 1
1
( n) 1
(We used FORMULA 110 with a 1, m 1, n
1 n
x
1 n
ln x
1 n
1 x
1 n 1 n
1 n
x
1 n
C
ln x 1 1n
n)
C
67. (a) Neither MAPLE nor MATHEMATICA can find this integral for arbitrary n.
(b) MAPLE and MATHEMATICA get stuck at about n 5.
(c) Let x
I
/2
dx
sin n x dx
n
0
I
u
2
n
sin x cos x
I
du; x
0
sin n
0
/2 sin
/2 sin n x cos n x
0
sin n cos n x
n
2
dx
u
u
2
, x
u du
2
cos
n
/2
0
Copyright
2
dx
/2
0
u
2
u
2
I
0;
cosn u du
n
n
cos u sin u
/2
0
4
2014 Pearson Education, Inc.
cosn x dx
cos n x sin n x
Section 8.7 Numerical Integration
8.7
1.
NUMERICAL INTEGRATION
2
1
x dx
I. (a) For n
b a
n
4, x
mf ( xi ) 12
f ( x)
(b)
1
f ( x) 1
0
ET
x dx
1
ET
True Value
II. (a) For n
x dx T
(c)
2.
3
1
2
1
x dx
M
3
2
Es
100
x
3
2 1 1
4
4
1 (18)
12
S
0
Es
True Value
m
mf ( xi )
x0
1
1
1
1
x1
5/4
5/4
2
5/2
x2
3/2
3/2
2
3
x3
7/4
7/4
2
7/2
x4
2
2
1
2
xi
f ( xi )
m
mf ( xi )
0%
mf ( xi ) 18
(b)
f ( xi )
0
b a
n
4, x
0
3
2
1
2
2
100
f (4) ( x)
f
1;
8
0
2
x2
2 1
2
ET
(c)
T
x
M
2
x
2 1 1
4
4
2
1 (12) 3 ;
8
2
xi
0
Es
2
1
1 ;
12
3;
2
x0
1
1
1
1
0
x1
5/4
5/4
4
5
x2
3/2
3/2
2
3
x3
7/4
7/4
4
7
x4
2
2
1
2
3
2
x dx S
3
2
0
0%
(2 x 1) dx
I. (a) For n
4, x
x
2
(c)
3
1
0
Es
True Value
II. (a) For n
4, x
(b)
3
1
( x)
(c)
Es
True Value
2
1
2
0
6;
f
0
0
x2
3
x
1
(9 3) (1 1)
b a
n
3 1 2
4
4
1 (36)
6
S
M
0
6
Es
Es
3
1
1
2
ET
f ( xi )
m
mf ( xi )
x0
1
1
1
1
x1
3/2
2
2
4
x2
2
3
2
6
x3
5/2
4
2
8
x4
3
5
1
5
3
1
(2 x 1) dx T
6 6
0
x
3
1;
6
xi
f ( xi )
m
mf ( xi )
6;
x0
1
1
1
1
0
x1
3/2
2
4
8
x2
2
3
2
6
x3
5/2
4
4
16
x4
3
5
1
5
(2 x 1) dx S
0
100
6
xi
0%
36
(2 x 1) dx
6 6
f ( x)
100
mf ( xi )
f
1 (24)
4
ET
(2 x 1) dx
(4)
2
4
T
24
2x 1
M
(b)
3 1
4
1;
4
mf ( xi )
f ( x)
b a
n
0%
Copyright
2014 Pearson Education, Inc.
607
608
3.
Chapter 8 Techniques of Integration
1
1
I. (a) For n
T
x2 1
f ( x)
M
2
1 ( 1)
4
1 (11)
4
b a
n
4, x
mf ( xi ) 11
f ( x)
2
4
x
2
1
2
2.75;
2x
f ( x)
1 ( 1) 1 2
(2)
12
2
ET
1;
4
2
1
12
or
0.08333
(b)
(c)
1
1
x3
3
x 2 1 dx
ET
1
12
ET
True Value
100
II. (a) For n
(b)
f ( x)
0
Es
0
1
1
x
2
1
12
8
3
(c)
2
1
Es
True Value
1
100
f (4)( x)
1
1
1 ( 1)
4
1 (16)
6
S
3
x
3
1 dx
Es
x
b a
n
4, x
3
0
f ( xi )
m
mf ( xi )
x0
1
2
1
2
x1
1/2
5/4
2
5/2
x2
0
1
2
2
x3
1/2
5/4
2
5/2
x4
1
2
1
2
1
3
1
3
1
1
8
3
1
ET
1
0
x
100
2
4
1
1
x
3
1
2
8
3
M
x 2 1 dx S
f ( x) x
M
1;
4
2
2
1;
6
2.66667;
0
8
3
8
3
8
3
0
0 ( 2)
4
mf ( xi )
3
f ( x)
2x
1
2
4
T
1
2
f ( x)
0 ( 2) 1 2
(2)
12
2
ET
3;
4
1 (3)
4
2
1
12
0.08333
(b)
(c)
0
2
x 2 1 dx
ET
True Value
11
4
1
12
xi
f ( xi )
m
mf ( xi )
x0
1
2
1
2
x1
1/2
5/4
4
5
x2
0
1
2
2
x3
1/2
5/4
4
5
x4
1
2
1
2
xi
f ( xi )
m
mf ( xi )
x0
2
3
1
3
x1
3/2
5/4
2
5/2
x2
1
0
2
0
x3
1/2
3/4
2
3/2
x4
0
1
1
1
0
x 2 1 dx T
2
3
0%
b a
n
4, x
x
2
8
3
3%
x 2 1 dx
I. (a) For n
x 2 1 dx T
0.08333
mf ( xi ) 16
4.
xi
x 2 1 dx
100
x3
3
1
12
2
3
x
0
2
0
8
3
2
2
3
ET
2
100 13%
Copyright
2014 Pearson Education, Inc.
3
4
1
12
ET
1
12
Section 8.7 Numerical Integration
II. (a) For n
mf ( xi )
f
(b)
(c)
5.
2
0
0
(4)
Es
True Value
M
0
2
3
f
Es
0
Es
x
3
1;
6
( x)
0
0
x 2 1 dx S
2
100
0%
x
b a
n
2 0
4
T
1 (25)
4
2
xi
f ( xi )
m
mf ( xi )
x0
2
3
1
3
x1
3/ 2
5/4
4
5
x2
1
0
2
0
x3
1/ 2
3/ 4
4
3
x4
0
1
1
1
t 3 t dt
f (t )
4,
t
M
(b)
(c)
2
0
25
3
t
mf (ti )
f (3) (t )
(b)
0
(c)
b a
n
S
6
0
22
2
1
2
1
2
0
6
x
3
1;
6
0
0
Es
0
t 3 t dt S
0
Es
True Value
100
2
f (ti )
m
mf (ti )
t0
0
0
1
0
t1
1/2
5/8
2
5/4
t2
1
2
2
4
t3
3/2
39/8
2
39/4
t4
2
10
1
10
t 3 t dt T
6 25
4
ti
f (ti )
m
mf (ti )
t0
0
0
1
0
t1
1/2
5/8
4
5/2
t2
1
2
2
4
t3
3/2
39/8
4
39/2
t4
2
10
1
10
ET
6;
M
2
Es
6t
ti
0
0%
ti
t 3 1 dt
I. (a) For n
4, x
mf (ti )
f (t )
M
(b)
(c)
1
1
8
t3 1
6
T
f (t )
f (1)
t 3 1 dt
ET
True Value
b a
n
100
t4
4
1 ( 1)
4
1 (8)
4
2
2;
ET
1 ( 1) 1 2
(6)
12
2
2
4
3t
t
1
1
1
4
1
4
ET
4%
2 0 2
4
4
1 (36)
6
f (4) (t )
t 3 t dt
24
4
100
6
36
6
6 6
1
4
f (t )
1;
4
2 0 1 2 (12)
12 2
2
t2
2 0
100
4, x
1
ET
t4
4
x
2 1
4
2
25 ;
4
3t
f (2)
t 3 t dt
ET
True Value
2
f (t )
12
II. (a) For n
1
0
1
2
(3)
0
mf (ti )
1
S
( x)
2
3
I. (a) For n
6.
4
x 2 1 dx
2
2
3
0 ( 2)
2
4
4
1 (4) 2 ;
6
3
b a
n
4, x
609
f (t )
14
4
x
2
1
2
1
1;
4
6t
( 1)4
4
1
4
( 1)
m
mf (ti )
t0
1
0
1
0
t1
1/2
7/8
2
7/4
t2
0
1
2
2
t3
1/2
9/8
2
9/4
t4
1
2
1
2
1
t 3 1 dt T
2
ET
0%
Copyright
f (ti )
2014 Pearson Education, Inc.
1
2 2
0
610
Chapter 8 Techniques of Integration
II. (a) For n
mf (ti ) 12
f (3) (t )
(b)
1
1
(c)
7.
2 1
1 s2
S
f (4) (t )
6
t 3 1 dt
2 2
2
2
4
1;
6
ti
M
1
1
0
Es
Es
True Valule
100
0%
b a
n
179,573
44,100
mf ( si )
6
s4
f (s)
2
ds
1
ET
(c)
1
0
t1
1/2
7/8
4
7/2
t2
0
1
2
2
t3
1/2
9/8
4
9/2
t4
1
2
1
2
si
f ( si )
m
mf ( si )
264,821
529,200
f (3) ( s )
4 1
2 ( s 1)2
1
32
s0
1
1
1
1
s1
5/4
16/25
2
32/25
s2
3/2
4/9
2
8/9
s3
7/4
16/49
2
32/49
s4
2
1/4
1
1/4
0.50899
0.00899
0.03125
2
1
s 1
100
0.00899
0.5
1
2
b a
n
264,821
44,100
100
2 1
4
S
1
1
1
2
ET
2 1
1 s2
f (4) ( s )
24
s5
2 1
ds 12
1 s2
Es
100
True Value
x
1 ;
3
12
1 264,821
12 44,100
120
s6
1
384
2 1
1 s2
Es
0.0004
0.5
si
f ( si )
m
mf ( si )
s0
1
1
1
1
s1
5/4
16/25
4
64/25
s2
3/2
4/9
2
8/9
s3
7/4
16/49
4
64/49
s4
2
1/4
1
1/4
1
4
M
0.00260
ds S
100
120
1
2
0.50042
0.00042
Es
0.00042
0.08%
si
f ( si )
m
mf ( si )
s0
2
1
1
1
s1
5/2
4/9
2
8/9
s2
3
1/4
2
1/2
s3
7/2
4/25
2
8/25
s4
4
1/9
1
1/9
ds
I. (a) For n
ET
b a
n
1269
450
2
4, x
mf ( si )
f (s)
1
2
ds T
2%
0.50042;
2 1 1 4 (120)
180 4
Es
(c)
2
s3
f (1)
s 2 ds
4, x
mf ( si )
(b)
6
1;
8
179,573
352,800
0.00899
ET
True Value
II. (a) For n
M
x
2
1 179,573
8 44,100
1
4
f (s)
2 1 1 2 (6)
12 4
ET
2 1
1 s2
T
1
s2
0.50899; f ( s )
(b)
2 1
4
mf (ti )
0
ds
4, x
m
1
0
t 3 1 dt S
f (ti )
t0
2;
0
Es
x
3
1
2
0
I. (a) For n
8.
1 ( 1)
4
1 (12)
6
b a
n
4, x
( s 1)
T
4 2 1
4
2
1 1269
4 450
f (s)
4 2 1 2 (6)
12 2
1
4
2
( s 1)3
0.25
Copyright
x
1;
2
4
1269
1800
0.70500;
f (s)
M
6
6
( s 1)4
2014 Pearson Education, Inc.
Section 8.7 Numerical Integration
(b)
4
1
2 ( s 1)2
ET
(c)
II. (a) For n
2
3
ET
4 1
2 ( s 1)2
2
3
ds T
x
4 2 1
4
2
3
S 16 1813
450
(3)
24
f (s)
( s 1)5
120
( s 1)6
(s)
M
4 1
2 ( s 1)2
(c)
Es
True Value
2
3
ds
1
12
0.08333
4 1
2 ( s 1)2
Es
0.00481
100
1;
6
120
4 2 1 4 (120)
180 2
(b)
2
3
ds S
f ( si )
m
mf ( si )
s0
2
1
1
1
s1
5/2
4/9
4
16/9
s2
3
1/4
2
1/2
s3
7/2
4/25
4
16/25
s4
4
1/9
1
1/9
0.67148
0.00481
b a
n
4
2 2 2
0
x
2
4
8
;
4.8284;
T
2 2 2 1.89612; f (t ) sin t
8
f (t ) cos t
f (t )
sin t
M 1
2
0
(1)
12 4
ET
(c)
0
sin t dt
ET
True Value
II. (a) For n
2.00456; f (3) (t )
(b)
0
1
2
0.00456
(c)
Es
True Value
100
Es
Es
100
5%
4
x
3
0
7.6569
0
4
S
(1)
ti
f (ti )
m
mf (ti )
t0
0
0
1
0
t1
/4
2
t2
/2
2/2
1
2
2
t3
3 /4
t4
12
sin t dt S
100
2
ET
0
;
2/2
0
2
2
1
2
0
sin t dt T
2 1.89612
0.10388
ti
f (ti )
m
mf (ti )
2 4 2
t0
0
0
1
0
sin t
t1
/4
2/2
4
2 2
t2
/2
1
2
2
t3
3 /4
2/2
4
2 2
0
1
0
0.00664
2 2.00456
t4
0.00456
0.00456
2
12
f (4) (t )
cos t
0
180 4
Es
sin t dt
( cos ) ( cos 0)
4
2 4 2
0.00481
0.16149
192
0.10388
2
b a
n
4, x
mf (ti )
M
3
cos t 0
100
Es
100 1%
2
3
4, x
mf (ti )
(b)
0.03833
si
sin t dt
I. (a) For n
0.705
6%
b a
n
1813
450
0.67148;
Es
0
1
2 1
100
2
3
4, x
1813
2700
9.
0.03833
100
mf ( si )
f
1
4 1
0.03833
ET
True Value
(4)
4
1
( s 1) 2
ds
611
0%
Copyright
2014 Pearson Education, Inc.
612
10.
Chapter 8 Techniques of Integration
1
0
ti
f (ti )
m
mf (ti )
t0
0
0
1
0
t1
1/4
2/2
2
t2
1/2
1
2
t3
3/4
2/2
2
t4
1
0
1
sin tdt
I. (a) For n
mf (ti )
1
8
T
(c)
1
2 2 2
1
2
0.03307
II. (a) For n
1
12
S
(3)
(b)
(c)
11. (a)
M
(b) M
12. (a) M
(b) M
13. (a) M
1
0
2 4 2
3
(t )
M
Es
2
100
1
cos
cos 0
2
0.63662
1
ET
0
2
2
2
0
sin t dt T
5%
x
3
1
4
1 ;
12
ti
f (ti )
m
mf (ti )
0
1
0
7.65685
t0
0
0.63807;
t1
1/4
2/2
4
2 2
sin t
t2
1/2
1
2
2
0.00211
t3
3/4
2/2
4
2 2
t4
1
0
1
0
cos t
4
sin t dt
Es
True Value
1 0
4
2 4 2
1
100
2
sin t
0.05140
0
0.03307
sin t
2
1
cos t
b a
n
4, x
mf (ti )
f
1 0 1 2
12 4
sin t dt
100
2
f (t )
ET
ET
True Value
1;
8
0.60355; f (t )
2
0.60355
x
2
1
4
4.828
cos t
M
0
1 0
4
2 2 2
f (t )
(b)
b a
n
4, x
f
(4)
(t )
1 0 1 4
180 4
0.63662
0.00145
2
4
4
1
Es
100
0
x 1
1 (1) 2 (0)
12
ET
2 (n must be even)
0 (see Exercise 2): Then n 1
0 (see Exercise 2): Then n
2 (see Exercise 3): Then
2
x
2
x
2
n
ET
2 2
12 n
1
2
2 (2) 2 (0)
12
ET
2 (n must be even)
x
0.63807
0.00145
Es
0.00145
0%
0 (see Exercise 1): Then n 1
0 (see Exercise 1): Then n
sin t dt S
2
x 1
(2)
4
3n 2
4
0 10
1 1 4 (0)
180 2
Es
0 10
Es
10 4
0 10
4
4
2 (1) 4 (0)
180
n2
4
3
104
10
4
n
4
3
10 4
4
3
10 4
n 115.4, so let n 116
(b) M
14. (a)
M
0 (see Exercise 3): Then n
2 (see Exercise 4): Then
2 (n must be even)
x
2
n
ET
2 2
12 n
2
x 1
(2)
4
3n 2
Es
10 4
2 (1) 4 (0)
180
n2
4
3
104
0 10 4
n
n 115.4, so let n 116
(b) M
0 (see Exercise 4): Then n
2 (n must be even)
Copyright
x 1
Es
2014 Pearson Education, Inc.
2 (1) 4 (0)
180
0 10 4
Section 8.7 Numerical Integration
15. (a)
M
12 (see Exercise 5): Then
n
282.8, so let n
(b) M
16. (a)
6 (see Exercise 6): Then
200, so let n
(b) M
17. (a)
70.7, so let n
(b) M
4 64
3
f ( x)
104
x
n2
8 10 4
n
8 104
10 4
4
n2
(6)
2 (1) 4 (0)
180
Es
n2
0 10 4
4 104
4 104
n
2
1 1
12 n
ET
10 4
1
2 n2
(6)
2 (1) 4 (0)
180
Es
0 10 4
n2
104
1
2
1
2
n
104
1
n
x
4
1 1
180 n
Es
10 4
2
3n 4
(120)
n4
104
2
3
9.04, so let n 10 (n must be even)
2
n
x
2
2 2
12 n
ET
10 4
4
n2
(6)
n2
4 104
4 104
n
201
ET
2
n
x
Es
21.5, so let n
1 (x
2
f ( x)
3
n
x 1
71
n
x 1
Then
f
1
n
x
120 (see Exercise 8): Then
n
(b)
n
200, so let n
(b) M
2
2 2
12 n
ET
x 1
2 (n must be even)
6 (see Exercise 8): Then
M
n
104
4 2
3
n
19. (a)
2
n
120 (see Exercise 7): Then
n
18. (a)
2 (n must be even )
x
6 (see Exercise 7): Then
n
10 4
8
n2
(12)
201
0 (Exercise 6): Then n
M
2
2 2
12 n
ET
283
0 (see Exercise 5): Then n
M
2
n
x
613
10 4
64
3n 4
(120)
n4
64
3
104
22 (n must be even)
1) 1/2
3 3 2 1
12 n
4
4
2 2
180 n
f ( x)
10 4
9
16 n 2
1 (x
4
1) 3/2
n2
9
16
1
4
104
x 1
4 1
104
9
16
n
1
M
3
n
1.
4
3
75, so let
76
(3)
( x)
x
3
n
3 (x
8
1) 5/2
f (4) ( x)
3 3 4 15
180 n
16
Es
1) 7/2
15 ( x
16
35 (15)
16(180) n
15
16
10 4
n4
f ( x)
3 (x
4
4
x 1
15
M
7
7
16 1
35 (15) 104
4
n
16(180)
15 .
16
Then
35 (15) 104
n 10.6, so let
16(180)
n 12 (n must be even)
20. (a)
x
(b)
1
x 1
f ( x)
3
n
f (3) ( x)
x
3
n
f ( x)
ET
15 ( x
8
Es
1 (x
2
3 3 2 3
12 n
4
1) 7/2
f
3 3 4 105
180 n
16
1) 3/2
34
48n 2
(4)
( x)
10 4
n2
105 ( x
16
35 (105)
16(180) n 4
1) 5/2
34 104
105
16
x 1
35 (105) 104
16(180)
2014 Pearson Education, Inc.
3.
4
5
Then
n 129.9, so let n 130
105
M
so let n 18 (n must be even)
Copyright
4 1
48
9
3
M
5
34 104
n
48
n4
x 1
4
1) 9/2
10 4
3
16 1
n
4
9
105 .
16
Then
35 (105) 104
16(180)
n 17.25,
614
Chapter 8 Techniques of Integration
21. (a)
f ( x)
f (3) ( x)
f ( x)
x
(b)
10 4
2
n
5
2
ET
f (3) ( x)
f ( x)
2 2
12 n
sin ( x
10 4
2
n4
4
4 32 10
180
n
sin ( x
)
32 104
10 4
cos ( x
4
4 32 10
180
180
n
12
n
f (4) ( x)
)
8 104
n2
M
8
12n 2
(1)
sin ( x 1)
sin ( x 1)
180
)
f ( x)
10 4
8
12 n2
32 104
n4
cos ( x
32
1804
23.
cos ( x 1)
f (4) ( x)
cos ( x 1)
32
180 n 4
22. (a)
f ( x)
2
2 2 (1)
| ET | 12
n
2
n
x
(b)
sin ( x 1)
1. Then
1. Then
8 104
2
n
x
cos ( x
8 104
n
12
)
M
1. Then
n
6.49, so let n
n
12
6.49, so let n
f ( x)
n2
M
81.6, so let n
2 2
180 n
Es
4
82
(1)
8 (n must be even)
)
M
8 104
12
2
n
x
1. Then
n
81.6, so let n
2 2
180 n
Es
4
82
(1)
8 (n must be even)
2(12.7) 13.0 (30) 15,990 ft 3 .
6.0 2(8.2) 2(9.1)
24. Use the conversion 30 mph 44 fps (ft per sec) since
time is measured in seconds. The distance traveled as the
car accelerates from, say, 40 mph 58.67 fps to 50 mph
73.33 fps in (4.5 3.2) 1.3 sec is the area of the
trapezoid (see figure) associated with that time interval:
1 (58.67 73.33)(1.3) 85.8 ft. The total distance
2
traveled by the Ford Mustang Cobra is the sum of all
these eleven trapezoids (using 2t and the table below):
v(mph)
0
30
v(fps)
0
44
t (sec)
0
2.2
3.2
t /2
0
1.1
0.5
s
40
50
60
70
80
90
100
110
120
130
88
102.67
117.33
132
146.67
161.33
176
190.67
4.5
5.9
7.8
10.2
12.7
16
20.6
26.2
37.1
0.65
0.7
0.95
1.2
1.25
1.65
2.3
2.8
5.45
58.67 73.33
(44)(1.1) (102.67)(0.5) (132)(0.65) (161.33)(0.7) (190.67)(0.95) (220)(1.2) (249.33)(1.25)
(278.67)(1.65) (308)(2.3) (337.33)(2.8) (366.67)(5.45)
25. Using Simpson s Rule,
myi
33.6
x
3
x 1
Cross Section Area
1;
3
1 (33.6)
3
11.2 ft 2 . Let x be the length of the tank. Then the
Volume V (Cross Sectional Area) x 11.2 x.
Now 5000 lb gasoline at 42 lb/ft
5000
42
3
3
V
119.05 ft
119.05 11.2 x
x 10.63 ft
Copyright
5166.346 ft
0.9785 mi
xi
yi
m
myi
x0
0
1.5
1
1.5
x1
1
1.6
4
6.4
x2
2
1.8
2
3.6
x3
3
1.9
4
7.6
x4
4
2.0
2
4.0
x5
5
2.1
4
8.4
x6
6
2.1
1
2.1
2014 Pearson Education, Inc.
Section 8.7 Numerical Integration
26.
24
2
0.019 2(0.020) 2(0.021)
27. (a)
Es
(b)
x
x
3
8
(c)
; f (4)
1
M
;
1
2
Es
0
180
4
8
(1)
0.00021
xi
f ( xi )
m
mf ( x1i )
1
1
1
x1
/8
0.974495358
4
3.897981432
x2
/4
0.900316316
2
1.800632632
x3
3 /8
0.784213303
4
3.136853212
x4
/2
0.636619772
1
0.636619772
(10.47208705) 1.37079
100
b a
n
0.015%
1 0
10
e0
erf (1)
2 0.1
3 3
4e 0.01 2e 0.04
4e 0.09
0.1
1 0 (0.1) 4 (12)
180
y0
2 y1 2 y2
y0
4 y1 2 y2
4 y3
4e 0.81 e 1
4 y9
y10
0.843
6.7 10 6
2 y3
b a y0 y1 y1 y2 y2
2
n
T
8
S
Es
x
2
0
4
0
2
30
29. T
2
x
x0
x
(b)
24
4
4.2 L
mf ( xi ) 10.47208705
24
0.00021
1.37079
28. (a)
x4 M ; n
b a
180
2(0.031) 0.035
615
2 yn 1
yn
1
yn
1
yn
yn where
b a
n
b a
n
x
f ( x0 ) f ( x1 )
2
and f is continuous on [a, b]. So
f ( xn 1 ) f ( xn )
2
f ( x1 ) f ( x2 )
2
f ( xk 1 ) f ( xk )
is always between
2
f ( xk 1 ) f ( xk )
; this is a consequence of
2
n
. Since f is
continuous on each interval xk 1 , xk , and
f ( xk 1 ) and f ( xk ), there is
a point ck in xk 1 , xk with f (ck )
the Intermediate Value Theorem.
n
Thus our sum is
k 1
b a
n
f (ck ) which has the form
xk f (ck ) with
xk
b a
n
for all k. This a Riemann
x
b a
n
and f is continuous on
k 1
Sum for f on [a , b].
x
3
30. S
y0
[a, b]. So S
b a
n
2
4 y1 2 y2
4 y3
b a y0 4 y1 y2
3
n
f ( x0 ) 4 f ( x1 ) f ( x2 )
6
f ( x2 k ) 4 f ( x2 k 1 ) f ( x2 k
6
2)
2 yn 2
y2 4 y3 y4
3
4 yn 1
yn where n is even,
y4 4 y5 y6
3
f ( x2 ) 4 f ( x3 ) f ( x4 )
6
yn
2
4 yn
3
f ( x4 ) 4 f ( x5 ) f ( x6 )
6
1
yn
f ( xn
2)
4 f ( xn 1 ) f ( xn )
6
is the average of the six values of the continuous function on the interval
x2k , x2k 2 , so it is between the minimum and maximum of f on this interval. By the Extreme Value
Theorem for continuous functions, f takes on its maximum and minimum in this interval, so there are xa and
xb with x2 k
xa , xb
x2 k 2 and f ( xa )
f ( x2 k ) 4 f ( x2 k 1 ) f ( x2 k
6
2)
f ( xb ).
By the Intermediate Value Theorem, there is ck in x2 k , x2 k 2 with f (ck )
n /2
So our sum has the form
xk f (ck ) with
k 1
Copyright
xk
b a ,
( n /2)
f ( x2 k ) 4 f ( x2 k 1 ) f ( x2 k
6
a Riemann sum for f on [ a, b].
2014 Pearson Education, Inc.
2)
.
616
Chapter 8 Techniques of Integration
1
2
31. (a) a 1, e
/2
2
Length
4 cos 2 t dt
0
/2
4
0
/2
0
f (t ) dt ; use the
Trapezoid Rule with n 10
b a
n
t
/2
0
2
.
20
10
10
2
4 cos t dt
0
mf ( xn )
t
2
(37.3686183)
40
(37.3686183)
2.934924419
Length 2(2.934924419)
(b)
37.3686183
n 0
T
f (t )
1
M
ET
5.870
1
2
b a
12
0
2
t M
12
xi
f ( xi )
m
mf ( xi )
x0
0
1.732050808
1
1.732050808
x1
/20
1.739100843
2
3.478201686
x2
/10
1.759400893
2
3.518801786
x3
3 /20
1.790560631
2
3.581121262
1 14 cos 2 t dt
2
20
1 0.0032
x4
/5
1.82906848
2
3.658136959
x5
/4
1.870828693
2
3.741657387
x6
3 /10
1.911676881
2
3.823353762
x7
7 /20
1.947791731
2
3.895583461
x8
2 /5
1.975982919
2
3.951965839
x9
9 /20
1.993872679
2
3.987745357
2
1
2
xi
f ( xi )
m
mf ( xi )
x0
0
1.414213562
1
1.414213562
x1
/8
1.361452677
4
5.445810706
x2
/4
1.224744871
2
2.449489743
x3
3 /8
1.070722471
4
4.282889883
x4
/2
1
2
2
x5
5 /8
1.070722471
4
4.282889883
x6
3 /4
1.224744871
2
2.449489743
x7
7 /8
1.361452677
4
5.445810706
1.414213562
1
1.414213562
x10
32.
x
8
0
mf ( xi )
S
24
x
3
8
24
/2
;
29.184807792
(29.18480779)
3.82028
x8
33. The length of the curve y
dy 2
dx
L
9 2
400
20
sin 320 x from 0 to 20 is: L
cos 2 320 x
20
L
0
0
1
dy 2
dy
dx; dx
dx
3
20
cos 320 x
2
1 9400 cos 2 320 x dx. Using numerical integration we find
21.07 in
34. First, we ll find the length of the cosine curve: L
dy 2
dx
2
4
sin 2 50x
L
25
25
1
2
4
25
25
1
dy 2
dx
dy
dx; dx
25
50
sin 50x
sin 2 50x dx. Using a numerical integrator we find L
Surface area is: A
73.1848 ft.
length width (73.1848)(300) 21,955.44 ft. Cost 2.35 A (2.35)(21,955.44)
$51,595.28. Answers may vary slightly, depending on the numerical integration used.
Copyright
2014 Pearson Education, Inc.
Section 8.8 Improper Integrals
35.
y
dy
dx
sin x
gives S
36.
cos 2 x
S
2 (sin x ) 1 cos 2 x dx; a numerical integration
0
14.4
x2
4
y
dy 2
dx
cos x
617
dy
dx
dy 2
dx
x
2
x2
4
2
S
0
x2
4
2
2
1 x4 dx; a numerical integration gives S
37. A calculator or computer numerical integrator yields sin 1 0.6
38. A calculator or computer numerical integrator yields
5.28
0.643501109.
3.1415929.
12
39. The amount of medication absorbed over a 12-hr period is given by
0
6 ln 2t 2
3t 3 dt . A numerical
integrator yields a value of 28.684 for this integral, so the amount of medication absorbed over a 12-hr period
is approximately 28.7 milligrams.
1 1
12.5 4ln t 2 3t 4 dt. A
6 0
numerical integrator yields a value of 6.078 for this integral, so the average concentration is approximately
6.1 grams per liter.
40. The average concentration of antihistamine over a 6-hr period is given by
8.8
1.
IMPROPER INTEGRALS
dx
0 x
2
1
2.
dx
1 x1.001
3.
1 dx
0 x
4.
5.
lim
b dx
0 x2 1
lim
b dx
1 x1.001
b
b
lim
b
4 dx
4 x
0
1
dx
1 x 2/3
0
1
b
4
b
0
1 dx
0 x 2/3
(0 3) (3 0)
6.
1
dx
8 x1/3
lim
b
0
b
lim
b
b
0
1
lim
b
3 x 2/3
2
8
b
2 4
lim 3 x1/3
0
c
1000
2 0
0
2 4 b
4
lim 3 x1/3
1000
b0.001
lim 2 2 b
b
0
b
1
c
0
2
1000
2
0 4
4
lim 3b1/3 3( 1)1/3
b
0
6
0 dx
8 x1/3
3 b 2/3
2
1
lim 2 x1/2
b
lim
1
b
2
b
b
1000 x 0.001
lim
(4 x) 1/2 dx
0 dx
1 x 2/3
lim tan 1 b tan 1 0
0
b
x 1/2 dx
lim
b
b
lim tan 1 x
1 dx
0 x1/3
3(
2
8)
2/3
b
0
lim
c
0
3 (1)
2
2/3
Copyright
lim
c
0
3 c 2/3
2
1
3 x 2/3
2
c
0 32 (4)
3
2
0
2014 Pearson Education, Inc.
9
2
lim 3(1)1/3 3c1/3
c
0
618
Chapter 8 Techniques of Integration
7.
1 dx
0 1 x2
8.
1 dr
0 r 0.999
x
b
11.
12.
2
1
2
4
2 v
2
v
0
2 x dx
1 b1
x2 4
1
2
lim
b
1
2
2 s 1
0 4 s2
lim
b
0
1
b
4 du
b 2 u
2
dx
0 (1 x ) x
2 2
1
c
;
;
2
( 1)
3/ 2
2
u
du
2(
c
lim
0
2
0
4
u
x
du
dx
2 x
0
1
c
lim
c
x2 1
2 ln 22 1
4 s
4 s
2
2 x dx
4
1) d
ds
du
2
;
u
du
0 12
0
b
u
4 s2
lim
du
u
0
b
2
ln xx 11
b
lim
0
3
4
0 2 ln 2
ln 4
0 ln 3 ln 3
1
1
u b
lim
b
4 du
2u 3/ 2
du
4 2u 3/ 2
c
1
u 1
lim
c
lim
b
4
1
u b
c
3 du
b 2 u
lim
b
u
0
1 0 du
2 4 u
c
lim sin 1 2s
0
c 2
lim
c
2
3
lim
b
b
c
ds
0
4 s2
lim 2
b
b
0
3
0
b
3 0
lim
b
b 2 du
0 u2 1
lim
b
b
2 tan 1 u
0
lim 2 tan 1 b 2 tan 1 0
b
2(0)
Copyright
2014 Pearson Education, Inc.
3
lim sin 1 2c sin 1 0
2
c
2
2 du
0 u2 1
c
1
u 4
lim
0
2s ds
4
b
du
1 u2
2 x dx
3 du
02 u
lim
b
1 du
u2
x2
2
0
x2 1
1
2
b
2
ln(1) ln 13
u
;
3/ 2
2
2 ln(1) 2 ln 12
0
x2 4
ds
4
( 1 0) (0 1)
1
2
2
ln x 1
ln 3 ln1 ln 3
ln 22 11
x dx
c
1 2 2 s ds
2 0 4 s2
ds
(2 0)
0
x2 4
d ;
2
lim 2 ln bb 1
2 x dx
b
lim
tan 1 1 tan 1 b2
lim
x dx
3/ 2
b
lim bb 11
lim ln bb 11
lim
0
b
b
c
2
ln x 1
b
2
x2 1
x dx
0
b
2 x dx
2
x2 1
1
2
tan 1 2x
b
lim
lim
ln 3 ln
b
lim ln tt 11
2
b
lim
17.
ln bb 11
b
lim 2ln vv 1
2
b
2 dt
2 t2 1
14.
2 dx
x 1
2
1000 0 1000
0
b
b
b
16.
b
2 dv
13.
15.
2 dx
x 1
lim 1000 1000b 0.001
b
0
b
2 dx
x
1
0
2
b 1
lim 1000r 0.001
lim ln 31
2
lim sin 1 b sin 1 0
0
b 1
2 2 dx
9.
10.
b
lim sin 1 x
Section 8.8 Improper Integrals
18.
2 dx
1 x x2 1
dx
1 x x2 1
1
lim sec
b 1
19.
0 1 v
b
lim
1
lim sec
1
c sec
c
b
lim ln 1 tan 1 v
dv
1 tan 1 v
2
1
2 sec
2 dx
b
x x2 1
b 1
dx
2 x x2 1
b
2
lim
c
3
b 1
2
2
0
3
lim ln 1 tan 1 b
c
lim sec 1 | x |
b
ln 1 tan 1 0
b
0
2
lim sec 1 | x |
c dx
2 x x2 1
619
2
c
ln 1 2
ln(1 0)
ln 1 2
20.
21.
16 tan 1 x
0 1 x2
0
2 b
lim 8 tan 1 x
dx
e d
b
b
2
2
lim 8 tan 1 b
8 tan 1 0
b
0
lim
e
0
e
b
b
8 2
0 e 0 e0
lim
beb
eb
1
b
2
b 1
e b
lim
2 2
8(0)
1
b
(l'Hopital's rule for
1 0
22.
sin d
0
25.
x
1
0
0
1
27.
2
0
e b
lim
lim
b
0
b
0
ln x dx
b
x2
2
1
b
4
b3
1
4
b
1
x2
4 b
b
2
( 1)
b
0
2
lim b4
b
1
4
0
1
lim x x ln x b
b 0
lim
b
b
lim
0
2
e x
0
( 1 0) (0 1)
0
b
0
(Formula 107 with a
lim
c
1
4
b2
2
2
e x
0
2
ln b b4
1
4
4 s
2
b
lim sin 1 2s
0
b 2
1 1ln1
lim sin 1 b2 sin 1 0
2
b
Copyright
ln b
lim
b
2
b2
0
0
1
4
b b ln b
1 0
0
ds
c
0
lim b 1 0 1
b
1, b 1 )
(1 0) 1
b
1 ln1
2
lim
cos )
1
lim 1 eb
2
2 xe x dx
e c
lim
ln x
0
0
c
2(0 1)
2
0
ex
lim
2
2 xe x dx
2
lim 2 e1 1 ( sin
b
sin d
2(sin 0 cos 0)
e x dx
0
1
x ln x dx
1
2e
2e 0
0
dx
lim
1
4
26.
b
2
2 xe x dx
b
0
2 eb
e
24.
b
lim
2(sin b cos b )
lim
b
23.
form)
1
2e
0
1
e b
lim
2
0
2
2014 Pearson Education, Inc.
lim
b
0
ln b
1
b
1
lim
b
0
1
b
1
b2
620
Chapter 8 Techniques of Integration
b
lim 2sin 1 r 2
28.
1 4 r dr
0 1 r4
29.
2 ds
1 s s2 1
30.
4 dt
2 t t2 4
31.
4 dx
1 x
2
lim sec 1 s
b
0
2
dx
x 1
lim
34.
2
1
d
5
6
dx
0 ( x 1) x 2 1
1 ln
2
lim
/2
0
b
c
0
( 1)
2 dx
x 1
2
3
lim
b
0
b
x
2 1 x
b 1
b
lim ln bb 23
2 c 1
1 tan 1 b
2
b
0
ln cos
lim
b
2
6
4
c
6
2
c
0 ln 12
1 tan 1 0
2
1 ln 1
2
1
0
0 2 2 0
b
1 tan 1 x
2
0
x2 1
1 ln
4
1
2
0 2 2 2 0
ln 11 23
b
0
c 1
lim 2 2 1
1
c
lim 2 x 1
0
c 1
b
1
2 3
lim 2 x
1
0
3
sec 1 b2
0
lim
b
b2 1
b
2
1
2
lim 2 4 2 c
lim 12 ln x 1
b 1
lim
c
2 1 0
b
tan d
2
4 dx
c x
1
lim ln
1 sec 1 4
2
2
lim
b
lim
2
2 1 b
b
35.
0
1 dx
0 1 x
b 1
33.
dx
1 x
2
b
0
b
lim
lim
32.
2
2 2
0
3
b 1
4
sec 1 2t
b
1
2
lim
2sin 1 0
b 1
lim sec 1 2 sec 1 b
b
b 1
b
lim 2sin 1 b 2
0
b 1
ln 2
x 1
lim 12 ln
x2 1
b
1 ln1
2
1
2 2
ln |cos b | ln1
1
2
lim
b
2
4
1 tan 1 x
2
ln1 12 0
b
0
4
ln |cos b |
, the integral
2
diverges
36.
/2
0
1
37.
cot d
ln x
0
1
1/3
x2
x
b
b
0
lim ln1 ln |sin b |
b
lim ln |sin b |
0
b
0
, the integral diverges
dx
ln x
2
/2
lim ln sin
1/3
dx is bounded, so convergence is determined by
0
On (0, 1/ 3], ln x
1
hence
0
ln x
x2
1 and
ln x
1
x2
x
1/3
. Since
2
0
1
x2
ln x
x2
dx .
1/3
dx diverges to
dx diverges.
Copyright
2014 Pearson Education, Inc.
, so does
0
ln x
x2
dx and
Section 8.8 Improper Integrals
2
1
dx ln ln x ,
x ln x
don t need a comparison test.)
38. Since
39.
ln 2
0
x 2 e 1/ x dx;
0 e
40.
41.
1e
x
1/ln 2
dx;
1
x
x
0
dt .
t sin t
1
1/ln 2 y 2 e
y
y
y
lim ln(ln 2) ln(ln a )
; the integral diverges. (In this case we
a 1
dy
3
1/ln 2
e y dy
b
e y
lim
e b
lim
1/ln 2
b
b
e 1/ln 2
e 1/ln 2 , so the integral converges.
y
0
1
dx
x ln x
621
1
y
2 e
x
Since for 0 t
2,
e
2
dy
0
so the integral converges.
1
t sin t
,0
1
t
and
dt
t
0
converges, then the original integral converges as
well by the Direct Comparison Test.
42.
1 dt
;
0 t sin t
Let f (t )
1 dt
Now,
0t
1
t sin t
and g (t )
1
1
2
2t b
lim
lim
3
0
b
1
2
0
b
1,
t3
f (t )
g
0 (t )
then lim
t
1
2b 2
t3
sin t
lim t
t
0
2
t
lim 1 3cos
t
t
6t
lim sin
t
0
t
0
1
dt
0 t sin t
, which diverges
6
lim cos
t
t
0
6.
diverges by the Limit
Comparison Test.
43.
2 dx
0 1 x2
1 dx
0 1 x2
2 dx
0 1 x2
44.
2 dx
01 x
45.
1
1
46.
1
1
2 dx
1 1 x
0
0
and
1 dx
01 x
1
1;
ln( x) dx
0
1
1 ln1
2
1
b 1
lim
b
1 ln 1 x
2
1 x 0
ln(1 x)
b 1
0
1
0
1
1
2
b
2
ln b
2
b
4
ln x dx
ln x dx
1
x ln( x) dx
1
1
4
1
ln x dx;
ln( x) dx
0
x ln | x | dx
lim
b
lim
1 ln 1 b
2
1 b
lim
b 1
0
, which diverges
b
0
lim
b 1
ln(1 b) 0
, which diverges
diverges as well.
ln x dx
1 0
1 dx
0 1 x2
and
diverges as well.
1 dx
01 x
2 dx
01 x
2 dx
1 1 x2
0
lim
c
0
lim x ln x x
b
0
1
b
lim
b
0
10 1
b ln b b
2 converges.
x ln x dx
1 ln1
2
1
4
lim
0
b
2
c
2
x2
2
ln c
ln x
1
x2
4 b
2
c
4
1
4
lim
c
0
0
1
4
x2
2
ln x
0
0
1
x2
4 c
the integral
converges (see Exercise 25 for the limit calculations).
47.
dx ;0
1 1 x3
1
x3 1
1
x3
for 1 x
and
dx
1 x3
converges
dx
1 1 x3
converges by the Direct Comparison
Test.
Copyright
2014 Pearson Education, Inc.
622
48.
Chapter 8 Techniques of Integration
4
dx
x 1
4
49.
2
51.
0
1
v 1
v
lim
1
v
1
1
lim
v 1
v
1
1 e
1
e
dx
4 x
1 and
b
lim 2 x
, which diverges
4
b
v
1
1 0
1
v
1 and
dv
v
2
b
lim 2 v
, which diverges
2
b
for 0
and
d
0 e
lim
b
e
b
e b 1
lim
0
b
d
0 e
1
converges
by the Direct Comparison Test.
dx
1
x6 1
0 x6 1
dx
0
1
x
1
1 0
1
x
diverges by the Limit Comparison Test.
; 0
d
0 1 e
x 1
x
1
lim
diverges by the Limit Comparison Test.
dv
v 1
d
0 1 e
x
lim
1
x
dv ; lim
v 1 v
2
50.
1
x 1
dx ; lim
x 1 x
x6 1
dx
1
dx
x6 1
0
x6 1
dx
1
dx
1 x3
dx
1 x3
and
lim
b
b
1
2
2x 1
lim
b
1
2b 2
1
2
1
2
converges by the Direct Comparison Test.
1
52.
dx
2
x
2
1
; lim
x
dx
2
x2 1
1
x
x2 1
lim
x
x
x
2
1
1
lim
1
x
1;
1
x2
1
2 x
dx
x
dx
b
lim ln b 2
b
diverges by the Limit Comparison Test.
x
53.
x 1
x2
1
2
x2
dx; lim
x 1
x2
x
lim
x 1
x2
x
1
, which diverges
x 1
x
1
1
lim
x
1
x
1;
1 x
2
dx
1 x3/ 2
lim
b
b
2 x 1/2
1
lim
b
2
b
2
dx converges by the Limit Comparison Test.
x
54.
x dx
2
x
4
1
; lim
x
x4 1
x
lim
x
x4
x
4
1
1
lim
1
x
x4
x dx
2
55.
x4 1
2 cos x
x
x dx
1;
1
x4
2
x
dx
2 x
4
b
lim ln x 2
x
, which diverges
diverges by the Limit Comparison Test.
dx; 0
1
x
2 cos x
x
for x
and
dx
x
lim ln x
b
b
, which diverges
diverges by the Direct Comparison Test.
Copyright
2014 Pearson Education, Inc.
2 cos x
x
dx
Section 8.8 Improper Integrals
1 sin x
56.
2 dt
4 t
3/ 2
x2
3/ 2
; lim 3/t 2
1
t
t
2
x2
x2
1 sin x
converges
57.
1 sin x
dx; 0
x2
for x
dx
2 b
x
lim
b
2
b
lim
b
2
2 dx
2
x2
dx converges by the Direct Comparison Test.
2 dt
1 and
1
2
x2
and
623
4 t
b
4t 1/2
lim
3/ 2
4
b
lim
4
b
b
2
2 dt
4 t 3/ 2
2
2 dt
4 t 3/ 2 1
converges
converges by the Limit Comparison Test.
58.
dx ;
2 ln x
59.
ex
1 x
60.
1
x
0
1
ln x
ex
x
1
x
dx; 0
ln y dy
e
dx
2 x
diverges
dx
2 ln x
for x 1 and
dx
1 x
diverges
e x dx
1 x
ey ]
ln (ln x) dx; [ x
ee
2 and
for x
(ln y )e y dy; 0
e
b
lim y ln y
ye
b
diverges by the Direct Comparison Test.
diverges by the Direct Comparison Test.
(ln y )e y for y
ln y
, which diverges
e
e and
ln y e y dy diverges
ee
ln (ln x) dx diverges by
the Direct Comparison Test.
1
61.
dx
1
e
x
x
ex x
; lim
x
ex
lim
x
1
e
x
1
lim
x
x
1
ex
2e b /2
lim
b
2e 1/2
2
e
1
1
1 0
x
ex
dx
1;
e x /2 dx converges
1
e
1
x
dx
1
e x /2 dx
lim
b
b
2e x /2
1
converges by the Limit Comparison
ex x
Test.
1
62.
ex 2x
dx ; lim
1 ex 2x x
dx
1 ex
x
4
dx
ex e
64.
2
1
1
b
lim
b
x
x
dx
0
1
2
x
4
ex
ex 2x
dx
1 ex 2x
converges
dx
63.
lim
1
ex
1
0
x
4
dx
1
x4 1
dx
0 ex e
x
x
; 0
2
e
1
1
1 0
x
dx
1 ex
1 and
lim
b
b
e x
lim
1
b
e b
e 1
1
dx
0 x
1
4
1
dx
1
1
x
4
1
0
dx
x
4
1
dx
1 x2
and
dx
1 x2
lim
b
1 b
x 1
converges by the Direct Comparison Test.
1
ex e
x
1
ex
for x
0;
dx
0 ex
converges
2
dx
0 ex e
x
converges by the Direct
Comparison Test.
65. (a)
2
dx ;
1 x (ln x ) p
t
ln x
1
e
converges by the Limit Comparison Test.
dx
;
1
lim
ln 2 dt
0 tp
lim
b
0
ln 2
1 t1 p
p 1
b
lim
b
0
b1 p
p 1
converges for p 1 and diverges for p 1
Copyright
2014 Pearson Education, Inc.
1 (ln 2)1 p
1 p
the integral
624
Chapter 8 Techniques of Integration
dx
2 x (ln x ) p
(b)
; [t
dt
ln 2 t p
ln x ]
and this integral is essentially the same as in Exercise 65(a): it converges
for p 1 and diverges for p 1
66.
2 x dx
0 x
2
1
b
x
b
lim (ln1)
e x dx
0
lim
e
b
68.
1
A 0
y
1
2A 0
69. V
70. V
71.
/2
A
e
0
0
2
e x dx
1
2 0
2
b
e x
xe x
b
b
b
e 2 x dx
0
1
b
b
lim
b
2
0
b
1 e 2x
2
0
1 e 2b
2
lim
b
ln |sec x tan x | ln |sec x | 0
2
0 1 1;
1e 20
2
1
2
be b
lim
b
b
lim
b
b
1 e 2b
2
lim 12
b
b
e x
xe x
lim
0 e 0 e 0
e b
b
1 e 2x
2
0
b
e 2 x dx
(sec x tan x) dx
b
lim 12
2
be b
lim
0
xe x dx
0
lim ln 1 sin b
b
b
2
lim ln b 1
ln b2 2 1
0 1 1
lim
2
e x dx
0
1
lim ln b2 1
dx
0
b
b
2 xe x dx
0
b
lim ln x 2 1
2x
x2 1
the integral
b
b
e x
lim
xe x dx
x
2
lim ln b2 1
0
0
b
A
lim ln b 2 1
b
0
b 2 x dx
diverges. But lim
67.
b
lim ln x 2 1
1
2
lim
b
e b
0 14
1
1
4
2
2
tan b
ln 1 sec
b
ln 1 0
sec2 x
sec2 x 1
2
ln 2
2
/2
72. (a) V
sec2 x dx
0
/2
0
0
lim
b
/2
0
/2
tan 2 x dx
0
sec 2 x tan 2 x dx
/2
0
dx
2
dx
2
/2
(b) Souter
/2
0
2 sec x 1 sec2 x tan 2 x dx
tan 2 b
0
2
lim
b
2 tan x sec2 x dx
0
2 sec x (sec x tan x ) dx
lim
b
tan 2 b
Souter diverges; Sinner
/2
0
tan 2 x
2
b
0
2 tan x 1 sec4 x dx
2
lim
b
/2
tan 2 x
2
b
0
lim
b
tan 2 b
0
2
diverges
Copyright
2014 Pearson Education, Inc.
lim
b
2
tan 2 b
Sinner
Section 8.8 Improper Integrals
1
625
1
dt
t (1 t )
73. (a)
0
With u
1
dt the limits of integration are unchanged.
2 t
1
1
dt
t (1 t )
0
1
t and du
0
2
du
1 u2
2 lim tan 1 1 tan
a
2
1
0
4
a
2
1
dt
t (1 t )
(b)
0
With u
1
dt the limits of integration are unchanged. We split the integral into two
2 t
integrals, the first of which was evaluated in (a).
t and du
1
1
dt
t (1 t )
0
0
2
1 u
2
2
du
1
2 lim tan
2
b
2
2
2
1
du
1 u2
b tan 1 1
2
74. Let c be any number in (3, ).
c
1
3
x x2 9
dx
3
1
x x2
1
dx
9
c
1
Formula 20 in Table 8.1 gives
dx
2
x x 9
Section 3.9.) Both integrals do converge:
c
3
1
x x2
dx
9
1
c
x x
2
3
dx
1
lim
1
sec
3
1
3
b
9
x x2
1
sec
3
a
1
Thus
lim
dx
9
6
dx provided both integrals on the right converge.
x x2
c
3
1
sec
3
1
b
3
1
sec
3
1
9
1
sec
3
1
x
. (The definition of the inverse secant is given in
3
a
3
1
sec
3
c
3
6
1c
3
1
sec
3
1c
3
.
Copyright
2014 Pearson Education, Inc.
626
Chapter 8 Techniques of Integration
75. (a)
3
e
3x
dx
b
1 e 3x
3
3
lim
b
x2
0.000042. Since e
3
replaced by
(b)
3
0
e
76. (a) V
x2
dx
e
x2
lim
1 e 3b
3
for x
3, then
3x
1 e 33
3
3
x2
e
dx
0
1
3
e
9
1e 9
3
0.0000411
0.000042 and therefore
0
e
x2
dx can be
dx without introducing an error greater than 0.000042.
0.88621
1
x
1
0
e
b
2
dx
lim
b
1 b
x 1
1
b
lim
b
1
1
(0 1)
(b) When you take the limit to , you are no longer modeling the real world which is finite. The comparison
step in the modeling process discussed in Section 4.2 relating the mathematical world to the real world
fails to hold.
77. (a)
(b)
78. (a)
(b)
int((sin(t))/t, t
0.. infinity);
answer is
f: 2*exp( t^2)/sqrt(Pi);
int(f , t 0..infinity); (answer is 1)
79. (a)
f ( x)
1
2
e
x 2 /2
f is increasing on (
, 0],
f is decreasing on [0, ),
f has a local maximum at 0, f (0)
0,
1
2
(b) Maple commands:
f: exp( x^2/2)(sqrt(2*pi);
int(f , x
1..1);
0.683
int(f, x
2..2);
0.954
int(f, x
3..3);
0.997
Copyright
2014 Pearson Education, Inc.
2
Section 8.8 Improper Integrals
(c) Part (b) suggests that as n increases, the integral approaches 1. We can take
we want by choosing n 1 large enough. Also, we can make
want by choosing n large enough. This is because 0
for x
n
1. ) Thus,
x /2
e
As n
dx
, 2e
f ( x) dx
n
c
lim
n
c
n /2
e
x /2
n
dx
c
n
b
(b)
a
f ( x) dx as close to 1 as
n
f ( x) dx and
e
n
x /2
f ( x) dx as small as we
for x 1. (Likewise, 0
f ( x)
e x /2
x /2 c
n
2e
lim
2e
c
c /2
n /2
2e
2e
n /2
f ( x) dx is as small as we want.
n
f ( x) dx is as small as we want.
f ( x) dx
a
b
f ( x) dx
a
f ( x ) dx,
f ( x) dx
b
a
f ( x) dx
b
a
f ( x) dx
f ( x) dx exists since f ( x) is integrable on every interval [a , b].
f ( x) dx
b
b
n
dx.
0, for large enough n,
80. (a) The statement is true since
a
x /2
lim
Likewise for large enough n,
and
e
f ( x)
n
627
a
f ( x ) dx
a
f ( x ) dx
a
b
f ( x ) dx
f ( x) dx
a
f ( x) dx
b
a
f ( x) dx
b
b
a
f ( x ) dx
f ( x) dx
b
a
f ( x) dx
f ( x) dx
81. Example CAS commands:
Maple:
f : (x,p) - x^p*ln(x);
domain : 0..exp(1);
fn_list : [seq( f (x,p), p -2..2 )];
plot( fn_list, x domain, y -50..10, color [red,blue,green,cyan,pink], linestyle [1,3,4,7,9],
thickness [3,4,1,2,0], legend ["p -2","p
-1","p
0","p 1","p
2"], title "#81 (Section 8.8)" );
q1: Int( f(x,p), x domain );
q2 : value( q1 );
q3 : simplify( q2 ) assuming p -1;
q4 : simplify( q2 ) assuming p -1;
q5 : value( eval( q1, p -1 ) );
i1: q1 piecewise( p -1, q4, p -1, q5, p -1, q3 );
82. Example CAS commands:
Maple:
f : (x,p) - x^p*ln(x);
domain : exp(1)..infinity;
fn_list : [seq( f(x,p), p -2..2 )];
plot( fn_list, x exp(1)..10, y 0..100, color [red,blue,green,cyan,pink], linestyle [1,3,4,7,9],
thickness [3,4,1,2,0], legend ["p -2","p
Copyright
-1","p
0","p 1","p
2014 Pearson Education, Inc.
2"], title "#82 (Section 8.8)" );
628
Chapter 8 Techniques of Integration
q6 : Int( f(x,p), x domain );
q7 : value( q6 );
q8 : simplify( q7 ) assuming p -1;
q9 : simplify( q7 ) assuming p -1;
q10 : value( eval( q6, p -1 ) );
i2 : q6
piecewise( p -1, q9, p -1, q10, p -1, q8 );
83. Example CAS commands:
Maple:
f : (x,p) - x^p*ln(x);
domain : 0..infinity;
fn_list : [seq( f(x,p), p -2..2 )];
plot( fn_list, x 0..10, y -50..50, color [red,blue,green,cyan,pink], linestyle [1,3,4,7,9],
thickness [3,4,1,2,0], legend ["p -2","p
-1","p
0","p 1","p
2"], title "#83 (Section 8.8)" );
q11: Int( f(x,p), x domain ):
q11 lhs(i1 i2);
`` rhs(i1 i2);
`` piecewise( p -1, q4 q9, p -1, q5 q10, p -1, q3 q8 );
`` piecewise( p -1, -infinity, p -1, undefined, p -1, infinity );
84. Example CAS commands:
Maple:
f : (x,p) - x^p*ln(abs(x));
domain : -infinity..infinity;
fn_list : [seq( f(x,p), p -2..2 )];
plot( fn_list, x 4..4, y -20..10, color [red,blue,green,cyan,pink], linestyle [1,3,4,7,9],
legend ["p -2","p
-1","p
0","p 1","p
2"], title "#84 (Section 8.8)" );
q12 : Int( f(x,p), x domain );
q12p : Int( f(x,p), x 0..infinity );
q12n : Int( f(x,p), x -infinity..0 );
q12
q12p q12n;
`` simplify( q12p q12n );
81-84.
Example CAS commands:
Mathematica: (functions and domains may vary)
Clear[x, f, p]
f[x_]: x p Log[Abs[x]]
int
Integrate[f[x], {x, e, 100)]
int / . p
2.5
Copyright
2014 Pearson Education, Inc.
Section 8.9 Probability
629
In order to plot the function, a value for p must be selected.
p 3;
Plot[f[x], {x, 2.72, 10}]
2/
1
dx
x
sin
85. Maple gives
0
by Ci(t )
t
2/
1
dx
x
x sin
0
t
0
8.9
1.
2.
3.
2
Ci
0.16462, where Ci is the cosine integral function defined
2
cos x
dx.
x
86. Maple gives
Si(t )
1
2
1
Si
2
2
2
4
0.10276, where Si is the sine integral function defined by
sin x
dx.
x
PROBABILITY
8
4
2
0
1
x dx
18
4
1; not a probability density.
3
1
(2 x ) dx 1; a probability density.
2
2x
ln 2
ln(1 ln 2)/ln 2 x
2 dx
0
ln(1 ln 2)/ln 2
0
1 ln 2
ln 2
1
ln 2
1
This is a probability density.
4.
x 1 is not nonnegative on [0,1
1
5.
1
x2
dx 1; a probability density.
8
6.
4 x2
0
3], so not a probability density.
dx
lim
b
4
2
4
x
tan 1
2
b
0
2
This is not a probability density.
7.
/4
0
2 cos 2 x dx 1; a probability density.
Copyright
2014 Pearson Education, Inc.
630
8.
Chapter 8 Techniques of Integration
e
0
1
dx diverges; not a probability density.
x
9. (a)
(b)
(c)
(d)
The probability that a tire lasts between 25,000 and 32,000 miles
The probability that a tire lasts more than 30,000 miles
The probability that a tire lasts less than 20,000 miles
The probability that a tire lasts less than 15,000 miles
3 /2
10. (a)
2
(b)
11.
3
1
2
15
14.
1
15
ln x 1
x
2
dx
x2
0.5
4e 3 2e 1 0.537
ln15 1
15 15
1
13
1 2
x ( x 3)
2
1/2
x(2 x ) dx
2
0
dx
0.682
3
( x 1)e x
ln x
2
1
/2 2
1
1
dx 1
2
xe x dx
12.
13.
1
dx
2
ln 2
2
11
16
1
2
0.599
0.688
Using software to evaluate the Sine Integral we find
sin 2 x
x2
200
1059
dx 1.00004780741 so the given function
is very nearly a probability density over the given interval. Again using software we find that
/6
sin 2 x
x2
200
1059
9
15.
4
18.
19.
x
/4
16.
17.
2
/6
c1
3
6
3
c
0
65
1296
sin x dx
x dx
c 11
c
dx
x
dx
4e 2 x dx
dx
0.6732.
0.0502
2
2
/4
cos x /6
1 2
c
12
3
2
3
1 2
. Solving
c
4
12
ln(c 1) ln c
2e 2c
ln
0.159
3
1, we find c
4
c 1
c 1
. Solving ln
c
c
2 . Solving
2e 2 c
Copyright
21 .
1, we find
2 1, we find c
c 1
c
1
ln 2 .
2
2014 Pearson Education, Inc.
e and thus c
1
e 1
.
Section 8.9 Probability
20.
5
0
1
c 25 x 2
3
cx 25 x 2 dx
3/2 5
3
.
125
125
c, so c
3
0
631
21. We will assume that the given function is to be a probability density over the whole real line.
c
1 x
22.
23.
24.
1
0
0
dx
2
c
so we take c
2 3/2
x
3
c
1
lim
cb
e cxdx
Var( X )
e bcx 1
2
X
( 2 X ) f ( X ) dX
4
mean
x
0
2
cx
( 2 X ) f ( X ) dX
2
X f ( X ) dX
2
f ( X ) dX
X 2 f ( X ) dX
1
x dx
8
3
mean
x
0
2
x
1 2
x dx
9
1
2
x3
dx
0
x
1
1
dx
x
1
2
1
4
1/2
1/4
15
x (1 x ) dx
4
0.10242 .
7
17
2
16
64
0.353
by c produces a probability density on [0, ).
f ( X ) dX
2
2
8
1 2
c
16
x dx
1
for c. Thus the median is
2
8.
9
4
lim
b
2
x
c1
0
9
x 2 dx
1 3
c
27
1
3
for c. Thus the median is 22/3
2
2
b
2
1
1
e
2
(1)
c1
c
mean
tan
.
To find the median we need to solve
28.
2
8
3
To find the median we need to solve
mean
x
1
15
. Then
4
4
c, so c
15
0
2
1
. Thus multiplying e
c
To find the median we need to solve
27.
1 x
dx
1
tan
f ( X ) dX
2
f ( X ) dX
Thus Var( X )
26.
1
. Then
1
2 5/2
x
5
X 2 f ( X ) dX
25.
2
1
c x (1 x ) dx
2
1
2
x
3
dx
1
c
2
1
1
for c. Thus the median is
2
2.
e 1 1.718
To find the median we need to solve
c
1
1
dx
x
Copyright
ln c
1
for c. Thus the median is
2
2014 Pearson Education, Inc.
e
1.649.
2.381.
632
Chapter 8 Techniques of Integration
29. The exponential density with mean 1 is e X . The probability that the food is digested in less than 30 minutes
1/2
is
0
X
e
dX
e
1/2
1 0.3935.
30. The exponential density with mean 4 is (1/ 4) e X /4 . The probability that a flower is pollinated within 5
minutes is
5
0
(1/ 4)e
X /4
dX
5/4
e
1 0.7135. Out of 1000 flowers we would expect 713 or 714 to be
pollinated within 5 minutes.
31. The exponential density with mean 1200 is (1/ 1200)e X /1200 .
1000
(a) The probability that a bulb will last less than 1000 hours is
0
(1/ 1200)e
X /1200
dX
e
5/6
1 0.5654.
(b) By Example 9 the median lifetime is 1200 ln 2 831.8 so the expected time until half the bulbs in a batch
fail is 832 hr.
1
3
. Then c
3
ln(3 / 2)
lifetime of the components is 7.4 years. The probability of failure within 1 year is
32. To find the density, solve
1
0
ln(3/2)
e
3
X ln(3/2)
3
3
0
(1/ c )e X / c dX
33. To find the density, solve
2
0
3/ c
1
2/ c
1
7.3989, so the mean
1/3
2
3
dX
e
1 0.1264.
(1/ c )e X / c dX
1/2
within 6 months, or half a year, is
0
e
ln(5/3)
e
2
X ln(5/3)
2
2
;c
5
2
. The probability that a hydra dies
ln(5 / 3)
3
5
dX
1/4
1 0.1199, so we would expect
(0.12)(500) 60 hydra to die within the first six months.
34. To find the density, solve
50
0
(1/ c )e X / c dX
e
50/ c
3
;c
10
1
80
risk driver is involved in an accident in the first 80 days is
0
50
. The probability that a highln(10 / 7)
X ln(10/7)
50
ln(10/7)
e
50
dX
8/5
7
1 0.4349, so we would expect 43 or 44 out of 100 high-risk drivers to be involved in an accident
10
in the first 80 days.
35. Using seconds as the time unit, the density is (1/ 30)e X /30 .
(a)
(b)
15
0
60
(1/ 30)e
X /30
dX
(1/ 30)e
X /30
dX
1/2
e
e
2
1 0.393
0.135
(c) In a continuous distribution the probability of a particular number is 0.
(d) The probability than a single customer waits less than 3 minutes is
e
6
1 0.997521. The probability
that at least one customer out of 200 waits longer than 3 minutes is 1 (0.997521)200
the most likely outcome is that all 200 are served within 3 minutes.
Copyright
2014 Pearson Education, Inc.
0.391
0.5, so
Section 8.9 Probability
633
36. For parts (a) and (b) the density is (1/ 16)e X /16 . For parts (c) and (d) the density is (1/ 32)e X /32 .
30
(a)
(1/ 16)e
X /16
dX
(1/ 16)e
X /16
dX
10
(b)
25
50
(c)
35
20
(d)
0
15/8
e
e
25/16
5/8
e
0.382
0.210
(1/ 32)e
X /32
dX
e
25/16
(1/ 32)e
X /32
dX
e
5/8
e
35/32
0.125
1 0.465
37. The expected payout per printer is 200
1
0
(1/ 2)e
X /2
dX 100
2
1
X /2
(1/ 2)e
dX
$102.56. Thus the
expected refund total for 100 machines is $10,256.
2
38. To find the density, solve
1
in the first year is
0
ln 2
e
2
1
e
c
0
X ln 2
2
X /c
dX
e
2/ c
1
1
, which gives c
2
2
. The probability of failure
ln(2)
2
1 0.293. We expect (150)(0.293) 43.934 or about 44 copiers
2
dX
to fail during the first year.
1
e
2
For Exercises 39 52, the density function is f ( X )
39.
162,
193
165
167
148
40.
41.
f ( X ) dX
0.323; about 323 children
f ( X ) dX
0.262; about 262 children
f ( X ) dX
55,
60
0
42.
with
and
4.7
1
2
17
f ( X ) dX
f ( X ) dX
0.74593
18,000
0.89435
4000
1
2
f ( X ) dX
18,000
(b) We want to find L such that
f ( X ) dX
L
0.84134; (4000)(0.84134) 3365 tires
f ( X ) dX
0.9. A CAS gives L 16,874, so 90% of tires will have a
lifetime of at least 16,874 miles.
43.
65.5,
(a)
(b)
68
64
61
as given in the solution.
4
22,000,
(a)
2
28
20.11,
17
1 x
2
2.5
f ( X ) dX
1
2
f ( X ) dX
0.23832
68
f ( X ) dX
0.159, or 16%.
Copyright
2014 Pearson Education, Inc.
634
Chapter 8 Techniques of Integration
44.
81,
85
7
0.520; one would expect 52 of the babies to live to between 75 and 85.
f ( X ) dX
75
45.
266,
280
16; (36)(7)
280
0.6184; we would expect 618 of the women to have pregnancies lasting between 36 and 40
f ( X ) dX
252
252, (40)(7)
weeks.
46.
1400,
1450
(a)
f ( X ) dX
1325
0.46484
1480
1
f ( X ) dX 0.21186
2
(500)(0.21186) 106; we would expect about 106 males to have a brain weight exceeding 1480 gm.
(b)
47.
100
1480
f ( X ) dX
80,
70
12
f ( X ) dX
0
0.20233
(300)(0.20233) 61; about 61 adults.
48.
4.4,
4.45
4.3
49.
0.2
f ( X ) dX
35,
0.29017
9
40
1
f ( X ) dX 0.28926
40
2
About 289 shafts would need more than 45 grams of added weight.
f ( X ) dX
50.
200,
260
170
0
30
f ( X ) dX
1
2
f ( X ) dX
0.15866
260
0.02275 0.15866
51.
(0.8)(2500)
(a)
(b)
(c)
1960
2000,
(0.4)(50)
1
2
f ( X ) dX
0.159
f ( X ) dX
0.840
1980
2020
0.02275
0.18141 or about 18%
f ( X ) dX
0
1940
f ( X ) dX
1960
20
f ( X ) dX
Copyright
0.977
2014 Pearson Education, Inc.
Section 8.9 Probability
52 .
20
10
2
To improve the approximation to the binomial distribution we will modify the interval of integration. We give
the true binomial values for comparison.
(0.5)(400)
209.5
(a)
189.5
169.5
(b)
0
(c)
220.5
200,
f ( X ) dX
0.68208; correct to 5 places
f ( X ) dX
0.00114; true value
f ( X ) dX
1
2
220.5
f ( X ) dX
0.00112
0.02018; true value
(d) The value is very close to 0, in fact about 10
53.
635
24
0.02012
.
(a) and (b)
Outcome
X
HHHH
0
THHH
1
HTHH
1
HHTH
1
HHHT
1
TTHH
2
THTH
2
THHT
2
HTTH
2
HTHT
2
HHTT
2
TTTH
3
TTHT
3
THTT
3
HTTT
3
TTTT
4
(c)
The probability of at least 2 heads is
P
0.4
0.3
0.2
0.1
0
1
2
3
4
X
Copyright
2014 Pearson Education, Inc.
1
1 4 6
16
11
.
16
636
Chapter 8 Techniques of Integration
54. (a) The first die is listed first in each pair, the second die second.
1+1=2
2+1=3
3+1=4
4+1=5
5+1=6
6+1=7
1+2=3
2+2=4
3+2=5
4+2=6
5+2=7
6+2=8
1+3=4
2+3=5
3+3=6
4+3=7
5+3=8
6+3=9
1+4=5
2+4=6
3+4=7
4+4=8
5+4=9
6 + 4 = 10
1+5=6
2+5=7
3+5=8
4+5=9
5 + 5 = 10
6 + 5 = 11
1+6=7
2+6=8
3+6=9
4 + 6 = 10
5 + 6 = 11
6 + 6 = 12
(b)
P
0.2
0.1
2 3 4 5 6 7 8 9 10 1112
P (8)
5
36
(d) P ( X
5)
(c)
X
1 2 3 4 5
36
18
3 2 1 1
9)
36
6
P( X
55. (a) {LLL, LLD, LDL, DLL, LLU, LUL, ULL, LDD, DLD, DDL,LUU, ULU, UUL, DDU, DUD, UDD,
DUU, UDU, UUD,LUD, LDU, ULD, UDL, DLU, DUL, DDD, UUU}
(b) We will assume that the three answers are equally likely, though other assumptions might be reasonable.
The plots for the X number of Ls, the number of Us and the number of Ds are identical.
P
0.5
0.4
0.3
0.2
0.1
0
1
2
3
X
1 3 3 7
0.26
27
27
(d) P (no more than one D) 1 P(at least two D)
7 20
1
0.74
27 27
(c)
P(at least two L)
Copyright
2014 Pearson Education, Inc.
Chapter 8 Practice Exercises
637
56. The probability that both systems fail is 0.0148. Since the two systems have the same performance
distribution, the failure probability for a single system is 0.0148 0.121655. The success probability for a
0.0148 so the probability that both succeed is 1
single system is 1
0.0148
2
0.771489. The
probability that one fails and one succeeds is 1 0.0148 0.771489 0.213711. Since the events main fails,
backup succeeds and main succeeds, backup fails have the same probability, the probability that only the
main fails is 0.213711/ 2 0.106856. Thus the probability that the main fails, either along with the backup or
by itself, is 0.0148 0.106856 0.121656.
CHAPTER 8
1. u
PRACTICE EXERCISES
dx ;
x 1
ln( x 1), du
ln( x 1) dx
dv
x
x ln( x 1)
x 1
( x 1) ln( x 1) x C1
2. u
x 2 ln x dx
tan
1
tan 1 3x dx
4. u
cos
1 x
2
x cos
x tan
1 x
2
cos
1 9 x2
1
dx
1 x
2
; dv
3x
dx
, du
4 x2
x cos
2 1
x ln( x 1)
dx
dx
x 1
x ln( x 1) x ln( x 1) C1
( x 1) ln( x 1) ( x 1) C , where C
3 x dx
1 9x
2
x3
9
C
y 1 9 x2
dy 18 x dx
;
dx, v
4 x
ln x
x;
x dx
1 x
2
x 2
2
x3
3
dx
dx, v
; dv
C1 1
1 x3 ;
3
1 x3 1
x
3
3 dx
3x, du
x;
dx
x 2 dx, v
dx ; dv
x
1 x3 ln x
3
ln x, du
3. u
dx, v
2
;
x tan
1
1
6
dy
y
1 x
2
1
2
3x
x tan 1 (3x)
( x 1) 2
( )
ex
2( x 1)
( )
ex
2
( )
ex
y
dy
4 x2
2 x dx
x cos
dy
y
x cos
1 x
2
C
( x 1) 2 e x dx
0
( x 1) 2
2( x 1) 2 e x
C
sin(1 x)
6.
2
( )
2x
( )
2
( )
x
0
1 9 x2
C
4 x2
C
x;
ex
5.
1 ln
6
cos(1 x)
sin(1 x)
cos(1 x)
x 2 sin(1 x) dx
Copyright
x 2 cos(1 x ) 2 x sin(1 x) 2 cos(1 x) C
2014 Pearson Education, Inc.
638
Chapter 8 Techniques of Integration
7. u
cos 2 x, du
e x cos 2 x dx
I
e x dx, v
2sin 2 x dx; dv
ex ;
e x cos 2 x 2 e x sin 2 x dx;
e x dx, v
2 cos 2 x dx; dv
ex ;
u
sin 2 x, du
I
e x cos 2 x 2 e x sin 2 x 2 e x cos 2 x dx
e x cos 2 x 2e x sin 2 x 4 I
e x cos 2 x
5
I
2e x sin 2 x
5
C
x sin x cos x dx
8.
u
x, du
dx, dv
u
cos 3 x, du
3sin 3x dx; dv
x sin x cos x dx
1 2
sin x
2
sin x cos x dx, v
uv
e 2 x dx, v
1
sin 2 x dx
2
x 2
sin x
2
v du
1 e 2x ;
2
1
x 2
sin x
1 cos 2 x dx
2
4
1
1
x 2
sin x
sin 2 x C
x
2
4
8
9.
10.
x dx
x dx
dx
x ( x 1)2
12.
x 1
x 2 ( x 1)
14.
15.
16.
17.
3
2
x2 4 x 3
11.
13.
2 dx
x 2
x2 3x 2
dx
x 3
1
x
2
1 ln cos
3
cos
1
2
cos d
sin
6
3x2 4 x 4
x3 x
dx
4 x dx
4 dx
x2 4
(v 3) dv
1
2
2v
8v
2 ln | x 2 | ln | x 1| C
dx
x 1
1
( x 1)2
2
x
1
x2
3 ln |
2
x 3|
dx
ln |x| ln | x 1|
dx
2ln
x 1
x
dy
y]
y2 y 2
1 ln |
2
x 1| C
1
x 1
C
1
x
C
2 ln |x|+ 1x 2ln | x 1| C
1
3
dy
y 1
1
3
dx
x 2
1
5
dx
x 3
dy
y 2
1 ln y 2
3
y 1
C
1 ln cos
3
cos
2
1
C
C
; [sin
x3 4 x
3
1
x 1
; [cos
cos
sin 2
1
2
2
x 1
dx
sin d
cos 2
dx
x 1
4
x
dx
2 tan
3
4v
dx
x2 x 6
x]
x 4
x2 1
1 x
2
5
8(v 2)
dx
1
5
4 ln | x |
1 ln
2
x2 1
1 ln sin
5
sin
4 tan
1
2
3
C
x C
C
1
8( v 2)
dv
Copyright
3 ln | v |
8
5
6
16
ln | v 2 |
1 ln | v
16
2014 Pearson Education, Inc.
2| C
5
1 ln (v 2) (v 2)
16
v
C
Chapter 8 Practice Exercises
18.
(3v 7) dv
(v 1)(v 2)(v 3)
19.
dt
t 4 4t 2 3
20.
1
2
t dt
1
3
t4 t2 2
( 2) dv
v 1
dt
t2 1
1
2
dt
t2 3
t dt
1
3
t2 1
t2 2
21.
x3 x 2
x2 x 2
22.
x3 1
x3 x
23.
x3 4 x 2
x2 4 x 3
24.
2 x3 x 2 21x 24 dx
x2 2 x 8
x 2 3 x 23 ln | x
25.
dx
dx
x3 x 1
2
x 1
x3 x
1
dx
x
;
26.
dx
x1 3x
;
ds
s
e 1
;
(b)
1 ln |
3
du
dx
2 x 1
dx
2u du
3
dx
x 1
dx
x 2
dx
x2
2
dx
x 1
dx
x 1
1 tan 1 t
2
C
3
6
tan
1 t
3
C
C
dx
3
2
x
x2 2 x 8
t
3
4
3
dx
x dx
1
t2 1
1 ln
6
2
3
1
x ( x 1)
tan
9
2
dx
x 3
4 ln |
3
2 ln |
3
x 2|
x 1| C
dx
x
x ln | x 1| ln | x | C
x2
2
9 ln |
2
dx
x 2
1
3
(2 x 3) dx
3 ln |
2
x 3|
2
3
x 1| C
dx
x 4
x 2| C
u du
2
3
1
3
u2 1 u
du
u 1
1
3
du
u 1
1 ln | u
3
1| 13 ln | u 1| C
1 ln x 1 1
31 1 x
x
3u 2 du
dx
3 x 2/3
2
3 u (1duu )
3
u (1 u )
3ln uu 1
C
3ln
3
1
x
3
x
C
3u du
du
e s ds
ds
du
u 1
du
ds
29. (a)
dx
4|
1
2 3
2
C
(v 1)2
es 1
u
28.
3x
x2 4 x 3
(v 2)( v 3)
ln
x dx
(2 x 3)
du
u
ds ;
es 1
dx
1
x 1
dx
27.
1 ln t 2
6
dx
u
u
dv
v 3
1 tan 1 t
2
t dt
2x
x x 2
x
dx
dv
v 2
2u du
s
u u
2 e 1
2u du
du
u
ln uu 1
C
s
ln e s 1
e
ln |1 es
C
| C
2
1
du
2 (u 1)(
u 1)
du
u 1
du
u 1
lns uu 11
C
s
ln e 1 1
e 1 1
u2 1
y dy
16 y 2
du
u 1
es 1
1
2
16 y 2
y dy
du
u (u 1)
e s ds
;[y
2 y dy
16 y 2
16 y 2
4sin x]
639
4
sin x cos x dx
cos x
Copyright
C
4cos x C
4 16 y 2
4
C
2014 Pearson Education, Inc.
16 y 2
C
C
C
640
Chapter 8 Techniques of Integration
30. (a)
x dx
(b)
x dx
4 x
4 x
x dx
4 x
2
t dt
32. (a)
4t
4t
x dx
9 x
2
dx
x 9 x2
35.
dx
9 x2
36.
dx
;
2
1
1 sec
2
1 sec
2
1
2
2 x dx
dx
x
dx
3 x
x
dx
du
u
1 dx
18 3 x
1 dx
18 3 x
dx
3 x
1 ln | 3
6
1
6
sin
3cos
3cos
3cos d
tan d
2
tan 12 sec d
tan
1 ln | u |
2
1 ln |
9
d
cos5 x sin 5 x dx
sin 5 x cos 4 x cos x dx
tan 3 x sec3 x dx
sec5 x
5
41.
sec3 x
3
u
C
sec 2
d
C
ln
tan
4
ln
C
1
C
4 x2
2
C
C
1 ln |
9
1 ln 9 x 2
x | 18
C
C
sin 1 3x C
sin 5 x 1 sin 2 x
sin 9 x cos x dx
4t 2 1
4
C
9 x2
1 ln x 3
6
x 3
cos 4 x sin x dx
sin 6 x
6
2
cos 6 x sin x dx
cos5 x
5
cos7 x
7
C
cos x dx
2sin8 x
8
sin10 x
10
C
C
sec 2 x 1 sec2 x sec x tan x dx
sec4 x sec x tan x dx
sec2 x sec x tan x dx
C
sin 5 cos 6 d
1 cos
2
tan 5 x
5
ln 1
C
d
38.
sin 5 x cos x dx 2 sin 7 x cos x dx
1
4
x | 16 ln | 3 x | C
cos 4 x 1 cos 2 x sin x dx
40.
ln | cos | C
1 ln | 3 x | 1 ln | 3 x | C
x | 18
18
sin 3 x cos 4 x dx
tan 4 x sec2 x dx
4 x2
C
37.
39.
2sec y C
4t 2 1 C
1
4
9 x2
1
9
1
6
4t
t
du
34.
2
1
u
;
9 x
2
4 cos
2 sec y tan y dy
C
8t dt
;
4 x2
2 sin 2 cos d
2sin ]
1
8
1
t dt
(b)
33.
2
1 ln
2
4 x2
4 x2
C
2 tan y 2sec2 y dy
2sec y
( 2 x) dx
; [x
1 ln
2
4 x2
2
2 tan y ]
1
2
4 x2
(b)
4 x
; [x
2
x dx
31. (a)
2 x dx
1
2
2
1 cos11
22
1
2
sin(
) sin(11 ) d
1
2
sin(
)d
1
2
sin(11 ) d
C
Copyright
2014 Pearson Education, Inc.
1 cos(
2
1 cos11
) 22
C
Chapter 8 Practice Exercises
42.
sec2 sin 3 d
43.
1 cos 2t dt
44.
et tan 2 et 1 dt
sin (1 cos2 )
cos2
sin
cos2
2 cos 4t dt
4 2 sin 4t
C
ln sec et
tan et
sec et et dt
3 1 ( x ) 4 M where
45. | Es | 180
f (4) ( x)
f (4) (1)
24
M
47.
x
1
x
f ( x)
24. Then | Es | 0.0001
n4
768
10, 000 180
46. | ET | 1120 ( x )2 M where
2
3n 2
2;
n
3n 2
2
10 3
b a
n
6
0
1 0
n
x
n2
1000
x
2
6
12
1;
n
0
2000
3
n
T
i 0
12
(12)
;
( cos ) C
x 1
x 2
f ( x)
i 0
mf ( xi ) 18 and 3x
S
18
(18)
3 1
180
2
n
4
(24)
0.0001
768
180
1
n4
n 16 (n must be even)
f ( x) 8
M
25.82
n
8. Then | ET | 10 3
C
2x 3
f ( x)
6x 4
0.0001
1 1
12 n
2
(8) 10 3
26
xi
f ( xi )
m
mf ( xi )
x0
0
0
1
0
x1
/6
1/2
2
1
x2
/3
3/2
2
3
x3
/2
2
2
4
x4
2 /3
3/2
2
3
x5
5 /6
1/2
2
1
0
1
0
xi
f ( xi )
m
mf ( xi )
x0
0
0
1
0
x1
/6
1/2
4
2
x2
/3
3/2
2
3
x3
/2
2
4
8
x4
2 /3
3/2
2
3
x5
5 /6
1/2
4
2
0
1
0
x6
Copyright
f ( x)
n 14.37
18
.
cos
maximum of f (4) ( x) on [1, 3] is
x6
6
sec
C
;
6
mf ( xi ) 12
cos 1
sin d
24 x 5 which is decreasing on [1, 3]
(0.0001) 180
768
1
n4
3 1
n
x
d
641
2014 Pearson Education, Inc.
642
48.
Chapter 8 Techniques of Integration
f (4) ( x)
n
49.
3
M
6.38
n
365
1
365 0 0
yav
3;
2 1
n
x
2 ( x 101)
37 sin 365
2 (264)
cos 365
37
2
(0.16705 0.16705) 25
675
1
675 20 20
mf ( xi ) 1211.8
1
12
55.
dx
9 x
0
1
2
2 (264)
cos 365
1
655
dT
ln x dx
dy
1 y 2/3
3
b
lim
b
0
0 dy
1 y 2/3
105
60
b
0
dx
9 x
2
1.87T 2
26T
283, 600
0
1 [2.5 2(2.4) 2(2.3)
hour. 24
x
3
x 15
0
25(0)
25
675
0.62333 T 3
5
10
20
5.434;
26
T
676 4(1.87)(283,600)
2(1.87)
29
12
2(2.4) 2.3]
5;
396.45 C
2.42 gal
6059 ft
2
xi
f ( xi )
m
mf ( xi )
x0
0
0
1
0
x1
15
36
4
144
x2
30
54
2
108
x3
45
51
4
204
x4
60
49.5
2
99
x5
75
54
4
216
x6
90
64.4
2
128.8
x7
105
67.5
4
270
x8
120
42
1
42
lim sin 1 b3
3
sin 1 30
2
(1211.8)(5)
2
6059 ft ;
$2.10/ft
b
lim sin 1 3x
0
b 3
1
x ln x x b
1 dy
0 y 2/3
365
25 x
2 ( 101)
cos 365
13 T 2
105
8.27T
2
the job cannot be done for $11, 000.
lim
n4
24.83 mi/gal
Area
The cost is Area $2.10/ft
$12, 723.90
10 5
1
60 n4
25 F
5.434
52. Using the Simpson s rule,
1
37
2
(5582.25 59.23125 1917.03194) (165.4 0.052 0.04987)
hours/gal
(b) (60 mph) 12
29
54.
(3) 10 5
2 (0 101)
37 365
cos 365
2
25(365)
8.27 10 5 26T 1.87T 2
51. (a) Each interval is 5 min
3
4
1
n
2 ( x 101)
37 365
cos 365
2
1
365
25 dx
2 ( 101)
25 237 cos 365
8.27 10 5 26T 1.87T 2
0
2 1
180
8 (n must be even)
37
2
50. av(Cv )
53.
Hence | Es | 10 5
2 (365 101)
37 365
cos 365
2
1
365
1
655
1.
n
2
lim
b
0
1 dy
0 y 2/3
(1 ln1 1)
2 3 lim
Copyright
b
0
b
b ln b b
y1/3
1
b
1
lim ln1b
b
0
6 lim 1 b1/3
b
2
0
2014 Pearson Education, Inc.
b
6
0
1
2
lim
b
0
1
b
1
b2
1 0
1
Chapter 8 Practice Exercises
56.
1
d
1)3/5
2(
d
and
57.
2 du
3 u 2 2u
58.
3v 1
1 4v3 v 2
du
3 u 2
dv
2(
1
d
1)3/5
d
1)3/5
2 (
lim ln u u 2
3
b
1
v2
4
4v 1
converges if each integral converges, but lim
3/5
1)3/5
(
1
diverges
b
du
3 u
1
v
d
1)3/5
1(
diverges
3/5
2
2
d
1)3/5
2(
643
lim ln b b 2
ln 3 32
b
b
lim ln v 1v ln(4v 1)
dv
ln 3
lim b ln 4b 1
1
b
0 ln 13
1
b
b
(ln1 1 ln 3)
ln 14 1 ln 3 1 ln 34
59.
x 2 e x dx
0
0
60.
dx
4 x2 9
1
2
63.
lim
64.
I
2
3 2
2
67.
1
dx
0 4 x2 9
0
2
ln z
1 z
t
t
x
1 and
2
b
b
b2e b
lim
0
b
b e3b
3
1
9
lim
2be b
1 e3b
9
b
2 tan 1 2 x
3
3
1 lim
2b
2I
b
2 lim tan 1 4x
2e b
1
9
( 2)
0 2
2
1
9
0
2 tan 1 2b
3
3
1 lim
2b
0
d
6
diverges
I
1
2
ln z
e z
dz
1
dz
2 dx
0 ex e x
d
2
0
0
b
1
1
2 tan 1 (0)
3
ex
2
0
2
diverges
e u sin u du 1
b
lim e u sin u
0
b
0
e u cos u du
converges
(ln z ) 2
2
e
lim
1
b
e t dt converges
4 dx
0
tan 1 (0)
b
6
b
e u cos u
lim
2 lim tan 1 b4
0
b
e t for t 1 and
2 dx
ex e
b
2e x
0
1 e3 x
9
dx
1
2 0 x2 9
4
4 dx
0 x 2 16
e ln z
1 z
dz
2 xe x
6
1 0 I
e
x e3 x
3
lim
e u cos u du
0
I
66. 0
b
2
4 dx
x 2 16
62.
65.
b
xe3 x dx
61.
x 2e x
lim
converges
Copyright
b
(ln z )2
2
e
1
t
t
2 dx
ex e
e
12
2
0
lim
b
dt converges
x
converges
2014 Pearson Education, Inc.
(ln b )2
2
1
2
diverges
644
Chapter 8 Techniques of Integration
1
dx
x 1 ex
68.
2
1
x2
lim
x
0
0
dx
x 1 ex
lim
x
1
x2 1 e x
lim 1 e x
x2
0
1
dx
0 x2 1 e x
dx
1 x2 1 ex
2
x
dx
1 x2 1 e x
1 dx
0 x2
2 and
0
;
1
dx
0 x2 1 e x
diverges
diverges
dx
x2 1 e x
x2 1 e x
diverges
69.
x dx
;
1 x
2 x3/ 2
3
70.
71.
x3 2
4 x2
u
x
du
dx
2 x
u 2 2u du
1 u
x 2 x 2 ln 1
dx
x
4x 2
x2 4
x
2u 2
u2
2u 2 ln |1 u | C
C
x dx 23 xdx2
dx
2x x 2 dx ; x 1 sin
2 u3
3
2u 2 1 2u du
dx
5
2
x2
2
dx
x 2
3 ln |
2
x 2 | 25 ln | x 2 | C
cos d
For the integrand to be nonnegative x must be between 0 and 2 so x 1 is between
between
2 and / 2, where cosine is nonnegative. Thus
2x x2
1 ( x 1)2
2 x x 2 dx
cos cos d
1
2
1
sin
2
72.
73.
dx
1
sin 2
4
1
dx
2 x x2
2 cos x sin x
sin 2 x
1 ( x 1)2
( x 1)
cos2
1
2
cos 2 d
1
2
C
cos .
1 cos 2
1
sin cos
2
1
( x 1) 2 x x 2
2
d
C
C
sin 1 ( x 1) C
2csc 2 x dx
dx
1 sin 2
cos x dx
2 cot x sin1 x ln | csc x cot x | C
csc x dx
sin 2 x
2cot x csc x ln | csc x cot x | C
74.
sin 2 cos5 d
sin 2 cos5 d
sin 2
1 sin 2
sin 2
2
1 and 1 and we can take
cos d
1 sin 2
2
cos d ; u sin
sin 2
u2
sin 3
3
2sin 4
2u 4
u 6 du
2 5
sin
5
Copyright
sin 6
u3
3
1 7
sin
7
du
cos d
cos d
2 5
u
5
1 7
u
7
C
C
2014 Pearson Education, Inc.
Chapter 8 Practice Exercises
9 dv
75.
dv
v2 9
1
2
81 v 4
dx
2 ( x 1)2
76.
1 b
1 x 2
lim
b
77.
cos(2
( )
1
12
dv
3 v
lim
1
1 b
b
1 cos(2
4
79.
80.
81.
sin 2 d
/4
84.
85.
2 x
dv; [v
1
(2 y ) dy
dx
y
x
2 2
3
1 v2
v2
1 v2
v
v
dy
( y 1)2 1
x dx
z 1
z2 z2 4
sin 2 x
4 y1/2
C
1) C
2 x 3ln | x 1|
1
x 1
C
/2
/4
2
2
x)3/2
2 (2
3
4(2 x)1/2
x
4
dz
1
2
1
4.
1 sin 2
cos cos d
sin
2
sin
d
2
csc 2
d
d
cot
C
dy
8 2x
2 y 3/2
3
2
2
x2
2
dx
( x 1) 2
1 sec 2
4
C
(2
C
C
sin ]
y2 2 y 2
2
/4
cos 2 x dx
1 cos
4
1)
3
2 x
sin
83.
y
dy
x dx
;
2 x
/2
2
sin (2
2
( x 2) dx 3 xdx1
1
2 1 cos 2
2
1 cos 2
1 cos 4 x dx
2
82.
1) d
dx
( 2 sin 2 ) d
1
2
2
1 cos 2
/2
3x 2
x2 2 x 1
x 2
x2 2 x 1
0 1 1
1)
cos(2
x3 dx
( 1)
C
1)
0
78.
1 tan 1 v
6
3
1 ln 3 v
12
3 v
1)
1 sin(2
2
( )
1
dv
3 v
1
12
tan 1 ( y 1) C
2 x dx
9
1
z
x2 1
1
z2
2
1 sin 1 x 2 1
2
3
z 1
z2 4
dz
1 ln |
4
Copyright
C
z|
1
4z
1 ln
8
z2
4
1 tan 1 z
8
2
2014 Pearson Education, Inc.
C
C
C
645
646
Chapter 8 Techniques of Integration
86.
x 2 ( x 1)1/3 dx ;
u
x 2 ( x 1)1/3 dx
u2
x 1 du
x2
dx
(u 1)2
2u 1 u1/3du
u 7/3 2u 4/3 u1/3 du
3 10/3 6 7/3 3 4/3
u
u
u
C
10
7
4
3
6
3
( x 1)10/3
( x 1)7/3
( x 1)4/3 C
10
7
4
t dt
87.
9 4t
1
x
et dt ;
e2t 3et 2
90.
tan 3 t dt
ln y dy
y
lim
b
92.
dx
1 x2
; dv
9 4t 2
C
dx , v
x2
1;
x
tan 1 x ln | x | 12 ln 1 x 2
89.
1
1
4
9 4t 2
tan 1 x, du
88. u
91.
8t dt
1
8
2
3
;
[et
x]
x
ln y
dx
dy
y
dy
e x dx
b
2e2 b
1
4e 2 b
y 3/2 (ln y )2 dy ;
0 14
u
Now we compute
dx
(ln y )2 , du
2x
dx
tan 2 t
2
ln xx 12
ln | sec t | C
lim
b
x e 2x
2
2ln y
dy , dv
y
y 3/2 dy, v
u
ln y , du
1
dy , dv
y
1 e 2x
4
b
0
2 5/2
y
5
y 3/2 dy , v
2 3/2
y dy
5
4 5/2
y
5
Copyright
x dx
dx
x
2
1 x
C
ln | x 1| ln | x 2 | C
tan t dt
1 tan 1 x
x
dx
x 1 x2
ln | x | ln 1 x 2
dx
x 2
xe
tan 1 x
4 3/2
y ln y dy
5
y 3/2 ln y dy :
2 5/2
y ln y
5
2 5/2
y ln y
5
0
1
x
1
4
2 5/2
y (ln y )2
5
y 3/2 (ln y )2 dy
y 3/2 ln y dy
x ex
0 e3 x
x dx
x2
dx
x 1
tan 2 t
2
(tan t ) sec 2 t 1 dt
1
tan 1 x
x
C
dx
( x 1)( x 2)
tan
2014 Pearson Education, Inc.
2 5/2
y
5
C
t
ln et 1
e
2
C
Chapter 8 Practice Exercises
2 5/2
y (ln y )2
5
2 5/2
y (ln y )2
5
y 3/2 (ln y )2 dy
2
(ln y )2
5
y 5/2
93.
eln
x
94.
e
3 4e d ;
95.
2
1 cos 5t
dv
e2v 1
dr
1
r
u
u
r
du
dr
2 r
x3
1 x2
dx
x
100.
x2
1 x3
dx
1 3x 2
3 1 x3
101.
1 x2
1 x3
dx;
du
x
1 x2
dx
2x
1
2 1 x2
x dx
x3
dx
1 ln 1
3
A
1 x
Bx C
1 x x2
(1/3) x 1/3
2/3
1 x
1 x2
1
3
dx
x
dx
1 2
2
u
3
4
3
2
2
u
du
1
3
x 1
1
3 1 x x2
1
3
u
u2
3
4
x
tan 1
1
2
3/2
dx
C
2 r
r
2 ln 1
C
C
1 x2
2
1 ln
2
A 1 x x2
A B 1,
2
1
3 1 x
dx
1 x x2
C
C
2u 2ln |1 u | C
dx
3/2
3 4e
1 tan 1 (cos 5t )
5
C
sec 1 ev
9
1
6
1 x2
C
C
( A B C) x ( A C)
1
2
2
1
3 1 x
ln x 4 10 x 2
x 4 10 x 2 9
1 x2
1 x3
2
dx
1 ln 3
6
4
2 1 2u du
4 x3 20 x dx
99.
x
2u du
1 u
dx
1 tan 1 u
5
sec 1 x C
x x2 1
u )3/2 C
1 2 (3
4 3
3 u du
du
1
5 1 u2
dx
4 x3 20 x
x 4 10 x 2 9
u
1
4
4e d
ev dv
dx
98.
1 x2
1 x3
4e
du
C
C
ev
x
( A B) x
8
16
ln y
25
125
u cos 5t
du
5sin 5t dt
;
;
;
2 x3/2
3
x dx
sin 5t dt
96.
97.
dx
4 3/2
y ln y dy
5
4 2 5/2
4 5/2
y ln y
y
5 5
5
2 ln |1
3
3
4
1 ln 1
6
x|
Copyright
A B C
x 1
1
3 1 x x2
dx
1
2
du
( Bx C )(1 x)
1
u2
du
2
1
3 1 x
dx
1
3
1 ln 3
6
4
u2
1
3
tan 1
x x2
1
3
3
4
tan 1 2 x 1
3
1
3
tan 1 2 x 1
3
2014 Pearson Education, Inc.
2,
3
A
x 1
dx
x x2
1 ln 1
6
0, A C 1
C
x
1 2
2
u
3/2
B
dx;
1, C
3
1;
3
647
648
Chapter 8 Techniques of Integration
1 x2
(1 x )3
102.
u 1 x
du dx
dx;
ln |1 x | 1 2x
x 1
103.
1 (u 1)2
u
1
(1 x )2
u 2 2u 2
u3
x
2 w2 1 w dw
1
u
du
2
u2
du
2
u3
du
ln | u | u2
32 (1
105
w)7/2 C
du
1
u2
C
C
w
x
w2
2 w dw dx
x dx;
du
3
1 w
( )
2
2w
w)3/2
2 (1
3
4w
( )
4 (1
15
w)5/2
4
( )
8 (1
105
w)7/2
2 w2 1 w dw
0
4
3
104.
x 1
x
3/2
16
15
w
1 x dx;
1
x 1
x
1 x
2w dw
4 w2 (1 w)3/2
3
7/2
5/2
32
105
w2
dx
1 x
x
1
C
2 w 1 w dw;
u
4 w(1 w)3/2
3
3/2
8
2 w 1 w dw
4
3
1 x 1
1
x 1 x
105.
2
1 u
106.
1/2
/6
15
d
x
2
2sec d
2
2
, dx
1 cos cos d
2 cos 6 1 cos 6
lim
3 1
c
0
u
2 ln sec
/6 1 cos 2
1 cos
1 cos
tan ,
1 w dw, v
2 (1
3
3
2
1/2
c
1/2
/6
c
cos d
2
sec2
, du
2 ln 1 u 2
2 1 cos
1/2
Copyright
0
0
u
C
4
3
1/2
1
2
0, x
sin
/6 sin cos
1 cos
d
sin
1 cos
sin d , dv
2cos c 1 cos c
2 cos c 1 cos c
w)3/2
w)5/2 C
8 (1
15
2
C
tan
cos , x
cos , du
/6
1/2
lim
0
du;
cos d , 1 x 2
0
2 cos
c
w)3/2
2dw, dv
d , 1 u2
sec
2 ln 1 x
x
1 x 2 dx;
lim
0
4 w(1
3
2 w, du
C
1 u2
u
c
5/2
1 x
1
x
u2
2u du dx
2sec 2
sec
du
sin ,
0
u
dx;
1
0
x
2
1 x
w)3/2 dw
4 (1
3
w)5/2
16 w(1
15
lim
c
0
c
d ,v
sin d
1/2
1 cos 6
4
3
1 cos
3/2
4
3
/6
3/2
c
1 cos c
2014 Pearson Education, Inc.
3/2
sin
/6 sin cos
1 cos
2 1 cos
6
d ;
1/2
C
Chapter 8 Practice Exercises
lim
0
c
107.
1/2
3
2
3 1
ln x
x x ln x
4
3
1
x ln x ln ln x
dx;
109.
x ln x ln x dx;
x
u
ln x 1
x
du
111.
x 1 x
cos
sin cos
1 x
x
dx; u
A
u 1
2
2
u2 1
113. (a)
a
0
ln u
C
dx;
u
ln x
ln x
ln x 1
x
dx
1 x
1
2
d
du
u2
0
sin x
sin x cos x
/2
1 x
1
u 1
2 du
f (a x) dx; u
/2
ln x
dx
0
cos x
cos x sin x
dx
0
/2 sin x cos x
sin x cos x
dx
1
u 1
a x
a
0
ln ln x
x
0
3
1
u
du
du
u ln | u | C
ln ln ln x
2 ln x dx
x
ln ln x
ln x
du
dx
2u 2
1 u2
( A B)u
A B
2u 2
u2 1
du
A B
u
0
a, x
a
sin
2
/2
0
/2
dx
x
x
2
cos
2
sin x
sin x cos x
x
x
dx
/2
0
Copyright
2
sin
/2
dx
4
1 ln 1 1 x
2
2
x
C
0, A B
2
0
0
2
u2 1
a
C
du;
A 1
1 ln
2
2 1 x
0
/2
0
2
sin
2
sin x
sin x cos x
/2
0
cos x cos
dx
sin x
sin x cos x
dx
sin x cos
2
/2
0
cos x cos
2
cos x
cos x sin x
/2
2
2014 Pearson Education, Inc.
dx
cos
2
u
ln ln x
x
C
du
2u ln | u 1| ln | u 1 | C
dx, x
ln x
1 x4
x2
1 ln 1
2
x2
C
2u du
ln x
ln x
x ln x
1 du
u
ln x ln ln x
u C
2 x ln x ln x dx
x
2u ln x dx
x
du
cos d , 1 x 4
, 2 x dx
cot
C
1 du
u
dx
2
du
2
B
1 x 1
1 x 1
C
a
f (u ) du
0
1;
f (u ) du, which
f ( x) dx.
sin
2
2
csc
du
/2
0
du
ln u
A(u 1) B(u 1)
2
is the same integral as
(b)
ln x
1 ln
2
3/2
1 cos c
C
sin , 0
csc d
1 x
B
u 1
x2
dx;
4
3
ln | u | C
ln xln x
4
3
3 2
u 1
u
1 du
u
1 x ln x
2
ln ln x
x
x
1
2
2
u2 1
xln x
4
3
3
dx
ln ln x
3/2
3
2
1
4
ln x ln 1 ln x
1 dx
x ln x
2
1
x
du
du
C
1/2
3
2
1
4
3
u 1 ln x
dx;
x
dx
4
u
ln ln x
x
1
x
u
1
112.
1u
2
du
ln x
110.
3/2
3
2
1
ln |1 ln x | C
108.
1/2
2 cos c 1 cos c
ln x
x 1 ln x
dx
1 ln x
1
2
1/2
3
2
3 1
649
0
2
2
sin x
cos x sin
2
sin x
dx
sin x
sin x cos x
dx
4
dx
650
Chapter 8 Techniques of Integration
sin x
sin x cos x
114.
sin x cos x cos x sin x sin x
sin x cos x
dx
cos x sin x
sin x cos x
dx
sin x
2 sin x cos x dx
1 cos x
1 cos x
dx
1 cos x
dx
tan 1
1
2
x
2
2
1 cos x
dx
csc 2 x dx 2 csc x cot x dx
CHAPTER 8
1. u
2
sin 1 x dx
sin
2.
1
x
cos x sin x
sin x cos x
sin x
dx
sin x cos x
sin x
sin x cos x
dx
2 tan x
1 2 cos x cos 2 x
sin 2 x
2
dx
x
2
1 ln
2
1
x dx
1 x2
1 2
2sin
x sin
1
x dx
1 x2
1 2
sin x cos x
dx
C
tan 2 x sec 2 x sec2 x
sec 2 x tan 2 x
dx
tan 2 x sec 2 x
sec 2 x tan 2 x
2 cos x
sin 2 x
dx
cos 2 x
sin 2 x
1
sin 2 x
dx
dx
dx
sec2 x
sec2 x tan 2 x
1
x
1 ,
x 1
x sin
1
dx, v
2 x sin
2 cot x 2 csc x x C
x;
x dx
1 x2
1 x2
x
1
2
2 sin 1 x
sin 1 x, du
; u
2 sin 1 x
2 dx
1
x ( x 1)( x 2)
1
2x
1
x ( x 1)( x 2)( x 3)
1
x 1
1
6x
1 x2
dx
1 x2
1 x2
; dv
1
x ( x 1)( x 2)( x 3)( x 4)
1
24 x
1
2( x 2)
1
6( x 1)
2x C
1 ,
6( x 3)
1
4( x 2)
the following pattern: x ( x 1)( x 12) ( x m)
dx
x ( x 1)( x 2) ( x m )
m
k 0
k
1
6( x 3)
m
k 0
( 1)
ln
( k !)( m k )!
Copyright
1
24( x 4)
( 1)k
;
( k !)( m k )!( x k )
x k
2 x dx
1 x2
,v
2 x C; therefore
1 ,
2( x 2)
1
2( x 1)
dx
dx
csc2 x 1 dx
cot x 2 csc x
1,
x
therefore
dx
C
cot x dx
; dv
x
2 sin 1 x
x
1
x ( x 1)
sin x
sin x cos x
dx
ADDITIONAL AND ADVANCED EXERCISES
2
sin 1 x , du
2 x sin
dx
x ln sin x cos x
tan 2 x
sec2 x tan 2 x
dx
sin 2 x
cos2 x
1
cos2 x
sec2 x
1 2 tan 2 x
dx
116.
cos2 x
dx
dx
x ln sin x cos x
sin 2 x
sin 2 x
1 sin 2 x
115.
sin x
sin x cos x
dx
sin x cos x
sin x cos x
dx
C
2014 Pearson Education, Inc.
2 1 x2 ;
Chapter 8 Additional and Advanced Exercises
sin 1 x, du
3. u
dx
1 x
x sin
dx cos d
x2
2
sin 1 x 12 2
dt
y
dz
dy
2 y
du
u 1
1
2
6.
du
u2 1
1
2
1 t2
1 ln
2
t
1
x4 4
dx
x
2
1 ln x 2 2 x 2
16
x2 2 x 2
7.
8.
x
lim
x
x
1 cos t
lim
x
x t
0
lim x
x
0
lim x
x
0
lim
n
k 1
2
1
t
;
1
x
1
dt is an indeterminate 0
dt
lim
x cos t
1 t2
1
x
0
x
n
lim
n
sin 1 x 12 sin 2
x2
2
sin 1 x
k 1
2
u ln u u 1
1
1
z sin
x 1 x 2 sin
4
z 2 sin
2
z dz
y y2
2
y
d
sin
1
1
1
y
2
x
z
C
z 1 z 2 sin
4
1
z
C
tan
sec2
d
du
u2 1
du
(u 1) u 2 1
d
1 ln tan 1
2
sec
C
dt
1
2
C
2x 2
x2 2 x 2
2
( x 1)2 1
2x 2
x2 2 x 2
2
( x 1)2 1
C
1 cos t
lim
x
x t
0
2
cos x cos x
lim
x
dt diverges since
1 dt
0 t2
lim 0
x
cos x
x2
ln 1 k 1n
lim cos x 1
1
x2
0
1
n
2 ln 2 2
Copyright
x
1
0
0
ln 1 x dx;
ln1 1
u 1 x, du
x
0
u 1, x 1
2 ln 2 1 ln 4 1
2014 Pearson Education, Inc.
0
diverges; thus
form and we apply l Hôpital s rule:
lim
x
1
16
dx
cos x cos( x)
lim
x
1
lim cos
t
t 0
cos t
t2
;
x2
2
du
1 tan 1 u
2
u2 1
1
x2 2 x 2 x2 2 x 2
dx
t2
0
y sin
2 1 x2
C
cos t x
lim
ln n 1 kn
ln u du
C
1
1 cos t
x t2
2
x
dt; lim
1 cos t
x t2
n
9.
2
y
tan 1 ( x 1) tan 1 ( x 1)
1
8
sin t dt
4x
1
d
tan
u 1
1 ln
2
u2 1
1 sin 1 t
2
2
C
u
cos d
sin cos
1
2
sin 1 x sin cos
4
x 2 dx
sin 1 x
sin 2 cos d
2 cos
sin 1 x
y 1 y sin
2
y
u du
1
2
x2
2
x2
2
x sin 1 x dx
;
2 z sin 1 z dz; from Exercise 3,
1
y sin
x2
2
x dx
t sin
dt cos d
;
1 t2
t
1
C
z
x2
2
x dx, v
sin 2
4
sin 1 y dy
5.
; dv
x sin
sin 1 y dy;
4.
2
651
dx
u
2
dx
C
652
Chapter 8 Techniques of Integration
n 1
10.
11.
lim
n
k 0 n
dy
dx
2
k
dy
dx
2x
1 x2
1/2
13. V
6
6
2
3
b
2
7
b
0
2 xe x dx
0
2
2
2
n
k 0
1 k
2 cos 2 x; L
1/2
4x2
1 2 x2 x 4
2
2
1 x
1 1 1x
0
1
dx
0
1
1 x
/4
0
1
n
1 1
0 1 x2
1
n
2
1
cos 2t
2
6
1
0
1
2 u 7/2
7
0
1 x2
1 x2
2
2
2
; L
1/2
2
ln 2
0
0
1/2
x ln 11 xx
0
dx
1
2
ln 3
dx, x 2
u1/2
70 84 30
105
2u 3/2 u 5/2 du
6
16
105
4 1
x
5
x2
1
ln 4 45
(1 u ) 2
32
35
1
5 x
dx
ln 14 5
2 ln 2
shell
height
xe x
2
dx
ex
1
2
0
ln 2 x e x 1 dx
ln 2 e x
ln 2 e x
1
sin 1 x
dx
dt
dy 2
dx
1
2 xy dx
1 x, du
4 25 dx
1 x 2 (5 x )
shell
radius
2
ln 2
16. V
1
lim
2 ln 4
a
1
2
1 x
4
5
x 1
ln 5 x x
15
4
6
y 2 dx
a
15. V
dx
4 u 5/2
5
4
5
k
2
1 cos 2 x
1 x2
b
shell
2 shell
radius height
a
1 2
x 1 x dx; u
0
0
6
(1 u )2 u du
1
2 u 3/2
3
14. V
dy 2
dx
dy 2
dx
2
1 x2
6
n
k 0
2
n 1
1
n
2
0
2
/4
0
cos 2 t dt
1
1
1
0
n
1
/4
n
lim
2
cos 2 x
2 sin t 0
12.
n 1
1
ln 2 xe x
ln 2 x xe x
ln 2
2
2 ln 2 2
x dx
ex
x2
2
ln 2
2
ln 2
0
2
Copyright
2
ln 2 1
2014 Pearson Education, Inc.
dx
1/2 1 x 2
1 x2
0
(0 ln1)
dx
ln 3 12
Chapter 8 Additional and Advanced Exercises
2
ln 2
2
2
ln 2 1
e
17. (a) V
1
1
2
ln x
x x ln x
dx
e
2
e
2
1
1
ln x dx
(FORMULA 110)
x x ln x
2
2
x x ln x
e
2 x ln x x
1
e
2 x ln x
1
e e 2e ( 1)
e
(b) V
2
1 ln x
1
e
dx
x 2 x ln x x
x ln x
2
x 2 x ln x x
x ln x
2
1
ey
0
1
(b) V
0
e2
2
ey 1
(b) V
2
0
2
x3
3
lim
b
0
8 ln 2
3
2
1
0
1
dy
ln x dx
2 x ln x x
e
1
0
e2 y
2
e 2 y 1 dy
e2 y
2e y 1 dy
y
e2 y
2
1
e2
2
0
2e y
y
e2 3
1
2
1
1
2
e2
2
0
1
2
2e 1
e2 4e 5
2
0
x 2 ln x
0
1
1
dx
(2e 5)
1 dy
2e 52
lim x ln x
x
2
e
2
2
1
(5e 4e e) (5)
18. (a) V
e
ln x
e
2
5 x 4 x ln x x ln x
19. (a)
1 2 ln x
1
lim f ( x)
x
2
0
dx;
ln x
16 ln 2
9
2
u
2
b
0
ln x
2 x3
0 3
f (0)
2
, du
f is continuous
2 ln x dx
; dv
x
2ln x dx
x
8
3
x 2 dx, v
ln 2
2
2
3
x3
3
lim
b
16
27
Copyright
2014 Pearson Education, Inc.
0
x3
3
ln x
x3
9
2
b
2
653
654
Chapter 8 Techniques of Integration
1
20. V
ln x
0
lim
b
ln x dx
e
e
x ln x
e
My
1
2
e
2
tan 1 e
/4
ey
du
y
2
1
2
e x2 1
x
1
dx
dx
dy
S
sec tan
e 1 e2
dx
e2 1 ;
1
2
1
4
Mx
M
e 2
2
;
1
2; therefore,
0 by symmetry
csc
2
ln 1e e
e dy
2);
0
tan sec
1
u
and y
1 x2
2
and y
1 e2
ln x
e2 1
4
1 ex
2 1
e2
2
e2
dx
1 (e
2
ln x dx
1
2
2
ln x
0
1
x2
1
1
1 e
2 1
1
2sin 1 x
1 2 x dx
0 1 x2
My
M
1
(e e) (0 1) 1;
e
x 2 ln x
2
1
e
x2
2 1
My
M
1 2 dx
0 1 x2
My
e
2
x ln x dx
therefore, x
y
e
1
x 2 ln x
22. M
x ln x x 1
ln x ln2x dx
1
1
2
24.
0
b
2
e
L
1
lim x ln x x b
0
1
dx
2 ln x dx
1
Mx
23.
2 1
2
ln x
0
b
21. M
x
1
dx
x ln x
0
2
1
2
2
ln
e
1
sec
S
2
1 u 2 du;
ln sec
1 e2 e
2 1
tan
2
d
c
x 1 x 2 dy
u
du
tan 1 e
/4
tan
sec
2
d
1 e2
e
ln
S
2
2
1
e
1 y
e
0
tan
/4
1
e
2 ln 1
2
1 e 2 y dy;
sec
1 e2 e ln 1 e2
2
Copyright
1
tan 1 e
/4
cot
1 e2
2
2
tan 1 e (sec ) tan
tan
/4
tan 1 e sec sec2 d
tan
/4
L
d
ln csc
2 ln 1
1 x2
2
tan
sec
dx
d
1
e
2
x
dx;
2014 Pearson Education, Inc.
sec 2
e
d
2 1 ln
2 1
d
Chapter 8 Additional and Advanced Exercises
25. S
1
2
1
4
0
1 x
2
5
6
26.
x
y
f ( x) 1
1
2/3 3/2
ax
x2 1
1
5/2 1
1
2x
dx
b2 1
lim 12 ln b
b
1;
2
diverges if a
1
2
1 ln 2
2
ln1
lim ln
b2 1
1
ax
x2 1
28. G ( x )
29.
A
1
2x
lim
b
dx
1 xp
2 dx
3 x1/3
du
x 1
16
L
1
b
lim
ax
x2 1
1
b
1
2x
a
b2 1
ln 2a ; lim
0
f ( x)
1
2
3/2
4 32
0
1
x 1 dx
1 u
1
1
1
x
2/3
S
1
1
6
du
2
for a
ln 2 ;
4
b
1
2
if a
lim
: 0
x
2
16
1
0
1
1 x 2/3
2
1 u
3/2
3/2
dx
x 2/3
( 1)du
16 4
x 1 dx
2a
lim bb
b
lim 1
b
b2 1
lim
b
b
b
1 ln x
2
1
1
a
b2 1
b
: lim
a ln
2
b
b
b
1
2
dx
1
x dx
16
4 x5/4
5
1
lim b
2a
1
2
1
b2
1
a
lim
b
lim
b
(b 1)2 a
b 1
lim
b
1
2
ln
a
b
x
1
1
2
if a
b
x2 1
1 ln
2
b2 1
b
the improper integral
1/ 2
ln 21/2
lim (b 1) 2a 1
b
0
a
the improper integral diverges if a
dx converges only when a
b
3/2
124
5
b
b
x 2/3
u
dx;
dy
dx
4 (1)5/4
5
1 x 2/3
dx; f ( x )
12
5
0
t 1 dt
1
4 (16)5/4
5
27.
1
x1/3
1 u
2
f ( x)
655
e xt dt
1 e xt
x
lim
b
b
1
2
b
1 0
x
in summary, the improper integral
ln 2
4
and has the value
xb
lim 1 ex
0
1;
2
1
x
if x
0
xG ( x)
x 1x
1 if x
0
converges if p 1 and diverges if p 1. Thus, p 1 for infinite area. The volume of the solid of
revolution about the x-axis is V
2 p 1. Thus we want p
1
2
1
1
xp
2
dx
dx
1 x2 p
which converges if 2 p 1 and diverges if
for finite volume. In conclusion, the curve y
volume for values of p satisfying 12
p 1.
Copyright
2014 Pearson Education, Inc.
x p gives infinite area and finite
656
Chapter 8 Techniques of Integration
1 dx
;
0 xp
30 . The area is given by the integral A
p 1: A
p 1: A
p 1: A
1
lim ln x b
b 0
lim ln b
b
lim
1
x1 p
1
lim
1
x1 p
1
b
b
b
0
b
0
, diverges;
0
lim b1 p
b
, diverges;
0
lim b1 p
b
1 0, converges; thus, p 1 for infinite area.
0
1 dx
0 x2 p
The volume of the solid of revolution about the x -axis is Vx
1.
2
and diverges if p
which converges if 2 p 1 or p
Thus, Vx is infinite whenever the area is infinite ( p 1). The volume of the solid of
revolution about the y -axis is V y
R( y)
1
2
dy
1 y 2/ p
dy
which converges if 2p
1
p
31. See the generalization proved in 32.
a
a
dx
2
f ( x) x
0
a
0
a
2
f ( x ) dx
0
a
(2 x a ) f ( x ) dx
1
x
The last integral is
3
3 a
0
Using integration by parts with u
f (a )
a
0
33.
2
f ( x ) dx
a
2
0
f ( x ) dx
( )
cos 3x
2e2 x
( )
1 sin 3x
3
4e2 x
( )
e2 x
3
a
2
sin 3x
2
dx
a3
.
12
2 x a , du
2dx , dv
f (0) b, the second integral is (2 x a ) f ( x )
e2 x
I
x
0
a
2
2ab
a
0
f ( x ), v
2
a
0
f ( x ) dx
f ( x ), and the fact that
2ab 2
a
0
f ( x ) dx. Thus
a3
.
12
1 cos 3x
9
2e 2 x
9
cos 3x
4
9
I
13 I
9
2 (see
x p gives infinite area and finite volume for values of p satisfying
Exercise 29). In conclusion, the curve y
1 p 2, as described above.
32. 0
1,
2
e2 x
9
Copyright
3sin 3x 2 cos 3x
I
e2 x
13
2014 Pearson Education, Inc.
3sin 3 x 2 cos 3x
C
Chapter 8 Additional and Advanced Exercises
34.
e3 x
( )
3e3 x
( )
1 cos 4x
4
9e3x
( )
1 sin 4x
16
e3 x
4
I
35.
36.
9
16
sin 4 x
( )
cos x
9sin 3x ( )
sin x
cos 5x
e3 x
16
5sin 5x
( )
1 cos 4x
4
25cos 5x
( )
1 sin 4x
16
8I
aeax
( )
1 cos bx
b
a 2 e ax
( )
1
b2
cos bx
3sin 4 x 4 cos 4 x
C
sin 3 x cos x 3cos 3 x sin x
8
I
C
9 I
16
1 cos 5 x cos 4 x
4
e ax
b2
a sin bx b cos bx
5 sin 5 x sin 4 x
16
sin bx
ae ax
b2
sin bx
( )
cos bx
aeax
( )
1 sin bx
b
a 2 e ax
( )
eax
b
sin bx
ln ax
( )
1
1
x
( )
x
1
b2
aeax
b2
1
x
a2
b2
sin bx
eax
x ln ax
e3 x
25
I
sin 3 x cos x 3cos 3x sin x
1 cos 5 x cos 4 x 5 sin 5 x sin 4 x 25 I
4
16
16
1 4 cos 5 x cos 4 x 5sin 5 x sin 4 x
C
9
e ax
b
3sin 4 x 4 cos 4 x
sin 4x
( )
( )
I
25 I
16
sin x
eax
I
I
sin 3x cos x 3cos 3x sin x 9 I
I
39.
3e3 x
16
3co