# chapter6 solution

C H A P T E R 6
Techniques of Integration
Section 6.1
Integration by Parts and Present Value..............................................406
Section 6.2
Integration Tables ...............................................................................422
Quiz Yourself .............................................................................................................432
Section 6.3
Numerical Integration.........................................................................436
Section 6.4
Improper Integrals ..............................................................................445
Review Exercises ........................................................................................................449
Test Yourself .............................................................................................................459
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 6
Techniques of Integration
Section 6.1 Integration by Parts and Present Value
Skills Warm Up
1. f ( x) = ln ( x + 1)
7. A =
1
f ′( x) =
x +1
(
)
2. f ( x) = ln x 2 − 1
f ′( x) =
2x
x −1
4. f ( x ) = e
=
 − 2 (− x
=
 − 2 (−2 x
=
x3
f ′( x) = 3 x 2e x
8. A =
3
− x2
f ′( x ) = −2 xe − x
 f ( x) − g ( x) dx
2
2
2
2
+ 4) − ( x 2 − 4) dx
+ 8) dx
= − 23 x3 + 8 x
2
3. f ( x) = e
b
a
2
(
b
+ 16 −
5. f ( x) = x e
f ′( x) = x 4e x + 4 x3e x
=
 −1 (− x
=
 −1 ( − x
1
1
= e x ( x 4 + 4 x3 )
=
(
=
4
3
9. A =
6. f ( x) = xe − 4 x
f ′( x) = − 4 xe− 4 x + e− 4 x
= e − 4 x (1 − 4 x)
− 13
b
a
2
2
+ 2) − 1 dx
+ 1) dx
1
−1
) ( 13 − 1)
+1 −
 f ( x) − g ( x ) dx
=
 −1 4 x − ( x
=
 −1 (− x
5
5
2
2
− 5) dx
+ 4 x + 5) dx
= − 13 x 3 + 2 x 2 + 5 x
=
(
64
3
 f ( x) − g ( x) dx
= − 13 x 3 + x
4 x
−2
) (163 − 16) =
− 16
3
a
2
− 125
3
5
−1
) ( 13 + 2 − 5)
+ 50 + 25 −
= 36
10. A =
b
a
 f ( x) − g ( x) dx
1
=
 −1  f ( x) − g ( x)
=
 −1 ( x
=
 −1 ( x
1
1
3
3
406
3
1
 g ( x) − f ( x) dx
− 3x 2 + 2) − ( x − 1) dx +
− 3 x 2 − x + 3) dx +
=  14 x 4 − x3 −
=
dx +
1 x2
2
1
3
1 ( − x
3
3
1 ( x − 1) − ( x
3
− 3 x 2 + 2) dx
+ 3 x 2 + x − 3) dx
+ 3x + − 14 x 4 + x3 +
−1
1 x2
2
3
− 3x
1
( 14 − 1 − 12 + 3) − ( 14 + 1 − 12 − 3) + (− 814 + 27 + 92 − 9) − (− 14 + 1 + 12 − 3) = 8
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
1.
 xe
9x
dx
7. Let u = ln x and dv = x3 dx. Then du =
u = x, dv = e9 x dx
2.
v =
x4
.
4
x
ln x dx =
x4
ln x −
4
 x ln 2 x dx
=
x4
ln x −
4
u = ln 2 x, dv = x dx
=
x
x4
+C
ln x −
4
16
 ln 4x dx
=
x4
(4 ln x − 1) + C
16
xe
2 3x
dx
u = x 2 , dv = e3 x dx
3.
4.
Integration by Parts and Present Value
3
u = ln 4x, dv = dx
Then du =
v = 13 e3x .
3x
 13 e
dx =
1 3x
xe
3
−
=
1 3x
xe
3
− 19 e3 x + C
=
1 3x
e
9
(3 x
3x
x
dx
12
=
− 16 xe −6 x
=
1 e −6 x
− 36
−
(6 x
 − 16 e
1 e −6 x
36
2 32
x ln x −
3

2 3 2 1 
x   dx
3
 x
2 32
2
x ln x −  x1 2 dx
3
3
2
4
= x3 2 ln x − x3 2 + C
3
9
− 1) + C
dx = − 16 xe −6 x −
4
=
v = − 16 e − x .
−6 x

x3
dx
4
1
2
dx and v = x 3 2 .
3
x
ln x dx =
6. Let u = x and dv = e−6 x dx. Then du = dx and
 xe
x4  1 
  dx
4  x
1
dx and
x
8. Let u = ln x and dv = x1 2 dx.
5. Let u = x and dv = e3 x dx. Then du = dx and
 xe

407
−6 x
9. Let u = ln 5 x and dv = dx. Then
5
1
du =
dx = dx and v = x.
5x
x
dx
+C
 ln 5x dx
+ 1) + C
1
= x ln 5 x −
 x x  dx
= x ln 5 x −
 dx
= x ln 5 x − x + C
= x(ln 5 x − 1) + C
10.
 ln (3 x)
2
dx =
2
 ln (9 x ) dx
=
 (ln 9) dx + 2  ln x dx
Use integration by parts on the second integral.
Let u = ln x and dv = dx. Then du =
 (ln 9) dx + 2  ln x dx
1
dx and v = x.
x
= (ln 9) x + 2  ln x dx

= (ln 9) x + 2  x ln x −

 x x  dx
= (ln 9) x + 2  x ln x −

 dx
1

= (ln 9) x + 2 x ln x − 2 x + C
= (ln 9) x + 2 x(ln x − 1) + C
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
408
Chapter 6
Techniques of Integration
11. Let u = x 2 and dv = e − x dx. Then du = 2 x dx and v = −e − x .
xe
2 −x
dx = − x 2e− x + 2  xe− x dx
Let u = x and dv = e− x dx. Then du = dx and v = −e − x .
xe
2 −x
dx = − x 2e − x + 2 − xe − x +

e
−x
dx

= − x 2e − x − 2 xe − x − 2e − x + C
= −e − x ( x 2 + 2 x + 2) + C
1 e7 x .
7
12. Let u = x 2 and dv = e7 x dx. Then du = 2 x dx and v =
xe
2 7x
dx =
1 x 2e7 x
7
2
7
−
 xe
7x
dx
1 e7 x .
7
Let u = x and dv = e7 x dx. Then du = dx and v =
xe
2 7x
dx =
1 2 7x
x e
7
−
2 1 7x
xe
7 7
=
1 2 7x
x e
7
−
2
xe7 x
49
=
1 7x
e
343
(49 x 2
−
+
12
2
x + 4 dx =
=
+ 4)
Let u = x and dv = ( x + 4)
x
2
32
32
(x
+ 4)
32
−
4 2 x
3 5
(x
+ 4)
52
−
2
5
=
2 x2
3
(x
+ 4)
32
−
4 2 x
3 5
(x
+ 4)
52
−
4
35
=
2 x2
3
(x
+ 4)
32
−
8
x
15
=
Then du = dx and v =
2
12
dx.
(x
− 3) .
− 3)
Let u = x and dv = ( x − 3)
x
(x
+ 4) .
32
(x
+ 4)
52
16
105
+
(x
+ 4) .
5 2
 ( x + 4)
(x
(x
+ 4)
+ 4)
52
72
72
dx

 +C

+C
32
32
 43 x( x − 3) dx
32
32
2 2
x ( x − 3) − 43  x( x − 3) dx
3
2 2
x
3
(x
2
3
2
5
dx. Then du = dx and v =
2 x2
3
x − 3 dx =
2
3
32
−
x + 4 dx =
Then du = 2 x dx and v =
2
+ C
dx. Then du = 2 x dx and v =
14. Let u = x 2 and dv = ( x − 3)
x
1 7x
e
49
 43 x( x + 4) dx
2 x 2 x + 4 3 2 − 4 x x + 4 3 2 dx
)
)
3 (
3 (
2 x2
3
(x
dx

1 7x
e
7
− 14 x + 2) + C
13. Let u = x 2 and dv = ( x + 4)
x

2
5
(x
32
32
−
dx.
− 3) .
52
x − 3 dx =
2 2
x
3
(x
− 3)
32
−
4 2
x
3 5
=
2 2
x
3
(x
− 3)
32
−
8
x
15
(x
(x
− 3)
− 3)
52
52
+
−
2
5
16
105
 ( x − 3)
(x
− 3)
52
7 2
dx

+C
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
Integration by Parts and Present Value
409
1
3
2 1 
15. Let u = (ln x) and dv = x 2 dx. Then du = 3(ln x)   dx and v = x 3 .
3
 x
3
2
 x (ln x) dx
1
2 1 
=
1 3
3
x (ln x) −
3
 3 x (3)(ln x)  x  dx
=
1 3
3
x (ln x) −
3
 x (ln x)
2
3
2
dx
1
2
1
Let u = (ln x) and dv = x 2 dx. Then du = 2(ln x)  dx and v = x 3.
x
3
 
3
2
 x (ln x) dx
=
1
1 3
3
2
x (ln x ) −  x 3 (ln x) −
3
3

1
1

 3 x (2)(ln x) x  dx
3
1 3
2
3
2
2
1

x (ln x ) −  x 3 (ln x) −  x 2 (ln x) dx
3
3
3

1
1
2
3
2
2
= x 3 (ln x ) − x 3 (ln x) +  x 2 (ln x) dx
3
3
3
=
1
2
1
Let u = (ln x) and dv = x 2 dx. Then du = 2(ln x)  dx and v = x3.
3
 x
3
2
 x (ln x) dx
=
1
1 3
3
2
x (ln x) −  x3 (ln x) −
3
3
1
1

 3 x (2)(ln x) x  dx
3
1 3
2
3
2
2
1

x (ln x ) −  x3 (ln x) −  x 2 (ln x) dx
3
3
3

1 3
1 3
2
3
2
2
= x (ln x ) − x (ln x) +  x 2 (ln x) dx
3
3
3
=
1
2
1
Let u = (ln x) and dv = x 2 dx. Then du = 2(ln x)  dx and v = x3.
3
 x
3
2
 x (ln x) dx
=
1
1 3
3
2
x (ln x) −  x3 (ln x) −
3
3

1
1

 3 x (2)(ln x) x  dx
3
1 3
2
3
2
1

x (ln x ) −  x3 (ln x) −  x 2 (ln x) dx
3
3
3


1 3
1 3
2
3
2
= x (ln x ) − x (ln x) +  x 2 (ln x) dx
3
3
3
=
1
1
Let u = ln x and dv = x 2 dx. Then du =   dx and v = x3 .
x
3
 
3
2
 x (ln x) dx
=
1 3
1
3
2
x (ln x) −  x3 (ln x) −
3
3
1
1

 3 x (2)(ln x) x  dx
3
1 3
2
3
2
1

x (ln x) −  x3 (ln x) −  x 2 (ln x) dx
3
3
3

1 3
1 3
2 2
3
2
= x (ln x) − x (ln x) +  x (ln x) dx
3
3
3
=
1 3
1
2 1
1  1 
3
2
x (ln x) − x3 (ln x) +  x3 ln x −  x3  dx 
3
3
3 3
3  x 
1
1
2
2
2
3
= x3 (ln x) − x3 (ln x) + x3 ln x −  x 2 dx
3
3
9
9
1
1
2
2 3
3
2
= x3 (ln x) − x3 (ln x) + x3 ln x −
x +C
3
3
9
27
=
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
410
16.
Chapter 6

Techniques of Integration
3x3
dx = 3  x3e− x dx
ex
Let u = x3 and dv = e− x dx. Then du = 3x 2 dx and v = − e− x .
3  x3e − x dx = 3 − x3e − x −

2
−x
 3x (− e ) dx
= 3 − x3e − x + 3  x 2e − x dx


Let u = x 2 and dv = e− x dx. Then du = 2 x dx and v = − e− x .
3  x3e − x dx = 3 − x3e − x −

2
−x
 3x (− e ) dx
= 3 − x3e − x + 3  x 2e − x dx


= 3 − x3e − x + 3 − x 2e − x −


−x
 (2 x)(− e ) dx
= 3 − x3e − x + 3 − x 2e − x + 2  xe − x dx


= 3 − x3e − x − 3x 2e − x + 6  xe − x dx
Let u = x and dv = e− x dx. Then du = dx and v = − e− x .
3  x3e − x dx = 3 − x3e − x −

2
−x
 3x (− e ) dx
= 3 − x3e − x + 3  x 2e − x dx


= 3 − x3e − x + 3 − x 2e − x −


−x
 (2 x)(− e ) dx
= 3 − x3e − x + 3 − x 2e − x + 2  xe − x dx



= 3 − x3e − x − 3 x 2e − x + 6  − xe − x −

 −e
−x
dx

= 3 − x3e − x − 3 x 2e − x − 6 xe − x + 6  e − x dx


= 3 − x3e − x − 3 x 2e − x − 6 xe − x − 6e− x  + C
= − 3e − x ( x3 + 3x 2 + 6 x + 6) + C
17.
e
4x
18.
e
−2 x
dx =
dx =
1
4
4x
 e (4) dx
− 12
=
1 e4 x
4
 e (− 2) dx
−2 x
=
20. Let u = x and dv = e−2 x dx. Then du = dx and
+C
− 12 e−2 x
+C
19. Let u = x and dv = e 4 x dx. Then du = dx and
v =
 xe
1 4x
e .
4
4x
v = − 12 e−2 x .
 xe
−2 x
dx = − 12 xe −2 x −
 − 12 e
−2 x
dx
= − 12 xe −2 x − 14 e −2 x + C
= − 14 e −2 x ( 2 x + 1) + C
1 4x
1
xe −  e 4 x dx
4
4
1 4x
1
= xe − e 4 x + C
4
16
dx =
=
e4 x
(4 x − 1) + C
16
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
21. Let u = x 2 and dv = e− x 4 dx.


+ 8  xe
2 −x 4
= − 4x e

22.
Then du = 2 x dx and v = − 4e− x 4 .
x 2e − x 4 dx = ( x 2 )( − 4 xe − x 4 ) −
Integration by Parts and Present Value
dx =
dx
3
2
xe
2 x 2
dx =
=
Then du = dx and v = − 4e− x 4 .
xe
dx = ( x
2
−x 4
3
2
xe
2 x2
dx =
 − 4e dx
= − 4 x 2e − x 4 − 32 xe− x 4 + 32  e− x 4 dx
2 −x 4
= − 4x e
= − 4e
−x 4
− 32 xe
−x 4
−x 4
− 128e
−x 4
dx
( x 2 )(2e x 2 ) −  (2 x)(2e x 2 ) dx
3
2
3 2 x2
2x e
2
− 4  xe x 2 dx

v = 2e x 2 .
= − 4 x 2e − x 4 + 8  xe − x 4 dx
= − 4 x 2e − x 4 + 8 − 4 xe − x 4 −

2 x2
Let u = x and dv = e x 2 dx. Then du = dx and
)(− 4 xe ) −  (2 x)(− 4e ) dx
−x 4
xe
v = 2e x 2 .
Let u = x and dv = e− x 4 dx.
2 −x 4
3
2
Let u = x 2 and dv = e x 2 dx. Then du = 2 x dx and
( 2 x)(− 4e− x 4 ) dx
−x 4
3 2 x2
xe
2
411
+C
( x + 8x + 32) + C
2
( x 2 )(2e x 2 ) −  (2 x)(2e x 2 ) dx
3
2
=
3 2 x2
2x e
2
− 4  xe x 2 dx
=
3 2 x2
2x e
2
− 4 2 xe x 2 −
=
3 2 x2
2x e
2
− 8 xe x 2 + 8  e x 2 dx

=
3 2 x2
2x e
2
− 8 xe x 2 + 16e x 2  + C
 2e
x2
dx

= 3e x 2 ( x 2 − 4 x + 8) + C
23. Let u = ln(t + 1) and dv = t dt. Then du =
 t ln(t
+ 1) dt =
=
t2
1
dt and v = .
2
(t + 1)
t2
t2
1
dt
ln (t + 1) − 
2
2 t +1
t2
1
ln (t + 1) − 
2
2
1 

t − 1 +
 dt
t + 1


t2
1 t 2
ln (t + 1) −  − t + ln t + 1  + C
2
2 2

1
= 2t 2 ln (t + 1) + t ( 2 − t ) − 2 ln t + 1  + C
4
=
1
25. Let u = .
t
24. Let u = 2 x − 1 and dv = e7 x .
Then du = 2 dx and v =
(2 x
1 e7 x .
7
− 1)e x dx =
1
7
(2 x
− 1)e7 x −
2
7
=
1
7
(2 x
− 1)e7 x −
2 e7 x
29
=
1 e7 x
49
(14 x
− 7 − 2) + C
=
1 e7 x
49
(14 x
− 9) + C
7x
e dx
+C
 1
Then du =  − 2  dt.
 t 
e1 t
u
u
1t
 t 2 dt = − e du = −e + C = −e + C
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412
26.
Chapter 6
1
Techniques of Integration
 x(ln x)3 dx
=
=
 (ln x)
(ln x)
= −
−3 1
x
1
and v = − .
x
−2
+C
−2
1
2(ln x)
2

+C
27. Let u = (ln x) and dv = x dx. Then du =
2
(2 ln x) dx
x
x2
.
and v =
2
 x(ln x)
2
dx =
x2
2
(ln x) −
2
 x ln x dx
Let u = ln x and v = x dx. Then du =
 x(ln x)
2
dx =
1
dx
x
 x2
x2
2
(ln x) −  ln x −
2
2

x 
dx
2 
x2
x2
x2
2
(ln x) − ln x + + C
2
2
4
x2 
2
2(ln x) − 2 ln x + 1 + C
=

4
=
28. Let u = ln 6 x and dv = dx. Then du =
6
1
dx = dx
x
6x
and v = x.
 ln (6 x) dx
= x ln 6x −
ln x
ln x
1 1
dx = −
−  − ⋅ dx
2
x
x
x x
ln x
1
= −
+  2 dx
x
x
ln x
1
=
− +C
x
x
ln x + 1
= −
+C
x
 dx
1
and v = − .
x

ln 2x
ln 2x
dx = −
x2
x
ln 2x
= −
x
ln 2x
= −
x
ln 2x
= −
x
and v =
= x[ln 6x − 1] + C

(ln x)
2
x
dx =
2 1 
 (ln x)  x  dx
=
1
30. Let u = ln 3 x. Then du = dx.
x
 (ln 3x)
−1  1 
  dx =
 x
u
−1
du
= ln u + C
= ln ln 3 x + C
(ln x)
3
3
+C
2
3
(x
1
−
 −x
+

1
+C
x
+1
+C
x − 3 dx. Then du = dx
− 3) .
x − 1 dx =
2
x
3
(x
− 3)
32
−
32
 23 ( x − 3) dx
=
2
x
3
(x
− 3)
32
−
4
15
(x
− 3)
(x
− 3)
=
2
15
=
2
5
3
32
32
(3 x
(x
(x
− 3)
(x
+C
+ 2) + C
13
3
4
52
+ 6) + C
x + 2 dx = ( x + 2)
Then du = dx and v =
3
1
dx
x2
32
34. u = x and dv =
x
1
dx
x
⋅
−
33. Let u = x and dv =
x
= x ln 6x − x + C
29.
1
1
dx. Then du = dx
x2
x
32. Let u = ln 2x and dv =
x2
.
2
and v =
1
1
dx. Then du = dx
x2
x
31. Let u = ln x and dv =
dx
dx.
+ 2) .
43
x + 2 dx dx =
3
x
4
( x + 2)
43
−
43
 34 ( x + 2) dx
=
3
x
4
( x + 2)
43
−
9
28
( x + 2)
73
3
28
( x + 2)
=
3
28
( x + 2) ( 4 x − 6) + C
=
3
14
( x + 2)4 3 (2 x − 3) + C
=
43
+C
7 x − 3( x + 2) + C
43
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
35. Let u = x and dv = ( x − 1)
v = 2 x − 1.
x
 x − 1 dx = 2 x
= 2x
=
1
3
−1 2
dx. Then du = dx and
x −1 −
2
x −1 −
4
32
( x − 1) + C
3
x − 1 dx
Integration by Parts and Present Value
1
39. Let u = ln x and dv = x 5 dx. Then du =   dx and
 x
x6
v =
.
6
e
1 x
5
ln x dx =
−1 2
=
40. Let u = ln x and dv = 2 x. Then du =
e
1
2 x ln x = x 2 ln x] 1 −
e
 (2 x + 1)2
= −
=
e
du = 2 xe

3 x2
xe
(x
2
+ 1)
2
(x
 (x e
2 x2
2
)
= −
)
−2
dx. Then
1
2( x 2 + 1)
2 x2
 xe
x2
2( x 2 + 1)
+C
1

2x
1
 0 1 + 2x
1
dx

 0 1 − 2 x + 1 dx
1
1


= ln 3 −  x − ln ( 2 x + 1)
2

0
.
= ln 3 − 1 +
1
ln 3
2
3
ln 3 − 1
2
≈ 0.648
=
dx
xe
1 2
+ ex + C
2
2( x + 1) 2
x2
1
= x ln (1 + 2 x) 0 −
= ln 3 −


x

 dx
 x2 + 1 2 
)
(
x 2e x
+
2( x 2 + 1)
e
2
dx and v = x.
1 + 2x
1
2
= −
=
du =
 0 ln(1 + 2 x) dx
+ 1) dx and v = −
dx =
41. Let u = ln(1 + 2 x) and dv = dx. Then
e2 x
+C
4( 2 x + 1)
(
2
=
xe2 x
1
+ e2 x + C
2( 2 x + 1) 4
38. Let u = x e and dv = x x + 1
x2
1 2 1
e +
2
2
≈ 4.195
1
.
2( 2 x + 1)
xe2 x
1 2x
+
e
2( 2 x + 1) 2 
2 x2
e
 1 x dx
 x2 
= e2 −  
 2 1
−2
dx = −
1
dx and
x
v = x2.
37. Let u = xe 2 x and dv = ( 2 x + 1) dx. Then
xe 2 x
x5
dx
6
e6
e6
1
−
+
6
36 36
5 6
1
=
e +
36
36
≈ 56.060
dx. Then du = dx
2
12
( 2 + 3x) .
3
x
2
2
12
12
dx = x( 2 + 3 x ) −  ( 2 + 3 x) dx
3
3
2 + 3x
2
4
12
32
= x( 2 + 3 x ) −
( 2 + 3x) + C
3
27
2
12
=
(2 + 3x) 9 x − 2(2 + 3x) + C
27
2
2 + 3 x ( 3 x − 4) + C
=
27
du = e 2 x ( 2 x + 1) dx and v = −
e
1
e
and v =

x6
e
ln x] 1 −
6
 x6 
e6
=
− 
6
 36  1
x − 1( x + 2)
36. Let u = x and dv = ( 2 + 3x)
413
42. Let u = x and dv = e− x 2 dx. Then du = dx and
v = −2e − x 2 .
4
0
4
4
x
dx = − 2 xe − x 2  + 2  e − x 2 dx
0
0
ex 2
4
= −8e −2 − 4e − x 2 
0
= −12e −2 + 4
≈ 2.376
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
414
Chapter 6
Techniques of Integration
43. Let u = x and dv = ( x + 1)
12
2
3
Then du = dx and v =
8
0 x
x + 1 dx =
2x
3
(x
(x
+ 1)
dx.
47. Area =
+ 1) .
32
32
−
=
2
3
4
15
(x
32
dx
 ( x + 4 )e
8
52
+ 1) 
0
 0 x ( x + 4)
dx = 2 x( x + 4)
12
2
 2( x + 4)
−
45. Let u = x and dv = e
and v =
2
1
4
3
(x
−3
2
=
1 x 2e 2 x 
2
1
48. Area = − 
dx. Then du = 2 x dx
2
2
−
−
2
1
2
1
(
)
xe
2x
−  12 xe 2 x −

=
1 x 2e 2 x
2
1 xe 2 x
2
−
(
5 4
e
4
 12 e
2x
dx
1
2
Area =
− 12 e 2 + 14 e 2
)
1
+ 2  xe x dx
2 x
 −1 (1 − x )e dx
1
1
−1
= (0 + 2e − 2e) − ( −2e −1 − 2e −1 )
4
e
≈ 1.472
=
3
2
−2
Let u = x and dv = e − 3 x dx.
1
Then du = dx and v = − e −3 x .
3
2
2 x
 −1 (1 − x )e dx
= (1 − x 2 )e x + 2 xe x − 2e x 
− 14 e 2 ≈ 66.400
46. Let u = x 2 and dv = e − 3 x dx.
1
Then du = 2 x dx and v = − e − 3 x .
3
2
1 2 −3x
2 −3 x
2 −3 x
 0 x e dx = − 3 x e −  − 3 xe dx
1
2
= − x 2e − 3 x +  xe −3 x dx
3
3
0
− 1)e x dx
= (1 − x 2 )e x + 2( xe x − e x ) ( Example 1)
2
) ( 12 e
( x2
2 x
2 x
 (1 − x )e dx = (1 − x )e
2
+ 14 e 2 x 
1
= 2e 4 − e 4 + 14 e 4 −
1
−1
2
and v = e x .
dx
1 x 2e 2 x 
1
2
1
dx = ( x + 4)e x − e x 
−2
Let u = 1 − x 2 and dv = e x dx. Then du = −2 x dx
2 x 12 e 2 x dx
=
=
0
=
1 2 2x 
xe 
2
1
x
12
32
+ 4) 
0
1 e2 x .
2
x 2e 2 x dx =
1
 − 2 ( x + 4)e
dx
64
3
2x
dx
x
14
12
12
= 2 x( x + 4) −

=
x
= (5e − e) − ( 2e − 2 − e − 2 ) = 4e − e − 2 ≈ 10.738
dx.
Then du = dx and v = 2( x + 4) .
−1 2
e
= ( x + 4 )e − e + C
12
12
dx = ( x + 4)e x −
x
x
Area =
−1 2
dx
and v = e x .
1192
15
44. Let u = x and dv = ( x + 4)
x
Let u = x + 4 and dv = e x dx. Then du = dx
 ( x + 1)
32
=  23 x( x + 1) −

1
− 2 ( x + 4)e
−3
1
2 1
1

x 2e − 3 x dx = − x 2e − 3 x + − xe − 3 x −  − e − 3 x dx
3
3 3
3

1 2 −3 x 2 −3 x 2 −3 x
= − xe
− xe
+  e dx
3
9
9
2
2
2 −3x 
 1
e 
= − x 2e − 3 x − xe − 3 x −
9
27
 3
0
=
2 − 50e − 6
27
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
49. Area =
3
0
1 −x 3
1 3
xe
dx =  xe − x 3 dx
9
9 0
Let u = x and dv = e − x 3 dx.
51. Area =
1
1
xe − x 3 dx = − 3 xe − x 3 + 3  e − x 3 dx

9
9
−x 3
−x 3
1
 + C
= − 3 xe
− 9e

9
x
4
x 4 ln x dx
ln x dx =
=
−x 3
−x 3
1

− 3 xe
− 9e
 0
9 
3
Area =
415
1
dx
x
x5
.
5
and v =
3
 1 −x 3
= − e
( x + 3)
 3
0
e
1
x5
ln x −
5

x4
dx
5
x5
x5
+ C
ln x −
5
25
e
 x5
x5 
x 4 ln x dx =  ln x −

25  1
5
 e5
e5  
1
4e5
1
=  −
+
 − 0 −
 =
5
25
25
25
25


 
≈ 23.7861
e − 2
=
e
≈ 0.2642
60
0.5
0
3
0
0
50. Area =
e
1
Then du =
−3
1
1
dx and v = − x − 2 .
2
x
1
1
1
ln x dx = − x − 2 ln x +  x − 2   dx
2
2
 x
ln x 1 − 3
= − 2 +  x dx
2x
2
ln x
1
= − 2 −
+C
2x
4 x2
e
1
ln x
dx
x2
1
1
dx. Then du = dx and
2
x
x
Let u = ln x and dv =
1
v = − .
x

ln x
1
dx = − ln x −
x2
x
Area =
e
1

−1
1
1
dx = − ln x −
x2
x
x
ln x
dx
x2
e
1
 1
= − ln x − 
x 1
 x
e
1 
1 − 3e − 2
 ln x
Area = − 2 −
=
≈ 0.1485
4 x 2  1
4
 2x
 1 1
=  − −  − ( −1)
 e e
2
=1−
e
≈ 0.264
0.2
−2
3
0
52. Area =
x − 3 ln x dx
Let u = ln x and dv = x − 3 dx.
x
e
1
Let u = ln x and dv = x 4 . Then du =
Then du = dx and v = − 3e− x 3.
Area =
Integration by Parts and Present Value
4
1
−0.1
0
0
3
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
416
Chapter 6
Techniques of Integration
1
53. Let u = ln x and dv = x n dx. Then du =   dx and
 x
x n +1
v =
.
(n + 1)
x
n
ln x dx =
=
x n +1
ln x −
n +1

56. Using n = 1 and a = −3,
 xe
x n +1
1
ln x −
x n dx
n +1
n + 1
x n +1
(n
+ 1)
2
xe −3 x
1
+  e −3 x dx
3
−3
1 −3 x
1
= − xe
− e −3 x + C
3
9
1 −3 x
= − e (3 x + 1) + C.
9
dx =
1 x n +1
⋅
dx
x n +1
57. Using n = − 4,
x n +1
1
x n +1
=
⋅
+C
ln x −
n +1
n +1 n +1
=
−3 x
x
−4
−
 1 + ( n + 1) ln x + C
1 ax
e .
a
xe
n ax
x neax
n
−  x n −1eax dx
a
a
xe
2 5x
dx =
xe
5
−
x
32
2
xe5 x dx.
5
2 5x
x
2 x5 2
( − 2 + 5 ln x) + C
25
3
379 −8
−
e ≈ 0.022
128 128
0 t e
=
e5 x
(25 x 2 − 10 x + 2) + C.
125
60.
2
1 ( x
61.
4
2
 0 x (25 − x )
62.
1 x
(b)
10
0
4
dt =
9
224 ln 2
− 19 ≈ 32.755
3
+ 4) ln x dx =
5
e
5


 −1 + ln x  + C
2


=
59.
3 −4 t
25
4
4 x5 2 
5

 −1 + ln x  + C
25 
2

x 2e5 x
2 xe5 x
2e 5 x
−
+
+ C
5
25
125
2
)
52
=
=
12,000
(2
=

x 2e5 x
2  xe5 x
1
x e dx =
− 
−  e5 x dx
5
5 5
5

63. (a)
−
 1 + ( − 4 + 1) ln x + C

x(3 2) +1 
3

−1 +  + 1 ln x + C
2
3 + 1 
2



ln x dx =
Now, using n = 1 and a = 5,

2
3
,
2
55. Using n = 2 and a = 5,
2 5x
+ 1)
x −3
(−1 − 3 ln x) + C
9
−1
=
(1 + 3 ln x) + C
9 x3
58. Using n =
dx =
(− 4
=
54. Let u = x n and dv = eax dx. Then du = nx n −1 dx and
v =
x − 4 +1
ln x dx =
32
ln x dx =
dx =
1,171,875
π ≈ 14,381.070
256
9 10
1
e +
≈ 1982.392
100
100
500( 20 + te −0.1t ) dt = 500  20 dt +
 0
10
= 500( 200) + 500
10
0
10
0
te −0.1t dt 

te −0.1t dt
Let u = t and dv = e−0.1t dt. Then du = dt and v = −10e −0.1t .
0
10,000
10
The company is forecasting
an increase in demand over
= 100,000 + 500 −10te −0.1t 
{
10
0
+ 10
10
0
e −0.1t dt
10
= 100,000 + 500 −100e −1 − 100e −0.1t 
0
= 100,000 + 500 ( −100e
−1
− 100e
= 100,000 + 50,000 (1 − 2e −1 )
−1
}
+ 100)
≈ 113,212 units
113,212
(c) Average =
≈ 11,321 per year
10
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
(
Integration by Parts and Present Value
417
)
64. I = 2000 375 + 68te− 0.2t , 0 ≤ t ≤ 5
(a)
1,050,000
0
750,000
5
The board of trustees expects the gift income to increase over the five-year period.
(b)
 0 2000(375 + 68te
5
−0.2t
) dt
5
5
= 2000  375 dt + 68  te −0.2t dt 

0
 0
5
= 2000(1875) + 136,000 te−0.2t dt
0
Let u = t and dv = e−0.2t dt. Then du = dt and v = −5e −0.2t .
5
5
= 3,750,000 + 136,000 −5te −0.2t  + 5  e −0.2t dt
0
0
{
5
= 3,750,000 + 136,000 − 25 e −1 − 25e −0.2t 
0
}
= 3,750,000 + 136,000( −25e −1 − 25e −1 + 25)
= 3,750,000 + 3,400,000(1 − 2e −1 )
= \$4,648,419.80
(c) Average annual gift income =
65.
 (1.6t ln t + 1) dt
=
=
5
1
1
2000(375 + 68te − 0.2t ) dt = ( 4,684,419.80) = \$936,883.96

0
5−0
5
 dt +  1.6t ln t dt
t + 1.6  t ln t dt
(a) Average value =
2
= t + 0.8t 2 ln t − 0.4t 2 
1
= 3.2 ln 2 − 0.2
2
Let u = ln t and dv = t dt. Then du =
t 2
= t + 1.6  ln t −
2
t
1
dt and v = .
2
t

t 
dt 
2 
≈ 2.018
(b) Average value =
t 2
t2 
= t + 1.6  ln t − 
4
2
66. Using Exercise 54 twice with n = 2 and a = −
2 − t 30
4
1
(1.6t ln t + 1) dt
4 − 3 3
4
= t + 0.8t 2 ln t − 0.4t 2 
3
= 12.8 ln 4 − 7.2 ln 3 − 1.8
= t + 0.8t 2 ln t − 0.4t 2
 (410.5t e
2
1
(1.6t ln t + 1) dt
2 − 1 1
≈ 8.035
1
1
and with n = 1 and a = − ,
30
30
+ 25,000) dt = 410.5−30t 2e −t 30 + 60 te−t 30 dt  + 25,000t


= 25,000t + 410.5−30t 2e −t 30 − 1800te −t 30 + 1800  e −t 30 dt 


= 25,000t + 410.5−30t 2e −t 30 − 1800te −t 30 − 54,000e−t 30  + C
= 25,000t − 12,315e −t 30 (t 2 + 60t + 1800) + C.
90
1
(410.5t 2e−t 30 + 25,000) dt

0
90 − 0
90
1
=
25,000t − 12,315e −t 30 (t 2 + 60t + 1800)

0
90
= \$167,068.28
(a) Average value =
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
418
Chapter 6
Techniques of Integration
365
1
(410.5t 2e−t 30 + 25,000) dt

274
365 − 274
365
1
=
25,000t − 12,315e −t 30 (t 2 + 60t + 1800)

274
91
= \$26,253.48
(b) Average value =
(c)
2 − t 30
 0 (410.5t e
365
t
67. V =
1
− rt
 0 c(t )e dt
68. V =
1
− rt
 0 c(t )e dt
t
69. Present value =
=
+ 25,000) dt = \$31,281,948.97
4
−0.04t
10
450e −0.05t dt =
=
 0 5000e
=
0
dt =
5000 −0.04t 4
 = \$18,482.03
e
0
−0.04
450 −0.04t 10

e
= \$3541.22
0
−0.05
10
−0.05t
 0 (100,000 + 4000t )e dt
10
0
100,000e−0.05t dt + 4000 
= −2,000,000e −0.05t 
10
0
10
0
+ 4000 
= 2,000,000(1 − e−0.5 ) + 4000 
10
0
10
0
te −0.05t dt
te −0.05t dt
te −0.05t dt
Let u = t and dv = e−0.05t dt. Then du = dt and v = −20e −0.05t .
{
) + 4000{ −200e
}
= 2,000,000(1 − e −0.5 ) + 4000 −20te −0.05t  + 20  e −0.05t dt
0
0
= 2,000,000(1 − e −0.5
10
−0.5
10
10
+ −400e −0.05t 
0
= 2,000,000(1 − e −0.5 ) + 4000( −600e −0.5 + 400)
}
= \$931,265.0973
70. Present value =
=
6
 0 (30,000 + 500t )e
6
 0 30,000e
−0.07 t
−0.07 t
dt
6
dt + 500  te −0.07 t dt
0
6
3,000,000 −0.07 t 6
 + 500  te −0.07 t dt
e
0
0
7
6
3,000,000 −0.42
= −
(e − 1) + 500  0 te−0.07t dt
7
= −
Let u = t and dv = e−0.07t dt. Then du = dt and v = −
100 −0.07 t
e
.
7
= −
  100 −0.07t  6
3,000,000 −0.42
100 6 −0.07t 
− 1) + 500  −
+
e
te
e
dt 
(

7
7 0
0
  7

= −
 600 −0.42  10,000 −0.07t  6 
3,000,000 −0.42
e
e
e
− 1) + 500 −
+ −
(
 
7
49

 0 
 7
= −
3,000,000 −0.42
10,000 −0.42 10,000 
e −0.42 −
e
+
(e − 1) + 500 − 600

7
7
49
49 
3,000,000 −0.42
10,000 
e −0.42 +
(e − 1) + 500 − 14,200

7
49
49 
= \$153,816.01
= −
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
71. Present value =
=
Integration by Parts and Present Value
419
t 2 −0.06t
 0 (1000 + 50te )e dt
4
 0 (1000e
4
−0.06 t
+ 50te0.44t ) dt
4
 1000 −0.06t
1 0.44t 
 1
= −
e
+ 50
te0.44t −
e

0.442
 0.44
 0
 0.06
 1000 −0.24
1 1.76   1000 0
1 0 
 1
 1
e 
e
e  − −
e + 50
= −
+ 50
(4)e1.76 −
(0)e0 −
2
2
0.44
0.06
0.44
0.44
0.06
0.44



 

≈ \$4955.34
72. Present value =
=
t 10 −0.06 t
 0 (5000 + 25te )e dt
10
10
0
5000e −0.06t dt + 25 
10
0
te 0.04t dt
10
5000 −0.06t 10
 + 25  te 0.04t dt
e
0
0
0.06
10
5000 −0.6
= −
(e − 1) + 25  0 te 0.04t dt
0.06
= −
Let u = t and dv = e 0.04t dt. Then du = dt and v = 25e 0.04t .
{
{
}
10
5000 −0.6
e
− 1) + 25 25te0.04t  − 25  e0.04t dt
(
0
0.06
10
5000 −0.6
e
= −
− 1) + 25 250e0.4 − 625 e0.04t 
(
0
0.06
5000 −0.6
= −
(e − 1) + 25(−375e0.4 + 625) = \$39,238.17
0.06
= −
73. Present value =
25
0
}
80,000e − 0.05t dt
29
0
74. Present value =
2,500,000e − 0.03t dt
=
80,000 25 − 0.05t
e
(− 0.05) dt
− 0.05  0
=
2,500,000 29 − 0.03t
e
(− 0.03) dt
− 0.03  0
=
80,000 − 0.05t 25
e

0
− 0.05 
=
2,500,000 − 0.03t 29
e

0
− 0.03 
≈ \$1,141,592.33
75. (a) Actual income =
(b) Present value =
=
≈ \$48,420,704.23
4
 0 (150,000 + 75,000t ) dt
4
 0 (150,000 + 75,000t )e
4
 0 150,000e
−0.04 t
dt +
= −3,750,000e−0.04t 
4
0
= 150,000t + 37,500t 2 
−0.04t
4
= \$1,200,000
0
dt
4
 0 75,000te
−0.04 t
dt
4
+ 75,000  te −0.04t dt
0
= −3,750,000(e −0.16 − 1) + 75,000  te −0.04t dt
4
0
Let u = t and dv = e−0.04t dt. Then du = dt and v = −25e −0.04t .
{
− 1) + 75,000 { −100e
}
= −3,750,000(e −0.16 − 1) + 75,000 −25te −0.04t  + 25  e −0.04t dt
0
0
= −3,750,000(e −0.16
4
−0.16
4
4
− 625 e −0.04t 
0
}
= −3,750,000(e −0.16 − 1) + 75,000( −725e −0.16 + 625) = \$1,094,142.27
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
420
Chapter 6
Techniques of Integration
3
 0 (2,400,000 + 600,000t ) dt
76. (a) Actual value =
3
= 2,400,000t + 300,000t 2 
0
= \$9,900,000
(b) Present value =
=
3
− 0.03t
 0 (2,400,000 + 600,000t )e dt
3
0
2,400,000e − 0.03t dt +
3
0
600,000te − 0.03t dt
3
3
 − 2,400,000 − 0.03t 
− 0.03t
= 
e
dt
 + 600,000  0 te
0.03

0
=
3
− 2,400,000 − 0.09
− 1) + 600,000  te − 0.03t dt
e
(
0
0.03
Let u = t and dv = e − 0.03t dt. Then du = dt and v = −
1 − 0.03t
.
e
0.03
=
3
1 3 − 0.03t 
− 2,400,000 − 0.09
 1 − 0.03t 
− 1) + 600,000−
+
e
te
e
dt 
(

0.03
0.03  0
0
 0.03

=
3
− 2,400,000 − 0.09
1
e − 0.03t  
(e − 1) + 600,000−100e−0.09 − 0.0009
0
0.03

1 
− 2,400,000 − 0.09
1
e − 0.09 +
(e − 1) + 600,000−100e−0.09 − 0.0009

0.0009 
0.03
≈ \$9,428,843.88
=
77. (a) Actual income =
4
 0 (3,000,000 + 750,000t ) dt
4
= 3,000,000t + 375,000t 2 
0
= \$18,000,000
(b) Present value =
=
4
− 0.05t
 0 (3,000,000 + 750,000t )e dt
4
0
3,000,000e − 0.05t dt +
4
4
0
750,000te − 0.05t dt
4
= − 60,000,000e − 0.05t  + 750,000  te − 0.05t dt
0
0
= − 60,000,000(e − 0.2 − 1) + 750,000  te − 0.05t dt
4
0
Let u = t and dv = e − 0.05t dt. Then du = dt and v = − 20e− 0.05t .
{
− 1) + 750,000{− 80e
}
= − 60,000,000(e − 0.2 − 1) + 750,000 − 20te − 0.05t  + 20  e − 0.05t dt
0
0
= − 60,000,000(e − 0.2
4
− 0.2
4
}
4
− 400 e − 0.05t 
0
= − 60,000,000(e − 0.2 − 1) + 750,000{− 80e − 0.2 − 400e − 0.2 + 400}
≈ \$16,133,083.71
78. Graph A shows the expected income and graph B shows the present value of the expected income. The present value of a
future amount is the amount to be deposited today to produce the future value (see Section 4.2). Because of inflation, a certain
amount today buys, or is worth more than, that same amount years from now. Therefore, for any specific year the present value
would be less than that of the actual income.
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
t1
79. Future value = e rt1 
0
f (t )e − rt dt
Integration by Parts and Present Value
80. Future value = e rt1 
t1
0
421
f (t )e − rt dt
5
= e(0.1)5  (3000e 0.05t )e −0.1t dt
10
= e(0.08)10  3000e −0.08t dt
0
0
5
10
= e0.5  3000e −0.05t dt
 3000 −0.08t 
= e0.8 
e

 −0.08
0
0
 3000 −0.05t  5
e
= e0.5 

 −0.05
0
≈ \$45,957.78
≈ \$21,881.75
81. (a) Future value = e (0.07)(10) 
= e 0.7 
10
0
10
0
1200e −0.07 t dt
1200e −0.07t dt
10
 1200 −0.07t 
= e 0.7 
e
0
 −0.07

≈ \$17,378.62
15
15
(b) Difference = e(0.10)(15)  1200e −0.10t dt  − e(0.09)(15)  1200e−0.09t dt 
 

0
0

15
15
= e 1.5  1200e −0.1t dt  − e 1.35  1200e−0.09t dt 

 

0
0
15
15
1200 −0.1t 
 1200 −0.09t 
e  − e 1.35 
e
= e 1.5 

 −0.1
0
 −0.09
0
≈ \$41,780.27 − \$38,099.01
= \$3681.26
18
82. Future value = e(0.07)(18)  450te − 0.07t dt
0
1.26
= 400e
18
0
te− 0.07t dt
Let u = t and dv = e− 0.07t dt. Then du = dt and v =
−1 − 0.07 t
e
.
0.07
 1 − 0.07 t  18
1 18 − 0.07t 
= 450e1.26 −
+
te
e
dt 

0.07  0
0
 0.07

18 
1
 −18 −1.26
e − 0.07t  
= 450e1.26 
−
e
0
0.07
0.0049


1
1 
 −18 −1.26
= 450e1.26 
−
e
e−1.26 +

0.07
0.0049
0.0049


≈ \$239,628.22
(a) Four years at The Pennsylvania State University costs \$30,452 &times; 4 = \$121,808. Yes, the fund would cover the costs.
(b) Four years at The Ohio State University costs \$26,537 &times; 4 = \$106,148. Yes, the fund would cover the costs.
83.
4
1
4
x +
3
x
dx = 4 
4
1
4
x +
 4 − 1
≈ 4
 f
 10  
3
x
dx
 23 
 29 
 35 
 41 
 47 
 53 
 + f + f + f + f + f 
 20 
 20 
 20 
 20 
 20 
 20 
 59 
 65 
 71 
 77 
+ f   + f  + f   + f   ≈ 4.254
 20 
 20 
 20 
 20 
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6
422
84.
Techniques of Integration
4
10
dx = 10 
x
1
xe
4
1
1
dx
xe x
 4 − 1  9 
 11 
 13
 15
 17 
 19 
 21
≈ 10
 f   + f   + f   + f   + f   + f   + f  
 12    8 
8
 8
 8
 8
 8
 8
 23
 25
 27
 29
 31
+ f   + f   + f   + f   + f   ≈ 2.691
 8
 8
 8
 8
 8 
Section 6.2 Integration Tables
Skills Warm Up
( x + 12 )
1.
(x
+ 4) = x 2 + 8 x + 16
3.
2.
(x
− 1) = x 2 − 2 x + 1
4. x −
2
2
(
1
3
)
2
2
= x2 + x +
= x2 −
2x
3
+
1
4
1
9
5. Let u = 2 x and dv = e x dx. Then du = 2 dx and v = e x .
 2 xe
x
dx = 2 xe x −
 2e
x
dx = 2 xe x − 2e x + C = 2e x ( x − 1) + C
1
dx and v = x 3 .
x
1
1

x 2 dx = x3 ln x − x3 + C = x3  ln x −  + C
3
3

6. Let u = ln x and dv = 3x 2 dx. Then du =
 3x
2
ln x dx = x 3 ln x −

1. Formula 4: u = x, du = dx, a = 2, b = 3
x
 (2 + 3x)2 dx
=
1 2

+ ln 2 + 3x  + C
9  2 + 3 x

2. Formula 11: u = x, du = dx, a = 3, b = 7
1
 x (3 + 7 x ) 2
dx =

1 1
1
x
+ ln
+C
3  3 + 7 x 3
3 + 7 x 
3. Formula 24: u = x 2 , du = 2 x dx, a = 4
8 x
x 2 + 16 dx = 8  x 2
2
x 2 + 16 dx = 8  18  x( 2 x 2 + 16)
 
= x( 2 x 2 + 16)
= 2 x( x 2 + 8)
= 2  x( x 2 + 8)

x 2 + 16 − 256 ln x +
x 2 + 16 − 256 ln x +
x 2 + 16 − 128 ln x +
4. Formula 21: u = x, du = dx, a = 3
4
1
4
x −3
ln
dx =
+C
2(3)
x −9
x +3
2
x 2 + 16 − 256 ln x
x 2 + 16   + C
 
x 2 + 16 + C
x 2 + 16 + C
x 2 + 16  + C

5. Formula 27: u = x 2 , du = 2 x dx, a = 3

2x
4
x −9
dx = ln x 2 +
x4 − 9 + C
2
x −3
= ln
+C
3
x +3
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
Integration Tables
423
6. Formula 23: u = 2 x 2 , du = 4 x dx, a = 5
x
4x
1
4
4 x 4 + 15 dx =
=
1 1
4
2
=
1
8
4 x 4 + 25 dx
(2 x
(2 x
2
2
4 x 4 + 25 + 25 ln 2 x 2 +
4 x 4 + 25 + 25 ln 2 x 2 +
3 x2
dx =
2 x2
xe
1
2
2 x dx =
1
2
4 x 4 + 25
)+C
15. Formula 42: u = 3 x, du = 3 dx
7. Formula 37: u = x 2 , du = 2 x dx
xe
)
4 x 4 + 25  + C

( x2
− 1)e x + C
2
 3x ln(3x) dx
8. Formula 39: u = x 2 , du = 2 x dx
x
 1 + e x2
1
2x
dx
2  1 + e x2
2
1
=  x 2 − ln 1 + e x  + C



2
dx =
)
(
 (ln 5 x)
1
x
dx = ln
+ C
1+ x
x(1 + x )
2
1

x
1 1
2
= −  + ln
 +C
5 x
5
5 + 2x 
1
x
2
x + 49
dx = −
1
7 +
ln
7
x 2 + 49
+C
x
12. Formula 27: u = x, du = dx, a = 1

1
2
x −1
dx = ln x +
1
x 4− x
2
dx = −
1
2+
ln
2
4− x
x
x2 − 9
dx = −
x2
x2 − 9
+ ln x +
x
=
3x 2
−1 + 2 ln (3 x) + C
4 
dx =
1
5
 (ln 5 x)
=
1
5
(5 x)2 −
2
5 dx
2
2 ln 5 x + (ln 5 x)  + C

(
dx = 3 x 2 − ln 1 + e3 x
2
)+C
18. Formula 38: u = 7 x 2 , du = 14 x dx
3 7 x2
 7x e
dx =
1
14
2 7 x2
 (7 x )e (14 x) dx
=
1  2 7 x2
14 
7 x e
2
−  e 7 x (14 x) dx

=
1  2 7 x2
14 
7 x e
2
− e7 x  + C

19. Formula 21: u = x 2 , du = 2 x dx, a = 6
2
+C
14. Formula 26: u = x, du = dx, a = 3

6x
x2 − 1 + C
13. Formula 33: u = x, du = dx, a = 2

2

1  (3 x )

−1 + 2 ln (3x )) + C
(
3 4


17. Formula 39: u = 3x 2 , du = 6 x dx
 1 + e3 x 2
11. Formula 28: u = x, du = dx, a = 7

=
2
= x 2 − 2 ln 5 x + (ln 5 x) 


10. Formula 12: u = x, du = dx, a = 5, b = 2
 x2 (5 + 2 x) dx
1
3 x ln (3x)(3) dx
3
16. Formula 44: u = 5 x, du = 5 dx
9. Formula 10: u = x, du = dx, a = b = 1

=
x2 − 9 + C
 x4
x
1
2x
dx =  4
dx
− 36
2 x − 36
=
1 1
x2 − 6 
ln 2

 +C

2  2(6)
x + 6 
=
1
x2 − 6
+C
ln 2
24
x + 6
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
424
Chapter 6
Techniques of Integration
20. Formula 24: u = 3t , du = 3 dt , a =
t
2
2
1 1
9t 2 2 + 9t 2 (3) dt
⋅
3 9
1 1 
2
2
=
 3t ( 2 ⋅ 9t + 2) 9t + 2 −
27  8 
2 + 9t 2 dt =
4
2 ln 3t +

9t 2 + 2   + C
 
1 
3t (18t 2 + 2) 9t 2 + 2 − 4 ln 3t + 9t 2 + 2  + C

216 
1 
6t (9t 2 + 1) 9t 2 + 2 − 4 ln 3t + 9t 2 + 2  + C
=

216 
1 
3t (9t 2 + 1) 9t 2 + 2 − 2 ln 3t + 9t 2 + 2  + C
=

108 
=
21. Formula 14: u = x, du = dx, a = 3, b = 1
x
2
3 + x dx =
=
2
(1)(2(2)
 x 2 (3 + x)3 2 − 2(3) x 3 + x dx


+ 3) 
2 2
32
12
x (3 + x) − 6  x(3 + x) dx

7
Formula 14: u = x, du = dx, a = 3, b = 1
=

2  2
2
32
32
x (3 + x ) − 3 
 x (3 + x ) − 6 
7 
1
2
1
3
+
)
 ( )( ( )
=
2 2
12
36
32
32
x (3 + x ) +
(3 + x)1 2 dx
 x (3 + x ) −
7
5
5 

=
2 2
12
24
32
32
x (3 + x ) +
(3 + x)3 2  + C
 x (3 + x ) −
7
5
5

=
2
32
(3 + x) (5 x 2 − 12 x + 24) + C
35
(

3 + x dx 

)
22. Formula 17: u = t , du = dt , a = 3, b = 4

3 + 4t
dt = 2 3 + 4t + 3 
t
t
1
dt
3 + 4t
Formula 15: u = t , du = dt , a = 3, b = 4

3 + 4t
3
dt = 2 3 + 4t +
ln
t
3
= 2 3 + 4t +
3 ln
23. Formula 23: u = x, du = dx, a =

3 + x 2 dx =
1
x
2

x 2 + 10
dx =
x
3
+C
3
3 + 4t −
3 + 4t +
3
+C
3
3
3 + x 2 + 3 ln x +
24. Formula 25: u = x, du = dx, a =

3 + 4t −
3 + 4t +
x 2 + 10 −
3 + x2  + C

10
10 ln
x 2 + 10
10 +
x
+C
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
25. Formula 6: u = x, du = dx, a = 1, b = 1
x2
1+ x
dx = −
Integration Tables
425
27. Formula 33: u = x, du = dx, a = 1
x
(2 − x) + ln 1 + x + C
2
 x2
26. Formula 40: u = x, du = dx, n = 2
1
1 − x2
dx = −
1 − x2
+C
x
28. Formula 30: u = x, du = dx, a = 2
1
1
2x
 1 + e2 x dx = x − 2 ln(1 + e ) + C
 x2
1
2
x −4
dx =
x2 − 4
+C
4x
29. Formula 9: u = x, du = dx, a = 9, b = − 2, n = 5
8x2
 (9 − 2 x )5
dx = 8 
x2
(9
− 2 x)
5
 1 

2(9)
−1
92

 + C
dx = 8 
+
−
3
5−3
5− 2
5 −1
 ( − 2)  (5 − 3)(9 − 2 x)

x
x
−
−
−
−
5
2
9
2
5
1
9
2
(
)(
)
(
)(
)

 

 1

18
81
−1
 + C
= 8 − 
+
−
2
3
4
 8  2(9 − 2 x)

x
x
3
9
2
4
9
2
−
−
(
)
(
)

 



18
81
−1
 +C
= −
+
−
2
3
4
 2(9 − 2 x)
3(9 − 2 x)
4(9 − 2 x) 

1
6
81
=
−
+
+C
2
3
4
2(9 − 2 x)
(9 − 2 x) 4(9 − 2 x)
30. Formula 4: u = x, du = dx, a = 1, b = −3
2
x
(1 − 3x)
2
dx =
2 1

+ ln 1 − 3 x  + C
9 1 − 3 x

31. Formula 43: u = 2 x, du = 2 dx, n = 2
 4x
ln ( 2 x) dx =
2
1
2
(2 x) ln (2 x)(2) dx
2
=
2 +1

1  (2 x)

−1 + ( 2 + 1) ln ( 2 x)  + C

2  ( 2 + 1)2 


=

1  8 x3
−
1 + 3 ln ( 2 x)  + C


2 9

=
4 x3
(−1 + 3 ln(2 x)) + C
9
32. Formula 36: u = x 2 , du = 2 x dx
 xe
x2
dx =
1
2
x2
 e (2 x) dx
=
1 e x2
2
+C
33. Formula 7: u = x, du = dx, a = −5, b = 3
x2
 (3x − 5)2
dx =
1
25

3x −
+ 10 ln 3 x − 5  + C
27 
3x − 5

34. Formula 13: u = x, du = dx, a = −1, b = 2
 4x − 1
1
1
x
− 4 ln
dx = −1
2  x 2 ( 2 x − 1)2
2x − 1
 x( 2 x − 1)

 +C

&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
426
Chapter 6
Techniques of Integration
35. Formula 3: u = ln x, du = (1 x) dx, a = 4, b = 3
ln x
dx =
x( 4 + 3 ln x)

=

37. Formula 19: u = x, du = dx, a = 1, b = 1
ln x  1 
 dx 
4 + 3 ln x  x 
1
0
2( 2 − x)
x
dx = −
3
1+ x
1
3 ln x − 4 ln 4 + 3 ln x  + C
9
= −
=
Formula 44: u = x, du = dx
3
dx = x(ln x) − 3 (ln x) dx
3
 4
2 − − 
 3
4− 2 2
3
≈ 0.391
36. Formula 45: u = x, du = dx, n = 3
 (ln x)
2
3
1

1 + x
0
2
38. Formula 19: u = x, du = dx, a = 5, b = 2
3
2
= x(ln x) − 3 x 2 − 2 ln x + (ln x)  + C


5
0
3
2
= x (ln x) − 3(ln x) + 6 ln x − 6 + C


−2(10 − 2 x)
x
dx =
12
5 + 2x
5

5 + 2x 
0
5 5
3
≈ 3.7
=
39. Formula 3: u = x, du = dx, a = − 7, b = 4
6
3
x
dx =
4x − 7
6
3
6
x
1

dx =  2 ( 4 x − ( − 7) ln − 7 + 4 x )
− 7x + 4x
4
3
6
1

=  ( 4 x + 7 ln − 7 + 4 x )
16
3
1
 1

= 
4(6) + 7 ln − 7 + 4(6)  − 
4(3) + 7 ln − 7 + 4(3) 
16
16

 

1
1
=
(24 + 7 ln 17 ) − 16 (12 + 7 ln 5 )
16
1
=
(12 + 7 ln 17 − 7 ln 5 )
16
1
=
12 + 7(ln 17 − ln 5)
16 
3
7 17
ln
=
+
≈ 1.2854
4 16
5
(
)
(
)
40. Formula 6: u = x, du = dx, a = −5, b = 3
x2
4
 2 3x − 5 dx =
4
1  3x

− ( −10 − 3 x) + 25 ln −5 + 3 x 
27  2
2
1
(132 + 25 ln 7) − 48
27 
1
=
(84 + 25 ln 7)
27
≈ 4.913
=
41. Formula 40: u = x, du = dx, n = 0.5
6
4
0
4
1
1


dx = 6  x −
ln (1 + e0.5 x )
1 + e0.5 x
0.5

0
{
}
= 6 4 − 2 ln (1 + e 2 ) − (0 − 2 ln 2) ≈ 6.795
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
42. Formula 43: u = x, du = dx, n = 3
2
1
Integration Tables
43. Formula 30: u = x, du = dx, a =
x3 ln ( x 2 ) dx = 2  x3 ln x dx
2
3
1 x 2
1
2
 x3 +1

1 + (3 + 1) ln x 
= 2 
−

2

 (3 + 1)
 1

dx = −
2

x +5
1

= −


2
 x4

= 2  ( 4 ln x − 1)
16

1
= −

1

= 2 ( 4 ln 2 − 1) −  ( −1) 
16



=
15 

= 2 4 ln 2 − 
16 

15
= 8 ln 2 −
≈ 3.6702
8
3
427
5
3
x2 + 5 

5x
 1
(3)2 +
5(3)
14
+
15
5  
− −
 
 
(1)2 +
5(1)
5 


6
5
6 − 14
≈ 0.2405
15
44. Formula 5: u = x, du = dx, a = 1, b = 3, n = 4
3
3
x
 0 (1 + 3x)4

1 
1
−1

+

dx =
2
4−2
4 −1
(3)  (4 − 2)(1 + 3x)
(4 − 1)(1 + 3x)  0
3
=

1
1
−1

+

2
3
9  2(1 + 3 x)
3(1 + 3x)  0

=
1  1
1   1 1 
+
−
 − − + 

9  200 3000   2 3 
=
9
500
45. Formula 35: u = x, du = dx, a = 4
2
 −2
1
(16 − x2 )
32

dx = 
 2
 ( 4)
2
x
(4)
2


2
− ( x) 
−2
2


x
= 

2
16 16 − x  − 2
=
2
16 12
−
(− 2)
16 12
0.05
1
=
4 12
=
1
8
3
Approximate area: 0.0722
−2
2
0
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
428
Chapter 6
Techniques of Integration
46. Formula 31: u = x, du = dx, a = 0.5
1

( x2
+ 0.25)
32

dx = 
 0.25
1


x 2 + 0.25  0
x

 
1
−
= 

 
2
 0.25 (1) + 0.25   0.25
1
=
−0
0.25 1.25
=



+ 0.25 
0
(0)
2
1
0.25 1.25
10
−2
0
2
Approximate area: 3.5777
47. Formula 12: u = x, du = dx, a = 2, b = 3
2
1
1
1  1 1
3
x
dx = −  + ln
2
9 x ( 2 + 3x)
9  2 x 2
2 + 3x
2


 1
= −
1  1 3  1   
3  1  
 + ln    − 1 + ln   
18  2 2  4   
2  5  
= −
1  1 3 1
1 
− +  ln − ln 
18  2 2  4
5 
 (1 4) 
3 ln

 (1 5) 
1
5
= − −1 + 3 ln 
36 
4
1
= − −1 +
36 
0.05
1
2
0
Approximate area: 0.0092
48. Formula 10: u = x, du = dx, a = 2, b = 5
2
1
2
10
x
1

dx = 10  ln
2 + 5 x  1
x( 2 + 5 x )
2

2
=  5 ln

2
5( 2)
+

= 5 ln

1
 − 5 ln

2
5(1)
+

2
1
− 5 ln
12
7
1
= 5 ln 6
1
7
7
= 5 ln
6
Approximate area: 0.7708
8
0
0
3
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
Integration Tables
429
49. Formula 24: u = x, a = 2, du = dx
0
A=
5
x2
x2 + 4
5
1
x( 2 x 2 + 4) x 2 + 4 − 16 ln x + x 2 + 4 
 0
8 
1
= 42 5 − 16 ln 5 + 3 + 16 ln 2

8
=
(
=
1

42 5 + 16 ln 
8

=
1

21 5 + 8 ln 
4

)
20
2 

5 + 3 
−1
2 

5 + 3 
5
−2
Approximate area: 9.8145
52. (a) Formula 37: u = 4 x, du = 4 dx
50. Formula 41: u = x 2 , du = 2 x dx
A =
4
0
x ln x 2 dx =
1
2
4
1
4
(−1 + ln x 2 )1
=
1 2
x
2
=
1
16
2
1
16
2
=
( −1 +
(
 4 xe
ln x 2 2 x dx
ln 16) − ( −1)
ln 16 − 15)
4x
dx =
1
4
4x
 4xe (4) dx
Then du = 4 dx and v =
 4 xe
4x
(4 x
− 1)e4 x + C
(
1 e4 x .
4
)  ( 14 e )(4) dx
dx = ( 4 x) 14 e 4 x −
= xe
=
0
1
4
(b) Let u = 4 x and dv = e 4 x dx.
= xe 4 x −
16
=
1
4
4x
(4 x
−
e
4x
1 e4 x
4
4x
dx
+C
− 1)e 4 x + C
5
−2
Approximate area: 14.6807
1
x
, du = dx
3
3
 x
 x  1 
ln   dx = 3  ln    dx
 3
 3  3 
51. (a) Formula 41: u =

x
 x  
= 3  −1 + ln    + C
3
 3  
 

 x 
= x −1 + ln    + C
 3 

 x
(b) Let u = ln   and dv = dx.
3
1 1
1
Then du =
  dx = dx and v = x.
x 3 3 
x
 x
 ln 3  dx
 x
= x ln   −
 3
 x
= x ln   −
 3
1
 x x  dx
 dx
 x
= x ln   − x + C
 3
  x

= x ln   − 1 + C
3
  

&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
430
Chapter 6
Techniques of Integration
53. (a) Formula 19: u = x, du = dx, a = − 3, b = 7

2( 2( − 3) − 7 x)
x
dx = −
2
−3 + 7x
3(7)
(b) Let u = x and dv = (7 x − 3)
Then du = dx and v =
 x(7 x − 3)
−1 2
−1 2
−3 + 7x + C =
dx.
2
12
(7 x − 3) .
7
2
2
12
12
x(7 x − 3) −  (7 x − 3) dx
7
7
2
4
12
32
= x(7 x − 3) −
(7 x − 3) + C
7
147
2
12
=
(7 x − 3) 21x − 2(7 x − 3) + C
147
2
7 x − 3 (7 x + 6) + C
=
147
dx =
55. Formula 19: u = x, du = dx, a = 4, b = 5
54. (a) Formula 42: u = 7 x, du = 7 dx
 7 x ln(7 x) dx
=
1
7 x ln (7 x)(7) dx
7
=
2

1  (7 x )

−1 + 2 ln (7 x)) + C
(
7 4


=
=

1  49 x 2
(−1 + 2 ln(7 x)) + C

7 4

b
1
= − (8 − 5 x) 4 + 5 x 
a
7
=
7 x2
(−1 + 2 ln(7 x)) + C
4
1
7
(7) dx and v = x 2 .
7x
2
 7x ln(7 x) dx
=
7 2
x ln (7 x) −
2

P( a ≤ x ≤ b) =
b
75 
 a 14 
x

 dx
4 + 5x 
75  2(8 − 5 x)
−
14 
75
b

4 + 5x 
a
0.8
1
(a) P(0.4 ≤ x ≤ 0.8) = − (8 − 5 x) 4 + 5 x 
0.4
7
1
= − 4 8 − 6 6 ≈ 0.483
7
(
(b) Let u = ln(7 x) and dv = 7 x dx.
Then du =
2
(7 x + 6) 7 x − 3 + C
147
7 2 1 
x   dx
2  x
)
0.5
1
(b) P(0 ≤ x ≤ 0.5) = − (8 − 5 x) 4 + 5 x 
0
7
1
= − 5.5 6.5 − 16 ≈ 0.283
7
(
)
7 2
7
x ln (7 x) −  x dx
2
2
7 2
7 2
= x ln (7 x) − x + C
2
4
7 x2
=
(2 ln(7 x) − 1) + C
4
=
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
Integration Tables
431
56. Formula 37: u = x 2 , du = 2 x dx
P( a ≤ x ≤ b ) =
b
3 x2
 a 2x e
b
2 x2
 a x e (2 x dx)
dx =
(
0.25
)
2
(a) P(0 ≤ x ≤ 0.25) =  x 2 − 1 e x 

0
(
= −0.9375e0.0625 − ( −1) ≈ 0.002
1
)
b
2
= ( x 2 − 1)e x 

a
(
)
2
(b) P(0.5 ≤ x ≤ 1) =  x 2 − 1 e x  = 0 − −0.75e0.25 ≈ 0.963

 0.5
57. Formula 39: u = 4.8 − 1.9t , du = −1.9 dt
2
1
5000
dt
2 − 0  0 1 + e4.8 −1.9t
2500 2
−1.9
= −
dt
1.9  0 1 + e 4.8 −1.9t
2
2500 
4.8 − 1.9t − ln (1 + e 4.8 −1.9t )
= −

0
1.9
2500 
= −
(1 − ln(1 + e)) − 4.8 − ln(1 + e4.8 ) 
1.9 
Average value =
(
6000
0
10
0
)
≈ 401.402
58. Formula 39: u = 4.20 − 0.25t , du = −0.25 dt
28
1
375
dt
28 − 21  21 1 + e 4.20 − 0.25t
28
−0.25
375
=
dt
7( −0.25)  21 1 + e 4.20 − 0.25t
Average =
400
28
1500 
4.20 − 0.25t − ln (1 + e 4.20 − 0.25t )
21
7 
1500 
= −
−2.8 − ln (1 + e −2.8 ) − −1.05 − ln (1 + e −1.05 ) 

7 
≈ 323.352
= −
(
59. R =
2

 0 10,0001 −

Formula 27: u = t
R = 20,000 −
) (

1
12
(1 + 0.1t )
2
0.1, du =
(
10,000 
ln t
0.1 

0
30
0
)
dt = 10,000t ]0 − 10,000
2
2
0
1
(1 + 0.1t 2 )
12
dt
0.1 dt , a = 1
0.1 +
)
2
0.1t 2 + 1  ≈ \$1138.43
 0
dS
, is approximately \$80 per week.
dt
dS
is positive over the entire interval 0 ≤ t ≤ 52.
(b) Sales are increasing over the entire interval, 0 ≤ t ≤ 52, since
dt
60. (a) After 16 weeks, t = 16, the rate of sales,
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
432
Chapter 6
Techniques of Integration
61. Equilibrium point:
60
2
=
x
3
x + 81
x
−
= 0
3
x 2 + 81
60
180 − x
x 2 + 81
x 2 + 81
3
x
= 0
x 2 + 81 = 180
x 2 ( x 2 + 81) = 32,400
x 4 + 81x 2 − 32,400 = 0
( x 2 − 144)( x 2 + 225) = 0
x 2 − 144 = 0
x 2 = 144  x = 12
2
x + 255 = 0
x 2 = 225  nonreal solution
p =
12
= 4
3
(a) Consumer surplus =
=
12
0 (demand function − price) dx



12
0

− 4  dx
x + 81

60
2
 12
15
dx −
= 4 
0
2
x
+ 81

Formula 27: u = x, du = dx
(
= 4 15 ln x +

x 2 + 81
)

 dx
12
− x
 0
≈ 17.92
(b) Producer surplus =
=
12
0 (price − supply function ) dx
12
0
x

 4 −  dx
3

12
1 2
x
6  0
= 4x −
= 24
Chapter 6 Quiz Yourself
1. Let u = x and dv = e5 x dx. Then du = dx and
v =
 xe
1 5 x.
e
5
5x
2.

ln x3 dx = 3 ln x dx
Let u = ln x and dv = dx. Then du =
e
dx =
1 5x
xe
5
−
1
5
=
1 5x
xe
5
−
1 5x
e
25
=
1 5x
e
5
5x
dx
+C
( x − 15 ) + C
1
dx and
x
v = x.

ln x 3 dx = 3 x ln x −  dx


= 3 x ln x − 3 x + C
= 3 x(ln x − 1) + C
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6
1

x dx = ln x x 2 + x  −
2

1

= ln x x 2 + x  −
2

433
1
1
dx and v = x 2 + x.
x
2
3. Let u = ln x and dv = ( x + 1) dx. Then du =
 ( x + 1) ln
Quiz Yourself
1 2
 1 
 x + x   dx
2
 x 

x
  2

+ 1 dx

1
 1
= ln x x 2 + x  − x 2 − x + C
2
 4
1 2
1
= x ln x + x ln x − x 2 − x + C
2
4
x
5.
3
x + 6 = ( x + 6)
13
3
4. Let u = x and dv =
dx. Then du = dx and v =
x + 6 dx =
3x
4
(x
+ 6)
43
−
3
4
=
3x
4
(x
+ 6)
43
−
9
28
(x
+ 6)
=
3
4
(x
+ 6)
43
x −

3
7
(x
+ 6) + C
=
3
4
(x
+ 6)
43
=
3
4
(x
+ 6)
43
( x − 73 x − 187 ) + C
( 74 x − 187 ) + C
=
3
14
(x
+ 6)
43
(2 x
 ( x + 6)
43
73
3
4
(x
+ 6) .
43
dx
+C
− 9) + C
1
x ln x dx
2
1
1
Let u = ln x and dv = x dx. Then du =
dx and v = x 2 .
x
2
1 1 2
1

 x ln x dx = 2  2 x ln x − 2  x dx
1
1
= x 2 ln x − x 2 + C
4
8
 x ln
x dx =
 x ln
x1 2 dx =
1
6. Let u = x 2 and du = e−2 x dx. Then du = 2 x dx and v = − e −2 x .
2
xe
2 −2 x
dx = −
x 2 −2 x
e
+
2
 xe
−2 x
dx
1
Let u = x and dv = e−2 x dx. Then du = dx and v = − e−2 x .
2
x 2 −2 x
x −2 x
1
−2 x
2 −2 x
 x e dx = − 2 e − 2 e + 2  e dx
1
x2
x
= − e −2 x − e −2 x − e −2 x + C
2
2
4
1 −2 x
2
= − e ( 2 x + 2 x + 1) + C
4
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
434
Chapter 6
Techniques of Integration
(
7. x = 1000 45 + 20te−0.5t
)
 0 1000(45 + 20te
5
(a) Total demand =
− 0.5t
) dt
5
5
0
0
= 1000  45 dt + 1000  20te − 0.5t dt
5
5
0
0
= 45,000  dt + 20,000  te − 0.5t dt
5
= 45,000[t ] 0 + 20,000 te − 0.5t dt
5
0
5
= 225,000 + 20,000 te − 0.5t dt
0
Let u = t and dv = e − 0.5t dt. Then du = dt and v = − 2e− 0.5t .
{
= 225,000 + 20,000{−10e
5
5
}
= 225,000 + 20,000 − 2te − 0.5t  + 2  e − 0.5t dt
0
0
− 2.5
}
5
− 4e − 0.5t 
0
= 225,000 − 200,000e − 2.5 − 80,000e − 2.5 + 80,000 ≈ 282,016 units
5
1
1000( 45 + 20te − 0.5t ) dt
5 − 0 0
1 5
=  1000( 45 + 20te − 0.5t ) dt
5 0
1
≈ ( 282,016) ≈ 56,403 units
5
(b) Average annual demand =
8. (a) Actual income =
(b) Present value =
7
7
0
32,000t dt = 16,000t 2  = \$784,000
0
7
32,000te − 0.033t dt = 32,000  te − 0.033t dt
0
7
0
Let u = t and dv = e − 0.033t dt. Then du = dt and v = −
1 − 0.033t
e
.
0.033
7
7
1
1



te − 0.033t  +
e − 0.033t dt 
= 32,000−

0
0.033
0.033
0


7

 1
 
1

− 0.033(7)
− 0.033t
 
e
= 32,000−
− 
( 7 )e
2
 (0.033)
 
 0.033
0

7 − 0.231
1
1 
e
e − 0.231 +
= 32,000−
−
2
2
(0.033)
(0.033) 
 0.033
≈ \$673,108.31
9. Formula 3: u = x, du = dx, a = 1, b = 2
x
 1 + 2x
dx =
1
(2 x − ln 1 + 2 x
4
)+C
10. Formula 10: u = x, du = dx, a = 0.1, b = 0.2

x
1
1
+ C
dx =
ln
x(0.1 + 0.2 x )
0.1
0.1 + 0.2 x
= 10 ln
x
+ C
0.1 + 0.2 x
11. Formula 26: u = x, du = dx, a = 4

x 2 − 81
dx = −
x2
x 2 − 81
+ ln x +
x
x 2 − 81 + C
12. Formula 15: u = x, du = dx, a = 4, b = 9

1
1
dx =
ln
2
x 4 + 9x
4 + 9x − 2
+C
4 + 9x + 2
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6
13. Formula 39: u = x 2 , du = 2 x dx, n = 4
2x
 1 + e 4 x2
14. Formula 37: u = x 2 + 1, du = 2 x dx
)
(
435
 2 x( x
2
1
ln 1 + e 4 x + C
4
2
+ 1)e x
2 +1
dx = x 2e x
2 +1
+C
144t 2 + 400
15. R =

dx = x 2 −
Quiz Yourself
144t 2 + 400 dt
Formula 23: u = 12t , du = 12 dt , a = 20

144t 2 + 400 dt =
1
12

144t 2 + 400 (12) dt
=
1 1
12 
2
(12t
=
1
t
2
144t 2 + 400 +
144t 2 + 400 + 400 ln 12t +
50
3
)
144t 2 + 400  + C

)
(
144t 2 + 400 + C
ln 12t +
(a) Total revenue over first three years:
144t 2 + 400 dt =  12 t

3
0
144t 2 + 400 +
50
3
(
ln 12t +
)
3
)
6
144t 2 + 400 
 0
≈ \$84.28112652 million or \$84,281,126.52
(b) Total revenue over first six years:
6
0
144t 2 + 400 dt =  12 t

144t 2 + 400 +
50
3
(
ln 12t +
144t 2 + 400 
 0
≈ \$257.39242972 million or \$257,392,429.72
16. Let u = x and dv = e x 2 dx. Then du = dx and
v = 2e x 2 .
0
 − 2 xe
x 2
18. Formula 19: u = x, du = dx, a = 8, b = 1
8
dx = 2 xe x 2 
0
−2
−
0
 − 2 2e
x2
dx
0
 −16 
= 
( 4)  −
 3

− 64 64
=
+
3
3
64
=
2 −1
3
≈ 8.8366
0
= 4e −1 − 4e x 2 
−2
= 4e −1 − ( 4 − 4e−1 )
8
− 4
e
≈ −1.057
=
8
 2(16 − x )
x
dx = −
3
8+ x

(

x + 8
0
 − 32

2 2 

3


(
)
2
)
17. Formula 42: u = x, du = dx
2
1
2
5 x ln x dx = 5  x ln x dx
1
2
 x2

= 5 ( −1 + 2 ln x)
4
1
1


= 5( −1 + 2 ln 2) − ( −1)
4


 3
= 5 − + 2 ln
 4
15
= 10 ln 2 −
4

2

≈ 3.1815
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
436
Chapter 6
Techniques of Integration
19. Formula 44: u = 6 x, du = 6 dx
e
2
 1 (ln 6x) dx
=
1 e
6 1
=
1
6
2
 (ln 6x) (6) dx
e
2
⋅ 6 x 2 − 2 ln (6 x) + (ln 6x) 

1
(
) (
)
2
2
= e 2 − 2 ln 6e + (ln 6e) − 2 − 2 ln 6 + (ln 6) 


= 2e − 2e ln 6e + e(ln 6e) − 2 + 2 ln 6 − (ln 6)
2
2
≈ 9.8182
20. Formula 34: u = x, du = dx, a = 3
1
3
2
x2
dx = −
9 − x2
3
9 − x2 
5
≈ 0.124
 =
9x
18
 2
21. Formula 21: u = x 2 , du = 2 x dx, a = 2
6
4
6
1
x2 − 2 
2x
dx =  ln 2

4
x − 4
x + 2  4
 4
1  17
7
ln
− ln 
4  19
9
≈ 0.035
=
Section 6.3 Numerical Integration
Skills Warm Up
1. f ( x) =
4. f ( x) = x3 − 2 x 2 + 7 x − 12
1
= x−2
x2
2
f ′( x) = − 2 x = − 3
x
6
−4
f ′′( x) = 6 x = 4
x
−3
2. f ( x) = ln ( 2 x + 1)
f (4) ( x) = −
+ 1)
3. f ( x) = 2 ln x
f ′( x) =
6. f ( x ) = e x
2
x
2
f ′′( x) = − 2
x
4
f ′′′( x) = 3
x
12
4
f ( ) ( x) = − 4
x
2
f ′( x) = 2 xe x
3
+ 1)
f ′( x) = 3e3 x
f (4) ( x) = 81e3 x
2
2
96
(2 x
5. f ( x) = e3 x
f ′′′( x) = 27e3 x
16
(2 x
f ′′( x) = 6 x − 4
f ′′( x) = 9e3 x
2
f ′( x) =
2x + 1
4
f ′′( x) = −
(2 x + 1)2
f ′′′( x) =
f ′( x) = 3 x 2 − 4 x + 7
4
f ′′( x) = 4 x 2e x + 2e x
2
7. f ( x ) = − x 2 + 6 x + 9,
[0, 4]
f ′( x) = −2 x + 6 = 0 when x = 3
f (0) = 9, f (3) = 18, and f ( 4) = 17, so (3, 18) is
the absolute maximum.
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
Numerical Integration
437
Skills Warm Up —continued—
8. f ( x ) =
8
,
x3
f ′( x) = −
[1, 2]
10.
24
x4
1
&lt; 0.0001
16n 4
1
&lt; 16n 4
0.0001
10,000 &lt; 16n 4
24
≠ 0 for any value of x, there are no
x4
critical numbers. f (1) = 8 and f ( 2) = 1, so (1, 8) is
Because −
625 &lt; n 4
4
625 &lt; n
the absolute maximum.
9.
5 &lt; n
So, n &gt; 5.
1
&lt; 0.001
4n 2
1
&lt; 4n 2
0.001
1000 &lt; 4n 2
250 &lt; n 2
n 2 − 250 = 0
250 &lt; n
5 10 &lt; n
So, n &gt; 5 10 or n &gt; 15.81.
1. Exact:
2
0
Trapezoidal Rule:
Simpson’s Rule:
2. Exact:
1
0
2
1
=
8
3
x 2 dx ≈
x 2 dx ≈
≈ 2.6667
1 0
4

1 0
6

( 12 )
+ 2
( 12 )
+ 4
2
2
+ 2(1) + 2
2
( 32 )
+ 2(1) + 4
2
( 23 )
2
2
2
+ ( 2)  =

2
+ ( 2)  =

8
3
11
4
= 2.75
≈ 2.6667
1
 x3

 x4

9
1

+ 1 dx =  + x =  + 1 − 0 =
= 1.125

2
8
8
8





0
Simpson’s Rule:
3
2
0
0
Trapezoidal Rule:
3. Exact:
2
1 x3 
3 0
x 2 dx =
1
 x3
 0  2
1
 x3
 0  2

1
145
 1

1

 27
 1

+ 1 dx ≈ (0 + 1) + 2
+ 1 + 2
+ 1 + 2
+ 1 +  + 1 =
≈ 1.133
8
128
16
128
2
128





 




1
9
 1

1

 27
 1

+ 1 dx ≈
(0 + 1) + 4 128 + 1 + 2 16 + 1 + 4 128 + 1 +  2 + 1 = 8 = 1.125
12





 



3
3


(4 − x 2 ) dx = 4 x − x3  = − 23 ≈ −0.6667

1
Trapezoidal Rule:
Simpson’s Rule:
2
 1 (4 − x ) dx
3
2
 1 (4 − x ) dx
3
≈
≈
1

7
 9
3 + 2  + 2(0) + 2 −  + ( −5) = −0.75
4 
 4
 4

1

2
7
 9
3 + 4  + 2(0) + 4 −  + ( −5) = − ≈ −0.6667
6 
3
 4
 4

&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
438
Chapter 6
4. Exact:
2
1
Techniques of Integration
2
dx = 2 ln x  12 = 2 ln 2 ≈ 1.3863
x
Trapezoidal Rule:
Simpson’s Rule:
5. Exact:
4
0
2
1
4
2
+ 2
2 + 4
5
2
+ 2
+ 2
1
1
dx ≈
1 +
2
x
12 
2
2
2
1
 4
 2
 4
4  + 2  + 4  +  ≈ 0.5004
4 
5
 3
7
1
4
3
7
2
3 + 2
3 + 4
7
2
+ 2 ≈ 5.2650

+ 2 ≈ 5.3046

1
1
 4
 2
 4  1
dx ≈ 1 + 2  + 2  + 2  +  ≈ 0.6970
x −2
8
5
3
 
 
 7  2
1
1
 4
 2
 4  1
dx ≈
1 + 4  + 2  + 4  +  ≈ 0.6933
x −2
12 
5
3
 
 
 7  2
4
3
2
1
3 x2
2
1 0
e
8
e − 4 x dx ≈
e− 4 x dx ≈
1 0
e
12 
1
2
dx =  16 e3 x  =

0
Simpson’s Rule:
x
2
0
0
Trapezoidal Rule:
2
3
2
5
2
2 + 2
2
2
2
1
 4
 2
4
2  + 2  + 2  +  ≈ 0.5090
4 
5
 3
7
e − 4 x dx = − 14 e − 4 x  = − 14 e −8 +
0
 0 xe
0
+ 2 1 + 4
+ 2
4
1
dx = ln x − 2  3 = ln 2 − ln 1 = ln 2 − 0 = ln 2 ≈ 0.6931
x − 2
Simpson’s Rule:
10. Exact:
1
2
+ 4
3
2
+ 2 1 + 2
1
1
dx ≈ 1 +
2
x
8 
2
1
2
Trapezoidal Rule:
9. Exact:
1
2
+ 2
2
Simpson’s Rule:
0
1
0
6

x dx ≈
≈ 5.3333
1
0
4

x dx ≈
4
Trapezoidal Rule:
8. Exact:
16
3
=
1
1
dx = −  = 0.5
x2
x 1
Simpson’s Rule:
3
4
0
0
Trapezoidal Rule:
7. Exact:
4
2 3 2
x 
3
0
x dx =
Simpson’s Rule:
6. Exact:
1
1

 16 
8
 16 
4
 16 
8
 16 
dx ≈
2 + 4  + 2  + 4  + 2  + 4  + 2  + 4  + 1 ≈ 1.3863
x
24 
9
5
11
3
13
7
15
 
 
 
 
 
 
 

2
1
Trapezoidal Rule:
1
1

 16 
8
 16 
4
 16 
8
 16 
dx ≈
2 + 2  + 2  + 2  + 2  + 2  + 2  + 2  + 1 ≈ 1.3882
x
16 
9
5
11
3
13
7
15
 
 
 
 
 
 
 

2
1
1
 0 xe
1
 0 xe
3 x2
3 x2
dx ≈
x 2 + 1 dx =
Trapezoidal Rule:
Simpson’s Rule:
2
0
2
0
x
x
dx ≈
1
3
( x2
≈ 0.2499
+ 2e −1 + 2e − 2 + 2e − 3 + 2e− 4 + 2e − 5 + 2e − 6 + 2e − 7 + e −8  ≈ 0.2704
+ 4e−1 + 2e − 2 + 4e− 3 + 2e − 4 + 4e− 5 + 2e− 6 + 4e− 7 + e−8  ≈ 0.2512
1 3
e
6
1
0
8
1
0
12 
1
4
1
6
≈ 3.1809
(
)
(
)
(
)
(
)
(
)
(
)
−
+ 2 14 e3 16 + 2 12 e3 4 + 2 34 e 27 16 + e3  ≈ 3.8643

+ 4 14 e3 16 + 2 12 e3 4 + 4 34 e 27 16 + e3  ≈ 3.3022

2
32
+ 1)  ≈ 3.3934
 0
x 2 + 1 dx ≈
x 2 + 1 dx ≈
1
0
4

1
0
6

( 12 )
1
4
+ 2
( 12 )
+ 4
1
4
( 32 )
+ 1 + 2(1) 1 + 1 + 2
( 32 )
+ 1 + 2(1) 1 + 1 + 4
9
4
9
4
+1 + 2
+1 + 2
4 + 1 ≈ 3.4567

4 + 1 ≈ 3.3922

&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
11. (a) Trapezoidal Rule:
(b) Simpson’s Rule:
(b) Simpson’s Rule:
439




1
1 1
2
2
2
1 
≈
+
+
+
+
≈ 0.7828
dx
2
2
2
1 + x2
8 1 + 0
1 + 12 
1
1
 3


1+ 
1+ 
1+ 

 4
 2
 4

1
0




1
1  1
4
2
4
1 
≈
+
+
+
+
≈ 0.7854
dx
2
2
2
1 + x2
12 1 + 0
1 + 12 
1
1
3


1+ 
1+ 
1+ 

 4
 2
4

1
0
12. (a) Trapezoidal Rule:
Numerical Integration
8
1
8
 8 
 8 
 8 
dx ≈ 0 + 2 2
 + 2 2
 + 2 2
 + 2
+
+
+
+
x2 + 3
2
1
3
2
3
3
3
4






4
0
8
1
8
 8 
 8 
 8 
dx ≈ 0 + 4 2
 + 2 2
 + 4 2
 + 2
+
+
+
+
x2 + 3
3
1
3
2
3
3
3
4






4
0

≈ 4.020
3 

≈ 4.458
3 
13 (a) Trapezoidal Rule:
2
0
1
4
9−
( 12 )
1
6
9−
( 12 )

3
9 − (0) + 2


9 − x3 dx ≈
3
 
 +
 
3 
9 − ( 2)  ≈ 5.090

3
 
+
 
3 
9 − ( 2)  ≈ 5.177

3


 + 2



3
9 − (1)  + 2


9−
( 32 )
3


 + 2



3
9 − (1)  + 4


9−
( 23 )
(b) Simpson’s Rule:
2
0

3
9 − (0) + 4


9 − x3 dx ≈
14 (a) Trapezoidal Rule:
1 3
0
1 3
16 
4 − x 2 dx ≈

2
4 − (0) + 2 3 4 −


( 18 )
2

3
 + 2 4 −


( 14 )

+ 2 3 4 −

( 85 )

2
4 − (0) + 4 3 4 −


( 18 )

+ 4 3 4 −

( 58 )
2
3

 + 2 4 −


( 83 )
2

3
 + 2 4 −


( 43 )

3
 + 2 4 −


( 14 )

3
 + 2 4 −


( 34 )
2

3
 + 2 4 −


2

3
 + 2 4 −


( 87 )
2

 +

3

 + 4 4 −


( 83 )

3
 + 4 4 −


( 78 )
3
( 12 )
2



2
4 − (1)  ≈ 1.540

(b) Simpson’s Rule:
1 3
0
1 3
24 
4 − x 2 dx ≈
2
2
2
2
2
2

3
 + 2 4 −



 +

3
( 12 )
2



2
4 − (1)  ≈ 1.541

15. (a) Trapezoidal Rule:
1
0 e
x2
dx ≈
1 e(0)
16 

2
2
2
2
2
2
2
2
+ 2e(1 8) + 2e(1 4) + 2e(3 8) + 2e(1 2) + 2e(5 8) + 2e(3 4) + 2e(7 8) + e(1)  ≈ 1.470

2
(b) Simpson’s Rule:
1
0 e
x2
dx ≈
1 e(0)
24 

16. (a) Trapezoidal Rule:
(b) Simpson’s Rule:
2
2
2
2
2
2
2
2
2
2
+ 4e(1 8) + 2e(1 4) + 4e(3 8) + 2e(1 2) + 4e(5 8) + 2e(3 4) + 4e(7 8) + e(1)  ≈ 1.463

0 e
2
0 e
− x2
− x2
dx ≈
dx ≈
1
6
1
4
(1 + 2e−1 4 + 2e−1 + 2e−9 4 + e−4 )
(1 + 4e−1 4 + 2e−1 + 4e−9 4 + e−4 )
≈ 0.881
≈ 0.882
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
440
Chapter 6
Techniques of Integration
17. (a) Trapezoidal Rule:
1
4
0
3
2
x +1


1
0 + 2

4


dx ≈

+ 2


1
(1 2)
3
2
1
3
(5 2)
2

 + 2



+ 1

 + 2



+ 1


+ 2


12 + 1 

1
3
(3 2)
3


+ 2


32 + 1 

1
3

 + 2



+ 1
1
2

+

+ 1 
1
3
(7 2)
2


2 + 1
1
3
2

 ≈ 2.540
42 + 1 

1
3
(b) Simpson’s Rule:
1
4
0
3
2
x +1


1
0 + 4

6


dx ≈

+ 4


18. (a) Trapezoidal Rule:
(b) Simpson’s Rule:
(b) Simpson’s Rule:
1
1+ x
3
3
2
1
3
(5 2)
2
dx ≈
3
dx ≈

 + 2



+ 1

 + 2



+ 1 
3
3
4
x
5
3
0
−x
3
x
 0 2 + x + x2
dx ≈
dx ≈ 3.6558.
3
2 − x +1
9x e
(7 2)
2

 ≈ 2.541
42 + 1 

1
3
1
3
2
1
6
1
 10 
0 + 4  + 2  + 4  + 2  + 4  +  ≈ 0.653
6 
11
4
23
4
43
14
 
 
 
 
 

24. Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 100,
6
3

 +

+ 1 
x
1
3
2
1
6
1
 10 
dx ≈ 0 + 2  + 2  + 2  + 2  + 2  +  ≈ 0.641
2
2+ x + x
4
 11 
 4
 23 
 4
 43  14 
23. Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 100
5
1
2
1
1 1
 4
 4
1
 4  1
dx ≈  + 2  + 2(1) + 2  + 2  + 2  +  ≈ 1.879
2
2 − 2x + x
4 2
5
5
 2
 13  5 
x + 6 dx ≈ 129.0692.
 2 10 xe



+
4


32 + 1 



2 + 1
1
3
1
 1 
 1 
 1  1
+ 4
1 + 4 9  + 2

 +  ≈ 1.405
6


 35  3
 2
8
8






x + 4 dx ≈ 19.5215.
x2
(3 2)
3
2
1
 1 
 1 
 1  1
1 + 2 9  + 2
 + 2 35  +  ≈ 1.397
4
2



 3


 8
 8 


22. Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 100,
1

 + 2



+ 1
1
1
21. Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 100
1


+ 4


12 + 1 

1
1
1 1
 4
 4
1
 4  1
dx ≈  + 4  + 2(1) + 4  + 2  + 4  +  ≈ 1.888
2 − 2x + x2
6 2
5
5
2
 
 
 
 13  5 
3
0
20. (a) Trapezoidal Rule:
(b) Simpson’s Rule:
1+ x
2
0
(1 2)
3
1
2
0
0
19. (a) Trapezoidal Rule:
1
dx ≈ 17.6742.
25. ΔR =
16
 14 5
8000 − x3 dx
Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 4, ΔR ≈ \$678.36.
26. ΔR ≈
16
 14 50
x
20 − x dx
Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 4, ΔR ≈ \$863.44.
27. P(0 ≤ x ≤ 1) =
1
0
1 − x2 2
e
dx
2π
Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 6, P ≈ 0.3413 = 34.13%.
28. P(0 ≤ x ≤ 2) =
2
0
1 − x2
e
2π
2
dx
Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 6, P ≈ 0.4772 = 47.72%.
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
3.5
29. P(0 ≤ x ≤ 3.5) =
0
1 − x2 2
e
dx
2π
Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 6, P ≈ 0.49998 = 49.98%.
31. A ≈
Numerical Integration
30. P(0 ≤ x ≤ 1.5) =
441
1 − x2 2
e
dx
2π
1.5
0
Using a program similar to the Simpson’s Rule program
in Appendix E, when n = 6, P ≈ 0.4332 = 43.32%.
1000
125 + 2(125) + 2(120) + 2(112) + 2(90) + 2(90) + 2(95) + 2(88) + 2(75) + 2(35) + 0
2(10) 
= 89,250 m 2
32. A ≈
120
75 + 2(81) + 2(84) + 2(76) + 2(67) + 2(68) + 2(69) + 2(72) + 2(68) + 2(56) + 2( 42) + 2( 23) + 0
2(12) 
= 7435 m 2
33. f ( x) = x 2 + 2 x
f ′( x) = 2 x + 2
f ′′( x) = 2
f ′′′( x) = 0
f (4) ( x) = 0
(a) Trapezoidal Rule: Because f ′′( x) is maximum
for all x in [0, 2] and f ′′( x ) = 2, you have
(2 − 0) 2
2 ( )
12( 4)
3
Error ≤
34. f ( x) =
1
≈ 0.0833.
12
=
1
x +1
f ′( x) = −
f ′′( x) =
+ 1)
+ 1)
f ′′′( x) = −
6
(x
+ 1)
= 0.
3
3
[0, 1] when
5
3
3
3
x = 1 and f ′′(1) = 15e, you have
(1 − 0) 15e
)
2 (
12( 4)
3
x = 0 and f ′′(0) = 2, you have
(1 − 0) 2
2 ( )
12( 4)
3
=
1
≈ 0.01.
96
(b) Simpson’s Rule: Because f (4) ( x) is maximum in
[0, 1] when
(0)
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
4
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
x = 0 and
(1 − 0) 24
)
4 (
180( 4)
f (4) (0) = 24, you have
5
Error ≤
180( 4)
5
4
f (4) ( x ) = 9(9 x8 + 36 x5 + 20 x 2 )e x
+ 1)
[0, 1] when
− 0)
f ′′′( x) = 3(9 x 6 + 18 x3 + 2)e x
3
24
Error ≤
(2
f ′′( x) = 3(3 x 4 + 2 x)e x
2
2
(x
Error ≤
f ′( x ) = 3 x 2e x
(x
f ( 4) ( x ) =
for all x in [0, 2] and f (4) ( x ) = 0, you have
35. f ( x) = e x
1
(x
(b) Simpson’s Rule: Because f (4) ( x) is maximum
=
Error ≤
=
5e
≈ 0.212.
64
(b) Simpson's Rule: Because f (4) ( x) is maximum in
[0, 1] when
4
x = 1 and f ( ) (1) = 585e, you have
(1 − 0) 585e
)
4(
180( 4)
5
Error ≤
=
13e
≈ 0.035.
1024
1
≈ 0.00052.
1920
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442
Chapter 6
36. f ( x) = e2 x
Techniques of Integration
2
38. f ( x) =
f ′( x) = 4 xe 2 x
2
f ′′( x) = (16 x 2 + 4)e2 x
2
f (4) ( x) = ( 256 x 4 + 384 x 2 + 48)e 2 x
2
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
x = 1 and f ′′(1) = 20e 2 , you have
(1 − 0)
2
12( 4)
Error ≤
1
x2
2
f ′′( x) = 3
x
6
f ′′′( x) = − 4
x
24
f ( 4) ( x ) = 5
x
f ′( x) = −
2
f ′′′( x) = 16 x( 4 x 2 + 3)e 2 x
[0, 1] when
1
x
3
20e2 ≈ 0.7697.
(b) Simpson’s Rule: Because f (4) ( x) is maximum in
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
[1, 3] when
Error ≤
x = 1 and f ′′(1) = 2, you have
(3 − 1)
3
12n 2
[0, 1] when x = 1 and f (4) (1) = 688e2 , you have
5
(1 − 0) 688e2 ≈ 0.1103.
Error ≤
4
180( 4)
( 2)
&lt; 0.0001
4
&lt; 0.0001
3n 2
n 2 &gt; 13,333.33
n &gt; 115.47.
Let n = 116.
37. f ( x) = x 4
(b) Simpson’s Rule: Because f (4) ( x) is maximum in
f ′( x) = 4 x3
[1, 3] when
f ′′( x) = 12 x 2
f ′′′( x) = 24 x
Error ≤
f (4) ( x) = 24
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
[0, 2] when
x = 2 and f ′′( 2) = 48, you have
x = 1 and f (4) (1) = 24, you have
(3 − 1)
180n 4
5
( 24)
&lt; 0.0001
64
&lt; 0.0001
15n 4
n 4 &gt; 42,666.67
n &gt; 14.4.
Error ≤
(2
− 0) ( 48)
Let n = 16. (n must be even.)
3
12n 2
&lt; 0.0001
32
&lt; 0.0001
n2
320,000 &lt; n 2
565.69 &lt; n.
Let n = 566.
(b) Simpson’s Rule: Because f (4) ( x) = 24 for all x in
[0, 2], you have
Error ≤
(2
− 0)
180n 4
5
(24)
&lt; 0.0001
64
&lt; 0.0001
15n 4
n 4 &gt; 42,666.67
n &gt; 14.37.
Let n = 16.
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
Numerical Integration
443
39. f ( x) = e2 x
f ′( x) = 2e2 x
f ′′( x) = 4e2 x
f ′′′( x) = 8e2 x
f (4) ( x) = 16e2 x
(b) Simpson’s Rule: Because f (4) ( x) is maximum
(a) Trapezoidal Rule: Because f ′′( x) is maximum
in [1, 3] when x = 3 and f ′′(3) = 4e6 , you have
Error ≤
(3 − 1)3
12n 2
in [1, 3] when x = 3 and f (4) (3) = 16e6 , you have
(4e6 ) &lt; 0.0001
Error ≤
(3 − 1)5
180n 4
8e6
&lt; 0.0001
3n 2
(16e6 ) &lt; 0.0001
128e6
&lt; 0.0001
45n 4
n 2 &gt; 10,758,101.16
n 4 &gt; 11,475,307.90
n &gt; 3279.95.
n &gt; 58.2.
Let n = 60. (n must be even.)
Let n = 3280.
40. f ( x) = ln x
f ′( x) =
1
x
1
x2
2
f ′′′( x) = 3
x
6
f (4) ( x) = − 4
x
f ′′( x) = −
(b) Simpson’s Rule: Because f (4) ( x) is maximum
(a) Trapezoidal Rule: Because f ′′( x) is maximum
in [3, 5] when x = 3 and f ′′(3) =
Error ≤
1
, you have
9
in [3, 5] when x = 3 and f (4) (3) =
(5 − 3)3  1 
Error ≤
  &lt; 0.0001
12n  9 
2
&lt; 0.0001
27 n 2
2
(5 − 3)5 
2
  &lt; 0.0001
180n 4  27 
16
&lt; 0.0001
1215n 4
n 2 &gt; 740.74
Let n = 28.
2
, you have
27
n 4 &gt; 131.69
n &gt; 27.2.
Let n = 4.
n &gt; 3.39.
41. Using a program similar to the Simpson’s Rule program in Appendix E, when n = 8, V ≈ \$21,831.20.
Using a graphing utility, V =
=
t
1
− rt
 0 c (t )e
dt
−0.07 t
 0 (6000 + 200 t )e
4
dt ≈ \$21,836.98.
42. Using a program similar to the Simpson’s Rule program in Appendix E, when n = 8, V ≈ \$1,215,971.94.
Using a graphing utility, V =
=
t
1
− rt
 0 c (t )e
dt
 0 (200,000 + 15,000
8
3
)
t e −0.10t dt ≈ \$1,218,045.50.
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444
Chapter 6
Techniques of Integration
43. (a) Average median age =
1
13 − 5
13
5 f ( x) dx
≈
1 1
 36.2 + 4(36.3) + 2(36.5) + 4(36.7) + 2(36.8) + 4(37.2)
8 3 

+ 2(37.3) + 4(37.5) + 37.6  ≈ 36.9 yr

13
1
(37.2 − 0.54t + 0.084t 2 − 0.0031t 3 ) dt
13 − 5 5
13
1
= 37.2t + 0.27t 2 + 0.028t 3 − 0.000775t 4 
5
8
≈ 36.9 yr
(b) Average median age =
(c) The results are approximately equal.
44. (a) Average residential price =
≈
1
14 − 6
14
6 f ( x) dx
1 8
10.40 + 4(10.65) + 2(11.26) + 4(11.51) + 2(11.54)

8  3(8) 

+ 4(11.72)+ 2(11.88) + 4(12.12) + 12.54 

≈ 11.5125 cents per killowatt-hour
(b) Average residential price =
1
14 − 6
16 (516.23254e
14
−t
+ 0.0182t 3 − 0.595t 2 + 6.58t − 12.9) dt
14
=
1
0.595 3

t + 3.29t 2 − 12.9t 
− 516.23254e − t + 0.00455t 4 −

8
3
6
≈ 11.49856 cents kilowatt-hour
(c) The results are approximately equal.
45. C =
12
0
8 − ln (t 2 − 2t + 4) dt


+ 4(6.82) + 2(6.05) + 4(5.28) + 2( 4.67) + 4( 4.19) + 2(3.80) + 4(3.46) + 3.18
≈ 58.915 mg
≈
1 6.61
2
46. (a) The Trapezoidal Rule tends to become more accurate as n increases, therefore using n = 16 should yield
a more accurate approximation.
(b) For a given value of n, Simpson’s Rule tends to be more accurate than the Trapezoidal Rule, therefore using
Simpson’s Rule with n = 8, should yield a more accurate approximation.
47. S =
6
 0 1000t e
2 −t
dt ≈
1
6
[0 + 4(151.63) + 2(367.88) + 4(502.04) + 2(541.34) + 4(513.03) + 2(448.08)
+ 4(369.92) + 2( 293.05) + 4( 224.96) + 2(168.45) + 4(123.62) + 89.24]
≈ 1878 subscribers
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Section 6.4
48.
Improper Integrals
445
P3 ( x) = ax3 + bx 2 + cx + d
P3′ ( x) = 3ax 2 + 2bx + c
P3′′ ( x) = 6ax + 2b
P3′′′ ( x) = 6a
P3(4) ( x) = 0
Error ≤
(b
− a) 
(b − a ) 0 = 0
max P3(4) ( x)  =
()

180n 4 
180n 4
5
5
So, Simpson’s Rule is exact when used to approximate a cubic polynomial.
1
0
x3 dx =
3
1 − 0 3
1
1
1
3
1

0 + 4  + (1)  = 0 + + 1 =
3( 2) 
6
2
4
 2


The exact value of this integral is
1
0 x
1
3
dx =
x4 
1
 = ,
4 0
4
which is the same as the Simpson approximation.
Section 6.4 Improper Integrals
Skills Warm Up
1. lim ( 2 x + 5) = 2( 2) + 5 = 9
x→2
x →1
3
3

2. lim  − x + 2  = − 3 + 2 = 0
x →3  x
3

3. lim
x → −4
x + 4
x + 4
= lim
x → −4 ( x + 4)( x − 4)
x 2 − 16
1
= lim
x → −4 x − 4
1
=
−4 − 4
1
= −
8
4. lim
x→0
1
= ∞
x −1
5. lim
Limit does not exist.
6. lim
x →−3
x →−3
= −3 − 1
= −4
7.
x ( x − 2)
x2 − 2 x
= lim 2
x → 0 x ( x + 3)
x3 + 3x 2
= lim
x→0
x − 2
x( x + 3)
( x + 3)( x − 1)
x2 + 2x − 3
= lim
x →−3
x +3
x +3
= lim x − 1
8.
4
3
(2 x
− 1)
3
(a)
4
3
(2b
− 1)
(b)
4
3
(2 ⋅ 0 − 1)
3
3
= − 43
1
3
+
x − 5 ( x − 2) 2
lim
x − 2
= ∞
x( x + 3)
(a)
1
3
+
b − 5 (b − 2) 2
lim
x − 2
= −∞
x( x + 3)
(b)
1
3
1
3
11
+
= − +
=
2
0 −5
5
4
20
( 0 − 2)
x → 0−
x → 0+
Limit does not exist.
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446
Chapter 6
Techniques of Integration
Skills Warm Up —continued—
(
)
9. ln 5 − 3x 2 − ln( x + 1)
2
10. e3 x + e−3 x
(
)
ln (5 − 3 ⋅ 0 ) − ln (0 + 1) = ln 5 − ln 1
2
(a) ln 5 − 3b 2 − ln(b + 1)
(b)
1
0
2
= 2
2
.
3
3
2x − 5
0 ( x − 3)( x − 2) dx
is proper because the function is continuous on [0, 1].
2
−2
ln ( x 2 ) dx is improper because the
function has an infinite discontinuity in
[− 2, 2] at x = 0.
5. The integral
∞
1
10. This integral converges because
∞ 5
∞
b
5
−2 x
e −2 x 
 0 e2 x dx = 5  0 e dx = − 2 blim
0
→∞ 
5
5
= − (0 − 1) = .
2
2
11. This integral diverges because
b
∞
1
 2 x − 1 = ∞ − 3 = ∞.
 5 2 x − 1 dx = blim
5
→∞ 
12. This integral diverges because
b
∞
x
 x 2 − 16  = ∞ − 3 = ∞.
 5 x 2 − 16 dx = blim
5
→∞ 
13. This integral diverges because
−1
−∞ e
= lim ( − e) − ( − e − a )
a → −∞
= − e + ∞ = ∞.
2
14. This integral converges because
−1
1
 1
dx = lim −  = 1 + 0 = 1.
b → −∞  x 
x2
b
−1
1
dx is improper because the
x2 + 3
integral has an infinite lower limit and infinite upper
limit of integration.
6. The integral
−1
dx = lim − e − x 
a
a → −∞
−x
e x dx is improper because the integral
has an infinite upper limit of integration.
∞
−∞
7. This integral converges because
∞ 1
∞
−4
1 x 4 dx = 1 x dx
 −∞
15. This integral diverges because
∞
1
e
x
dx = lim 2e
b →∞ 
x
= lim 2e
b →∞ 
= ∞.
b
 1 
= lim − 3 
b → ∞  3x 
1
 1   1 
1
1
= lim  − 3  −  −  = 0 +
= .
b → ∞  3b 
3
3
 3 

8. This integral diverges because
b
∞ 1
2 x  = ∞ − 2 = ∞.
 1 x dx = blim
1
→∞ 
9. This integral diverges because
∞
0
b
e x 3 dx = lim 3e x 3  = ∞.
0
b →∞
)
2
=1+1
1
dx is proper because the function is
x2
continuous on [1, 3].
4. The integral
2
≈ 1.609
1
1
(
(b) e3(0) + e−3(0) = e0 + e0
1
dx is improper because the
3x − 2
3. The integral
1
2x − 5
0 x 2 − 5x + 6 dx =
2
= ln 5
function has an infinite discontinuity in [0, 1] at x =
2. The integral
2
(a) e3b + e −3b = e −3b e 6b + 1
2
1. The integral
2
x
b
1
b
− 2e

16. This integral diverges because
0
 − ∞ x2
0
x
1

dx = lim  ln ( x 2 + 1)
b → −∞  2
+1
b
= 0 − ∞
= −∞.
17. This integral converges because
∞
1
 4 x(ln x)3
b

1 
1
 =
.
dx = lim −
2
b →∞  2 ln x 2 
2
ln
) 4
( 4)
 (
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.4
Improper Integrals
447
18. This integral converges because
7x
∞
0
(3 x
2
)
+ 5
2
7 ∞
6 x 3x 2 + 5
6 0
(
dx =
)
−2
dx
∞


7

= lim −
b → ∞  6 3x 2 + 5 

0
(
)


 
7
7

 − −
= lim  −
2


b → ∞  6 3b 2 + 5 
  6 3(0) + 5 

7
7
.
= 0 +
=
30
30
(
(
)
)
27. μ = 1.63, σ = 0.076
19. This integral converges because
∞
 −∞ 2 xe
−3 x 2
0
 −∞ 2 xe
dx =
−3 x 2
∞
0
dx +
2 xe
−3 x 2
dx
2 ( 2σ 2 )
1
e−( x − μ )
σ 2π
2
1
=
e − ( x −1.63)
0.076 2π
f ( x) =
b
0
2
2
= lim − 13 e −3 x  + lim − 13 e −3 x 
a →−∞ 
 a b →∞ 
0
(
) (
= − 13 + 0 + 0 +
1
3
)
Using a graphing utility:
= 0.
20. This integral diverges because
∞
2 − x3
 −∞ x e
dx =
0
2 − x3
 −∞ x e
= lim − 3 e
b → −∞ 
dx +
1 − x3 
=
(
− 13
0
 − ∞
) (
+ ∞ + 0+
∞
0
+ lim − 3 e
c→∞ 
21. A =
0
22. A =
 −∞ e
0
e − x dx =
x 4
)
 −∞ − ( x − 1)3
24. A =
 −∞
25. A =
26. A =
∞
∞
0
−1

7 
7
= lim −
 =
3
a → −∞ 
8
 ( x − 1)  a
0
5
dx = lim −10 4 − x  = ∞
a
a →−∞
4−x
0
0
=1
0
23. A =
2
1
e − ( x −1.63)
0.076 2π
1.50
(b)
1.70 0.076
(c)
1.80 0.076
∞
∞
1
2π
1
2π
0.011552
dx ≈ 0.9438
e − ( x −1.63)
2
0.011552
dx ≈ 0.1785
e − ( x −1.63)
2
0.011552
dx ≈ 0.0126
28. (a) The probability of choosing a car at random that
gets between 26 and 28 kilometers per gallon is
greater because the area of the region bounded by
26 ≤ x ≤ 28 is greater than that bounded by
22 ≤ x ≤ 24.
dx = lim 4e x 4  = 4
a
a → −∞
7
−1
b
lim − e − x 
0
b →∞ 
c
 0
= ∞.
∞
1.80
(a)
3
x 2e − x dx
1 − x3 
1
3
0.011552
b
6x
dx = lim 3 ln ( x 2 + 1) = ∞
0
b →∞
x +1
2
b
16 x
dx = lim 8 ln ( x 2 + 4) = ∞
2
0
→∞
b
x + 4
(b) The probability of choosing a car at random that gets
at least 30 kilometers per gallon is greater because
the area of the region for x ≥ 30 is greater than that
bounded by 20 ≤ x ≤ 22.
29. μ = 36, σ = 0.2
f ( x) =
2 ( 2σ 2 )
2
1
e−( x − μ )
= 1.99471e−( x − 36)
σ 2π
0.08
Using a graphing utility:
∞
− ( x − 36)2 0.08
dx ≈ 0.9938
∞
− ( x − 36)2 0.08
dx ≈ 0.6915
(a)
 35.5 1.99471e
(b)
 35.9 1.99471e
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
448
Chapter 6
Techniques of Integration
30. μ = 9000, σ = 500
38. (a) Present value =
2
1
f ( x) =
e−( x − μ )
σ 2π
2σ 2
=
1
2π
500
e
− ( x − 9000)2 500,000
∞
1
8000 500
(b)
 11,000 500
∞
2π
e − ( x − 9000)
1
31. Present value =
2π
2 500,000
e − ( x − 9000)
20
≈ \$748,222.01
dx ≈ 0.9772
2 500,000
(b) Present value =
33. Present value =
P
5000
=
≈ \$66,666.67
r
0.075
34. Present value =
P
3500
=
≈ \$100,000
r
0.035
35. Present value =
p
18,000
=
= \$360,000
r
0.05
 75,000 −0.08t 
= lim −
e

b →∞ 
0.08
0
The amount the foundation needs to fund the donation is
\$437,500. Yes, the foundation has enough money to start
the fund.
500,000e
dt
(a) For n = 5, you have
C = 650,000 − 250,000(e −0.50 − 1)
≈ \$748,367.34.
(c) For n = ∞, you have
n
C = 650,000 − lim 250,000e −0.10t 
0
n→∞
= 650,000 − 250,000(0 − 1) = \$900,000.
40. C = 800,000 +
n
0
30,000e − 0.04t dt
n
= 800,000 − 750,000e − 0.04t 
0
(b) For n = 10, you have
≈ \$4,637,228.40
0
n
= 650,000 − 250,000e −0.10t 
0
C = 800,000 − 750,000(e− 0.2 − 1)
≈ \$935,951.94.
20
(b) Present value =
25,000e −0.10t dt
(a) For n = 5, you have
−0.09 t
500,000 −0.09t 
e

0.09
0
∞
n
0
C = 650,000 − 250,000(e−1 − 1) ≈ \$808,030.14.
P
35,000
36. Present value =
=
= \$437,500
r
0.08
= −
39. C = 650,000 +
(b) For n = 10, you have
The amount you need to start the scholarship fund is
\$360,000. Yes, you have enough money to start the
scholarship fund.
0
75,000e −0.08t dt
= \$937,500.00
P
10,000
32. Present value =
=
≈ \$142,857.14
r
0.07
37. (a) Present value =
∞
0
b
dx ≈ 0.00003
P
12,000
=
= \$200,000
r
0.06
20
75,000e −0.08t dt
 75,000 −0.08t 
e
= −

 0.08
0
Using a graphing utility:
(a)
20
0
500,000e
−0.09 t
C = 800,000 − 750,000(e− 0.4 − 1)
= \$1,047,259.97.
dt
b
 500,000 −0.09t 
= lim −
e

b →∞ 
0.09
0
= \$5,555,555.56
(c) For n = ∞, you have
n
C = 800,000 − lim 750,000e − 0.04t 
0
n →∞
= 800,000 − 750,000(0 − 1)
= \$1,550,000.
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
41. C = 300,000 +
n
 0 15,000te
− 0.06 t
42. C = 650,000 +
dt
n
n
0
449
25,000(1 + 0.08t )e −0.12t dt
n
= 650,000 + 25,000 − 19 (125 + 6t )e −0.12t 
0
 50
2500  − 0.06t 
= 300,000 − 15,000  t +
e

3
9 


0
(a) For n = 5, you have
(a) For n = 5, you have
(
125
9
)
(
125
9
)
C = 650,000 + 25,000 − 155
e −0.6 +
9
2500 
 3250 − 0.3
C = 300,000 − 15,000
e
−

9
9 

≈ \$453,901.30.
≈ \$760,928.32.
(b) For n = 10, you have
(b) For n = 10, you have
C = 650,000 + 25,000 − 185
e −1.2 +
9
2500 
 4000 − 0.6
C = 300,000 − 15,000
e
−

9
9 

≈ \$807,922.43.
≈ \$842,441.86.
(c) For n = ∞, you have
n
C = 650,000 + 25,000 lim − 19 (125 + 6t )e−0.12t 
0
n →∞
(c) For n = ∞, you have
(
n
= 650,000 + 25,000 0 +
 50
2500  − 0.06t 
C = 300,000 − 15,000 lim  t +
e

n →∞  3
9 

0
125
9
)
≈ \$997,222.22.
2500 

= 300,000 − 15,000 0 −

9 

≈ \$4,466,666.67.
Review Exercises for Chapter 6
1. Let u = x − 1 and dv = e x dx. Then du = dx and
v = ex.
4. Let u = ln 4 x and dv = x dx.
Then du =
 ( x − 1)e
x
dx = ( x − 1)e x −
e
x
dx
= ( x − 1)e − e + C
x
x
 x ln 4 x dx
= ( x − 2)e x + C
 − 13e dx
+ 13  e − 3 x dx
dx = − 13 xe − 3 x −
= − 13 xe − 3 x
−3 x
= − 13 xe − 3 x − 19 e − 3 x + C
(
du = (1 x) dx and v = 2 x .

12
Then du = dx and v =
1
x   dx
 x
x ln x −
2
= 2
x ln x −
 2x
= 2
x ln x − 4
−1 2
dx
x +C
2
3
(x
dx.
− 5).
x − 5 dx =
2x
3
(x
− 5)
32
−
32
 23 ( x − 5) dx
=
2x
3
(x
− 5)
32
−
4
15
6. Let u = x and dv = ( 2 x + 7)
−1 2
x
x dx. Then
ln x
dx = 2
x
1 2 1 
x   dx
2  x

5. Let u = x and dv = ( x − 5)
)
3. Let u = ln x and dv = 1
1 2
x ln 4 x −
2
1 2
1
x ln 4 x −  x dx
2
2
1 2
1 2
= x ln 4 x − x + C
2
4
Then du = dx and v = − 13 e−3 x .
−3x
=
=
2. Let u = x and dv = e − 3 x dx.
 xe
1
1
1
(4) dx = dx and v = x 2 .
4x
x
2
(x
− 5)
52
+ C
dx.
Then du = dx and v = ( 2 x + 7) .
12

x
dx = x 2 x + 7 −
2x + 7
= x 2x + 7 −
 ( 2 x + 7)
12
dx
1
32
( 2 x + 7) + C
3
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
450
Chapter 6
Techniques of Integration
7. Let u = 2 x 2 and dv = e2 x . Then du = 4 x dx and v =
 2x e
2 2x
(
)  12 e
dx = ( 2 x 2 ) 12 e 2 x −
2x
(4 x
dx) = x 2e 2 x −
Let u = 2 x and dv = e 2 x . Then du = 2 dx and v =
 2x e
2 2x
(
)  12 e
dx = x 2e 2 x − ( 2 x) 12 e 2 x −

e
= x 2e 2 x − xe 2 x +
2x
2x
1 e2 x .
2
 2 xe
2x
dx
1 e2 x .
2
(2) dx
dx
= x 2e 2 x − xe 2 x + 12 e 2 x + C
8. Let u = (ln x) and dv = dx. Then du = 3(ln x )
3
3
 (ln x)
2
1
dx and v = x.
x
dx = x(ln x) −  3(ln x) dx
3
2
1
2
Let u = (ln x) and dv = 3 dx. Then du = 2(ln x ) and v = 3 x.
x
3
 (ln x)
3
2
dx = x(ln x) − 3 x(ln x) −

 6 ln
Let u = ln x and dv = 6 dx. Then du =
3
 (ln x)
x dx

1
dx and v = 6 x.
x
3
2
dx = x(ln x) − 3x(ln x) + 6 x ln x −
 6 dx
9. Let u = ln x and dv = 6 x dx.
1
Then du = dx and v = 3 x 2 .
x
e
1
6 x ln x dx = 3 x 2 ln x −

= x(ln x) − 3x(ln x) + 6 x ln x − 6 x + C
3
2
11. Let u = x and dv = e − x 4 dx.
Then du = dx and v = − 4e− x 4 .
1
 0 xe
1
3 x 2   dx
 x
−x 4
dx = − 4 xe − x 4 + 4  e− x 4 dx
1
= − 4 xe − x 4 − 16e − x 4 
0
e
3 

= 3 x 2 ln x − x 2 
2 1

= ( − 4e −1 4 − 16e −1 4 ) − (0 − 16)
3  
3

=  3e 2 − e 2  −  0 − 
2  
2

3 2
3
= e +
2
2
≈ 12.584
= 16 − 20e −1 4
≈ 0.4240
10. Let u = ln(1 + 3x) and dv = dx.
Then du =
4
1
(3) dx and v = x.
1 + 3x
 0 ln(1 + 3x) dx
3x
= x ln (1 + 3x) −
 1 + 3x dx
= x ln (1 + 3x) −
 1 − 1 + 3x  dx

1

4
1


=  x ln (1 + 3x) − x + ln (1 + 3x)
3

0
1


=  4 ln 13 − 4 + ln (13)  − 0
3


13
ln (13) − 4
=
3
≈ 7.1148
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
451
12. Let u = x 2 and dv = e3 x dx.
Then du = 2 x dx and v = 13 e3x .
1
 −1 x e
2 3x
dx =
1 2 3x
xe
3
2
3
−
 xe
3x
dx
Let u = x and dv = e3 x dx. Then du = dx and v = 13 e3 x .
=
1 x 2e3 x
3
− 23  13 xe3 x −

=  13 x 2e3 x −
2 3x
xe
9
2
=  13 (1) e3(1) −

=
( 13 e
=
1
27
3
2
9
(5e
− 17e

2 e3
27
−3
)
e3 x dx

1
2 3x 
e 
27
−1
+
(1)e3(1)
− 92 e3 +
3
1 2
3 0
+
2 e3(1) 
27

) − ( 13 e
−3
2
−  13 ( −1) e3(−1) −

− 92 e − 3 +
2 e−3
27
2
9
(−1)e3(−1)
+
2 e3( −1) 
27

)
≈ 3.6882
5
5
13. Present value =
 0 20,000e
14. Present value =
−0.07 t
 0 (1500t )e dt
=
−0.04 t
dt = −500,000e −0.04t  = \$90,634.62
0
10
10
0
1500te −0.07 t dt
= 1500 
10
0
te −0.07 t dt
Let u = t and dv = e−0.07t dt. Then du = dt and v = −
1 −0.07 t
e
.
0.07
10
t −0.07t
1


e
e −0.07 t dt 
= 1500  −
+
0.07  0
 0.07

10
t −0.07t
1


e
e −0.07t 
= 1500 −
−
0.07
0.0049

0
 10 −0.7
1
1 
 
e
e −0.7  −  0 −
= 1500  −
−

0.0049
0.0049 
 
 0.07
≈ \$47,695.40
15. Present value =
10
0
24,000te −0.05t dt = 24,000
10
0
e −0.05t dt
Let u = t and dv = e−0.05t dt. Then du = dt and v = −20e −0.05t .
{
= 24,000{−200e
10
10
}
= 24,000 −20te −0.05t  + 20  e −0.05t dt
0
0
−0.5
−
}
10
400 e −0.05t 
0
= 24,000 −200e −0.5 − 400(e −0.5 − 1)
= 24,000(400 − 600e −0.5 )
= \$865,958.50
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
452
Chapter 6
Techniques of Integration
16. Present value =
=
t 2 −0.05t
 0 (20,000 + 100te )e dt
10
10
0
20,000e −0.05t dt +
10
0
100te0.45t dt
10
= −
10
20,000 − 0.05t 
e
+ 100  te0.45t dt

0
0.05
0
= −
10
20,000 − 0.5t
(e − 1) + 100  0 te0.45t dt
0.05
Let u = t and dv = e0.45t dt. Then du = dt and v =
1 0.45t
e .
0.45
= −
10
20,000 −0.5
t 0.45t
1

e
e0.45t dt 
−
( e − 1) + 100 0.45
0.05
0.45  0

= −
t 0.45t
20,000 −0.5
1

e
e0.45t 
−
( e − 1) + 100 0.45
0.05
0.2025
0
10
20,000 −0.5
10 4.5
1
1 
 
e −
e 4.5  −  0 −
( e − 1) + 100 0.45

0.05
0.2025
0.2025
 


≈ \$313,466.73
= −
17. (a) Actual income =
4
 0 (200,000 + 50,000t ) dt
4
= 200,000t + 25,000t 2 
0
= \$1,200,000
(b) Present value =
=
4
− 0.06 t
 0 (200,000 + 50,000t )e dt
4
0
4
0
200,000e − 0.06t dt +
50,000te − 0.06t dt
4
4
 200,000 − 0.06t 
− 0.06 t
e
dt
= 
 + 50,000  0 te
 − 0.06
0
4
= 711,240.46 + 50,000  te − 0.06t dt
0
Let u = t and dv = e − 0.06t dt. Then du = dt and v = −
1 − 0.06t
e
.
0.06
1
 1 − 0.06t

= 711,240.46 + 50,000 −
te
+
e − 0.06t dt 
0.06 
 0.06

4
− 0.06 t
− 0.06t 
1
 1
te
e
= 711,240.46 + 50,000 −
−

0.0036
 0.06
0
≈ \$1,052,649.52
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
18. (a) Actual income =
453
7
 0 (400,000 + 175,000t ) dt
7
= 400,000t + 87,500t 2 
0
= \$7,087,500
(b) Present value =
=
7
− 0.04 t
 0 (400,000 + 175,000t )e dt
7
0
400,000e − 0.04t dt +
7
7
 0 175,000te
= −10,000,000e − 0.04t  +
0
7
0
− 0.04 t
dt
175,000te − 0.04t dt
7
= 2,442,162.59 + 175,000  te − 0.04t dt
0
Let u = t and dv = e − 0.04t dt. Then du = dt and v = − 25e− 0.04t .
= 2,442,162.59 + 175,000 − 25te − 0.04t +

7
0
25e − 0.04t dt 

7
− 0.04 t
− 0.04 t

= 2,442,162.59 + 175,000 − 25te
− 625e
 0

= \$6,007,438.79
19. Formula 6: u = x, du = dx, a = 2, b = 3

x2
1  3x

dx =
− ( 4 − 3 x) + 4 ln 2 + 3x  + C
3
2 + 3x
2

(3) 
1 1

− (12 x − 9 x 2 ) + 4 ln 2 + 3 x  + C
27  2

1
=
(9 x2 − 12 x + 8 ln 2 + 3x ) + C
54
=
20. Formula 40: u = x, du = dx, n = 6
1
 1 + e6 x
dx = x −
22. Formula 43: u = x, du = dx, n = 5
1
ln (1 + e6 x ) + C
6
x
5
ln x dx =
21. Formula 28: u = x, du = dx, a = 8

1
x
2
x + 64
dx = −
1
8+
ln
8
=
x 2 + 64
+C
x
x5 + 1
(5 + 1)2
−1 + (5 + 1) ln x + C
x6
(−1 + 6 ln x) + C
36
23. Formula 11: u = x, du = dx, a = 2, b = 3
1
 x( 2 + 3 x ) 2
dx =

1 1
1
x
+ ln

+C
2  2 + 3x
2
2 + 3x 
24. Formula 8: u = x, du = dx, a = 7, b = 2
x2
 (7 + 2 x)3 dx
=

1  14
49

−
+ ln 7 + 2 x  + C
2
8  7 + 2 x

2(7 + 2 x)
25. Formula 23: u = x, du = dx, a = 5

x 2 + 25
dx =
x
x 2 + 25 − 5 ln
5+
x 2 + 25
+C
x
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
454
Chapter 6
Techniques of Integration
26. Formula 29: u = x, du = dx, a =
x

2
dx =
x2 − 6
(
1
x
2
6
x 2 − 6 − 6 ln x +
x2 − 6
)+C
27. Formula 21: u = 2 x, du = 2 dx, a = 7

1
1
2
dx = 
dx
4 x 2 − 49
2 4 x 2 − 49
28. Formula 44: u = 3x, du = 3 dx
 (ln 3x)
=
1 1
2x − 7 
 ln
 + C
2  14
2x + 7 
=
1
2x − 7
ln
+ C
28
2x + 7
2
dx =
1
(ln 3x)2 3 dx
3
2
= x 2 − 2 ln 3x + (ln 3x)  + C


29. Formula 17: u = x, du = dx, a = b = 1

1+ x
dx = 2 1 + x +
x
x
1
dx
1+ x
Formula 15: u = x, du = dx, a = b = 1

1+ x
dx = 2 1 + x + ln
x
1+ x −1
+C
1+ x +1
30. Formula 22: u = x, du = dx, a = 4, n = 2

1
( x2
− 16)
2
dx =


x
1

dx
+ ( 4 − 3)  2
2
−
1
2( 4) ( 2 − 1)  ( x 2 − 16)
( x − 16) 

−1
2
= −
1  x
+
32  x 2 − 16

1

dx
x 2 − 16 
Formula 21: u = x, du = dx, a = 4
= −
1  x
1
x − 4 
+ ln
+ C
32  x 2 − 16
8
x + 4 
31. Formula 10: u = x, du = dx, a = 4, b = 3
2
2
1
1 x(4 + 3x) dx
1
x

=  ln
4 + 3 x  1
4
P ( a ≤ x ≤ b) =
1
1  1
1 
=  ln
 −  ln

5  4
7 
4
1
1
1
7
= ln 5 = ln
1
4
4
5
7
≈ 0.0841
32. Formula 31: u = x, du = dx, a =
3
−1
1
( x2 + 2)
32

dx = 
 2
2
3


2
x + 2  −1
x
 3   −1 
= 
−

 2 11   2 3 
3
1
=
+
2 11 2 3
=
33. Formula 19: u = x, du = dx, a = 9, b = 16
=
b
a
96 
11 
x

 dx
9 − 16 x 
96  2(18 − 16 x )
−
11 
768
= −
b

9 + 16 x 
a
b
1
(9 − 8 x) 9 + 16 x  a

22
1  9 − 8x
(a) P(0 ≤ x ≤ 0.8) = − 22
) 9 + 16 x 
(
=
1
− 22
(2.6
21.8 − 27
0.8
0
)
≈ 0.675
1  9 − 8x
(b) P(0 ≤ x ≤ 0.5) = − 22
) 9 + 16 x 
(
(
1 5 17 − 27
= − 22
0.5
0
)
≈ 0.290
3 3 + 11
2 33
≈ 0.7409
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
455
34. Formula 37: u = x1.5 , du = 1.5 x 0.5 dx
b
P( a ≤ x ≤ b) =
2 x1.5
 a 1.5 x e
b
a
=
1.5
x1.5e x
dx
(1.5x0.5 dx)
b
1.5
= ( x1.5 − 1)e x 

a
(
0.6
)
1.5
(a) P(0.4 ≤ x ≤ 0.6) =  x1.5 − 1 e x  ≈ 0.110

 0.4
(
0.5
)
1.5
(b) P(0 ≤ x ≤ 0.5) =  x1.5 − 1 e x 

0
35. Trapezoidal Rule:
Simpson’s Rule:
Exact value:
3
1
3
1
≈ 0.079
1
1 1
 1 
 1 
 1   1 
dx ≈  2 + 2
+ 2 2  + 2
+
= 0.705

2
2
 (3 2) 
 ( 2) 
 (5 2)2   (3)2 
x
4  (1)





 


1
1 1
 1 
 1 
 1   1 

dx
≈
+
4
+
2
+
4
+
≈ 0.6715





2
 (3 2)2 
 ( 2)2 
  2 
x2
6  (1)2




 (5 2)   (3) 

3
1
3
1
2
 1
dx = −  =
x2
3
 x1
36. Trapezoidal Rule:
2
1
1
1 1
 1 
 1 
 1 
 1 
 1 

dx ≈
+ 2
+ 2
+ 2
+ 2
+ 2
3
 (9 8)3 
 (5 4)3 
 (11 8)3 
 (3 2)3 
 (13 8)3 
16  (1)3
x











 1 
 1   1 
+ 2
+ 2
+
≈ 0.3786

3
 (7 4) 
 (15 8)3   ( 2)3 



 

Simpson’s Rule:
2
1
1
1 1
 1 
 1 
 1 
 1 
 1 

dx ≈
+ 4
+ 2
+ 4
+ 2
+ 4
3
 (9 8)3 
 (5 4)3 
 (11 8)3 
 (3 2)3 
 (7 4)3 
24  (1)3
x











 1 
 1   1 
+ 2
+ 4
+
≈ 0.3751

3
 (15 8) 
 (15 8)3   ( 2)3 



 

Exact value:
2
1
2
1
3
 1 
dx = − 2  =
= 0.375
x3
2
x
8

1
37. Trapezoidal Rule:
2
 0 (x
2
+ 1) dx ≈
((0)

1
8
2
) (( ) + 1) + 2(( ) + 1) + 2((1) + 1)
+ 2(( ) + 1) + 2(( ) + 1) + 2(( ) + 1) + (( 2) + 1)

) (( )
1
4
2
5
4
2
+1 + 2
+1 + 2
1
2
2
3
4
2
3
2
2
7
4
2
2
2
= 4.6875
Simpson’s Rule:
2
 0 (x
2
+ 1) dx ≈
((0)

1 
12 
2
)
+ 1 + 4

+ 4

=
Exact value:
2
 0 (x
2
14
3
( 14 )
( 54 )
2
+ 1 + 2


2
+ 1 + 2


( 12 )
( 32 )
2
+ 1 + 4


2
+ 1 + 4


( 34 )
( 74 )
(
)
2
2
+ 1 + 2 (1) + 1

2
+ 1 +

((2)
2
)

+1

= 4.6
+ 1) dx =  13 x3 + x =
0
2
8
3
+ 2 =
14
3
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
456
Chapter 6
Techniques of Integration
38. Trapezoidal Rule:
0 (2 − x ) dx
1
3
≈
(2 − (0) ) + 2 2 − ( )  + 2 2 − ( )  + 2 2 − ( )  + (2 − (1) ) =
111
64
(2 − (0) ) + 4 2 − ( )  + 2 2 − ( )  + 4 2 − ( )  + (2 − (1) ) =

7
4
1
8
3
3
1
4

1
2
3
3
4
3
3
= 1.734375
Simpson’s Rule:
0 (2 − x ) dx
1
3
Exact value:
≈
3
3
0 (2 − x ) dx
1
39. Trapezoidal Rule:
4
0
4
0
4
0
Simpson’s Rule:
Exact value:
1 
12 
1 4
x
4 0
1 0
e
2
e − x 2 dx ≈
1 0
e
3
1
2
(
1
= 2 x −
e − x 2 dx ≈
3
1
4
= 2−
3
4
)−0=
7
4
3
3
= 1.75
= 1.75
+ 2e −1 2 + 2e − 2 2 + 2e − 3 2 + e − 4 2  ≈ 1.7652
+ 4e −1 2 + 2e − 2 2 + 4e − 3 2 + e − 4 2  ≈ 1.7299
4
e − x 2 dx = − 2e − x 2  = 2 − 2e − 2 ≈ 1.7293
0
40. Trapezoidal Rule:
8
0
x + 3 dx ≈
1
2
0+3 + 2 1+3 + 2
+2
Simpson’s Rule:
8
0
x + 3 dx ≈
1
3
5+3 + 2
8
0
5+3 + 2
8
32
x + 3 dx =  23 ( x + 3)  =

0
2
3
(11
2+3 + 2
6+3 + 2
0+3 + 4 1+3 + 2
+4
Exact value:
1
4
3
4+3
8 + 3  ≈ 20.8464
7 +3 +
2+3 + 4
6+3 + 4
3+3 + 2
3+3 + 2
4+3
8 + 3  ≈ 20.8577
7 +3 +
)
11 − 3 3 ≈ 20.8578
41. (a) Trapezoidal Rule:
2
1
1
1 
1
1
1
1
1






 

dx ≈ 
+ 2
+ 2
+ 2
+



 ≈ 0.741
1 + ln x
8  1 + ln (1) 
1
ln
5
4
1
ln
3
2
1
ln
7
4
1
ln
2
 + ( )
 + ( )
 + ( )   + ( ) 
(b) Simpson’s Rule:
2
1
1
1 
1
1
1
1
1






 

dx ≈
+ 4

 + 2 1 + ln 3 2  + 4 1 + ln 7 4  +  1 + ln 2  ≈ 0.737
1 + ln x
12  1 + ln(1) 
1
ln
5
4
+
( ) 
( ) 
( ) 
( ) 

42. (a) Trapezoidal Rule:
2
0
1
2
x +1
dx ≈

1 
8 



 + 2
+ 1 



1
( 0)
2

+ 2





 + 2
1
 


  + 1

 4




 + 2
1
 


  + 1

 2




 + 2

3
 


  + 1
 4



2

+
1
1
()



 + 2
5


  + 1

 4




 + 4
 3


  + 1

 2


 
+
7
 
  + 1 
 4

1
1
2
1
2
1
2
1
2
1
1
2
1
2
( 2)
2

 ≈ 1.443
+ 1 


(b) Simpson’s Rule:
2
0
1
x2 + 1
dx ≈

1 
12 

1
( 0)
2


 + 4

+ 1





 + 2
1


  + 1

 4




 + 4
1


  + 1

2




 + 2

 3


  + 1
 4


+ 4





 + 2
5


  + 1

 4




 + 4
 3


  + 1

 2


 
+
7
 
  + 1
 4

1
2
1
2
1
2
1
2
1
2
1
2
1
(1)
2
1
( 2)
2



+ 1

 ≈ 1.444
+ 1 


&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
1
0
43. (a) Trapezoidal Rule:
(b) Simpson’s Rule:
1
0
(b) Simpson’s Rule:
 (1 4)3 2 
 (1 2)3 2 
 (3 4)3 2   (1)3 2 
x3 2
1






 +
 ≈ 0.305
dx
0
2
2
2
≈
+
+
+
2
2
2
2




2 − x2
8
 2 − (1 4) 
 2 − (1 2) 
 2 − (3 4)   2 − (1) 

 (1 4)3 2 
 (1 2)3 2 
 (3 4)3 2   (1)3 2 
1
x3 2






+
 ≈ 0.289
≈
+
+
+
0
4
2
4
dx
 2 − (1 4) 2 
 2 − (1 2)2 
 2 − (3 4)2   2 − (1)2 
2 − x2
12 





 


1
 −1 e
44. (a) Trapezoidal Rule:
1
 −1 e
8
0
45. (a) Trapezoidal Rule:
457
x4
x4
dx ≈
dx ≈
1 e(−1)
4

1  ( −1)
e
6

4
4
4
4
4
4
+ 2e(− 0.5) + 2e(0) + 2e(0.5) + e(1)  ≈ 2.924

4
4
4
4
+ 4e(− 0.5) + 2e(0) + 4e(0.5) + e(1)  ≈ 2.659

3
1 
3
3
3
3
3









dx ≈  2
+ 2 2
+ 2 2
+ 2 2
+ 2 2
 (1) + 2 
 ( 2) + 2 
 (3) + 2 
 ( 4) + 2 
2  (0) + 2 
x + 2










2
3
3
3
3





 

+ 2 2
+ 2 2
+ 2 2
+ 2
≈ 2.961



 (5) + 2 
 (6) + 2 
 (7) + 2   (8) + 2 





 

(b) Simpson’s Rule:
3
1 
3
3
3
3
3









dx ≈  2
+ 4 2
+ 2 2
+ 4 2
+ 2 2
 (1) + 2 
 ( 2) + 2 
 (3) + 2 
 ( 4) + 2 
3  (0) + 2 
x + 2










8
0
2
3
3
3
3





 

+ 4 2
+ 2 2
+ 4 2
+ 2
≈ 2.936





 ( 6) + 2 
 (7) + 2   (8) + 2 
 (5) + 2 



 

1
0
46. (a) Trapezoidal Rule:
(b) Simpson’s Rule:
1
0
1 − x dx ≈
1 − x dx ≈
1
8

1
12 

1−0 + 2 1−
1−0 + 4 1−
47. f ( x) = e3 x
+ 2 1−
1
2
1
2
3
4
+ 2 1−
+ 4 1−
3
4
+
1 − 1 ≈ 0.643

+
1 − 1 ≈ 0.657

1
x −1
1
f ′( x) = −
( x − 1)2
f ′′( x) = 9e3 x
f ′′′( x) = 27e3 x
f ′′( x) =
f (4) ( x) = 81e3 x
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
[0, 1] when x = 1 and f ′′( x) = 9e3 , you have
(1 − 0)
2
12( 4)
3
3e3
9e3 =
≈ 0.9415.
64
(b) Simpson’s Rule: Because
[0, 1] when x
f (4) ( x) is maximum in
= 1 and f (4) ( x) = 81e3 , you have
(1 − 0) 81e3
4
180( 4)
3
Error ≤
1
4
+ 2 1−
48. f ( x) =
f ′( x) = 3e3 x
Error ≤
1
4
=
9e3
≈ 0.0353.
5120
2
(x
f ′′′( x) = −
f (4) ( x) =
− 1)
3
6
(x
− 1)
4
24
(x
− 1)
5
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
[2, 4] when
x = 2 and f ′′( x ) = 2, you have
(4 − 2) 2
2 ( )
12( 4)
3
Error ≤
=
1
≈ 0.0833.
12
(b) Simpson’s Rule: Because f (4) ( x) is maximum in
[2, 4] when
x = 2 and f (4) ( x) = 24, you have
(4 − 2) 24
)
4 (
180( 4)
5
Error ≤
=
1
≈ 0.0167.
60
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
458
Chapter 6
Techniques of Integration
50. f ( x) = e x 5
49. f ( x) = x5
f ′( x) = 5 x 4
1 x5
e
5
1 x5
f ′′( x) =
e
25
1 x5
f ′′′( x) =
e
125
1 x5
f (4) ( x) =
e
625
f ′( x) =
f ′′( x ) = 20 x3
f ′′′( x) = 60 x 2
f (4) ( x ) = 120 x
f (5) ( x) = 120
f ( 6) ( x ) = 0
(a) Trapezoidal Rule: Because f ′′( x) = 20 x3 is a
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
maximum in [0, 3] when x = 3
Error ≤
( 3 − 0)
12n 2
3
[0, 5] when
(20(3) ) ≤ 0.0001
3
(5 − 0 )
12,150,000 ≤ n 2
n ≥ 3485.6.
Let n = 3486.
(b) Simpson’s Rule: Because f (4) ( x) = 120 x is a
( 3 − 0)
4
180n
486
≤ 0.0001
n4
3
(120(3))
≤ 0.0001
≤ 0.0001
≤ n2
&gt; 106.42.
Let n = 107.
(b) Simpson’s Rule: Because f (4) ( x) is maximum in
maximum in [0, 3] when x = 3
Error ≤
1
e, you have
25
3
1 
 e
12n 2  25 
5
e
12n 2
12,500
e
3
n
Error ≤
1215
≤ 0.0001
n2
x = 5 and f ′′( x) =
[0, 5] when
≤ 0.0001
x = 5 and f (4) ( x) =
(5 − 0)
Error ≤
180n
4,860,000 ≤ n 4
n ≥ 46.95.
Let n = 48.
4
5
 1 
e

 625 
e
36n 4
2500
e
9
n
1
e, you have
625
≤ 0.0001
≤ 0.0001
≤ n4
&gt; 5.24.
Let n = 6.
51. This integral converges because
52. This integral diverges because
−1
1
1
1
 1 
dx = lim − 4  = − + 0 = − .
5
a
→−∞
4
4
x
 4x  a
−1
 −∞
∞
1
4
1
dx =
x
∞
1
( ( b) ) −
b
x −1 4 dx = lim  43 x3 4  = lim 
1
b →∞
b →∞ 
4
3
34
4
3
(1)
3 4
 = ∞ −
4
3
= ∞
0
53. This integral diverges because
0
 −∞
54. This integral converges because
3
∞
0
 3(8 − x)2 3 
1
 = − 6 − ∞ = − ∞.
dx = lim −
a → −∞ 
2
8− x

 a
b
1
 1

 1
e − 2 x dx = lim − e − 2 x  = 0 −  −  = .
b →∞  2
2
2
0
 
b
55. This integral diverges because
∞
1
 (ln x)2 
ln x
dx = lim 
 = ∞ − 0 = ∞.
b →∞ 
x
2 

1
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6
∞
0
56. This integral diverges because
459
b
ex
dx = lim ln(1 + e x ) = ∞ − ln 2 = ∞.
x
0
b →∞
1+ e
∞
e − x 4 dx = lim − 4e − x 4  = 0 − ( − 4) = 4
0
b→∞
∞
b
2x
dx = lim ln ( x 2 + 2) = ∞ − ln 2 = ∞
0
b→∞
x2 + 2
b
57. A =
0
58. A =
0
59. A =
0
60. A =
 − ∞ (1 − 3x)2 3
∞
Test Yourself
2
4 xe−2 x dx = lim −
b →∞
3
0
61. Present value =
∞
0
b
−2 x2
 0 e (− 4 x dx)
dx = lim −
b → −∞
b
2
= lim −e2 x  = 1
b →∞ 
0
0
−2 3
 b (1 − 3x) (−3 dx)
8000e − 0.03t dt
0
13
= lim −3(1 − 3 x)  = −3 + ∞ = ∞
b
b → −∞ 
64. (a) Present value =
b
15
0
100,000e − 0.06t dt
15
 8000 − 0.03t 
e
= lim −

b → ∞  0.03
0
 100,000 − 0.06t 
e
= −

 0.06
0
= 0 − ( − 266,666.67)
≈ \$989,050.57
= \$266,666.67
62. Present value =
∞
0
(b) Present value =
∞
0
100,000e − 0.06t dt
b
15,000e − 0.05t dt
 100,000 − 0.06t 
= lim −
e

b→∞ 
0.06
0
b
= lim − 300,000e − 0.05t 
0
b→∞
 100,000 
= 0 − −

0.06 

≈ \$1,666,666.67
= 0 − ( − 300,000)
= \$300,000
P
21,000
=
= \$300,000
r
0.07
The amount you need to start the scholarship fund is
\$300,000. Yes, you have enough money to start the fund.
63. Present value =
Chapter 6 Test Yourself
1. Let u = x and dv = e x +1 dx. Then du = dx and
v = e x + 1.
 xe
x +1
v = −3e − x 3 .
dx = xe x + 1 −
e
x +1
dx
= xe x + 1 − e x + 1 + C
= ( x − 1)e x + 1 + C
2. Let u = ln 9 x and dv = x 2 dx.
Then du =

3. Let u = x 2 and dv = e− x 3 dx. Then du = 2 x dx and
9
1
1
dx = dx and v = x3 .
9x
x
3
1
1
x 2 ln 9x dx = x3 ln 9x −  x 2 dx
3
3
1 3
1
= x ln 9 x − x3 + C
3
9
1 3
= x (3 ln 9 x − 1) + C
9
xe
2 −x 3
dx = −3x 2e− x 3 +
 6 xe
−x 3
dx
Let u = 6 x and dv = e− x 3 dx. Then du = 6 dx and
v = −3e − x 3 .
xe
2 −x 3
dx = −3x 2e − x 3 − 18 xe − x 3 +
2 −x 3
= −3x e
= −3e
−x 3
(x
− 18 xe
2
−x 3
 18e
− 54e
−x 3
−x 3
dx
+C
+ 6 x + 18) + C
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
460
Chapter 6
Techniques of Integration
4. S = 2.1 t ln t − 2, 8 ≤ t ≤ 13 and t = 8 → 2008
(a) Total revenue =
8 (2.1
13
)
t ln t − 2 dt = 2.1 
Let u = ln t and dv = t1 2 dt. Then du =
2
= 2.1  t 3 2 ln t −
3

13
8
(
)
13
8
t ln t dt −
2 dt
1
2
dt and c = t 3 2 .
t
3
2 12 
t dt  −
3

 2 dt
13
8.4 3 2
 4.2

t − 2t 
=  t 3 2 ln t −
9
 3
8
≈ \$69.8128 billion
(b) Average revenue =
1
13 − 8
13
8 S (t ) dt
1
(69.8128) ≈ \$13.9625 billion
5
=
5. Formula 4: u = x, du = dx, a = 7, b = 2
x
 (7 + 2 x ) 2
dx =
6. Formula 39: u = x 3 , du = 3x 2 dx
1 7

+ ln 7 + 2 x  + C

4 7 + 2x

3x 2
 1 + e x3
(
dx = x3 − ln 1 + e x
3
)+C
7. Formula 24: u = x 2 , du = 2 x dx, a = 3
 2x
5
x 4 − 9 dx =
=
2
2
 (x ) (x )
2
1 x2
8

(2 x 4
2
− (3) ( 2 x) dx
− 9)
2
x 4 − 9 − 81 ln x 2 +
x4 − 9  + C

8. Formula 41: u = 3 − 2 x, du = −2 dx
1
 0 ln(3 − 2x) dx
9. Let u = x and dv = ( x − 2)
1
= − 12 (3 − 2 x) {−1 + ln (3 − 2 x)}
0
−1 2
dx.
Then du = dx and v = 2( x − 2) .
12
= − 12 −1 + ln 1 − 3( −1 + ln 3)
6
−1 2
 3 x( x − 2) dx
= − 12 ( 2 − 3 ln 3)
= 2 x( x − 2)
12
 2( x − 2)
12
−
12
= 2 x( x − 2) −

≈ 0.6479
(
= 24 −
=
4
3
(8))
(
4
3
(x
− 6−
dx
6
32
− 2) 
3
4
3
)
26
3
10. Formula 25: u = x, du = dx, a = 7
−1
 −3
1

x 2 + 49
7 +
dx =  x 2 + 49 − 7 ln
x


= 


=
=
(
( −1)2
+ 49 − 7 ln
50 − 7 ln − 7 −
50 −
x 2 + 49 

x

−1
7 +
50
(−1)2
( −1)

) − 
58 − 7 ln − 7 −

 
+ 49  
 −
 
 
58 − 7 ln
50 + 7 ln
( − 3)2
7 + 58
−3
+ 49 − 7 ln
7 +
(− 3)2
( − 3)

+ 49 







7 + 58
−3
≈ − 7.9691
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6
Test Yourself
461
11. Trapezoidal Rule:
2
 2 (x
5
((2)

− 2 x) dx ≈
3
8
=
585
32
Exact value:
2
 2 (x
)
− 2( 2) + 2

2
(114 )
2
(114 )  + 2 ( 72 )
− 2
1
3x
− 2
2
(174 ) + ((5)
− 2
2
− 2(5)
)
≈ 18.28
− 2 x) dx =  13 x3 − x 2  = 18
2
5
5
12. Exact: Let u = 9 x and dv = e3 x dx. Then du = 9 dx and v =
 0 9 xe
( 72 ) + 2 (174 )
2
1
dx = 3 xe3 x  −
0
 3e
3x
1 e3 x .
3
dx
1
= 3e3 − e3 x 
0
= 3e3 − (e3 − 1)
= 2e3 + 1
≈ 41.1711
Trapezoidal Rule:
1
 0 9 xe
3x
dx =
1
0
12 
(
)
(
13. f ( x) = 2 x 6 + 1
f ′( x) = 12 x
94
∞
0
4
∞
f (4) ( x) = 720 x 2
(a) Trapezoidal Rule: Because f ′′( x) is maximum in
[0, 1] when
3
≈ 41.3606
b
e −3 x dx = lim − 13 e −3 x  = 0 +
0
b →∞
1
2
dx = lim 4
b→∞
x
(1 − 0)
3
12n 2
5
60 = 2 ≤ 0.01
n
5
≤ 0.01
n2
5
≤ n2
0.01
b→∞
(
= 13 .
1
b − 4
)
16. This integral diverges because
0
1
 − ∞ (4 x − 1)2 3
3
13 0
lim ( 4 x − 1) 
a
4 a → −∞ 
dx =
3
+ ∞
4
= ∞.
= −
n ≥ 22.36.
17. This integral converges because
Let n = 23.
(b) Simpson’s Rule: Because f (4) ( x) is maximum in
x = 1 and f (4) ( x) = 720 you have
(1 − 0)
180n
4
≤ 0.01
n4
4
≤ n4
0.01
b
= ∞.
500 ≤ n 2
Error ≤
x 
= lim 4
x = 1 and f ′′( x) = 60, you have
[0, 1] when
1
3
15. This integral diverges because
f ′′′( x) = 240 x3
Error ≤
( 274 e ) + 9e 
14. This integral converges because
5
f ′′( x) = 60 x
)
+ 4 94 e3 4 + 2 92 e3 2 + 4
4
5
720 =
4
≤ 0.01
n4
∞
0
b
4
4
2 x3e − x dx = lim − 12 e − x 
b →∞ 
0
4
4
= lim  − 12 e − (b) + 12 e − (0) 
b →∞ 

= 0+
1
2
= 12 .
400 ≤ n 4
n ≥ 4.47.
Let n = 6.
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
462
Chapter 6
Techniques of Integration
18. (a) Present value for 20 years
Present value =
20
0
600,000e − 0.05t dt
20
= −12,000,000e − 0.05t 
0
= −12,000,000e − 0.05(20) + 12,000,000e − 0.05(0)
≈ \$7,585,446.71
(b) Present value forever
Present value =
∞
0
600,000e − 0.05t dt = lim
b →∞
b
0
600,000e − 0.05t dt
b
= lim −12,000,000e − 0.05t 
0
b →∞
(
= lim −12,000.000e− 0.05(b) + 12,000,000e − 0.05(0)
b →∞
)
= 0 + 12,000,000 = \$12,000,000
&copy; 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.