234 Section 5.1
Chapter 5
The Definite Integral
Section 5.1 Estimating with Finite Sums
(pp. 263-273)
Exploration 1 Which RAM is the
Biggest?
1.
4. LRAM > MRAM > RRAM, because the heights of the
rectangles decrease as you move toward the right under a
decreasing function.
Quick Review 5.1
1. 80 mph i 5 hr = 400 mi
2. 48 mph i 3 hr = 144 mi
3. 10 ft/sec 2 i 10 sec = 100 ft/sec
100 ft/sec i
1 mi 3600 sec
i
≈ 68.18 mph
5280 ft
1h
4. 300, 000 km / sec i
3600 sec 24 hr 365 days
i
i
i 1 yr
1 yr
1 hr
1 day
≈ 9.46 × 1012 km
5. (6 mph)(3 h) + (5 mph)(2 h) = 18 mi + 10 mi = 28 mi
6. 20 gal/min i 1 h i
60 min
= 1200 gal
1h
7. (−1°C/h)(12 h) + (1.5°C)(6 h) = −3°C
8. 300 ft 3 /sec i
3600 sec 24 h
i
i 1 day = 25, 920,0000 ft 3
1h
1 day
9. 350 people/mi 2 i 50 mi 2 = 17, 500 people
10. 70 times/sec i
3600 sec
i 1 h i 0.7 = 176, 400 times
1h
Section 5.1 Exercises
LRAM > MRAM > RRAM
2.
1. Since v(t) = 5 is a strait line, compute the area under the
curve.
x = (t ) v(t ) = (4)(5) = 20
2. Since v(t ) = 2t + 1 creates a trapezoid with the x-axis,
compute the area of the curve under the trapezoid.
h
A = (a + b)
2
a = t = 0 = v(0) = 2(0) + 1 = 1
b = t = 4 = v(4) = 2(4) + 1 = 9
h=4
4
A = (9 + 1) = 20
2
3. Each rectangle has base 1. The height of each rectangle is
Found by using the points t = (0.5, 1.5, 2.5, 3.5) in the
equation v(t ) t 2 + 1. The area under the curve is
⎛ 5 13 29 53 ⎞
+ ⎟ = 25, so the particle is
approximately 1⎜ + +
⎝4 4 4
4⎠
MRAM > RRAM > LRAM
3. RRAM > MRAM > LRAM, because the heights of the
rectangles increase as you move toward the right under an
increasing function.
close to x = 25.
4. Each rectangle has base 1. The height of each rectangle is
found by using the points y = (0.5, 1.5, 2.5, 3.5, 4.5) in the
equation v(t ) = t 2 + 1. The area under the curve is
⎛ 5 13 29 53 85 ⎞
+ + ⎟ = 46.25, so the
approximately 1⎜ + +
⎝4 4 4
4 4⎠
particle is close to x = 46.25.
Section 5.1
y
5. (a)
(b)
2
235
y
2
R
2
(b)
x
2
x
y
2
⎡ ⎛ 1⎞ ⎛ 1⎞ 2 ⎤⎛ 1⎞ ⎡ ⎛ 3⎞ ⎛ 3⎞ 2 ⎤⎛ 1⎞
MRAM: ⎢ 2 ⎜ ⎟ − ⎜ ⎟ ⎥ ⎜ ⎟ + ⎢ 2 ⎜ ⎟ − ⎜ ⎟ ⎥ ⎜ ⎟
⎢⎣ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎥⎦ ⎝ 2 ⎠ ⎢⎣ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎥⎦ ⎝ 2 ⎠
⎡ ⎛ 5 ⎞ ⎛ 5 ⎞ 2 ⎤ ⎛ 1 ⎞ ⎡ ⎛ 7 ⎞ ⎛ 7 ⎞ 2 ⎤ ⎛ 1 ⎞ 11
+ ⎢ 2 ⎜ ⎟ − ⎜ ⎟ ⎥ ⎜ ⎟ + ⎢ 2 ⎜ ⎟ − ⎜ ⎟ ⎥ ⎜ ⎟ = = 1.375
⎢⎣ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎥⎦ ⎝ 2 ⎠ ⎢⎣ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎥⎦ ⎝ 2 ⎠ 8
x
2
Δx =
7.
n
10
1
2
2
⎛ 1⎞ ⎡ ⎛ 1⎞ ⎛ 1⎞ ⎤⎛ 1⎞
LRAM: [2(0) − (0)2 ] ⎜ ⎟ + ⎢ 2 ⎜ ⎟ − ⎜ ⎟ ⎥ ⎜ ⎟
⎝ 2⎠ ⎢ ⎝ 2⎠ ⎝ 2⎠ ⎥⎝ 2⎠
⎣
⎦
2⎤
⎡
⎛ 3⎞ ⎛ 3⎞ ⎛ 1⎞ 5
⎛ 1⎞
+[2(1) − (1)2 ] ⎜ ⎟ + ⎢ 2 ⎜ ⎟ − ⎜ ⎟ ⎥ ⎜ ⎟ = = 1.25
⎝ 2⎠ ⎢ ⎝ 2⎠ ⎝ 2⎠ ⎥⎝ 2⎠ 4
⎣
⎦
6. (a)
LRAM n
1.32
MRAM n
RRAM n
1.34
1.32
50
1.3328
1.3336
1.3328
100
1.3332
1.3334
1.3332
500
1.333328
1.333336
1.333328
4
8. The area is 1.333 = .
3
9.
n
LRAM n
MRAM n
RRAM n
y
2
10
12.645
13.4775
14.445
50
13.3218
13.4991
13.6818
100
13.41045
13.499775
13.59045
500
13.482018
13.499991
13.518018
Estimate the area to be 13.5.
10.
2
x
⎡ ⎛ 1⎞ ⎛ 1⎞ 2 ⎤⎛ 1⎞
⎛ 1⎞
RRAM: ⎢ 2 ⎜ ⎟ − ⎜ ⎟ ⎥ ⎜ ⎟ + [2(1) − (1)2 ] ⎜ ⎟
⎝ 2⎠
⎢⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎝ 2 ⎠
2
⎡ ⎛ 3⎞ ⎛ 3⎞ ⎤⎛ 1⎞
⎛ 1⎞ 5
+ ⎢ 2 ⎜ ⎟ − ⎜ ⎟ ⎥ ⎜ ⎟ + [2(2) − (2)2 ] ⎜ ⎟ = = 1.25
⎝ 2⎠ 4
⎢⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎝ 2 ⎠
n
LRAM n
MRAM n
10
1.16823
1.09714
1.03490
50
1.11206
1.09855
1.08540
100
1.10531
1.09860
1.09198
500
1.09995
1.09861
1.09728
1000
1.09928
1.09861
1.09795
Estimate the area to be 1.0986.
RRAM n
236 Section 5.1
11.
n
LRAM n
MRAM n
RRAM n
10
0.98001
0.88220
0.78367
50
0.90171
0.88209
0.86244
100
0.89190
0.88208
0.87226
500
0.88404
0.88208
0.88012
1000
0.88306
0.88208
0.88110
Estimate the area to be 0.8821.
12.
n
LRAM n
MRAM n
RRAM n
10
1.98352
2.00825
1.98352
50
1.99934
2.00033
1.99934
100
1.99984
2.00008
1.99984
500
1.99999
2.00000
1.99999
15. LRAM:
Area
≈ f (2) i 2 + f (4) i 2 + f (6) i 2 + + f (22) i 2
= 2 i (0 + 0.6 + 1.4 + + 0.5)
= 44.8 (mg/L ) i sec
RRAM:
Area
≈ f (4) i 2 + f (6) i 2 + f (8) i 2 + + f (24) i 2
= 2(0.6 + 1.4 + 2.7 + + 0)
= 44.8 (mg/L) i sec
Patient's cardiac output:
5 mg
60 sec
i
≈ 6.7 L /min
44.8 (mg/L) i sec 1 min
Note that estimates for the area may vary.
16. (a) LRAM:1 i (0 + 12 + 22 + 10 + 5 + 13 + 11 + 6
+ 2 + 6) = 87 in. = 7.25 ft
(b) RRAM:1 i (12 + 22 + 10 + 5 + 13 + 11 + 6 + 2
+ 6 + 0) = 87 in. = 7.25 ft
Estimate the area to be 2.
13. Use f (x ) = 25 − x and approximate the volume using
2
π r 2 h = π ( 25 − ni2 )2 Δx , so for the MRAM program, use
π (25 − x 2 ) on the interval ⎡⎣ −5, 5 ⎤⎦ .
n
MRAM
10
526.21677
20
524.25327
40
523.76240
80
523.63968
160
523.60900
14. V =
17. 5 min = 300 sec
(a) LRAM: 300 i (1 + 1.2 + 1.7 + + 1.2) = 5220 m
(b) RRAM: 300 i (1.2 + 1.7 + 2.0 + + 0) = 4920 m
18. LRAM:10 i (0 + 44 + 15 + + 30) = 3490 ft
RRAM:10 i (44 + 15 + 35 + + 35) = 3840 ft
3490 ft + 3840 ft
= 3665 ft
Average =
2
19. (a) LRAM: 0.001(0 + 40 + 62 + + 137) = 0.898 mi
RRAM: 0.001(40 + 62 + 82 + + 142) = 1.04 mi
Average = 0.9
969 mi
4
500π
π (5)3 =
≈ 523.59878
3
3
n
error
% error
10
2.61799
0.5
20
0.65450
0.125
40
0.16362
0.0312
80
0.04091
0.0078
160
0.01023
0.0020
(b) The halfway point is 0.4845 mi. The average of LRAM
and RRAM is 0.4460 at 0.006 h and 0.5665 at 0.007 h.
Estimate that it took 0.006 h = 21.6 sec. The car was
going 116 mph.
20. (a) Use LRAM with π (16 − x 2 ).
S8 ≈ 146.08406
S8 is an overestimate because each rectangle is below
the curve.
V − S8
(b)
≈ 0.09 = 9%
V
21. (a) Use RRAM with π (16 − x 2 ).
S8 ≈ 120.95132
S8 is an underestimate because each rectangle is below
the curve.
V − S8
(b)
≈ 0.10 = 10%
V
22. (a) Use LRAM with π (64 − x 2 ) on the interval [4, 8], n = 8.
S ≈ 372.27873 m 3
(b)
V − S8
V
≈ 0.11 = 11%
Section 5.1
23. (a) (5)(6.0 + 8.2 + 9.1 + + 12.7)(30) ≈ 15, 465 ft 3
(b) (5)(8.2 + 9.1 + 9.9 + + 13.0)(30) ≈ 16, 515 ft 3
24. Use LRAM with π x on the interval [ 0, 5], n = 5.
1(0 + π + 2π + 3π + 4π ) = 10π ≈ 31.41593
25. Use MRAM with π x on the interval [0, 5]. n = 5.
⎛1
3
5
7
9 ⎞ 25
1⎜ π + π + π + π + π ⎟ = π ≈ 39.26991
2
2
2
2 ⎠ 2
⎝2
26. (a) LRAM5 :
32.00 + 19.41 + 11.77 + 7.14 + 4.33 = 74.65 ft/sec
(b) RRAM5 :
19.41 + 11.77 + 7.14 + 4.33 + 2.63 = 45.28 ft/sec
(c) The upper estimates for speed are 32.00 ft/sec for the
first sec, 32.00 + 19.41 = 51.41 ft/sec for the second sec,
and 32.00 + 19.41 + 11.77 = 63.18 ft/sec for the third
sec. Therefore, an upper estimate for the distance fallen
is 32.00 + 51.41 + 63.18 = 146.59 ft.
27. (a) 400 ft /sec – (5 sec)(32 ft/sec 2 ) = 240 ft/sec
(b) Use RRAM with 400 – 32x on [0, 5], n = 5.
368 + 336 + 304 + 272 + 240 = 1520 ft
28. (a) Upper = 70 + 97 + 136 + 190 + 265 = 758 gal
Lower = 50 + 70 + 97 + 136 + 190 = 543 gal
(b) Upper = 70 + 97 + 136 + 190 + 265 + 369 + 516 + 720
= 2363 gal
Lower = 50 + 70 + 97 + 136 + 190 + 265 + 369 + 516
= 1693 gal
(c) 25,000 – 2363 = 22,637 gal
22, 637
≈ 31.44 h (worst case)
720
25,000 − 1693 = 23, 307 gal
23,307
≈ 32.37 h (best case)
720
29. (a) Since the release rate of pollutants is increasing, an
upper estimate is given by using the data for the end of
each month (right rectangles), assuming that new
scrubbers were installed before the beginning of
January. Upper estimate:
30(0.20 + 0.25 + 0.27 + 0.34 + 0.45 + 0.52)
≈ 60.9 tons off pollutants
A lower estimate is given by using the data for the end
of the previous month (left rectangles). We have no data
for the beginning of January, but we know that
pollutants were released at the new-scrubber rate of
0.05 ton/day, so we may use this value.
Lower Estimate:
30(0.05 + 0.20 + 0.25 + 0.27 + 0.34 + 0.45)
≈ 46.8 tons off pollutants
(b) Using left rectangles, the amount of pollutants
released by the end of October is
30(0.05 + 0.20 + 0.25 + 0.27 + 0.34 + 0.45
+ 0.52 + 0.63 + 0.70 + 0.81) ≈ 126.6 tons
Therefore, a total of 125 tons will have been released
into the atmosphere by the end of October.
237
30. The area of the region is the total number of units sold, in
millions, over the 10-year period. The area units are
(millions of units per year)(years) = (millions of units).
31. True. Because the graph rises from left to right, the lefthand rectangles will all lie under the curve.
32. False. For example, all three approximations are the same if
the function is constant.
33. E. y = 4 x − x 2 = 0
4x = x2
x = 0, 4
Use MRAM on the interval [0, 4], n = 4.
1(1.75 + 3.75 + 3.75 + 1.75) = 11
34. D.
35. C.
π⎛
⎛π⎞
⎛π⎞
⎛ 3π ⎞ ⎞
sin(0) + sin ⎜ ⎟ + sin ⎜ ⎟ + sin ⎜ ⎟ ⎟
⎜
⎝ 4⎠
⎝ 2⎠
⎝ 4 ⎠⎠
4⎝
2⎞
2
π⎛
+ 1+
⎟
⎜0+
2
4⎝
2 ⎠
36. D.
37. (a) The diagonal of the square has length 2, so the side
length is 2 . Area = ( 2 )2 = 2
(b) Think of the octagon as a collection of 16 right triangles
with a hypotenuse of length 1 and an acute angle
2π π
measuring
= .
16 8
⎛ 1⎞ ⎛ π ⎞ ⎛
π⎞
Area = 16 ⎜ ⎟ ⎜ sin ⎟ ⎜ cos ⎟
8⎠
⎝ 2⎠ ⎝ 8 ⎠ ⎝
π
= 4 sin
4
= 2 2 ≈ 2.828
(c) Think of the 16-gon as a collection of 32 right triangles
with a hypotenuse of length 1 and an acute angle
2π π
measuring
= .
32 16
π ⎞⎛
π⎞
⎛ 1⎞ ⎛
Area = 32 ⎜ ⎟ ⎜ sin ⎟ ⎜ cos ⎟
⎝ 2 ⎠ ⎝ 16 ⎠ ⎝
16 ⎠
π
= 8 sin ≈ 3.061
8
(d) Each area is less than the area of the circle, π . As n
increases, the area approaches π .
38. The statement is false. We disprove it by presenting a
counterexample, the function f ( x ) = x 2 over the interval
0 ≤ x ≤ 1, with n = 1. MRAM1 = 1 f (0.5) = 0.25
LRAM1 + RRAM1 1 f (0) + 1 f (1)
=
2
2
0 +1
=
= 0.5 ≠ MRAM1
2
238 Section 5.2
39. RRAM n f = ( Δx )[ f ( x1 ) + f ( x 2 ) + + f ( x n−1 ) + f ( x n )]
= ( Δx )[ f ( x 0 ) + f ( x1 ) + f ( x 2 ) + + f ( x n−1 )]
+ ( Δx )[ f ( x n ) − f ( x 0 )]
= LRAM n f + ( Δx )[ f ( x n ) − f ( x 0 )]
But f (a) = f (b) by symmetry, so f (xn) – f (x0) = 0.
Therefore, RRAM n f = LRAM n f.
40. (a) Each of the isosceles triangles is made up of two right
triangles having hypotenuse 1 and an acute angle
2π π
measuring
= . The area of each isosceles triangle
2n n
π ⎞ 1 2π
⎛ 1⎞ ⎛ π ⎞ ⎛
is A T = 2 ⎜ ⎟ ⎜ sin ⎟ ⎜ cos ⎟ = sin .
⎝ 2⎠ ⎝ n ⎠ ⎝
n⎠ 2
n
π
4. 2π + 2. The same area as ∫ sin x dx sits above a rectangle of
0
area π × 2.)
5. 4. (Each rectangle in a typical Riemann sum is twice as tall
π
as in ∫ sin x dx.)
0
(b) The area of the polygon is
n 2π
AP = nAT = sin
, so
2
n
n 2π
lim AP = lim sin
=π
n→∞
n→∞ 2
n
π
6. 2. (This is the same region as in ∫ sin x dx , translated 2
(c) Multiply each area by r 2 :
0
1
2π
A T = r 2 sin
n
2
n 2 2π
A P = r sin
n
2
lim A P = π r 2
units to the right.)
n→∞
Section 5.2 Definite Integrals (274–284)
Exploration 1 Finding Integrals by
Signed Areas
7. 0. (The equal areas above and below the x-axis sum to
zero.)
π
1. −2. (This is the same area as ∫ sin x dx , but below the
0
x-axis.)
8. 4. (Each rectangle in a typical Riemann sum is twice as
π
wide as in ∫ sin x dx.)
0
2. 0. (The equal areas above and below the x- axis sum to
zero.)
9. 0. (The equal areas above and below the x-axis sum to
zero.)
π
3. 1. (This is half the area of ∫ sin x dx.)
0
Section 5.2
10. 0. (The equal areas above and below the x-axis sum to zero,
since sin x is an odd function.)
239
25
5.
∑ 2k
k= 0
500
6.
∑ 3k 2
k=1
50
50
50
x =1
x =1
x =1
7. 2∑ x 2 + 3∑ x = ∑ (2 x 2 + 3x )
Exploration 2 More Discontinuous
Integrands
8.
1. The function has a removable discontinuity at x = 2.
8
20
20
k =0
k =9
k =0
∑ xK + ∑ xk = ∑ xk
n
9.
∑ (−1)k = 0 if n is odd.
k =0
n
10.
∑ (−1)k = 1 if n is even.
k =0
2. The thin strip above x = 2 has zero area, so the area under
Section 5.2 Exercises
3
the curve is the same as ∫ ( x + 2), which is 10.5.
n
1. lim ∑ (ck 2 Δx k =
0
n→∞
k =1
2 2
∫0 x
dx where n is any partition of [0, 2].
n
5
2. lim ∑ (ck 2 − 3ck )Δx k =
n→∞
∫−7 ( x
k =1
2
− 3x ) dx where n is any
partition of [−7, 5] .
n
3. The graph has jump discontinuities at all integer values, but
the Riemann sums tend to the area of the shaded region
shown. The area is the sum of the areas of 5 rectangles (one
of them with height 0):
5
1
Δx k =
n→∞
k =1 ck
3. lim ∑
n
1
Δx =
−
ck k
1
k =1
4. lim ∑
n→∞
∫ 0 int( x ) dx = 0 + 1 + 2 + 3 + 4 = 10.
4
∫1
1
dx where n is any partition of [1, 4].
x
1
3
∫2 1 − x dx
where n is any partition of
[2, 3].
n
5. lim ∑ 4 − ck 2 Δx k =
n→∞
k =1
1
∫0
4 − x 2 dx where n is any partition
of [0, 1].
n
6. lim ∑ (sin 3 ck )Δx k =
n→∞
5
1. ∑ n = (1) + (2) + (3) + (4) + (5) = 55
2
2
2
2
2
4
2. ∑ (3k − 2) = [3(0) − 2] + [3(1) − 2] + [3(2) − 2]
k =0
+[3(3) − 2] + [3(4) − 2] = 20
3.
∑ 100( j + 1)
j=0
2
= 100[(1) + (2) + (3) + (4) + (5) ]
2
= 5500
x dx where n is any partition of
2
2
2
2
1
7.
∫−2 5 dx = 5[1− (−2)] = 15
8.
∫3 (−20) dx = (−20)(7 − 3) = −80
9.
∫0 (−160) dt = (−160)(3 − 0) = −480
n =1
4
3
[ −π , π ] .
Quick Review 5.2
2
k =1
π
∫−π sin
7
3
−1 π
π
3π
2
10.
∫−4 2 dθ = 2 [−1 − (−4)] =
11.
∫−2.1 0.5 ds = 0.5[3.4 − (−2.1)] = 2.75
12.
∫
3.4
99
4.
∑k
k=1
18
2
2 dr = 2( 18 − 2) = 4
240 Section 5.2
13. Graph the region under y =
x
+ 3 for − 2 ≤ x ≤ 4.
2
The region is one quarter of a circle of radius 4.
0
1
2
2
∫−4 16 − x dx = 4 π (4) = 4π
1
⎛x ⎞
∫−2 ⎜⎝ 2 + 3⎟⎠ dx = 2 (6)(2 + 5) = 21
4
14. Graph the region under y = −2 x + 4 for
16. Graph the region under y = 16 − x 2 for − 4 ≤ x ≤ 0.
1
3
≤x≤ .
2
2
17. Graph the region under y = x for − 2 ≤ x ≤ 1.
1
1
1
5
∫−2 x dx = 2 (2)(2) + 2 (1)(1) = 2
3/ 2
1
∫1/ 2 (−2 x + 4) dx = 2 (1)(3 + 1) = 2
18. Graph the region under y = 1 − x for − 1 ≤ x ≤ 1.
15. Graph the region under y = 9 − x 2 for − 3 ≤ x ≤ 3.
1
1
∫−1(1 − x ) dx = 2 (2)(1) = 1
This region is half of a circle radius 3.
3
1
9π
2
2
∫−3 9 − x dx = 2 π (3) = 2
Section 5.2
19. Graph the region under y = 2 − x for − 1 ≤ x ≤ 1.
1
1
22. Graph the region under y = r for 2 ≤ r ≤ 5 2.
1
∫−1(2 − x ) dx = 2 (1)(1 + 2) + 2 (1)(1 + 2) = 3
20. Graph the region under y = 1 + 1 − x 2 for − 1 ≤ x ≤ 1.
1
∫−1(1 +
1
π
1 − x 2 ) dx = (2)(1) + π (1)2 = 2 +
2
2
21. Graph the region under y = θ for π ≤ θ ≤ 2π
∫
5 2
2
1
r dr = (5 2 − 2 )( 2 + 5 2 ) = 24
2
1
b
1
23.
∫0 x dx = 2 (b)(b) = 2 b
24.
∫0 4 x dx = 2 (b)(4b) = 2b
25.
∫a 2s ds = 2 (b − a)(2b + 2a) = b
26.
∫a 3t dt = 2 (b − a)(3b + 3a) = 2 (b
27.
∫a
28.
∫a
29.
∫8
2
1
b
2
1
b
1
b
3
− a2
2
30.
3a
1
1
x dx = ( 3a − a)( 3a + a) = (3a 2 − a 2 ) = a 2
2
2
87 dt = 87 t
60
∫0
11
8
25 dt = 25 t
60
0
25(60) − 25(0) = 1500 gallons
31.
7.5
∫6
300 dt = 300 t
7.5
6
calories
300(7.5) − 300(6) = 450
2π
1
3π 2
θ dθ = (2π − π )(2π + π ) =
2
2
32.
11
− a2 )
1
3a 2
x dx = (2a − a)(2a + a) =
2
2
2a
11
2
87(11) − 87(8) = 261 miles
∫π
241
∫8.5 0.4 dt = 0.4t 8.5
11
0.4(11) − 0.4(8.5) = 1liter
⎛ x
⎞
, x , 0, 5⎟ ≈ 0.9905
33. NINT ⎜ 2
⎝ x +4
⎠
34. 3 + 2 ⋅ NINT(tan x , x , 0,
3
) ≈ 4.3863
35. NINT(4 − x 2 , x , − 2.2) ≈ 10.6667
36. NINT(x 2e − x , x , − 1, 3) ≈ 1.8719
242 Section 5.2
37. (a) The function hasa discontinuity at x = 0.
(b)
42. True. All the products in the Riemann sums are positive.
5
43. E. ∫ ( f ( x ) + 4) dx
2
5
5
∫2 f ( x ) dx + ∫2 4 dx
=
= 18 + 4 x
5
2
= 30
4
3
∫−2
44. D. ∫ (4 − x ) dx
x
dx = −2 + 3 = 1
x
−4
4
38. (a) The function has discontinuities at
x = −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5.
4
= 4x
(b)
0
∫−4 4 dx + ∫0 x dx + ∫−4 − x dx
=
4
−4
+
x2
2
4
0
−
x2
2
0
−4
= 16
45. C.
46. A.
47. Observe that the graph of f ( x ) = x 3 is symmetric with
respect to the origin. Hence the area above and below the
x-axis is equal for −1 ≤ x ≤ 1.
5
∫−6 2 int( x − 3) dx = (−18) + (−16) + (−14)
+ (−12) + (−10) + (−8) + (−6) + (−4) + (−2)
+ 0 + 2 = −88
39. (a) The function hasa discontinuity at x = −1.
3
48. The graph of f ( x ) = x 3 + 3 is three units higher than the
graph of g( x ) = x 3 . The extra area is (3)(1) = 3.
1
∫0 ( x
(b)
4
1
∫−1 x dx = −(area below x -axis) + (area above x -axis) = 0
x2 − 1
1
1
3
+ 3) dx =
1
13
+3=
4
4
49. Observe that the region under the graph of f ( x ) = ( x − 2)3
for 2 ≤ x ≤ 3 is just the region under the graph of
g( x ) = x 3 for 0 ≤ x ≤ 1 translated two units to the right.
3
1 3
1
3
∫2 ( x − 2) dx = ∫0 x dx = 4
7
∫−3 x + 1 dx = − 2 (4)(4) + 2 (3)(3) = − 2
3
40. (a) The function hasa discontinuity at x = 3.
50. Observe that the graph of f ( x ) = x is symmetric with
respect to the y-axis and the right half is the graph of
g( x ) = x 3.
(b)
1
∫−1 x
3
1
dx = 2 ∫ x 3 dx =
0
1
2
51. Observe from the graph below that the region under the
6
∫−5
graph f ( x ) = 1 − x 3 for 0 ≤ x ≤ 1 cuts out a region R from
the square identical to the region under the graph of
9 − x2
1
1
77
dx = (2)(2) − (9)(9) = −
x−3
2
2
2
g( x ) = x 3 for 0 ≤ x ≤ 1.
y
41. False. Consider the function in the graph below.
1
y
y = f(x)
a
b
R
x
1
1
∫0 (1 − x
3
) dx = 1 − ∫
x
1 3
x dx = 1 −
0
1 3
=
4 4
Section 5.2
52. Observe from the graph of f ( x ) = ( x − 1)3 for − 1 ≤ x ≤ 2
that there are two regions below the x-axis and one region
above the axis, each of whose area is equal to the area of
the region under the graph of g( x ) = x 3 for 0 ≤ x ≤ 1.
243
56.Observe from the graph below that the region between the
graph of f ( x ) = 3 x and the x-axis for 0 ≤ x ≤ 1 cuts out a
region R from the square identical to the region under the
graph of g( x ) = x 3 for 0 ≤ x ≤ 1.
y
1
R
y = f(x)
1
1
∫0
1
⎛ 1⎞ ⎛ 1⎞
3
∫−1( x − 1) dx = 2 ⎜⎝ − 4 ⎟⎠ + ⎜⎝ 4 ⎟⎠ = − 4
2
3
(b) Using right endpoints we have
⎛ 1 ⎞
n
2 ⎟ ⎛ 1⎞
⎜
dx
=
lim
∫0 x 2 n→∞ ∑
⎜ ⎛ k ⎞ ⎟ ⎜⎝ n ⎟⎠
k =1 ⎜ ⎜ ⎟ ⎟
⎝ ⎝ n⎠ ⎠
n
n
= lim ∑ 2
n→∞
k =1 k
1
1⎞
⎛
= lim n ⎜ 1 + 2 + + 2 ⎟ .
n→∞ ⎝
2
n ⎠
1
3
⎛ x⎞
a factor of 2. Thus the area under f ( x ) = ⎜ ⎟ for 0 ≤ x ≤ 2
⎝ 2⎠
is twice the area under the graph of g( x ) = x 3 for 0 ≤ x ≤ 1.
2⎛
x⎞
3
1
dx = 2 ∫ x 3 dx =
0
1
2
respect to the orgin. Hence the area above and below the
x-axis is equal for −8 ≤ x ≤ 8.
∫−8 x
3
58. (a) Δx =
dx = − (area below x -axis) + (area above x -axiis) = 0
2
y
x
n
(b)
(c)
y = f(x)
R
–1
1
∫0 ( x
3
− 1) dx = −1 +
1
3
=−
4
4
1
k
,x =
n k n
2
2
1
1 ⎛ 2⎞
1
⎛ 1⎞
⎛ n⎞
RRAM = ⎜ ⎟ i + ⎜ ⎟ i + + ⎜ ⎟ i
⎝ n⎠ n ⎝ n⎠ n
⎝ n⎠ n
2
n ⎛
1⎞
⎛ k⎞
= ∑⎜ ⎜ ⎟ i ⎟
⎜ ⎝ ⎠ n ⎟⎠
k =1 ⎝ n
55. Observe from the graph below that the region between the
graph of f ( x ) = x 3 − 1 and the x-axis for 0 ≤ x ≤ 1 cuts out a
region R from the square identical to the region under the
graph of g( x ) = x 3 for 0 ≤ x ≤ 1.
1
1
1
1⎞
⎛
Note that n ⎜ 1 + 2 + + 2 ⎟ > n and n → ∞, so
⎝ 2
n ⎠
1⎞
1
⎛
n ⎜ 1 + 2 + + 2 ⎟ → ∞.
⎝ 2
n ⎠
54. Observe that the graph of f ( x ) = x 3 is symmetric with
8
1 3
=
4 4
57. (a) As x approaches 0 from the right, f (x) goes to ∞.
⎛ x⎞
53. Observe that the graph of f ( x ) = ⎜ ⎟ for 0 ≤ x ≤ 2 is just a
⎝ 2⎠
horizontal stretch of the graph of g( x ) = x 3 for 0 ≤ x ≤ 1 by
∫0 ⎜⎝ 2 ⎟⎠
x dx = 1 −
x
⎛⎛ k⎞2
∑ ⎜⎜ ⎜⎝ n ⎟⎠
k =1 ⎝
1
n
3
n
i
1 ⎞ n ⎛ k2 ⎞ 1 n 2
= ∑k
⎟ =∑
n ⎟⎠ k =1 ⎜⎝ n 3 ⎟⎠ n 3 k =1
1
∑ k 2 = n3 i
k =1
n(n + 1)(2n + 1) n(n + 1)(2n + 1)
=
6
6n 3
2
∞ ⎛
n(n + 1)(2n + 1)
1⎞
⎛ k⎞
(d) lim ∑ ⎜ ⎜ ⎟ i ⎟ = lim
⎜
⎟
n→∞
n→∞
⎝
⎠
n
n
6n 3
k =1 ⎝
⎠
= lim
2n 3 + 3n 2 + n
n→∞
=
2 1
=
6 3
6n 3
244 Section 5.3
58. Continued
(e) Since ∫
1 2
x dx
0
equals the limit of any Riemann sum over
the interval [0, 1]as n approaches ∞, part (d) proves that
1 2
∫0 x
1
dx = .
3
Exploration 2 Finding the Derivative of
an Integral
Pictures will vary according to the value of x chosen.
(Indeed, this is the point of the exploration.) We show a
typical solution here.
Section 5.3 Definite Integrals and
Antiderivatives (pp. 285–293)
Exploration 1 How Long is the Average
Chord of a Circle?
1. The chord is twice as long as the leg of the right triangle in
the first quadrant, which has length
Pythagorean Theorem.
r 2 − x 2 by the
1. We have chosen an arbitray x between a and b.
2. We have shaded the region using vertical line segments.
3. The shaded region can be written as
x
∫a
f (t ) dt using the
definition of the definite integral in Section 5.2. We use t as
a dummy variable because x cannot vary between a and
itself.
4. The area of the shaded region is our value of F(x).
2. Average value =
r
1
2 r 2 − x 2 dx.
∫
−
r − (− r ) r
2 r
r 2 − x 2 dx
2r ∫− r
1
= i (area of semicircleof radius r )
r
3. Average value =
1 πr2
= i
r 2
πr
=
2
4. Although we only computed the average length of chords
perpendicular to a particular diameter, the same
computation applies to any diameter. The average length of
πr
a chord of a circle of radius r is .
2
5. The function y = 2 r 2 − x 2 is continuous on [−r , r ], so the
Mean Value Theorem applies and there is a c in [a, b] so
πr
that y(c) is the average value
.
2
5. We have drawn one more vertical shading segment to
represent ΔF.
6. We have moved x a distance of Δx so that it rests above the
new shading segment.
Section 5.3
7. Now the (signed) height of the newly-added vertical
segment is f (x).
8. The (signed) area of the segment is Δ F = Δ x i f ( x ), so
F ′(x ) = lim
Δx →0
9
9
1
1
2. (a) ∫ −2 f ( x ) dx = −2 ∫ f ( x ) dx = −2(−1) = 2
(b)
ΔF
= f (x)
Δx
9
9
(c)
9
9
dy sec x tan x
=
= tan x
dx
sec x
4.
dy cos x
=
= cot x
dx sin x
5.
dy sec x tan x + sec x
=
= sec x
dx
sec x + tan x
6.
dy
⎛ 1⎞
= x ⎜ ⎟ + 1n x − 1 = 1n x
⎝ x⎠
dx
7.
dy (n + 1) x
=
= xn
dx
n +1
(e)
∫1
(f)
=
∫1
9
f ( x ) dx − ∫ f ( x ) dx
9
9
7
1
2
∫1 3 f ( x ) dx = 3∫1
f ( x ) dx = 3(−4) = −12
1
5
∫2 f ( x ) dx = ∫2 f ( x ) dx + ∫1
2
5
1
(c)
∫2 f (t ) dt = − ∫1
9
7
(d)
∫1 [− f ( x )] dx = − ∫1
0
g(t ) dt = − ∫ g(t ) dt = − 2
−3
0
(c)
∫−3[− g( x )] dx = − ∫−3 g( x )dx = −
(d)
∫−3
5. (a)
2
0
0
4
∫3
0
g(r )
2
1
0
∫ g(r ) dr = 1
2 −3
dr =
f ( z ) dz =
2
0
4
∫3
f ( z ) dz + ∫ f ( z ) dz
0
3
4
∫0 f (z) dz + ∫0 f (z ) dz
= −3 + 7 = 4
5
4 f ( x ) dx − ∫ g( x ) dx
1
5
= 4 ∫ f ( x ) dx − ∫ g( x ) dx
1
f ( x ) dx = −5
∫−3 g(u) du =
3
1
0
3
∫4 f (t ) dt = ∫4 f (t ) dt + ∫0 f (t ) dt
4
3
0
0
= − ∫ f (t ) dt + ∫ f (t ) dt
1
= 4(6) − 8 = 16
2
−3
∫0
5
5
f (t ) dt = −5
2
f ( x ) dx − ∫ g( x ) dx
∫1
2
(b)
(b)
5
2
1
1
= −7 + 3 = −4
= 6 − 8 = −2
[4 f ( x ) − g( x )] dx =
9
7
3 f ( z ) dz = 3 ∫ f ( z ) dz = 5 3
=
= 4 + 6 = 10
5
7
f (u) du = 5
2
f ( x ) dx
1
∫1
2
∫1
= − ∫ f ( x ) dx + ∫ f ( x ) dx
(e) ∫ [ f ( x ) − g( x )] dx =
7
(b)
5
∫5 g( x ) dx = − ∫1 g( x ) dx = −8
2
7
∫9 [h( x ) − f ( x )] dx = ∫9 h( x ) dx − ∫9 f ( x ) dx
∫1
2
∫2 g( x ) dx = 0
∫1
f ( x ) dx + ∫ f ( x ) dx
3. (a)
4. (a)
Section 5.3 Exercises
5
7
∫1
= −4 + 5 = 1
dy
1
=
dx x 2 + 1
(f)
9
f ( x ) dx =
= − ∫ h( x ) dx + ∫ f ( x ) dx
dy
9.
= xe x + e x
dx
1
7
f ( x ) dx = 1
= −1− 5 = −6
2 x1n 2
1
dy
x
i
(
n
)
8.
=− x
1
2
2
=
−
dx
(2 + 1)2
(2 x + 1)2
5
9
∫9 f ( x ) dx = − ∫1
n
5
1
(d)
2
(d)
9
7
= 2(5) − 3(4) = −2
3.
(c)
9
7
= 2 ∫ f ( x ) dx − 3∫ h( x ) dx
dy
2.
= cos x
dx
(b)
9
∫7 [2 f ( x ) − 3h( x )] dx = ∫7 2 f ( x ) dx + ∫7 3h( x ) dx
dy
1.
= sin x
dx
1. (a)
9
∫7 [ f ( x ) + h( x )] dx = ∫7 f ( x ) dx + ∫7 h( x ) dx
= 5+ 4 = 9
Quick Review 5.3
10.
245
6. (a)
3
−1
∫1 h(r ) dr = ∫1
3
h(r ) dr + ∫ h(r ) dr
−1
1
3
−1
−1
= − ∫ h(r ) dr + ∫ h(r ) dr = 6
246 Section 5.3
6. Continued
−1
1
1
(b) − ∫ h(u) du = − ∫ h(u) du − ∫ h(u) du
3
−1
3
=
3
1
∫−1 h(u) du − ∫−1 h(u) du = 6
7. max sin (x2) = sin (1) on [0, 1]
1
∫0 sin( x
2
) dx ≤ sin (1) < 1
8. max x + 8 = 3 and min x + 8 = 2 2 on [0, 1]
2 2≤∫
1
0
x + 8 dx ≤ 3
9. (b − a) min f (x ) ≥ 0 on [a, b]
b
0 ≤ (b − a) min f ( x ) ≤ ∫ f ( x ) dx
a
10. (b − a) max f ( x ) ≤ 0 on [a, b]
b
∫a f ( x )dx ≤ (b − a) max f ( x ) ≤ 0
11. An antiderivative of x 2 − 1 is F ( x ) =
av =
1
∫
3 0
3
1 3
x − x.
3
( x 2 − 1) dx
1
14. An antiderivative of ( x − 1)2 is F ( x ) = ( x − 1)3 .
3
1 3
1
1 ⎛ 8 1⎞
av = ∫ ( x − 1)2 dx = [ F (3) − F (0)] = ⎜ + ⎟ = 1
0
3
3
3 ⎝ 3 3⎠
2
Find x = c in [0, 3] such that (c − 1) = 1.
c − 1 = ±1
c = 2 or c = 0.
Since both are in [0, 3], x = 0 or x = 2.
15. The region between the graph and the x-axis is a triangle of
height 3 and base 6, so the area of the region
1
is (3)(6) = 9.
2
1 2
9 3
av( f ) = ∫ f ( x ) dx = = .
6 −4
6 2
16. The region between the graph and the x-axis is a rectangle
with a half circle of radius 1 cut out. The area of the region
1
4−π
is 2(1) − π (1)2 =
.
2
2
av( f ) =
17. There are equal areas above and below the x-axis.
1 2π
1
av( f ) =
f (t ) dt =
i0=0
2π ∫ 0
2π
1 ⎡
F ( 3 ) − F (0) ⎤
⎦
3⎣
1
=
(0 − 0) = 0
3
=
Find x = c in [0, 3 ] such that c 2 − 1 = 0
c2 = 1
c = ±1
18. Since tan θ is an odd function, there are equal areas above
and below the x-axis.
1 π /4
2
av( f ) =
f (θ ) dθ = i 0 = 0
π / 2 ∫−π / 4
π
19.
Since 1 is in [0, 3 ], x = 1.
12. An antiderivative of −
av =
2π
∫π
sin x dx = − cos(2π ) + cos(π )
= −2
x2
x3
is F ( x ) = − .
2
6
1 3⎛ x2 ⎞
1
1⎛ 9⎞
3
−
dx = [ F (3)] − F (0)] = ⎜ − ⎟ = −
3 ∫0 ⎜⎝ 2 ⎟⎠
3
3⎝ 2⎠
2
Find x = c in [0, 3] such that −
1 1
1⎛ 4−π ⎞ 4−π
f (t ) dt = ⎜
.
=
2 ∫−1
2 ⎝ 2 ⎟⎠
4
3
c2
=− .
2
2
c2 = 3
c=± 3
Since 3 is in [0, 3], x = 3.
13. An antiderivative of − 3x 2 − 1 is F ( x ) = − x 3 − x.
1 1
av = ∫ (−3x 2 − 1) dx = F (1) − F (0) = −2
1 0
Find x = c in [0, 1] such that − 3c 2 − 1 = −2
−3c 2 = −1
1
c2 =
3
1
c=±
3
1
1
is in [0, 1], x =
.
Since
3
3
π /2
⎛π⎞
cos x dx = sin ⎜ ⎟ − sin (0) = 1
⎝ 2⎠
20.
∫0
21.
∫0
22.
∫0
23.
∫1 2 x dx = x
24.
∫−13x
25.
∫−2 5 dx =5 x −2 = 5(6) − 5(−2) = 40
26.
∫3 8 dx = 8 x 3 = 8(7) − 8(3) = 32
27.
∫−1 1 + x 2 dx = tan
28.
∫0
π /1 x
e dx = e1 − e 0 = e − 1
π /4
⎛π⎞
sec 2 x dx = tan ⎜ ⎟ − tan 0 = 1
⎝ 4⎠
4
2
2
2
dx = x 3
6
6
7
7
1
1/ 2
4
1
1
= 4 2 − 12 = 15
2
−1
= 23 − (−1)3 = 9
−1
(1) − tan −1 (−1) =
π
2
π
⎛ 1⎞
dx = sin −1 ⎜ ⎟ − sin −1 (0) =
⎝
⎠
2
6
1− x
1
2
Section 5.3
29.
30.
e
b
1
39. Yes, ∫ av( f ) dx =
∫1 x dx = ln e − ln1 = 1
4
∫1
a
b
∫a av ( f ) dx = ⎡⎣ av( f ) i x ⎤⎦ a
=
2e 1
1
1
dx = (ln 2e − ln e)
∫
e
2e − e
x
e
ln2
=
e
∫
π
−0
4
4
=
π
34. av ( f ) =
=
⎛π⎞
x dx = tan ⎜ ⎟ − tan(0)
⎝ 4⎠
(b)
150 mi 150 mi
+
=8h
30 mph 50 mph
(c)
300 mi
= 37.5 mph
8h
(d) The average speed is the total distance divided by the
d + d2
. The driver computed
total time. Algebraically, 1
t1 + t2
1 ⎛ d1 d 2 ⎞
+
. The two expressions are not equal.
2 ⎜⎝ t1 t2 ⎟⎠
1 1 1
dx = tan −1 (1) − tan −1 (0)
1 − 0 ∫0 1 + x 2
π
4
41. Time for first release =
(
2
1
1
3x 2 + 2 x dx = x 3 + x 2
∫
−
1
2 − (−1)
3
=4
35. av ( f ) =
36. av ( f ) =
1
π
−0
3
3
=
π
π
3⎛
⎛π⎞
)
2
1
and max f = 1
2
1 1
1
dx ≤ 1
≤
2 ∫0 1 + x 4
16
17
0.5
1
⎛ 1 ⎞ ⎛ 16 ⎞
⎛ 1⎞
dx ≤ ⎜ ⎟ (1)
⎜⎝ 2 ⎟⎠ ⎜⎝ 17 ⎟⎠ ≤ ∫ 0
4
⎝ 2⎠
1+ x
0.5
8
1
1
≤
dx ≤
17 ∫ 0 1 + x 4
2
1
1
⎛ 1⎞ ⎛ 1⎞
⎛ 1 ⎞ ⎛ 16 ⎞
dx ≤ ⎜ ⎟ ⎜ ⎟
⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ≤ ∫ 0 . 5
4
⎝ 2 ⎠ ⎝ 17 ⎠
1+ x
1
1
1
8
dx ≤
≤
17
4 ∫0 .5 1 + x 4
1
8 1
1
1 8
+ ≤
dx ≤ +
17 4 ∫ 0 1 + x 4
2 17
1
49
1
33
≤
dx ≤
68 ∫ 0 1 + x 4
34
1000 m 3
10 m 3 /min
Time for second release =
= 100 min
100 m 3
= 50 min
20 m 3 /min
total released 2000 m 3
1
=
= 13 m 3 /min
Average rate =
total time
150 miin
3
−1
⎞
∫03 sec x tan x dx = π ⎜⎝ sec ⎜⎝ 3 ⎟⎠ − sec 0⎟⎠
37. min f =
38. f (0. 5) =
b
∫a f ( x ) dx
40. (a) 300 mi
32. av ( f ) =
π
4 sec 2
0
b
= av ( f ) i b − av ( f ) i a
= (b − a) av ( f )
b
⎡ 1
⎤
= (b − a) ⎢
f ( x ) dx ⎥
∫
a
⎣b − a
⎦
π
1
31. av ( f ) =
sin x dx
π − 0 ∫0
1
2
= (− cos π − (− cos 0) =
π
π
33. av ( f ) =
b
∫a f ( x ) dx.
This is because av(f) is a constant, so
4
1
1 1
3
− x dx =
= − =−
4
x1 4 1
−2
1
247
42.
1
1
⎡1
∫0 sin x dx ≤ ∫0 x dx = ⎢⎣ 2 x
1
2⎤
1
⎥ =2
⎦0
1
⎡
7
x2 ⎞
x3 ⎤
43. ∫ sec x dx ≥ ∫ ⎜ 1 + ⎟ dx = ⎢ x + ⎥ =
0
0⎝
2⎠
6
6
⎥⎦ 0
⎢⎣
1
1⎛
44. Let L(x) = cx + d. Then the average value of f on [a, b] is
b
1
av( f ) =
(cx + d )dx
b − a ∫a
⎞⎤
⎞ ⎛ ca 2
1 ⎡⎛ cb 2
+ da ⎟ ⎥
=
+ db ⎟ − ⎜
⎢⎜
b − a ⎢⎣⎝ 2
⎠ ⎥⎦
⎠ ⎝ 2
2
2
⎤
1 ⎡ c( b − a )
+ d (b − a) ⎥
=
⎢
b − a ⎢⎣
2
⎥⎦
c( b + a ) + 2 d
=
2
(ca + d ) + (cb + d )
=
2
L (a) + L (b)
=
2
45. False. For example, sin 0 = sin π = 0, but the average value
of sin x on [0, π] is greater than 0.
46. False. For example,
3
∫−3 2 x dx = 0 but 2(−3) ≠ 2(3)
248 Section 5.4
47. A. There is no rule for the multiplication of functions.
48. D. There is no rule for the negation of the bounds.
1
1
cos xdx = (sin 5 − sin 1)
5 − 1 ∫1
4
⫽ – 0.450.
5
49. B. av( f ) =
50. C. 10 =
b
1
F ( x )dx
∫
a
b−a
10(b − a) =
51. (a) Area =
(b)
(c)
∫0
+ 12 x + c
y = 4x − 5
m= f′=4
3(0)2 + 12(0) + c = 4
c=4
2
2
+ 12 x + 4)dx
f (x) = x + 6x + 4 x + c
f (0) = (0)3 + 6(0)2 + 4(0) + c = −5
c = −5
f (x) = x3 + 6x 2 + 4 x − 5
3
b
∫a f ( x ) dx
1
bh
2
(b) av( f ) =
b
hb 2 1
⎡h
⎤
y( x ) dx = ⎢ x 2 ⎥ =
= bh
⎣ 2b ⎦ 0 2b 2
k
52. av( x k ) =
∫ f ′′ ( x )dx = 3x
∫ f ′( x )dx = ∫ (3x
h 2
x +C
2b
b
4. (a) f ′′ ( x ) = 6 x + 12
1 k k
1 ⎡ 1 k +1 ⎤
k k +1
x dx = ⎢
x ⎥ =
∫
0
k
k ⎣k +1
⎦ 0 k (k + 1)
x x +1
and y2 = x on a graphing calculator and
x ( x + 1)
find the point of intersection for x > 1.
Graph y1 =
2
1
1
( x 3 + 6 x 2 + 4 x − 5)dx
1 − (−1) ∫−1
⎞
1 ⎛ x4
+ 2 x 3 + 2 x 2 − 5x ⎟
2 ⎜⎝ 4
⎠
1
= −3
−1
Section 5.4 Fundamental Theorem of
Calculus (pp. 294–305)
Exploration 1 Graphing NINT f
2. The function y ⫽ tan x has vertical asymptotes at all odd
π
multiples of . There are six of these between –10 and 10.
2
−10
Thus, k ≈ 2.39838
53. An antiderivative of F′ (x) is F(x) and an antiderivative of
G′(x) is G(x).
b
∫a F ( x ) dx = F (b) − F (a)
b
∫a G ′( x ) dx = G (b) − G (a)
b
b
Sincee F ′( x ) = G ′( x ), ∫ F ′( x ) dx = ∫ G ′( x ) dx , so
a
a
3. In attempting to find F( – 10) ⫽ ∫ tan(t ) dt + 5, the
3
calculator must find a limit of Riemann sums for the
integral, using values of tan t for t between – 10 and 3. The
large positive and negative values of tan t found near the
asymptotes cause the sums to fluctuate erratically so that no
limit is approached. (We will see in Section 8.3 that the
“areas” near the asymptotes are infinite, although NINT is
not designed to determine this.)
4. y ⫽ tan x
F (b) − F (a) = G (b) − G (a).
Quick Quiz Sections 5.1–5.3
b
b
a
a
1. D. ∫ ( F ( x ) + 3) dx = a + 2b + ∫ 3dx
a + 2b + 3b − 3a = 5b − 2a
2. B.
3. C.
5. The domain of this continuous function is the open interval
⎛ π 3π ⎞
⎜⎝ 2 , 2 ⎟⎠ .
6. The domain of F is the same as the domain of the
2 2
∫2 x
dx =
3
x
3
2
=
2
3
3
2
2
−
= 0.
3
3
⎛ π 3π ⎞
continuous function in step 4, namely ⎜ , ⎟ .
⎝2 2 ⎠
Section 5.4
7. We need to choose a closed window narrower than
⎛ π 3π ⎞
⎜⎝ 2 , 2 ⎟⎠ to avoid the asymptotes.
4.
dy 3
7
=
−
=0
dx 3x 7 x
5.
dy
= 2 x ln 2
dx
6.
dy 1 −1/ 2
1
= x
=
dx 2
2 x
7.
x sin x + cos x
dy (− sin x )( x ) − (cos x )(1)
=
=−
2
dx
x
x2
8. The graph of F looks the graph in step 7. It would be
⎛π ⎤
⎡ 3π ⎞
decreasing on ⎜ , π ⎥ and increasing on ⎢π , ⎟ , with
⎝2 ⎦
⎣ 2⎠
vertical asymptotes at x ⫽
π
3π
and x ⫽
.
2
2
Exploration 2 The Effect of Changing
a in
x
∫a
8.
f (t) dt
1.
9. Implicitly differentiate:
dy
dy
x + (1) y + 1 = 2 y
dx
dx
dy
( x − 2 y) = − ( y + 1)
dx
y +1
dy
y +1
=
=−
x − 2y 2y − x
dx
10.
3. Since NINT (x , x, 0, 0) ⫽ 0, the x-intercept is 0.
2
4. Since NINT (x2, x, 5, 5) ⫽ 0, the x-intercept is 5.
d x
5. Changing a has no effect on the graph of y =
f (t ) dt.
dx ∫ a
It will always be the same as the graph of y ⫽ f (x).
6. Changing a shifts the graph of y ⫽
x
∫a
from a1 to a2, the distance of the vertical shift is
a1
∫a
2
f (t ) dt.
dy
= 2(sin x )(cos x ) = 2 sin x cos x
dx
3. dy = 2(sec x )(sec x tan x ) − 2(tan x )(sec 2 x )
dx
= 2 sec 2 x tan x − 2 tan x sec 2 x = 0
)
dy d x
=
sin 2 t dt = sin 2 x
dx dx ∫0
2.
dy d x
=
3t + cos t 2 dt = 3x + cos x 2
dx dx ∫ 2
3.
5
dy d x 3
=
t − t dt = x 3 − x
∫
0
dx dx
4.
dy d x
=
1 + e5t dt = 1 + e5x
dx dx ∫−2
5.
dy d x
=
tan 3 u dt = tan 3 x
dx dx ∫ 0
6.
dy d x u
=
e sec u du = e x sec x
dx dx ∫ 4
7.
dy d x 1 + t
1+ x
=
dt =
∫
2
7
dx dx 1 + t
1+ x2
8.
dy d x 2 − sin t
2 − sin x
=
dt =
dx dx ∫− 3 + cos t
3 + cos x
9.
2 du
2
dy d x 2 t 3
=
e dt = e x
= 2x ex
dx dx ∫ 0
dx
Quick Review 5.4
dy
= cos( x 2 ) i 2 x = 2 x cos( x 2 )
dx
(
1.
f (t ) dt vertically in
such a way that a is always the x-intercept. If we change
2.
dy
1
1
=
=
dx dx / dy 3x
Section 5.4 Exercises
2.
1.
dy
dy
= cos t ,
= − sin t
dt
dt
dy dy / dt cos t
=
=
= − cot t
dx dx / dt − sin t
(
(
(
)
)
(
)
5
)
10.
dy d x 2
du
=
cot 3 t dt = cot x 2
= 2 x cot 3x 2
∫
6
dx dx
dx
11.
dy d 5 x 1 + u 2
1 + 25 x 2
=
du =
∫
dx dx 2
u
x
249
250 Section 5.4
−x
1 + sin 2 u
1 + sin 2 ( − x )
12.
dy d
=
dx dx ∫
13.
dy d 6
d x
=
ln 1 + t 2 dt = − ∫ ln 1 + t 2 dt
∫
x
dx dx
dx 6
1 + cos u
2
(
(
= ln 1 + x 2
14.
du =
1 + cos ( − x )
)
)
27.
2
(
18.
−1
⎡⎛ 1 ⎞ x ⎤
3 dx = ⎢⎜
⎟3 ⎥
⎣⎝ ln 3 ⎠ ⎦ 2
−1 x
∫2
⎛ 1 ⎞⎛ 1
⎞
=⎜
−9
⎝ ln 3 ⎟⎠ ⎜⎝ 3 ⎟⎠
26
=−
≈ −7.889
3 ln 3
3x 2 cos x 3
x6 + 2
25 x 4 − 10 x 2 + 9 du 250 x 5 − 100 x 3 + 90 x
=
dx
125 x 6 + 6
125 x 6 + 6
( )
( )
x
dy d 0
d
=
sin r 2 dr = − ∫ sin r 2 dr
dx dx ∫ x
dx 0
du
sin x
= − sin x
=−
dx
2 x
(
)
(
)
dy d 10
d 3x 2
2
=
ln 2 + p2 dp
2 ln 2 + p dp = −
∫
3
x
dx dx
dx ∫10
du
= − ln 2 + p2
= −6 x ln 2 + 9 x 4
dx
(
19.
28.
dy d 25 t 2 − 2t + 9
d 5 x 2 t − 2t + 9
dt = − ∫
=
dt
2
∫
3
5
x
dx dx
dx 25
t +6
t 3+ 6
)
(
)
dy d x 3
du
du
=
cos 2t dt = cos2 x 3
+ cos 2 x 2
dx dx ∫ x 2
dx
dx
29.
1
⎛ 1 2⎞
⎡ 1 3 2 3/ 2 ⎤
2
∫0 ( x + x ) dx = ⎢⎣ 3 x + 3 x ⎥⎦ = ⎜⎝ 3 + 3 ⎟⎠ − (0 + 0) = 1
0
1
5 3/ 2
x
0
5
2
⎡2
⎤
dx = ⎢ x 5/ 2 ⎥ = (25 5 − 0) = 10 5 ≈ 22.361
⎣5
⎦0 5
30.
∫
31.
∫1
32.
∫−2 x 2 dx = 2 ∫−2 x
33.
∫0 sin x dx = ⎡⎣ − cos x ⎤⎦ 0 = 1 − (−1) = 2
34.
∫0 (1 + cos x ) dx = ⎡⎣ x + sin x ⎤⎦ 0
32 −6 / 5
x
−1
dx = ⎡⎣ −5 x −1/ 5 ⎤⎦
dy d cos x 2
du
du
=
− sin 2 x
t dt = cos 2 x
dx dx ∫ sin x
dx
dx
−1 −2
2
π
21. y =
∫5 sin
22. y =
x −t
e
8
∫
3
35.
24. y =
∫−3
25. y =
x
∫7
26. y =
x
3 − cos t dt + 4
cos 5t dt − 2
∫0 e
2
t
dt + 1
−1
−2
⎡ 1⎤
= 2 ⎢1 − ⎥ = 1
⎣ 2⎦
π
π
π /3
∫0
π /3
2 sec 2 θ dθ = 2 ⎡⎣ tan θ ⎤⎦ 0
= 2( 3 − 0)
= 2 3 ≈ 3.464
36.
5π / 6
∫π / 6
5π / 6
csc 2 θ dθ = ⎡⎣ − cot θ ⎤⎦ π / 6
= 3 − (− 3 )
= 2 3 ≈ 3.464
23. Es
= 10 −4 Esn
10n
x
dx = 2 ⎡⎣ − x −1 ⎤⎦
π
t dt
tan t dt
⎛1 ⎞ 5
= −5 ⎜ − 1⎟ =
⎝2 ⎠ 2
= (π + 0) − (0 + 0)
= π ≈ 3.142
= − sin x cos 2 x − cos x sin 2 x
x
32
1
= 3x 2 cos 2 x 3 + 2 x cos 2 x 2
20.
3
1⎞
⎛
= (6 − ln 3) − ⎜ 1 − ln ⎟
⎝
2⎠
1
= 5 − ln 3 + ln
2
= 5 − ln 3 − ln 2
= 5 − ln 6 ≈ 3.208
dy d 5 cos t
d x 3 cos t
cos x 3 du
dt
=
=
−
=
dt
3
∫
∫
dx dx x t 2 − 2
dx 5 t 2 − 2
x 6 − 2 dx
=
17.
1⎞
dy d 7
d x
=
2t 4 + t + 1 dt = − ∫ 2t 4 + t + 1 dt
dx dx ∫ x
dx 7
=−
16.
⎛
)
= − 2x 4 + x + 1
15.
3
∫1/ 2 ⎜⎝ 2 − x ⎟⎠ dx = ⎡⎣ 2 x − In x ⎤⎦1/ 2
3π /4
37.
∫π / 4
38.
∫0
39.
∫−1(r + 1)
π /3
1
3π / 4
csc x cot x dx = ⎡⎣ − csc x ⎤⎦ π / 4 = (− 2 ) − (− 2 ) = 0
4 sec x tan x dx = 4 ⎡⎣sec x ⎤⎦ π0 / 3 = 4(2 − 1) = 4
2
1
8
8
3⎤
⎡1
dr = ⎢ ( r + 1) ⎥ = − 0 =
3
⎦ −1 3
⎣3
Section 5.4
40.
4 1−
u
∫0
u
4
∫0
du =
(u
)
−1/ 2
44. Graph y = x 3 – 4 x.
− 1 du
= ⎡⎣ 2u −1/ 2 − u ⎤⎦
251
4
0
= (4 − 4) − (0 − 0) = 0
41. Graph y = 2 − x.
Over [–2, 0]:
∫−2 ( x
0
)
⎡1
⎤
− 4 x dx = ⎢ x 4 − 2 x 2 ⎥ = 0 − (−4) = 4
4
⎣
⎦ −2
Over [0, 2]:
0
Over [0, 2]: ∫
2
0
( 2 − x ) dx = ⎡⎢ 2 x − 12 x 2 ⎤⎥
⎣
⎦0
=2
2
)
⎡1
⎤
x 3 − 4 x dx = ⎢ x 4 − 2 x 2 ⎥ = −4 − 0 = −4
⎣4
⎦0
Total area = 4 + −4 = 8
3
1 ⎤
3
1
⎡
Over [2, 3]: ∫ ( 2 − x ) dx = ⎢ 2 x − x 2 ⎥ = − 2 = −
2
2
2
2
⎣
⎦2
3
∫0 (
2
2
3
45. First, find the area under the graph of y ⫽ x 2 .
1
1
⎡1 3 ⎤
∫ dx = ⎢⎣ 3 x ⎥⎦ = 3
0
Next find the area under the graph of y ⫽ 2 – x.
1 2
x
0
1 5
Total area = 2 + − =
2 2
42. Graph y = 3x2 – 3.
1 2⎤
⎡
∫1 ( 2 − x ) dx = ⎢⎣ 2 x − 2 x ⎥⎦
2
2
= 2−
1
Area of the shaded region =
3 1
=
2 2
1 1 5
+ =
3 2 6
46. First find the area under the graph of y ⫽
Over [–2, –1]:
−1
(
2
)
2
)
−1
3
∫−2 3x − 3 dx = ⎡⎣ x − 3x ⎤⎦−2 = 2 − ( −2) = 4
Over [–1, 1]:
∫−1(3x
1
1
− 3 dx = ⎡⎣ x 3 − 3x ⎤⎦
Over [1, 2]: ∫
2
1
(
)
−1
= −2 − 2 = −4
2
3x − 3 dx = ⎡⎣ x 3 − 3x ⎤⎦ = 2 − (−2) = 4
1
2
Total area = 4 + −4 + 4 = 12
43. Graph y = x 3 – 3x 2 – 2 x.
∫
1 1/ 2
x dx
0
x.
1
2
⎤
⎡2
= ⎢ x 3/ 2 ⎥ =
⎦0 3
⎣3
Next find the area under the graph of y ⫽ x 2.
2 2
x
1
∫
2
8 1 7
⎡1 ⎤
dx = ⎢ x 3 ⎥ = − =
⎣ 3 ⎦1 3 3 3
2 7
+ =3
3 3
47. First, find the area under the graph of y = 1 ⫹ cos x.
Area of the shaded region =
π
π
∫0 (1 + cos x ) dx = ⎡⎣ x + sin x ⎤⎦0 = π
The area of the rectangle is 2π.
Area of the shaded region ⫽ 2π – π ⫽ π.
48. First, find the area of the region between y ⫽ sin x and the
⎡ π 5π ⎤
x-axis for ⎢ ,
⎥.
⎣6 6 ⎦
Over [0, 1]:
(
)
1
)
2
1
1
⎡1 4
3
2⎤
∫0 x − 3x + 2 x dx = ⎢⎣ 4 x − x + x ⎥⎦ = 4 − 0 = 4
0
Over [1, 2]:
1
3
∫1 ( x
2
3
2
1
1
⎡1
⎤
− 3x 2 + 2 x dx = ⎢ x 4 − x 3 + x 2 ⎥ = 0 − = −
4
4
⎣4
⎦1
Total area =
1
1 1
+− =
4
4 2
5π / 6
∫π / 6
5π / 6
sin x dx = ⎡⎣ − cos x ⎤⎦ π / 6 =
3 ⎛
3⎞
−⎜−
⎟= 3
2 ⎝ 2 ⎠
⎛ π ⎞ ⎛ 2π ⎞ π
The area of the rectangle is ⎜ sin ⎟ ⎜ ⎟ =
⎝ 6⎠⎝ 3 ⎠ 3
Area of the shaded region = 3 −
π
3
1
⎛
⎞
, x , 0, 10 ⎟ ≈ 3.802
49. NINT ⎜
⎝ 3 + 2 sin x
⎠
252 Section 5.4
⎛ 2x 4 − 1
⎞
, x , − 0.8, 0.8⎟ ≈ 1.427
50. NINT ⎜ 4
⎝ x −1
⎠
51.
52.
(
1
NINT
2
)
cos x , x , − 1, 1 ≈ 0.914
8 − 2 x ≥ 0 between x ⫽ ⫺2 and x ⫽ 2
NINT( 8 − 2 x , x , − 2, 2) ≈ 8.886
2
)
53. Plot y1 = NINT e − t , t , 0, x , y2 = 0.6 in a [0, 1] by [0, 1]
2
window, then use the intersect function to find x ≈ 0.699.
54. When y ⫽ 0, x ⫽ 1.
y3 = 1 − x 3
y = 3 1− x3
3
x
∫a
3
f (t ) dt + K =
x
∫b
x
(b) s ′′(t ) = f ′(t ) < 0 at t = 5 since the graph is decreasing, so
acceleration at t = 5 is negative.
x
a
a
b
x
f (t ) dt + ∫ f (t ) dt
∫x
b
x
= ∫ f (t ) dt
b
−1
K = ∫ (t 2 − 3t + 1) dt
2
(c) s(3) =
−1
56. To find an antiderivative of sin 2 x , recall from trigonometry
that cos 2 x = 1 − 2 sin 2 x , so sin 2 x =
=
0
∫2 sin
2
1 1
− cos 2 x.
2 2
t dt
0⎡1
∫2
1
⎤
⎢ 2 − 2 cos (2 x ) ⎥ dx
⎣
⎦
0
1
⎡1
⎤
= ⎢ x − sin (2 x ) ⎥
4
2
⎣
⎦2
0
0
∫0 f (t ) dt = 0
d ⎛ x
⎞
⎜ f (t ) dt ⎟⎠ = f ( x )
dx ⎝ ∫ 0
H ′( x ) > 0 when f ( x ) > 0.
H is increasing on [0, 6].
(b) H ′( x ) =
1
(e) The acceleration is zero when s ′′(t ) = f ′(t ) = 0. This
occurs when t = 4 sec and t = 7sec.
(f) Since s(0) = 0 and s ′(t ) = f (t ) > 0 on (0, 6), the particle
moves away from the origin in the positive direction on
(0, 6). The particle then moves in the negative direction,
towards the origin, on (6, 9) since
s′(t ) = f (t ) < 0 on (6, 9) and the area below the x-axis is
smaller than the area above the x-axis.
(g) The particle is on the positive side since
s(9) =
9
∫0 f ( x ) dx > 0 (the area below the x-axis is
smaller than the area above the x-axis).
59. (a) s ′(3) = f (3) = 0 units / sec
(b) s ′′(3) = f ′(3) > 0 so acceleration is positive.
1
⎡1
⎤
= ⎢ x − sin x cos x ⎥
2
2
⎣
⎦2
⎛ sin 2 cos 2 ⎞ sin 2 cos 2 − 2
= 0 − ⎜1−
≈ −1.189
⎟⎠ =
⎝
2
2
57. (a) H (0) =
3
∫0 f ( x ) dx = 2 (3)(3) = 4.5 units
(d) s has its largest value at t = 6 sec since
s ′(6) = f (6) = 0 and s ′′(6) = f ′(6) < 0.
3
⎡1
⎤
= ⎢ t3 − t2 + t ⎥
3
2
⎣
⎦2
3
⎡ 1 3
⎤ ⎡8
⎤
= ⎢ − − + (−1) ⎥ − ⎢ − 6 + 2 ⎥ = −
3
2
3
2
⎣
⎦ ⎣
⎦
K=
the x-axis than below the x-axis.
H(12) is positive.
(e) H ′( x ) = f ( x ) = 0 at x = 6 and x = 12. Since
H ′( x ) = f ( x ) > 0 on [0, 6), the values of H are
increasing to the left of x = 6, and since
H ′( x ) = f ( x ) < 0 on (6, 12], the values of H are
decreasing to the right of x = 6. H achieves its
maximum value at x = 6.
58. (a) s ′(t ) = f (t ). The velocity at t = 5 is f (5) = 2 units/sec.
f (t ) dt
K = − ∫ f (t ) dt + ∫ f (t ) dt
=
f (t )dt > 0 because there is more area above
(f) H ( x ) > 0 on (0, 12]. Since H (0) = 0, H achieves its
minimum value at x = 0.
NINT( 1 − x , x , 0, 1) ≈ 0.883
55.
12
∫0
(d) H (12) =
2
(
(c) H is concave up on the open interval where
H ′′( x ) = f ′( x ) > 0.
f ′( x ) > 0 when 9 < x ≤ 12.
H is concave up on (9, 12).
3
1
6
1
(c) s(3) =
∫0 f ( x ) dx = 2 (−6)(3) = −9 units
(d) s(6) =
∫0 f ( x ) dx = 2 (−6)(3) + 2 (6)(3) = 0, so the
1
particle passes through the origin at t = 6 sec.
(e) s ′′(t ) = f ′(t ) = 0 at t = 7 sec
(f) The particle is moving away from the origin in the
negative direction on (0, 3) since s(0) = 0 and
s ′(t ) < 0 on (0, 3). The particle is moving toward the
origin on (3, 6) since s′(t ) > 0 on (3, 6) and s(6) = 0.
The particle moves away from the origin in the positive
direction for t > 6 since s′(t ) > 0.
Section 5.4
59. Continued
69. E. f (a) + f 1 (a)( x − π )
f (π ) = 0
f (π ) = −1
− 1( x − π ) = π − x
(g) The particle is on the positive side since
s(9) =
9
∫0
f ( x ) dx > 0 (the area below the x-axis is
smaller than the area above the x-axis).
60. f ( x ) =
(
)
d ⎛ x
⎞ d x 2 − 2x + 1 = 2x − 2
⎜ f (t ) dt ⎟⎠ =
dx ⎝ ∫1
dx
70. E.
71. (a) f (t) is an even function so
x 10
d ⎛
10
⎞
dt =
2+ ∫
⎜
0
dx ⎝
1 + t ⎟⎠ 1 + x
61. f ′( x ) =
10
dt = 2
1+ t
L ( x ) = 2 + 10 x
f ′(0) = 2 + ∫
0
0
d ⎛ x
⎞
⎜ f (t ) dt ⎟⎠
dx ⎝ ∫ 0
(b) Si(0) =
64. (a)
∫−3(
2
π /k
π
.
k
−x
0 sin t
∫0
t
dt = 0
[⫺20, 20] by [⫺3, 20, 203]
1 ⎛ 1⎞ 2
⎡ 1
⎤
sin kx dx = ⎢ − cos kx ⎥ = − ⎜ − ⎟ =
k ⎝ k⎠ k
⎣ k
⎦0
72. (a) c(100) − c(1) =
2
)
=
22 ⎛ 27 ⎞
− −
3 ⎜⎝ 2 ⎟⎠
125
=
6
=
dc ⎞
⎜⎝ dx ⎟⎠ dx
100
∫1
(b) c(400) − c(100) =
b
.)
2a
⎛ 1 ⎞ 25
25
y ⎜ − ⎟ = , so the height is
.
⎝ 2⎠ 4
4
(c) The base is 2 − (−3) = 5.
2
2 ⎛ 25 ⎞ 125
(base)(height) = (5) ⎜ ⎟ =
6
3
3 ⎝ 4⎠
65. True. The Fundamental Theorem of Calculus guarantees
that F is differentiable on I, so it must be continuous on I.
66. False. In fact, ∫ e x dx is a real number, so its derivative is
2
a
always 0.
1
100
400 ⎛
dc ⎞
∫100 ⎜⎝ dx ⎟⎠ dx
400
1
400
dx = ⎡ x ⎤
⎣ ⎦100
x
= 20 − 10 = 10 or $10
=
− (−1)
1
= − . (Recall that the vertex
2(−1)
2
of a parabola y = ax 2 + bx + c is at x = −
100 ⎛
∫1
dx = ⎡ x ⎤
⎣ ⎦1
2 x
= 10 − 1 = 9 or $9
1
1 ⎤
⎡
6 − x − x 2 dx = ⎢6 x − x 2 − x 3 ⎥
2
3 ⎦ −3
⎣
b
sin(t )
dt.
t
∫0
π /k
(b) The vertex is at x =
x
∫0
(d)
63. One arch of sin kx is from x ⫽ 0 to x ⫽
0
sin(t )
dt =
t
(c) Si′( x ) = f (t ) = 0 when t = π k , k a nonzero int eger.
d
( x cos π x )
dx
= x ( −π sin π x ) + 1 i cos π x
= −π x sin π x + cos π x
f (4) = −4π sin 4π + cos 4π = 1
=
Area ⫽ ∫
0
∫− x
sin(t )
dt
t
0 sin(t )
= −∫
dt
−x
t
x sin(t )
= −∫
dt = −Si( x )
0
t
Si(− x ) =
f ′(0) = 10
62. f ( x ) =
253
73.
3⎛
2
∫100 2
⎞
∫0 ⎜⎝ 2 − ( x + 1)2 ⎟⎠ dx = ⎡⎣ 2 x + 2( x + 1)
−1 ⎤
3
⎦0
1⎤
9
⎡
= ⎢6 + ⎥ − 2 =
2
2
⎣
⎦
= 4.5 thousand
The company should expect $4500.
30
30 ⎛
1
x2 ⎞
1 ⎡
x3 ⎤
dx
x
450
450
−
=
−
⎥
⎢
⎜
⎟
74. (a) 30 − 0 ∫0
2⎠
30 ⎢⎣
6 ⎦⎥
⎝
0
= 300 drums
(b) (300 drums)($ 0 . 02 per drum ) = $ 6
75. (a) True, because h ′( x ) = f ( x ) and therefore h ′′( x ) = f ′( x ).
67. D.
(b) True because h and h′ are both differentiable by part (a).
68. D. See the Fundamental Theorem of Calculus.
(c) True, because h ′(1) = f (1) = 0.
Section 5.5
254
75. Continued
(d) True, becaue h′(1) = f (1) = 0 and h′′(1) = f ′(1) < 0.
this interval and x ≈ 1.065 is a solution. Furthermore,
Si( x ) = 1 has no solution on the interval [π , 2π ] because
(e) False, because h′′(1) = f ′(1) < 0.
(f) False, because h′′(1) = f ′(1) ≠ 0
(g) True, because h′( x ) = f ( x ), and f is a decreasing
function that includes the point (1,0).
x
0
∫− x f (t )dt = − ∫0
76. Since f (t)is odd,
f (t )dt because the area
between the curve and the x-axis from 0 to x is the opposite
of the area between the curve and the x-axis from –x to 0,
but it is on the opposite side of the x-axis.
−x
x
0
⎡ x
⎤
∫0 f (t ) dt = − ∫− x f (t ) dt = − ⎣⎢ − ∫0 f (t )dt ⎦⎥ = ∫0 f (t ) dt
Thus
x
∫0
f (t ) dt is even.
77. Since f (t) is even, ∫
0
−x
∫0
f (t ) dt = − ∫
∫0
f (t ) dt =
f (t ) dt because the area
Thus
x
∫0
x
0
−x
f (t ) dt = − ∫ f (t ) dt
Section 5.5 Trapezoidal Rule (pp. 306–315)
Exploration 1 Area Under a Parabolic
Arc
1. Let y = f ( x ) = Ax 2 + Bx + C
y1 = f (0) = A(0)2 + B(0) + C = C , and
y2 = f (h) = Ah 2 + Bh + C.
2. y0 + 4 y1 + y2 = Ah 2 − Bh + C + 4C + Ah 2 + Bh + C
= 2 Ah 2 + 6C.
0
f (t ) dt is odd.
3. Ap =
x
78. If f is an even continuous function, then ∫ f (t ) dt is odd,
0
but
Si(x) is decreasing on this interval and Si(2π ) ≈ 1.42 > 1.
Thus, Si( x ) = 1 has exactly one solution in the interval
[0, ∞). Also, there is no solution in the interval (−∞, 0]
because Si(x) is odd by Exercise 56 (or 62), which means
that Si( x ) ≤ 0 for x ≤ 0 ( since Si( x ) ≥ 0 for x ≥ 0 ).
Then y0 = f (− h) = Ah 2 − Bh + C ,
x
between the curve and the x-axis from 0 to x is the same as
the area between the curve and the x-axis from –x to 0.
−x
Si( x ) ≠ 1) for x ≥ 2π . Now, Si( x ) = 1 has exactly one
solution in the interval [0, π ] because Si(x) is increasing on
d x
f (t ) dt = f ( x ). Therefore, f is the derivative of the
dx ∫0
odd continuous function
x
∫0
f (t ) dt.
Similarly, if f is an odd continuous function, then f is the
0
⎛ sin t
⎞
, t , 0, x ⎟ = 1 graphically, the solution is
79. Solving NINT ⎜
⎝ t
⎠
x ≈ 1.0648397. We now argue that there are no other
solutions, using the functions Si(x) and f(t) as defined in
Exercise 56. Since d Si( x ) = f ( x ) = sin x , Si( x ) is
x
dx
increasing on each interval ⎡⎣ 2k π ,(2k + 1)π ⎤⎦ and decreasing
on each interval ⎡⎣(2k + 1)π ,(2k + 2)π ⎤⎦ , where K is a
nonnegative integer. Thus, for x > 0, Si( x ) has its local
minima at x = 2kπ , where k is a positive integer. Furthermore, each arch of y = f ( x ) is smaller in height than the
previous one, so ∫
( 2 k +1)π
2 kπ
f ( x ) dx > ∫
means that Si(2k + 2)π ) − Si(2kπ ) =
( 2 k + 2 )π
( 2 k +1)π
( 2 k + 2 )π
∫2kπ
f ( x ) dx. This
f ( x ) dx > 0, so
each successive minimum value is greater than the previous
⎛ sin x
⎞
, x , o, 2π ⎟ ≈ 1.42 and Si ( x )
one. Since f (2π ) ≈ NINT ⎜
⎝ x
⎠
is continuous for x > 0, this means Si( x ) > 1.42 (and hence
2
+ Bx + C )dx
h
⎤
⎡ x3
x2
= ⎢ A + B + Cx ⎥
2
⎥⎦ − h
⎢⎣ 3
3
2
⎛
⎞
h
h
h3
h2
= A + B + Ch − ⎜ − A + B − Ch ⎟
3
2
3
2
⎝
⎠
= 2A
x
derivative of the even continuous function ∫ f (t ) dt.
h
∫− h ( Ax
=
h3
+ 2Ch
3
h
(2 Ah 2 + 6C )
3
4. Substitute the expression in step 2 for the parenthetically
enclosed expression in step 3:
h
Ap = (2 Ah 2 + 6C )
3
h
= ( y0 + 4 y1 + y2 ).
3
Quick Review 5.5
1. y′ = − sin x
y′′ = − cos x
y′′ < 0 on [−1, 1], so the curve is concave down on [−1, 1].
2. y′ = 4 x 3 − 12
y′′ = 12 x 2
y′′ > 0 on [8, 17], so the curve is concave up on [8, 17].
3. y′ = 12 x 2 − 6 x
y′′ = 24 x − 6
y′′ < 0 on [−8, 0], so the curve is concave down on [−8, 0].
Section 5.5 255
1
x
cos
2
2
1
x
y′′ = − sin
4 2
y′′ ≤ 0 on [48π , 50π ], so the curve is concave down on
[48π , 50π ].
4. y′ =
(c)
∫0
2. (a) f ( x ) = x 2 , h =
2−0 1
=
4
2
x
0
1
2
1
3
2
2
f (x)
0
1
4
1
9
4
4
5. y′ = 2e 2 x
y′′ = 4e 2 x
y′′ > 0 on [−5, 5], so the curve is concave up on [−5, 5].
6. y′ =
1
x
1
x2
y′′ < 0 on [100, 200], so the curve is concave down on
[100, 200].
y′′ =
⎞
1⎛
⎛ 1⎞
⎛ 9⎞
0 + 2 ⎜ ⎟ + 2(1) + 2 ⎜ ⎟ + 4 ⎟ = 2.75
⎝ 4⎠
⎝ 4⎠
4 ⎜⎝
⎠
T=
y′′ = −
7. y′ = −
2
⎡1 ⎤
x dx = ⎢ x 2 ⎥ = 2
⎣2 ⎦0
2
(b) f ′( x ) = 2 x , f ′′( x ) = 2 > 0 on [0, 2]
The aproximation is an overestimate.
(c)
1
x2
2
2 2
∫0 x
3. (a) f ( x ) = x 3 , h =
x3
y′′ > 0 on [3, 6], so the curve is concave up on [3, 6].
8. y′ = − csc x cot x
y′′ = (− csc x )(− csc 2 x ) + (csc x cot x )(cot x )
= csc3 x + csc x cot 2 x
y′′ > 0 on [0, π ], so the curve is concave up on [0, π ].
9. y′ = −100 x
T=
9
y′′ = −900 x 8
y′′ < 0 on [10, 1010 ], so the curve is concave down on
[10, 1010 ].
10. y′ = cos x + sin x
y′′ = − sin x + cos x
y′′ < 0 on [1, 2], so the curve is concave down.
1. (a) f ( x ) = x , h =
2−0 1
=
4
2
x
0
1
2
1
3
2
2
f (x)
0
1
2
1
3
2
2
⎞
1⎛
⎛ 1⎞
⎛ 3⎞
0 + 2 ⎜ ⎟ + 2(1) + 2 ⎜ ⎟ + 2⎟ = 2
⎜
⎝ 2⎠
⎝ 2⎠
4⎝
⎠
(b) f ′( x ) = 1, f ′′( x ) = 0
The approximation is exact.
2−0 1
=
4
2
x
0
1
2
1
3
2
2
f (x)
0
1
8
1
27
8
8
1⎛
⎛ 1⎞
⎛ 27 ⎞ ⎞
0 + 2 ⎜ ⎟ + 2(1) + 2 ⎜ ⎟ + 8⎟ = 4.25
⎝ 8⎠
⎝ 8⎠ ⎠
4 ⎜⎝
(b) f ′( x ) = 3x 2 , f ′′( x ) = 6 x > 0 on [0, 2]
The aproximation is an overestimate.
(c)
∫
2 3
x dx
0
4. (a) f ( x ) =
Section 5.5 Exercises
T=
2
8
⎡1 ⎤
dx = ⎢ x 3 ⎥ =
⎣3 ⎦0 3
2
⎡1 ⎤
= ⎢ x4 ⎥ = 4
⎣4 ⎦0
1
2 −1 1
,h=
=
x
4
4
x
1
5
4
3
2
7
4
2
f (x)
1
4
5
2
3
4
7
1
2
1⎛
⎛ 4⎞
⎛ 2⎞
⎛ 4⎞ 1⎞
T = ⎜ 1 + 2 ⎜ ⎟ + 2 ⎜ ⎟ + 2 ⎜ ⎟ + ⎟ ≈ 0.697
⎝ 5⎠
⎝ 3⎠
⎝ 7⎠ 2⎠
8⎝
(b) f ′( x ) = −
1
, f ′′( x ) =
2
> 0 on [1, 2]
x
x3
The approximation is an overestimate.
21
2
(c) ∫ dx = ⎡⎣ ln x ⎤⎦ = ln 2 ≈ 0.693
1
1 x
2
Section 5.5
256
5. (a) f ( x ) = x , h =
T=
4−0
=1
4
x
0
1
f (x)
0
1
11. Sum the trapezoids and multiply by
2
3
4
3
2
2
(
)
1
0 + 2 (1) + 2 2 + 2 3 + 2 ≈ 5.146
2
1
1
(b) f ′( x ) = − x −1/ 2 , f ′′( x ) = − x −3/ 2 < 0 on [0, 4]
2
4
The approximation is an underestimate.
(c)
4
∫0
12. Sum the trapezoids and multiply by
π −0 π
6. (a) f ( x ) = sin x , h =
=
4
4
x
0
f (x)
T=
0
π
4
π
2
3π
4
π
2
2
1
2
2
0
⎞
⎛ 2⎞
⎛ 2⎞
π⎛
⎜ 0 + 2⎜
⎟ + 2 (1) + 2 ⎜
⎟ + 0 ⎟ ≈ 1.896
8⎝
⎝ 2 ⎠
⎝ 2 ⎠
⎠
π
6
∫0
∫2
14. (a)
h
( y + 2 y1 + 2 y2 + + 2 yn−1 + yn )
2 0
(b)
∫0
x dx =
2 2
∫0 x
x2
2
h
( y + 2 y1 + 2 y2 + + 2 yn−1 + yn )
2 0
1
f ( x ) dx ≈ (16 + 2(19) + 2(17) + 2(14) + 2(13) + 2(16) + 20) = 97
2
5
9. (6.0 + 2(8.2) + 2(9.1) + + 2(12.7) + 13.0)(30)
2
= 15.990 ft
(b)
200
(0 + 2(520) + 2(800) + 2(1000) + + 2(860)
2
0
22 0 2
+
=2
2
2
8
3
x3
=
3
2
23 0 3 8
+
=
3
3 3
∫
∫
2 3
x dx
0
3
3
⎞
⎛ 1/ 2⎞ ⎛ 3
⎛ 1⎞
⎛ 3⎞
3
0
4
2
1
4
=⎜
+
+
+
(
)
+ 23 ⎟ = 4
⎜
⎟
⎜
⎜
⎟
⎟
⎟⎠
⎝ 3 ⎠ ⎜⎝
⎝ 2⎠
⎝ 2⎠
∫
2 3
x dx
0
=
16. (a) ∫
2
x4
4
=
0
2
=
0
24 0 4
−
=4
4
4
1
⎛ 1 ⎞ 1⎞
⎛ 1/ 4⎞ ⎛ 1
⎛ 1 ⎞
⎛ 1 ⎞
+
dx = ⎜
+ 4⎜
+ 2⎜ ⎟ + 4⎜
⎟
⎟
⎜
⎝ 1.75 ⎟⎠ 2 ⎟⎠
⎝ 3 ⎠ ⎝1
⎝ 1.25 ⎠
⎝ 1.5 ⎠
x
≈ 0.69325
(b)
3
(b) You plan to start with 26,360 fish. You intend to have
(0.75)(26, 360) = 19, 770 fish to be caught. Since
=
2 2
x dx
0
1
3
+ 0)(20) = 26, 360, 000 ft
15. (a)
2
2
2
⎞
⎛ 1/ 2⎞ ⎛ 2
⎛ 1⎞
⎛ 3⎞
2
2
dx = ⎜
⎜ 0 + 4 ⎜ ⎟ + 2(1) + 4 ⎜ ⎟ + 2 ⎟
⎟
⎟⎠
⎝ 3 ⎠ ⎜⎝
⎝ 2⎠
⎝ 2⎠
1
2
10. (a)
2
=
π
⎞
⎛ 1/ 2⎞ ⎛
⎛ 1⎞
⎛ 3⎞
0 + 4 ⎜ ⎟ + 2(1) + 4 ⎜ ⎟ + 2⎟ = 2
⎝ 2⎠
⎝ 2⎠
3 ⎟⎠ ⎜⎝
⎠
2
∫0 x dx = ⎜⎝
f ( x )dx ≈ (12 + 2(10) + 2(9) + 2(11) + 2(13) + 2(16) + 188) = 74
8. T =
8
(b)
∫0 sin x dx = ⎡⎣ − cos x ⎤⎦0 = 2
7. T =
1
to change
3600
seconds to hours.
1 ⎛ 1⎞
(0 + 2(3) + 2(7) + 2(12) + 2(17) + 2(25) + 2(33)
3600 ⎜⎝ 2 ⎟⎠
+2(41) + 48) = 0.045 mi ≈ 238 feet.
13. (a)
(b) f ′( x ) = cos x , f ′′( x ) = − sin x < 0 on [0, π ]
The approximation is an underestimate.
(c)
seconds to hours
1
(2.0(0 + 30) + (3.2 − 2.0)(30 + 40) + (4.5 − 3.2)(40 + 50)
2
+ (5.8 − 4.5)(50 + 60) + (7.7 − 5.8)(60 + 70)
+ (9.5 − 7.7)(70 + 80) + (11.6 − 9.5)(80 + 90)
+ (14.9 − 11.6)(90 + 100) + (17.8 − 14.9)(100 + 110)
+ (21.7 − 17.8)(110 + 120) + (26.3 − 21.7)(120 + 130))
1
≈ 0.633 mi ≈ 3340 feet.
3600
4
16
⎡2
⎤
xdx = ⎢ x 3/ 2 ⎥ =
≈ 5.333
⎣3
⎦0 3
1
to change
3600
17. (a)
2
1
2
∫1 x dx = ln x 1 = ln 2 − ln1 ≈ 0.69315
4
∫0
⎛ 1⎞
x dx = ⎜ ⎟ ( 0 + 4( 1) + 2( 2 ) + 4( 3 ) + ( 4 ))
⎝ 3⎠
≈ 5.2522
19, 770
= 988.5, the town can sell at most 988 licenses.
20
(b)
4
∫0
xdx =
2 3/ 2 4 2 3/ 2 2 3/ 2 16
x
= (4) − (0) =
0
3
3
3
3
Section 5.5 257
25. S50 = 1.37066, S100 = 1.37066 using a = 0.0001 as lower
⎛ ⎛ π ⎞⎞
⎛ π / 4⎞ ⎛
sin ( 0 ) + 4 ⎜ sin ⎜ ⎟ ⎟
⎜
⎟
3 ⎠⎝
⎝ ⎝ 4⎠⎠
π
∫0 sin x dx = ⎜⎝
18. (a)
⎛ ⎛ π ⎞⎞
⎛ ⎛ 3π ⎞ ⎞
+2 ⎜ sin ⎜ ⎟ ⎟ + 4 ⎜ sin ⎜ ⎟ ⎟ + sin π ≈ 2.00456
⎝ ⎝ 2⎠⎠
⎝ ⎝ 4 ⎠⎠
π
∫0 sin x dx = − cos x 0 = − cos π − ( − cos 0 ) = 2
(b)
π
19. (a) f ( x ) = x 3 − 2 x , h =
(b)
27. (a) T10 ≈ 1.983523538
T100 ≈ 1.999835504
T1000 ≈ 1.999998355
–1
0
1
2
3
f (x)
1
0
–1
4
21
(
(b)
)
1
1 + 4 ( 0 ) + 2 ( −1) + 4 ( 4 ) + 21 = 12
3
∫−1(
3
lower limit
26. S50 ≈ 0.82812, S100 ≈ 0.82812
3 − (−1)
=1
4
x
S=
limit
S50 = 1.37076, S100 = 1.37076 using a = 0.000000001 as
3
)
⎡1
⎤
x 3 − 2 x dx = ⎢ x 4 − x 2 ⎥
⎣4
⎦ −1
⎛ 81 ⎞ ⎛ 1 ⎞
= ⎜ − 9⎟ − ⎜ − 1⎟
⎝ 4
⎠ ⎝4 ⎠
= 12
Es = 0
22. Sketch a graph of 4 line segments joined at sharp corners.
One example:
y
10
0.016476462 = 1.6476462 × 10 −2
100
1.64496 × 10 −4
1000
1.645 × 10 −6
10 n
≈ 10 −2 ET
n
π2
,M = 1
n2
π ⎛ π2 ⎞
π3
ET ≤ ⎜ 2 ⎟ =
n 12 ⎝ n ⎠ 12n 2
(d) b − a = π , h 2 =
(d) Simpson’s Rule for cubic polynomials will always give
exact values since f ( 4 ) = 0 for all cubic polynomials.
(b) We are approximating the area under the temperature
graph. By doubling the endpoints, the error in the first
and last trapezoids increases.
ET = 2 − Tn
(c) ET
4
(c) For f ( x ) = x 3 − 2 x , M ( 4 ) = 0 sin ce f ( ) = 0.
f
20. The average of the 13 discrete temperatures gives equal
weight to the low values at the end.
1
21. (a) (126 + 2 i 65 + 2 i 66 + + 2 i 58 + 110 ) = 841
2
1
av( f ) ≈ i 841 ≈ 70.08
12
n
ET
10 n
≤
π3
12 (10 n )
23. S50 ≈ 3.13791, S100 ≈ 3.14029
24. S50 ≈ 1.08943, S100 ≈ 1.08943
= 10 −2 ET
n
28. (a) S10 ≈ 2.000109517
S100 ≈ 2.000000011
S1000 ≈ 2.000000000
(b)
n
Es = 2 − S n
10
1.09517 × 10–4
100
1.1 × 10–8
1000
(c) Es
= 10
10n
0
−4
Esn
π4
,M = 1
n4
π ⎛ π4 ⎞
π5
Es ≤
=
⎜
⎟
n 180 ⎝ n 4 ⎠ 180 n 4
π5
= 10 −4 Esn
Es
≤
10n
180(10 n)4
(d) b − a = π , h 4 =
x
2
Section 5.5
258
24 in.
= 4 in.
6
Estimate the area to be
4
[0 + 4(18.75) + 2(24) + 4(26) + 2(24) + 4(18.75) + 0]
3
≈ 466.67 in 2
29. h =
30. Note that the tank cross-section is represented by the shaded
area, not the entire wing cross-section. Using Simpson’s
Rule, estimate the cross-section area to be
1
[ y + 4 y1 + 2 y2 + 4 y3 + 2 y4 + 4 y5 + y6 ]
3 0
1
= [1.5 + 4(1.6) + 2(1.8) + 4(1.9) + 2(2.0)
3
+ 4(2.1) + 2.1] = 11.2 ft 2
1
⎛
⎞⎛ 1 ⎞
Length ≈ (5000 lb) ⎜
≈ 10.63 ft
⎝ 42 lb / ft 3 ⎟⎠ ⎜⎝ 11.2ft 2 ⎟⎠
31. False. The Trapezoidal Rule will over estimate the integral
if it is concave up.
(e) For 0 < h ≤ 0.1, ET ≤
(f) n ≥
34. B.
4
∫−2
ex
dx =
2
e0
e2
e4 ⎞
1 ⎛ e −2
2
+4 +4 +2 ⎟
⎜
2⎝ 2
2
2
2⎠
= e 4 + 2e 2 + 2e 0 + e −2
35. C.
π
∫0
sin x dx =
π /4⎛
⎛ π⎞
sin 0 + 4 ⎜ sin ⎟
⎝ 4⎠
2 ⎜⎝
⎞
⎛ π⎞
⎛ 3π ⎞
+ 2 ⎜ sin ⎟ + 4 ⎜ sin ⎟ + sin π ⎟
⎝ 2⎠
⎝
4⎠
⎠
⎞
⎛ 2⎞
⎛ 2⎞
π /4⎛
⎜ 0 + 4⎜
⎟ + 2(1) + 4 ⎜
⎟ + 0⎟
2 ⎝
⎝ 2 ⎠
⎝ 2 ⎠
⎠
π
= 1+
+ 2
4
=
(
)
36. C.
37. (a) f ′( x ) = 2 x cos ( x 2 )
f ′′( x ) = 2 x i − 2 x sin ( x 2 ) + 2 cos ( x 2 )
= −4 x 2 sin ( x 2 ) + 2 cos ( x 2 )
(b)
1 − (−1) 2
≥
= 20
0.1
h
38. (a) f ′′′( x ) = −4 x 2 i 2 x cos ( x 2 ) − 8 x sin ( x 2 ) − 4 x sin ( x 2 )
(4)
f
= −8 x 3 cos ( x 2 ) − 12 x sin ( x 2 )
( x ) = −8 x 3 i − 2 x sin ( x 2 ) − 24 x 2 cos ( x 2 )
− 12 x i 2 x cos ( x 2 ) − 12 sin ( x 2 )
= (16 x 4 − 12) sin ( x 2 ) − 48 x 2 cos ( x 2 )
(b)
[−1, 1] by [−30, 10]
(c) The graph shows that −30 ≤ f ( 4 ) ( x ) ≤ 10 so
32. False. For example, the two approximations will be the
same if f is constant on [a, b].
33. A. LRAM < T < RRAM, so RRAM < 16.4.
h 2 0.12
≤
= 0.005 < 0.01
2
2
f ( 4 ) ( x ) ≤ 30 for −1 ≤ x ≤ 1.
(d) ES ≤
1 − (−1) 4
h4
(h )(30) =
180
3
(e) For 0 < h ≤ 0.4, ES ≤
(f) n ≥
h 4 0.4 4
≤
≈ 0.00853 < 0.01
3
3
1 − (−1)
2
≥
=5
h
0.4
h
39. Tn = [ y0 + 2 y1 + 2 y2 + + 2 yn−1 + yn ]
2
h[ y0 + y1 + + yn−1 ] + h[ y1 + y2 + + yn ]
=
2
LRAM n + RRAM n
=
2
h
40. S2 n = [ y0 + 4 y1 + 2 y2 + 4 y3 + + 2 y2 n− 2
3
+ 4 y2 n−1 + y2 n ]
1
= [h( y0 + 2 y1 + 2 y2 + + 2 y2 n−1 + y2 n )
3
+ (2h)( y1 + y3 + y5 + + y2 n−1 )]
2T2 n + MRAM n
b−a
=
, where h =
.
3
2n
Quick Quiz Sections 5.4 and 5.5
(d) ET ≤
1 − (−1) 2
h2
(h )(3) =
12
2
1
f ( x ) dx = ((4 − 1)(10 + 30) + (6 − 4)(30 + 40)
2
+ (7 − 6)(40 + 20)) = 160
∫1
2. D.
∫ sin x dx = ⎜⎝
[−1, 1] by [−3, 3]
(c) The graph shows that −3 ≤ f ′′( x ) ≤ 2 so f ′′( x ) ≤ 3
for − 1 ≤ x ≤ 1.
7
1. C.
3
⎛ − (sin 2 x )
3
2⎞
− ⎟ cos x
3⎠
⎛ − (sin 2 (8)) 2 ⎞
⎛ − (sin 2 (1)) 2 ⎞
− ⎟ cos 8 − ⎜
− ⎟ cos 1 = 0.632
⎜
3
3⎠
3
3⎠
⎝
⎝
Chapter 5 Review 259
3. C. df ( x ) =
d x 2 −3 x t 2
e dt
dx ∫−2
df ( x )
= (2 x − 3)e
dx
2x − 3 = 0
3
x= .
2
4. (a)
( x 2 − 3 x )2
y
4.
4
=0
2
2−0
(sin 0 + 2 sin(0.52 ) + 2 sin(1.0 2 )
2(4)
1
x
2
+ 2 sin(1.52 ) + 2 sin(22 )) = 0.744
1 ⎛ 15
21 ⎞ 15
RRAM 4 : ⎜ + 3 + + 0 ⎟ =
= 3.75
⎠ 4
2⎝ 8
8
(b) F increases on [0, π ] and [ 2π , 3] because sin(t 2 ) > 0
(c) f (t ) = k =
d 3
sin (t 2 ) dt = 3K − 0 K = 3K
dx ∫0
y
5.
4
Chapter 5 Review (315–319)
y
1.
4
2
2
1
T4 =
1
2
x
y
2.
6.
4
2
⎡ 2 1 4⎤
3
∫0 (4 x − x ) dx = ⎢⎣ 2 x − 4 x ⎥⎦ = 8 − 4 = 4
0
2
2
x
15
21⎞ 15
1⎛
= 3.75
LRAM 4 : ⎜ 0 + + 3 + ⎟ =
8
8⎠ 4
2⎝
y
3.
8.
4
x
1
( LRAM4 + RRAM4 ) = 12 ⎛⎜⎝ 154 + 154 ⎞⎟⎠ = 3.75
2
2
7.
1
2
n
LRAMn
MRAMn
RRAMn
10
1.78204
1.60321
1.46204
20
1.69262
1.60785
1.53262
30
1.66419
1.60873
1.55752
50
1.64195
1.60918
1.57795
100
1.62557
1.60937
1.59357
1000
1.61104
1.60944
1.60784
51
5
∫1 x
9. (a)
dx = ⎡⎣ ln x ⎤⎦ = ln 5 − ln 1 = ln 5 ≈ 1.60944
1
2
5
∫5 f ( x ) dx = − ∫2 f ( x ) dx = −3
The statement is true.
2
(b)
1
2
x
1 ⎛ 63 165 195 105 ⎞
MRAM 4 : ⎜ +
+
+
= 4.125
2 ⎝ 64 64 64 64 ⎟⎠
5
∫-2 [ f ( x ) + g( x )] dx
5
5
∫−2 f ( x ) dx + ∫−2 g( x ) dx
2
5
5
= ∫ f ( x ) dx + ∫ f ( x ) dx + ∫ g( x ) dx
2
−2
−2
=
= 4 + 3+ 2 = 9
The statement is true.
260
Chapter 5 Review
9. Continued
(c) If f ( x ) ≤ g( x ) on [–2, 5], then
5
5
∫−2 f ( x ) dx ≤ ∫−2 g( x ) dx ,
13. The graph is above the x-axis for 0 ≤ x < 4 and below the
x-axis for 4 < x ≤ 6
Total area =
but this is not true since
5
2
5
∫−2 f ( x ) dx = ∫−2 f ( x ) + ∫2 f ( x ) = 4 + 3 = 7
5
∫−2 g( x ) dx = 2.
The statement is false.
n
Total volume: V = lim ∑ π sin 2 (mi ) Δx
n→∞
i =1
Total area =
(b) Use π sin x on [0, π ].
(b)
30
Position (m)
10
10
(b)
∫0
(c)
∫0
(d)
∫0
(e)
10
10
10
10
∫0
cos x dx − ∫
π
cos x dx
π /2
π /2
π
= (1 − 0) − (0 − 1) = 2
2
2
15.
∫−2 5 dx = ⎡⎣5x ⎤⎦ −2 = 10 − (−10) = 20
16.
∫2 4 x dx = ⎢⎣ 2 x
17.
∫0
18.
∫−1 (3x
5
⎡
π /4
5
2⎤
⎥⎦ =
2
50 − 8 = 42
π /4
cos x dx = ⎡⎣sin x ⎤⎦ 0 =
1
2
2 −0= 2
2
2
− 4 x + 7) dx = ⎡⎣ x 3 − 2 x 2 + 7 x ⎤⎦
x 3 dx
x sin x dx
x (3x − 2)2 dx
1 dx
1+ x2
(
1
1
−1
1
− 12s 2 + 5) ds = ⎡⎣ 2s 4 − 4 s 3 + 5s ⎤⎦ = 3 − 0 = 3
19.
∫0 (8s
20.
∫1
21.
∫1
22.
∫1
23.
∫0
24.
∫1
25.
∫0 (2 x36+ 1)3 dx = ∫0 36(2 x + 1)
t
The curve is always increasing because the velocity is
always positive, and the graph is steepest when the
velocity is highest, at t = 6.
∫0
π /2
∫0
= 6 − (−10) = 16
Time (sec)
12. (a)
6
= ⎡⎣sin x ⎤⎦ 0 − ⎡⎣sin x ⎤⎦ π /2
NINT (π sin 2 x , x , 0, π ) ≈ 4.9348
s
4
14. The graph is above the x-axis for 0 ≤ x < π and below the
2
π
x-axis for < x ≤ π
2
2
11. (a) Approximations may vary. Using Simpson’s Rule, the
area under the curve is approximately
1
[0 + 4(0.5) + 2(1) + 4(2) + 2(3.5) + 4(4.5) +
3
2(4.75) + 4(4.5) + 2(3.5) + 4(2) + 0] = 26.5
The body traveled about 26.5 m.
6
⎤
⎡
⎤
⎡
= ⎢4 x − 1 x 2 ⎥ − ⎢4 x − 1 x 2 ⎥
⎢⎣
2 ⎥⎦ 4
2 ⎥⎦ 0 ⎢⎣
= [8 − 0] − [6 − 8] = 10
and
10. (a) Volume of one cylinder: π r 2 h = π sin 2 (mi ) Δx
4
∫0 (4 − x ) dx − ∫4 (4 − x ) dx
2
2
27
y −4 /3 dy = ⎡⎢ −3 y −1/3 ⎤⎥ = −1 − (−3) = 2
⎣
dt =
t t
π /3
e
0
4 dx = ⎡ − 4 ⎤ = −2 − (−4) = 2
⎢⎣ x ⎥⎦ 1
x2
27
4
3
⎦1
4 −3/ 2
∫1 t
4
dt = ⎡⎢ −2t −1/2 ⎤⎥ = −1 − (−2) = 1
⎣
⎦1
π /3
sec 2 θ dθ = ⎡⎣ tan θ ⎤⎦ 0 = 3 − 0 = 3
1 dx = ⎡ ln | x | ⎤ e = 1 − 0 = 1
⎣
⎦1
x
1
1
−3
dx
1
)
π 9 − sin 2 π x dx
10
= ⎡⎢ −9(2 x + 1)−2 ⎤⎥
⎣
= −1 − (−9) = 8
⎦0
Chapter 5 Review 261
26.
2⎛
∫1 ⎜⎝ x +
1 ⎞ dx =
⎟
x2 ⎠
2
∫1 ( x + x
−2
⎤
⎡
33. (a) Note that each interval is 1 day = 24 hours
Upper estimate:
24(0.020 + 0.021 + 0.023 + 0.025 + 0.028
+ 0.031 + 0.035) = 4.392 L
Lower estimate:
24(0.019 + 0.020 + 0.021 + 0.023 + 0.025
+ 0.028 + 0.031) = 4.008 L
) dx
2
= ⎢ 1 x 2 − x −1 ⎥
⎢⎣ 2
⎥⎦ 1
( )
= 3− −1 =2
2
2
0
0
27.
∫−π /3 sec x tan x dx = ⎡⎣sec x ⎤⎦ −π /3 = 1 − 2 = −1
28.
∫−1 2 x sin(1 − x
29.
1
2
∫0
2
(b)
1
) dx = ⎡⎣ cos(1 − x 2 ) ⎤⎦ = 1 − 1 = 0
−1
2
2 dy = ⎡ 2 ln | y + 1 | ⎤ = 2 ln 3 − 0 = 2 ln 3
⎣
⎦0
y +1
30. Graph y = 4 − x 2 on [0, 2].
34. (a) Upper estimate:
3(5.30 + 5.25 + 5.04 + + 1.11) = 103.05 ft
Lower estimate:
3(5.25 + 5.04 + 4.71 + + 0) = 87.15 ft
(b)
The region under the curve is a quarter of a circle of
radius 2.
2
2
2
∫0 4 − x dx = 41 π (2) = π
24
[0.019 + 2(0.020) + 2(0.021) + + 2(0.031)
2
+ 0.035] = 4.2 L
3
[5.30 + 2(5.25) + 2(5.04) + + 2(1.11) + 0]
2
= 95.1 ft
35. One possible answer:
The dx is important because it corresponds to the actual
physical quantity Δx in a Riemann sum. Without the Δx, our
integral approximations would be way off.
36.
4
0
4
∫−4 f ( x ) dx = ∫−4 f ( x ) dx + ∫0 f ( x ) dx
=
31. Graph y =| x | dx on [−4, 8] .
0
4 2
∫−4 ( x − 2) dx + ∫0 x
0
dx
4
⎡1
⎤
⎡1 ⎤
= ⎢ x 2 − 2x ⎥ + ⎢ x3 ⎥
2
⎣
⎦ −4 ⎣ 3 ⎦ 0
⎡ 64
⎤ 16
= [0 − 16] + ⎢ − 0 ⎥ =
3
⎣
⎦ 3
37. Let f ( x ) = 1 + sin 2 x
max f = 2 since max sin 2 x = 1
min f = 1 since min sin 2 x = 0
The region under the curve consists of two triangles.
8
∫−4 | x | dx = 12 (4)(4) + 12 (8)(8) = 40
(min f )(1 − 0) ≤ ∫
32. Graph y = 64 − x 2 on [−8, 8] .
0 <1≤ ∫
1
0
1
0
2
1 + sin 2 x dx ≤ (max f )(1 − 0)
1 + sin x dx ≤ 2
4
38. (a) av( y) =
4
1
1 ⎡2
1 ⎛ 16
⎞ 4
⎤
x dx = ⎢ x 3/ 2 ⎥ = ⎜ − 0 ⎟ =
⎠ 3
⎝
4 − 0 ∫0
4 ⎣3
4
3
⎦0
a
(b) av( y) =
The region under the curve y = 64 − x 2 is half a circle of
radius 8.
8
8
⎡
2
2
2⎤
∫−8 2 64 − x dx = 2 ∫−8 64 − x dx = 2 ⎢⎢⎣ 12 π (8) ⎥⎥⎦ = 64π
a
1
1 ⎡2
2
⎤
a x dx = ⎢ ax 3/ 2 ⎥ = a 3/ 2
a − 0 ∫0
a ⎣3
⎦0 3
39.
dy
= 2 + cos3 x
dx
40.
dy
d
= 2 + cos3 (7 x 2 ) i (7 x 2 ) = 14 x 2 + cos3 (7 x 2 )
dx
dx
41.
dy d ⎛ x 6
6
⎞
=
−
dt = −
dx dx ⎜⎝ ∫1 3 + t 4 ⎟⎠
3 + x4
Chapter 5 Review
262
42.
x 1
dy d ⎛ 2 x 1
⎞
=
dt − ∫ 2
dt
0 t +1 ⎟
⎠
dx dx ⎜⎝ ∫0 t 2 + 1
1
=
i2 −
1
x
∫25
2
t
25
= 4 x − 20 + 50
= 4 x + 30
c(2500) = 4 2500 + 30 = 230
The total cost for printing 2500 newsletters is $230.
1
(600 + 600t ) dt
14 ∫0
14
1
= [600t + 300t 2 ⎤⎦ = 4800
0
14
Rich’s average daily inventory is 4800 cases.
c(t ) = 0.04 I (t ) = 24 + 24t
14
1 14
1
av(c) =
(24 + 24t ) dt = ⎡⎣ 24t + 12t 2 ⎤⎦ = 192
0
14 ∫0
14
14
Rich’s average daily holding cost is $192.
We could also say (0.04)4800 = 192.
x
⎡1 4 2
⎤
3
∫0 (t - 2t + 3) dt = ⎢⎣ 4 t - t + 3t ⎥⎦
0
x
=
x
∫5
y′′ = 2 −
dt + 50
x
45.
1 + x 4 dx = F (1) − F (0)
49. y′ = 2 x +
= ⎡⎣ 4t1/ 2 ⎤⎦ + 50
44. av( I ) =
1
∫0
48. y( x ) =
(2 x ) 2 + 1
x2 + 1
2
1
= 2
−
4x + 1 x2 + 1
43. c( x ) =
47.
1 4
x − x 2 + 3x
4
1 4
x − x 2 + 3x = 4
4
1 4
x − x 2 + 3x − 4 = 0
4
x 4 − 4 x 2 + 12 x − 16 = 0
Using a graphing calculator, x ≈ −3.09131 or x ≈ 1.63052.
46. (a) True, because g′( x ) = f ( x ).
(b) True, because g is differentiable.
(c) True, because g′(1) = f (1) = 0.
(d) False, because g′′(1) = f ′(1) > 0.
sin t
dt + 3
t
1
x
1
x2
Thus, it satisfies condition i.
11
y(1) = 1 + ∫ dt + 1 = 1 + 0 + 1 = 2
1t
1
y ′(1) = 2 + = 2 + 1 = 3
1
Thus, it satisfies condition ii.
50. Graph (b).
y=
x
∫1 2t dt + 4 = ⎡⎣t
2⎤
x
2
2
⎦ 1 + 4 = ( x − 1) + 4 = x + 3
51. (a) Each interval is 5 min =
1
h.
12
1
[2.5 + 2(2.4) + 2(2.3) + + 2(2.4) + 2.3]
24
29
=
≈ 2.42 gal
12
⎛ 12
⎞
(b) (60 mi/h) ⎜ h/gal ⎟ ≈ 24.83 mi/gal
⎝ 29
⎠
1 2
gt from Section 3.4,
2
1
the distance A falls in 4 seconds is (32)(4 2 ) = 256 ft.
2
When her parachute opens, her altitude is 6400 – 256 =
6144 ft.
52. (a) Using the freefall equation s =
(b) The distance B falls in 13 seconds is
1
(32)(132 ) = 2704 ft. When her parachute opens, her
2
altitude is 7000 – 2704 = 4296 ft.
(c) Let t represent the number of seconds after A jumps. For
t ≥ 4 sec, A’s position is given by
S A (t ) = 6144 − 16(t − 4) = 6208 − 16t , so A lands at
(f) False, because g′′(1) = f ′(1) ≠ 0.
6208
= 388 sec. For t ≥ 45 + 13 = 58 sec, B’s position
16
is given by S B (t ) = 4296 − 16(t − 58) = 5224 − 16t , so B
(g) True, because g′( x ) = f ( x ), and f is an increasing
function which includes the point (1, 0).
lands at t =
(e) True, because g′(1) = f (1) = 0 and g′′(1) = f ′(1) > 0.
t=
5224
= 326.5 sec. B lands first.
16
Chapter 5 Review 263
1
(2h)( y1 + y3 ) = h( y1 + y3 )
2
Area of the rectangle = (2h) y2 = 2hy2
53. (a) Area of the trapezoid =
57. (a) V 2 = (vmax )2 sin 2 (120 π t )
Using NINT:
2
1 1
av(V 2 ) = ∫ (Vmax ) sin 2 (120 π t ) dt
1 0
h( y1 + y3 ) + 2(2hy2 ) = h( y1 + 4 y2 + y3 )
b−a
.
2n
h
= [ y0 + 4 y1 + 2 y2 + 4 y3 + 2 y4 + + 2 y2 n− 2
3
+ 4 y2 n−1 + y2 n ]
= (Vmax )2 ∫ sin 2 (120 π t ) dt = (Vmax )2
1
(b) Let h =
S2 n
1
= [h( y0 + 4 y1 + y2 ) + h( y2 + 4 y3 + y4 ) + 3
+ h( y2 n− 2 + 4 y2 n−1 + y2 n )]
Since each expression of the form
h( y2i − 2 + 4 y2i −1 + y2i ) is equal to twice the area of the
ith of n rectangles plus the area of the ith of n
2 i MRAM n + Tn
trapezoids, S2 n =
.
3
54. (a) g(1) =
(b) g(3) =
1
∫1 f (t ) dt = 0
1
f (t ) dt = − (2)(1) = −1
2
3
∫1
−1
1
1
f (t ) dt = − ∫ f (t ) dt = − π (2)2 = −π
−1
4
(d) g′( x ) = f ( x ); Since f ( x ) > 0 for −3 < x < 1 and
f ( x ) < 0 for 1 < x < 3, g( x ) has a relative maximum at
x = 1.
(e) g′(−1) = f (−1) = 2
The equation of the tangent line is
y − (−π ) = 2( x + 1) or y = 2 x + 2 − π
(c) g(−1) =
∫1
(f) g′′( x ) = f ′( x ), f ′( x ) = 0 at x = –1 and f ′( x ) is not defined
at x = 2. The inflection points are at x = –1 and x = 2.
Note that g′′( x ) = f ′( x ) is undefined at x = 1 as well, but
since g′′( x ) = f ′( x ) is negative on both sides of x = 1,
x = 1 is not an inflection point.
(g) Note that the absolute maximum is g(1) = 0 and the
absolute minimum is
−3
1
1
g(−3) = ∫ f (t ) dt = − ∫ f (t ) dt = − π (2)2 = −2π .
−3
1
2
The range of g is ⎡⎣ −2π , 0 ⎤⎦ .
55. (a) NINT(e − x
2 /2
NINT(e − x
2
/2
, x , −10, 10) ≈ 2.506628275
, x , − 20, 20) ≈ 2.506628275
(b) The area is 2π .
56. First estimate the surface area of the swamp.
20
[146 + 2(122) + 2(76) + 2(54) + 2(40) + 2(30)
2
+ 13] = 8030 ft 2
1 yd 3
(5 ft )(8030 ft 2 ) i
≈ 1500 yd 3
27 ft 3
0
Vrms =
(Vmax )2 = Vmax
2
1 (Vmax )
=
2
2
2
2
(b) Vmax = 240 2 ≈ 339.41 volts
58. (a)
24
∫0
⎛ 4⎞
R(t )dt ≈ ⎜ ⎟ (9.6 + 2 (10.3 + 10.9 + 11.1
⎝ 2⎠
+ 10.9 + 10.5) + 9.6) ≈ 253.2,
which is the total number of gallons of water that
flowed through the pipe during the 24 hour period.
(b) Yes, because R(0) = R(24), the Mean Value Theorem
guarantees that there is a number c between 0 and 24
such that R′(c) = 0.
(c) Q(t ) = 0.01(950 + 25(4) − (4)2 ) = 10.58 gal / hr
59. f ′(1) = a(1)2 + b(1) = −6
f ′′( x ) = 2ax + b
f ′′(1) = 2a(1) + b = 6
2a + b = 6
− (a + b = −6)
a = 12
b = −18
f ′( x ) = 12 x 2 − 18 x
f (x) = 4 x3 − 9x 2 + c
2
∫1
2
f ( x )dx = ∫ 4 x 3 − 9 x 2 + c dx = 14
1
2
( x 4 − 3x 3 + cx ) = 14
1
c = 20
f ( x ) = 4 x 3 − 9 x + 20
2
1
(1 (3 + 1) + 2 (1 + (−1))) = 2
2
1
9
g (−2) = (−3(3 + 0)) = −
2
2
60. (a) g (4) =
(b) g (2) = f (2) = 1
9
2
(d) g has a point of inflection at x = 1. It is the only place
where the slope goes from positive to negative.
(c) The minimum value is g (−2) = −