C H A P T E R
Integration
4
Section 4.1
Antiderivatives and Indefinite Integration . . . . . . . . . 350
Section 4.2
Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
Section 4.3
Riemann Sums and Definite Integrals . . . . . . . . . . . 376
Section 4.4
The Fundamental Theorem of Calculus . . . . . . . . . . 385
Section 4.5
Integration by Substitution . . . . . . . . . . . . . . . . . 396
Section 4.6
Numerical Integration
Review Exercises
. . . . . . . . . . . . . . . . . . . 412
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433
C H A P T E R
Integration
Section 4.1
4
Antiderivatives and Indefinite Integration
1.
d 3
d
9
C 3x3 C 9x4 4
dx x3
dx
x
3.
d 1 3
x 4x C x2 4 x 2x 2
dx 3
2.
d 4 1
1
x C 4x3 2
dx
x
x
4.
d 2 32
d 2x2 3
C x 2x12 C
dx
dx 3
3x
x12 x32 5.
dy
3t2
dt
6.
dr
d
r C
y t3 C
Check:
7.
d 3
t C 3t2
dt
Check:
dy
x32
dx
8.
2
y x52 C
5
Check:
Given
9.
10.
11.
dy
2x3
dx
y
d 2 52
x C x32
dx 5
Rewrite
2x2
1
C 2 C
2
x
Check:
Integrate
Simplify
x 43
3 43
x C
4
C
d 1
C 2x3
dx x 2
3 x dx
x13 dx
1
dx
x2
x2 dx
x1
C
1
1
C
x
x32 dx
x12
C
12
x3 3x dx
x2
x4
C
3
4
2
1 4 3 2
x x C
4
2
1
xx
dx
43
2
x
C
12.
xx2 3 dx
13.
1
dx
2x3
1 3
x dx
2
1 x2
C
2 2
14.
1
dx
3x2
1 2
x dx
9
1 x1
C
9 1
1
C
9x
350
d
C d
1
C
4x2
x2 1
x32
Section 4.1
15.
x 3 dx Check:
17.
d x2
3x C x 3
dx 2
18.
d 2
x x3 C 2x 3x 2
dx
20.
d 1 4
x 2x C x3 2
dx 4
x2
C
2
d 4
x 2x 3 x C 4x 3 6x 2 1
dx
x 3 4x 2 dx x4
2x 2 2x C
4
Check:
22.
x2
d
5x C 5 x
dx
2
4x3 6x 2 1 dx x 4 2x 3 x C
Check:
2
x32 2x 1 dx x52 x2 x C
5
Check:
5 x dx 5x Check:
1
x3 2 dx x 4 2x C
4
Check:
21.
16.
2x 3x 2 dx x 2 x3 C
Check:
19.
x2
3x C
2
Antiderivatives and Indefinite Integration
d x4
2x 2 2x C x3 4x 2
dx 4
x 21 x dx x
12
d 2 52
x x2 x C x32 2x 1
dx 5
1
x12 dx
2
x32 1 x12
C
32 2 12
2
x32 x12 C
3
Check:
d 2 32
1
x x12 C x12 x12
dx 3
2
x 23.
3 x2 dx Check:
25.
27.
x3 dx 4 x3 1 dx Check:
1
x2
C 2C
2
2x
d
1
1
C 3
dx 2x2
x
26.
d 4 74
4 x3 1
x x C x34 1 dx 7
1
dx x4
Check:
4
x34 1 dx x74 x C
7
x4 dx 1
x3
C 3C
3
3x
d
1
1
C 4
dx 3x3
x
2
2
2
x32 x12 x12 dx x52 x32 2x12 C x123x2 5x 15 C
5
3
15
x2 x 1
d 2 52 2 32
x x 2x12 C x32 x12 x12 dx 5
3
x
x 2 2x 3
dx x4
Check:
24.
x2 x 1
dx x
Check:
28.
x53
3
C x53 C
53
5
d 3 53
3 x2
x C x23 dx 5
1
dx x3
Check:
x23 dx x2 2x3 3x4 dx 1
x1 2x2 3x3
1
1
C
2 3C
1
2
3
x
x
x
1
d
1
1
x 2 2x 3
2 3 C x2 2x3 3x4 dx
x
x
x
x4
1
2x
351
352
29.
Chapter 4
Integration
x 13x 2 dx 3x2 x 2 dx
30.
2t2 12 dt 4
4
t5 t3 t C
5
3
1
x3 x2 2x C
2
Check:
4t4 4t2 1 dt
d 3 1 2
x x 2x C 3x2 x 2
dx
2
Check:
d 45 43
t t t C 4t 4 4t2 1
dt 5
3
2t2 12
x 13x 2
31.
y2y dy Check:
33.
35.
1 dx x C
Check:
d
x C 1
dx
2 sin x 3 cos x dx 2 cos x 3 sin x C
Check:
34.
36.
d
3t C 3
dt
d 13
t cos t C t2 sin t
dt 3
1
2 sec2 d 3 tan C
3
Check:
40.
d 13 34
t t C t2 3t3 1 3tt2
dt 3
4
1
t 2 sin t dt t3 cos t C
3
Check:
38.
1
3
t2 3t3 dt t3 t4 C
3
4
3 dt 3t C
Check:
d
t csc t C 1 csc t cot t
dt
sec2 sin d tan cos C
1 3tt2 dt Check:
d
2 cos x 3 sin x C 2 sin x 3 cos x
dx
1 csc t cot t dt t csc t C
Check:
39.
32.
d 2 72
y C y52 y2y
dy 7
dx Check:
37.
2
y52 dy y72 C
7
d 1 3
tan C 2 sec2 d 3
sec ytan y sec y dy sec y tan y sec2 y dy
sec y tan y C
d
tan cos C sec2 sin d
Check:
d
sec y tan y C sec y tan y sec2 y
dy
sec ytan y sec y
41.
tan2 y 1 dy Check:
sec2 y dy tan y C
42.
cos x
dx 1 cos2 x
d
tan y C sec2 y tan2 y 1
dy
Check:
cos x
dx sin2 x
x
dx
sin1 xcos
sin x csc x cot x dx csc x C
d
1
csc x C csc x cot x dx
sin x
cos x
1 cos2 x
cos x
sin x
Section 4.1
43. f x 2
Antiderivatives and Indefinite Integration
44. fx x
f x 2x C
f x f )x)
2x
45. fx 1 x2
x2
C
2
f x x y
2
f ( x) = x + 2
2
x3
3
f )x)
5
f )x)
x
y
x2
8
4
f′
x3
C
3
2
y
3
353
f ( x) = 2
f )x)
x3
3
x
3
4
6
2x
3
f′
4
2
x
3
2
1
2
3
x
−4
3
x
−2
2
2
3
1
4
f
Answers will vary.
2
Answers will vary.
46. fx y
1
x2
47.
f′
f (x) = − 1x + 1
1
f x C
x
−4
2x 1 dx x2 x C
y
x
−2
4
−2
1 12 1 C ⇒ C 1
f (x) = − 1x
y x2 x 1
−4
48.
dy
2x 1, 1, 1
dx
dy
2x 1 2x 2, 3, 2
dx
2x 1 dx x2 2x C
y
2 32 23 C ⇒ C 1
y x2 2x 1
50. (a)
49. (a) Answers will vary.
y
5
y
5
x
−4
4
x
−3
5
−5
−3
dy 1
(b)
x 1, 4, 2
dx 2
(b)
6
2
x
y xC
4
2
4
2 4C
4
2C
y
x2
x2
4
−4
dy
x2 1, 1, 3
dx
y
x3
xC
3
3
13
1 C
3
8
−2
1
3 1C
3
C
7
3
y
x3
7
x
3
3
5
(−1, 3)
−4
4
−5
354
Chapter 4
51. (a)
Integration
(b)
y
dy
cos x, 0, 4
dx
7
6
cos x dx sin x C
y
−6
4 sin0 C ⇒ C 4
x
−4
3
−2
52. (a)
y sin x 4
(b)
y
(1, 3)
dy 1
2 , x > 0,
dx
x
y
1
7
y
53. (a)
(b)
9
−3
5
x2 dx 1
x1
C C
1
x
−1
8
−1
1
2
x
dy
2 x, 2, 2
dx
y
3
(c)
2x dx x 2 C
12
−15
15
2 22 C 4 C ⇒ C 6
−9
y
20
(b)
x2
−8
6
dy
2x,
dx
y
0
1
dx x2
1, 3
1
3 C ⇒ C2
1
x
54. (a)
6
−1
4, 12
(c)
20
4
2x12 dx x 3 2 C
3
6
0
4
4
32
4
12 432 C 8 C C ⇒ C
3
3
3
3
0
6
0
4
4
y x 32 3
3
55. fx 4x, f 0 6
f x 4x dx 2x 2 C
56. gx 6x 2, g0 1
gx 6x 2 dx 2x3 C
f0 6 202 C ⇒ C 6
g0 1 203 C ⇒ C 1
f x 2x 2 6
gx 2x3 1
57. ht 8t3 5, h1 4
ht 8t3 5 dt 2t4 5t C
58. fs 6s 8s 3, f 2 3
f s 6s 8s 3 ds 3s 2 2s 4 C
h1 4 2 5 C ⇒ C 11
f 2 3 322 224 C 12 32 C ⇒ C 23
ht 2t4 5t 11
f s 3s 2 2s 4 23
Section 4.1
Antiderivatives and Indefinite Integration
60. f x x2
59. f x 2
f2 5
f0 6
f 2 10
f 0 3
fx fx 2 dx 2x C1
f0 0 C1 6 ⇒ C1 6
f2 4 C1 5 ⇒ C1 1
fx 2x 1
f x 1
x 2 dx x3 C1
3
1
fx x3 6
3
2x 1 dx x2 x C2
13x
f x f 2 6 C2 10 ⇒ C2 4
3
f x 61. f x x32
1 4
x 6x 3
12
62. f x sin x
f4 2
f0 1
f 0 0
f 0 6
x32 dx 2x12 C1 2
x
fx C1
fx f x x
sin x dx cos x C1
f0 1 C1 1 ⇒ C1 2
2
f4 C1 2 ⇒ C1 3
2
2
1 4
x 6x C2
12
f 0 0 0 C2 3 ⇒ C2 3
f x x2 x 4
fx 6 dx fx cos x 2
3
f x cos x 2 dx sin x 2x C2
f 0 0 0 C2 6 ⇒ C2 6
2x12 3 dx 4x12 3x C2
f x sin x 2x 6
f 0 0 0 C2 0 ⇒ C2 0
f x 4x12 3x 4x 3x
63. (a) ht 1.5t 5 dt 0.75t 2 5t C
h0 0 0 C 12 ⇒ C 12
ht 0.75t 5t 12
2
(b) h6 0.7562 56 12 69 cm
64.
dP
kt, 0 ≤ t ≤ 10
dt
Pt 2
kt12 dt kt32 C
3
P0 0 C 500 ⇒ C 500
2
P1 k 500 600 ⇒ k 150
3
2
Pt 150t32 500 100t32 500
3
P7 100732 500
2352 bacteria
355
356
Chapter 4
Integration
65. f 0 4. Graph of f is given.
(a) f4
1.0
(b) No. The slopes of the tangent lines are greater than 2
on 0, 2 . Therefore, f must increase more than 4 units
on 0, 4 .
(c) No, f 5 < f 4 because f is decreasing on 4, 5 .
(d) f is a maximum at x 3.5 because f3.5
First Derivative Test.
(e) f is concave upward when f is increasing on , 1 and
5, . f is concave downward on 1, 5. Points of inflection
at x 1, 5.
(f) f is a minimum at x 3.
(g)
y
6
0 and the
4
2
x
−2
2
4
6
8
−6
66. Since f is negative on , 0, f is decreasing on
, 0. Since f is positive on 0, , f is increasing
on 0, . f has a relative minimum at 0, 0. Since f is
positive on , , f is increasing on , .
y
2
f ′′
1
−3
x
−2
vt 1
2
3
32 dt 32t C1
v0 60 C1
st 3
f′
67. at 32 ftsec2
32t 60 dt 16t 2 60t C2
s0 6 C2
st 16t2 60t 6, Position function
−2
f
The ball reaches its maximum height when
−3
vt 32t 60 0
32t 60
t 15
8 seconds.
15
15
15
s 8 16 8 60 8 6 62.25 feet
2
68. f t at 32 ftsec2
st 16t2 v0t
f0 v0
f 0 s0
ft vt 69. From Exercise 68, we have:
32 dt 32t C1
v0
st 32t v0 0 when t time to reach
32
maximum height.
f0 0 C1 v0 ⇒ C1 v0
ft 32t v0
f t st 32t v0 dt 16t2 v0t C2
f 0 0 0 C2 s0 ⇒ C2 s0
f t 16t 2 v0t s0
s
32 1632
v0
v0
2
v0
v02 v02
550
64
32
v02 35,200
v0
187.617 ftsec
32 550
v0
Section 4.1
Antiderivatives and Indefinite Integration
357
70. v0 16 ftsec
s0 64 ft
st 16t2 16t 64 0
(a)
vt st 32t 16
(b)
16 t 4 0
t2
v
1 ± 17
t
2
1 2 17 321 2 17 16
1617
Choosing the positive value,
t
1 17
2
2.562 seconds.
72. From Exercise 71, f t 4.9t2 1800. (Using the
canyon floor as position 0.)
71. at 9.8
vt 9.8 dt 9.8t C1
f t 0 4.9t2 1800
v0 v0 C1 ⇒ vt 9.8t v0
f t 65.970 ftsec
9.8t v0 dt 4.9t 2 v0t C2
4.9t2 1800
t2 1800
⇒ t
4.9
9.2 sec
f 0 s0 C2 ⇒ f t 4.9t 2 v0t s0
73. From Exercise 71, f t 4.9t2 10t 2.
74. From Exercise 71, f t 4.9t2 v0t 2. If
v t 9.8t 10 0 (Maximum height when v 0.)
9.8t 10
t
10
f
9.8
f t 200 4.9t2 v0t 2,
then
vt 9.8t v0 0
10
9.8
for this t value. Hence, t v09.8 and we solve
7.1 m
4.9
9.8
v0
2
v0
9.8 2 200
v0
v02
4.9 v02
198
2 9.8
9.8
4.9 v02 9.8 v02 9.82 198
4.9 v02 9.82 198
v02 3880.8 ⇒ v0
75.
a 1.6
vt 1.6 dt 1.6t v0 1.6t, since the stone was dropped, v0 0.
st 1.6t dt 0.8t2 s0
s20 0 ⇒ 0.8202 s0 0
s0 320
Thus, the height of the cliff is 320 meters.
vt 1.6t
v20 32 msec
62.3 msec.
358
Chapter 4
76.
v dv GM
Integration
1
dy
y2
0 ≤ t ≤ 5
77. xt t3 6t2 9t 2
(a) vt xt 3t 12t 9
2
1 2 GM
v C
2
y
3t2 4t 3 3t 1t 3
When y R, v v0.
at vt 6t 12 6t 2
1 2 GM
v C
2 0
R
(b) vt > 0 when 0 < t < 1 or 3 < t < 5.
(c) at 6t 2 0 when t 2.
1
GM
C v02 2
R
v2 311 3
1 2 GM 1 2 GM
v v0 2
y
2
R
v2 2GM
2GM
v02 y
R
v2 v02 2GM
1y R1 0 ≤ t ≤ 5
78. xt t 1t 32
79. vt t3 7t2 15t 9
xt (a) vt xt 3t2 14t 15 3t 5t 3
at vt 6t 14
(b) vt > 0 when 0 < t <
t12
t > 0
vt dt 2t12 C
x1 4 21 C ⇒ C 2
5
and 3 < t < 5.
3
7
(c) at 6t 14 0 when t .
3
v
1
t
73 373 573 3 2 32 43
80. (a) at cos t
vt at dt cos t dt sin t C1 sin t since v0 0
f t vt dt sin t dt cos t C2
f 0 3 cos0 C2 1 C2 ⇒ C2 4
f t cos t 4
(b) vt 0 sin t for t k, k 0, 1, 2, . . .
xt 2t12 2 position function
1
1
at vt t32 32 acceleration
2
2t
Section 4.1
81. (a)
v0 25 kmhr 25 1000 250
msec
3600
36
v13 80 kmhr 80 1000 800
msec
3600
36
Antiderivatives and Indefinite Integration
st a
(b)
s13 t 2 250
t
2
36
359
s0 0
275 132 250
13
234 2
36
189.58 m
at a constant acceleration
vt at C
v0 v13 250
250
⇒ vt at 36
36
250
800
13a 36
36
550
13a
36
1.175 msec2
v0 45 mph 66 ftsec
(a) 16.5t 66 44
30 mph 44 ftsec
t
15 mph 22 ftsec
at a
s
vt at 66
22
16.5
s
132 when a 33
16.5.
2
at 16.5
vt 16.5t 66
st 8.25t 2 66t
83. Truck: vt 30
st 30t Let s0 0.
Automobile: at 6
vt 6t Let v0 0.
st 3t2 Let s0 0.
At the point where the automobile overtakes the truck:
30t 3t2
0 3t2 30t
0 3tt 10 when t 10 sec.
ec
2.667
44
mp
h=
66
h=
mp
44
16.5
117.33 ft
ft/s
66
66
a
45
(c)
44
16.5
30
66
at 66 0 when t .
a
73.33 ft
t
vt 0 after car moves 132 ft.
2
1.333
(b) 16.5t 66 22
a
st t2 66t Let s0 0.
2
a 66
66
s
a
2 a
22
16.5
ft/s
ec
mp
0m h=
ph 22 f
t/se
c
82.
550 275
468 234
0
15
a
132
73.33
feet
117.33
feet
It takes 1.333 seconds to reduce the speed from 45 mph to
30 mph, 1.333 seconds to reduce the speed from 30 mph
to 15 mph, and 1.333 seconds to reduce the speed from
15 mph to 0 mph. Each time, less distance is needed to
reach the next speed reduction.
(a) s10 3102 300 ft
(b) v10 610 60 ftsec
41 mph
360
84.
Chapter 4
Integration
1 mihr5280 ftmi 22
ftsec
3600 sechr
15
(a)
t
0
5
10
15
20
25
30
V1ftsec
0
3.67
10.27
23.47
42.53
66
95.33
V2ftsec
0
30.8
55.73
74.8
88
93.87
95.33
(c) S1t S2t V1t dt (b) V1t 0.1068t2 0.0416t 0.3679
V2t 0.1208t2 6.7991t 0.0707
0.1068 3 0.0416 2
t t 0.3679t
3
2
V2t dt 0.1208t3 6.7991t2
0.0707t
3
2
In both cases, the constant of integration is 0 because S10 S20 0.
S130
953.5 feet
S230
1970.3 feet
The second car was going faster than the first until the end.
85. at k
vt kt
k
st t2 since v0 s0 0.
2
At the time of lift-off, kt 160 and k2t2 0.7. Since k2t2 0.7,
1.4k
1.4
1.4
160
v k
k
k
t
1.4k 1602 ⇒ k 1602
1.4
18,285.714 mihr2
7.45 ftsec2.
86. Let the aircrafts be located 10 and 17 miles away from the airport, as indicated in the figure.
vAt kAt 150
vB kB t 250
1
sAt kAt2 150t 10
2
1
sB kB t2 250t 17
2
(a) When aircraft A lands at time tA you have
vAtA kA tA 150 100 ⇒ kA 1
sAtA kA tA2 150tA 10 0
2
1 50 2
t 150tA 10
2 tA A
125tA 10
tA —CONTINUED—
10
.
125
50
tA
Airport
A
B
0
10
17
Section 4.1
Antiderivatives and Indefinite Integration
86. —CONTINUED—
kA 125
50
625 2
t 150t 10
50
625 ⇒ sAt tA
10
2
Similarly, when aircraft B lands at time tB you have
vBtB kB tB 250 115 ⇒ kB 135
tB
1
sBtB kB tB2 250tB 17 0
2
1 135 2
t 250tB 17
2 tB B
365
t 17
2 B
tB kB (b)
34
hr .
365
135
49,275
49,275 2
365
135
⇒ sBt t 250t 17
tB
34
34
68
28,025 2
t 100t 7
68
Yes, d < 3 for t > 0.0505 hr.
(c) d sBt sAt 20
sB
20
sA
0.1
0
0
d
3
0.1
0
0
87. True
91. False. For example,
88. True
x x dx
x dx 89. True
x dx because
92. False. f has an infinite number of antiderivatives, each
differing by a constant.
x3
C
3
90. True
x2 C x2 C .
2
2
1
2
93. f x 2x
fx x 2 C
f2 0 ⇒ 4 C 0 ⇒ C 4
f x x3
4x C1
3
f 2 0 ⇒
8
16
8 C1 0 ⇒ C1 3
3
Answer: f x x3
16
4x 3
3
361
362
Chapter 4
1,
94. fx 2 ,
0 ,
Integration
0 ≤ x < 2
2 < x < 3
3 < x ≤ 4
0 ≤ x < 2
2 < x < 3
3 < x ≤ 4
x C1,
f x 2x C2 ,
C3
,
y
2
1
x
1
2
3
4
f 0 1 ⇒ C1 1
f continuous at x 2 ⇒ 2 1 4 C2 ⇒ C2 5
f continuous at x 3 ⇒ 6 5 C3 1
0 ≤ x < 2
2 ≤ x < 3
3 ≤ x ≤ 4
x 1,
f x 2x 5 ,
1
,
95. fx 1, 0 ≤ x < 2
3x, 2 ≤ x ≤ 5
x C1,
0 ≤ x < 2
f x 3x2
C2, 2 ≤ x ≤ 5
2
f 1 3 ⇒ 1 C1 3 ⇒ C1 2
f is continuous: Values must agree at x 2:
4 6 C2 ⇒ C2 2
0 ≤ x < 2
x 2,
f x 3x 2
2, 2 ≤ x ≤ 5
2
The left and right hand derivatives at x 2 do not agree. Hence f is not differentiable at x 2.
96.
d
sx 2 cx
dx
2
2sxsx 2cxcx
2sxcx 2cxsx
0
Thus, sx 2 cx 2 k for some constant k. Since, s0 0 and c0 1, k 1.
Therefore, sx 2 cx 2 1.
[Note that sx sin x and cx cos x satisfy these properties.]
97. f x y f xf y gxg y
gx y f xg y gxf y
f 0 0
Note: f x cos x and gx sin x satisfy these conditions
f x y f xf y gxg y Differentiate with respect to y
g x y f xg y gxf y Differentiate with respect to y
Letting y 0, f x f xf 0 gxg0 gxg0
g x f xg 0 gxf0 f xg0
—CONTINUED—
Section 4.2
Area
363
97. —CONTINUED—
Hence, 2 f xfx 2 f xgxg0
2gxgx 2gxf xg0.
Adding, 2 f x fx 2g xgx 0.
Integrating, f x2 gx2 C.
Clearly C 0, for if C 0, then f x2 gx2 ⇒ f x gx 0, which contradicts that f, g are nonconstant.
Now, C f x y2 gx y2 f x f y gxg y2 f xg y gx f y2
f x2f y2 g x2g y2 f x2g y2 gx2f y2
f x2 gx2 f y2 g y2
C2
Thus, C 1 and we have f x2 gx2 1.
Section 4.2
1.
Area
5
5
5
i1
i1
i1
2i 1 2 i 1 21 2 3 4 5 5 35
6
2.
4
kk 2 31 42 53 64 50
3.
k3
5
4.
1
1
1
1
4
1
1 1
1
1
158
1 1
2 5 10 17
85
4
47
j 3 4 5 60
i 1
2
k0
5.
j3
6.
k
c c c c c 4c
k1
i 13 0 8 1 27 4 64 9 125 238
2
i1
9
7.
15
1
3i
i1
8.
n 2 n
n i1
15.
2i 2 i 2
2i
20
i1
3
2in
11.
20
i1
12.
9.
2
13.
3i
3 n
2 1
n i1
n
2
16.
2i 3 2 i 315
14.
15
15
i1
i1
15216
20
i 1
2
i1
19
i
2
i1
192039
2470
6
10
18.
i
2
1 i1
10
i
i1
j
2
j1
2
17.
1 4
4
10.
j1
2i
2 n
1
1
n i1
n
2021
420
2
j
5 8 3
8
5
1
i
i1
2
1 n1
n i0
45 195
10
1
i1
1011621
10 375
1
ni 2
364
Chapter 4
15
19.
ii 1
Integration
15
i
2
i1
2
3
i1
15
i
2
10
15
i
i1
20.
ii
2
1 i1
i1
10
i
3
i1
151631 1516
152162
2
4
6
2
10
i
i1
102112
1011
3080
4
2
14,400 2480 120
12,040
20
i
2
2 3, x, 1, 20, 1 2930 (TI-82)
3 i1
22. sum seqx
2020 1220 1
320
6
15
i
3
>
>
21. sum seqx
3 2x, x, 1, 15, 1 14,160 (TI-82)
2i i1
202141
60 2930
6
9
33
23. S 3 4 2 51 2 16.5
s 1 3 4 1 9
2
25
2
s 4 4 2 01 10
12.5
s4 0
28. S8 14 1
1
2 2
4
5
1
2 4
4
s5 2
8
2
8
2 1 1
9
4.5
3 2 3
2
32
3
3
1
1
2 1 2
4
4
4
3
1
2 2
4
7
1
1
2 2 2
4
4
4
2
1
1
2
4
4
0.768
0.518
1
1 . . .
2
2
4
6.038
7
1
2
4
4
5.685
15 61515 71515 81515 91515 51 61 71 81 91
0.746
1 1
1 1
1 1
1 1
1 1
1 1 1 1
1
6 5 5
7 5 5
8 5 5
9 5 5
2 5
6 7 8 9 10
30. S5 1
14 1 3 1
1
4 4
1 1
2 4
1
1
2 4
4
1
1
1
2
3
5
6
7
16 1
4
2
2
2
2
2
2
29. S5 1
s5 3 1
4 4
1 1
4 4
1
s8 0 2 4
2 1
55
3 2
6
s 21
1 1
2 4
26. S 5 2 1 s 2 2 31 7
1 1
4 4
15 1
1
5
1
1
24
5
15 15 2
21
5
15 15 2
152162
1516 14,160
4
24. S 5 5 4 21 16
25. S 3 3 51 11
27. S4 1515 1
15 215 1 2
2
4
2
1
16
9
5
5
1
25 15 2
35 15 2
1
45 15
2
0.859
25 15 2
1
0.646
1
35 15 2
1
45 15 0
2
0.659
Section 4.2
81 n2n 12
81
n4 2n3 n2
81
81
lim
1 4
4
4 n→
n4
4
4
31. lim
n 32. lim
64n nn 16 2n 1
n→ n→ n→
3
64
64
64
2n3 3n2 n
2 lim
n→
6 n3
6
3
n→
1 nn 1
n2 n
1
1
1
lim
1 2
2
2 n→
n2
2
2
n 34. lim
1 n
1 nn 1
n2
2
2i 1
2
2i 1 2 2
n 1 Sn
2
n
n
n
2
n
n
i1
i1
n
35.
18 nn 1
18
n2 n
18
lim
1 9
2
2
2 n→
n2
2
n 33. lim
S10 12
1.2
10
S100 1.02
S1000 1.002
S10,000 1.0002
4j 3
1 n
1 4nn 1
2n 5
3n Sn
2
4j 3 2
2
n j1
n
2
n
j1 n
n
36.
S10 25
2.5
10
S100 2.05
S1000 2.005
S10,000 2.0005
6 n 2
6 nn 12n 1 nn 1
6kk 1
3
k k 3
3
n
n k1
n
6
2
k1
n
37.
6 2n2 3n 1 3n 3
1
2
22n2 2 2 2 Sn
n2
6
n
n
S10 1.98
S100 1.9998
S1000 1.999998
S10,000 1.99999998
4i2i 1
4 n 3
4 n2n 12 nn 12n 1
4
i i2 4
4
n
n
n
4
6
i1
i1
n
38.
Area
S10 1.056
S100 1.006566
S1000 1.00066567
S10,000 1.000066657
4 n3 2n2 n 2n2 3n 1
n3
4
6
1
3n3 6n2 3n 4n2 6n 2
3n3
1
3n3 2n2 3n 2 Sn
3n3
365
366
Chapter 4
Integration
16
16 nn 1
16i
n n
n lim n i lim n 2 lim 8 n n
39. lim
n
2
n→ i1
n n n
40. lim
n→ 2i
2
lim
n→ i1
n
n→
1
n→ i1
2
2
n→ i1
n→ lim
2
3
1 2n3 3n2 n
1 2 3 n 1 n2
lim
3
n→ 6
n→ 6
n
1
1 n n n
2i
2
2
n→ i1
lim
n→ n→ n
n
2
i1
i1
i1
2
4
26
3
3
1 nn 2 lim n 1 n i
2
n
1
n→
i1
2
1 2 n2 34 n2 3n2
1
i1
n
1
1 nn 1
n
n→ n
n
2
2 lim
i1
1 n 2n n
2
2 lim
n→
2
2 1
1
3
2
1 n n 2 lim n n 2i
n
2i
44. lim
n→
n
2 12
n→
n
i 4 i
4n
n→ i
1
3
nn 1
2 3
4nn 12n 1
n 4n
n3
2
6
2 lim
n
2
n→ n3
lim
2 n
n 2i2
n3i1
lim
43. lim
1
8
n
4 n
4 nn 1
4
1
i lim 2
1
2
lim
n→ n
n→ 2
n2i1
2
n
lim
42. lim
8 lim 1 2
n→ 1 n1 2
1 n 1n2n 1
i lim 3
n→ n3 i1
n→ n
6
i 1
n
41. lim
2
3
n
1
2
n→
i1
3
4
i1
1 n 3
n 6n2i 12ni 2 8i 3
n→ n4 i1
2 lim
2 lim
1 4
nn 1
nn 12n 1
n2n 12
n 6n2
12n
8
n4
2
6
4
2 lim
1 3 n3 4 n6 n2 2 n4 n2 n→ n→ 45. (a)
2
2 lim 10 n→ 3
2
13
4
2 20
n
n
(b) x y
20 2
n
n
Endpoints:
2
0 < 1
1
x
1
—CONTINUED—
3
2n < 22n < . . . < n 12n < n2n 2
Section 4.2
Area
367
45. —CONTINUED—
(d) f Mi f xi on xi1, xi
(c) Since y x is increasing, f mi f xi1 on xi1, xi.
n
sn n
Sn f xi1 x
i1
i1
f
n
i1
(e)
x
sn
Sn
f xi x 2i 2
n
2n i 12n 2n
n
n→ i1
i1
i1
2
i 1
n
4 n
2
lim 2 i 1
n→ n i1
n
50
100
1.6
1.8
1.96
1.98
2.04
lim
4 nn 1
n
n2
2
lim
2n n 1 n4
n→ 10
2.2
(f) lim
n
f
i1
5
2.4
n
2i 2
2
2
n n i n n n
n→ 2.02
i
n n
n
lim
n→
i1
2
2
(b) x lim
n4 nn 2 1
lim
2n 1
2
n
n→ y
2
4 n
i
n→ n2 i1
lim
n→ 46. (a)
2
31 2
n
n
4
Endpoints:
3
1 < 1
2
1
1 <11
x
2
4
2
4
2n
< 1 <...< 1
3
n
n
n
2n < 1 22n < . . . < 1 n 12n < 1 n2n
(c) Since y x is increasing, f mi f xi1 on xi1, xi.
n
f x
sn x
i1
i1
f 1 i 1n n 1 i 1n n
n
2
n
2
i1
2
2
i1
(d) f Mi f xi on [xi1, xi
f x x f 1 in n 1 in n
n
Sn n
i1
(e)
2
2
x
i1
2
2
i1
5
10
50
100
sn
3.6
3.8
3.96
3.98
Sn
4.4
4.2
4.04
4.02
1 i 1n n n
(f) lim
n
i
2
2
n→ i1
lim
n→ lim
n→ 1 i n n n
lim
n→ i1
2
2
lim
n→ lim
n→ 2nn n2nn 2 1 n
2 2n n 2 n
4
lim
n→ 4 n2
4
2
2 nn 1
n
n
n
2
2 2n n 1
lim
n→ 4 n2
4
368
Chapter 4
Integration
Note: x 1 n 0 1n
i 1
i
1
sn f 2 3 n n
n
n
y
47. y 2x 3 on 0, 1.
n
n
i1
i1
3
2
2 n
2n 1n
1
i3
2
n2i1
2n2
n
1
x
1
Area lim sn 2
2
3
n→ Note: x 5 n 2 3n
48. y 3x 4 on 2, 5.
3i
4
n
n
3i
Sn f 2
n
i1
n
3 2
i1
6
y
12
3
n
10
8
3
3
18 3
n
n
27 n 1n
27
1
6
1
n2
2
2
n
Area lim Sn 6 n→ 2 n
6
i 12
4
i1
2
x
4
6
8
10
12
27 39
2
2
Note: x n1
i 1
i
1
Sn f 2 n n
n
n
y
49. y x2 2 on 0, 1.
n
n
i1
i1
3
2
nn 12n 1
1
3
1 n 2
1
3
i 2
2 2 2 2
n i1
6n3
6
n n
1
x
1
7
Area lim Sn n→ 3
50. y x 2 1 on 0, 3. Note: x Sn n
3i
n
f
l1
n
n
3
n
l1
3i
n
2
30 3
n
n
1
3
n
n
27
3
l2 1
n 3 l1
n l1
27 nn 12n 1 3
9 2n2 3n 1
3
n 3
n
6
n
2
n2
9
Area lim Sn 2 3 12
n →
2
2
3
y
12
10
8
6
4
2
x
1
2
3
Section 4.2
51. y 16 x 2 on 1, 3. Note: x sn n
f 1
i1
2i
n
2
n
Area
y
18
n
2
2i
16 1 n
n
i1
2
14
12
10
8
6
4
2
2
n
2 n
4i2 4i
15 2 n i1
n
n
−1
2
4 nn 12n 1 4 nn 1
15n 2
n
n
6
n
2
x
1
2
3
8
4
n 12n 1 n 1
6n2
n
30 Area lim sn 30 n →
8
70
1
4
23
3
3
3
Note: x n1
52. y 1 x2 on 1, 1. Find area of region over the interval 0, 1.
n
i
sn f
n
i1
y
3
n
1
i
1
n
n
i1
2
1
n
2
1 n 2
nn 12n 1
1
3
1
i 1
1 2 2
n3 i1
6n3
6
n n
1
x
−2
2
−1
1
1 2
Area lim sn 1 n→ 2
3 3
Area 4
3
53. y 64 x3 on 1, 4. Note: x 41 3
n
n
y
70
60
n
3i
sn f 1
n
i1
n
n
3
3i
64 1 n
n
i1
27i3
3
3
n
50
40
30
20
27i2
3
9i
63 3 2 n i1
n
n
n
10
−1
x
1
2
3
5
6
3
27 n2n 12 27 nn 12n 1 9 nn 1
2
63n 3
n
n
4
n
6
n
2
189 81
81
27 n 1
n 12 2n 12n 1 4n2
6n
2
n
Area lim sn 189 n →
54. y 2x x3 on 0, 1.
81
27 513
27 128.25
4
2
4
Note: x 1 n 0 1n
y
2.0
Since y both increases and decreases on 0, 1, Tn is neither an upper nor lower sum.
n
i
Tn f
n
i1
369
n
1
i
i
2
n
n
n
i1
3
1
n
1.0
0.5
2
1
nn 1
1 n n 1
i 4
i3 4
n2 i1
n i1
n2
n
4
n
1
n
1 1
2
1
n 4 4n 4n2
Area lim Tn 1 n→ 1 3
4 4
2
1.5
2
x
0.5
1.0
2.0
370
Chapter 4
Integration
Note: x 1 n1 n2
55. y x2 x3 on 1, 1.
y
Again, Tn is neither an upper nor a lower sum.
2
f 1 n n 1 n n
Tn 2i
n
2
i1
n
2
1 i1
1
i1
2 n
2i
i1
4i 4i2
6i 12i 2 8i3
2 1 2 3
n
n
n
n
n
10i 16i 2 8i3
2 3
n
n
n
4
20
n 2
n
n
nn 1 32
3
2
n
4 10 1 2i
n
2n
3
x
2
n
n
2
i1
n
3
i1
nn 12n 1 16
4
6
n
1
16
3
1
2
1
2 2 4 1 2
n
3
n n
n n
1
1
2n n4 1 20n i 32n i
n
1
2
i1
16 n 3
i
n4 i1
n2n 12
4
32
2
4
3
3
Area lim Tn 4 10 n→ Note: x 0 n1 n1
i 1
i
i
1
sn f 1 1 1 n n
n
n
n
y
56. y x2 x3 on 1, 0.
n
n
i1
i1
5i 4i
2 n n
n
2
2
i1
2
2
i3
n3
1n 2 n5 i n4 i
n
2
i1
n
3
i1
2
2
1
1 n 3
i
n4 i1
−2
x
−1
1
−1
5
1
5
4 2
2
1
1
2 2n 3 n 3n3 4 2n 4n2
Area lim sn 2 n→ 5 4 1
7
2 3 4 12
Note: y 2 n 0 2n
57. f y 3y, 0 ≤ y ≤ 2
y
4
n
n
n
2
2i
3
n
n
i1
2i
Sn f mi y f
n
i1
i1
12 n
12
2
i 2
n i1
n
n→ n→ x
2
4
6
2
y
5
4
g2 n n
2i
2
6
6
n
1
42 2
58. gy y, 2 ≤ y ≤ 4. Note: y 2
n
n
n
2
n
6
nn 1 6n 1
6
n
n
2
Area lim Sn lim 6 Sn 3
3
2
3
i1
2
n
1
2i
2
2
n
i1
2
2 n
i
1
n
n i1
n
2
1 nn 1
n1
n
2
n
n
2
n
Area lim Sn 2 1 3
n →
1
x
1
2
3
4
5
2
Section 4.2
Note: y 3 n 0 3n
27
3i 3
3i 3
Sn f i
n n
n
n
n n
2
6
n
4
2
i1
27
n3
3
i1
2
9 2n 3n 1
nn 12n 1
27
9
2
9
6
n
2
2n 2n2
Area lim Sn lim 9 n→ n→ −4
21 1
n
n
f 1 nn
n
i
x
2
−2
27
9
9
2n 2n2
60. f y 4y y 2, 1 ≤ y ≤ 2. Note: y Sn i1
2
y
5
1
3
i1
1 n
i
i
4 1
1
n i1
n
n
2
2
1 n
4i
2i
i2
4 1 2
n i1
n
n
n
1
x
1
1 n
2i
i2
3 2
n i1
n
n
1
2 nn 1
1 nn 12n 1
3n 2
n
n
2
n
6
3
4
5
3
n 1 n 12n 1
n
6
Area lim Sn 3 1 n →
61. gy 4y2 y3, 1 ≤ y ≤ 3.
2
n
2i
Sn g 1
n
i1
2
n
i1
2i
4 1
n
1 11
3
3
Note: y 3 n 1 2n
6
y
10
8
2
n
2i
1
n
3
2
2
n
−4 −2
−2
−4
2 n
4i 4i2
6i 12i2 8i3
4 1 2 1 2 3
n i1
n
n
n
n
n
2 n
10i 4i2 8i3
2
10 nn 1
4 nn 12n 1
8 n2n 12
3
2 3 3n 2
2
n i1
n
n
n
n
n
2
n
6
n
4
Area lim Sn 6 10 n →
44
8
4
3
3
371
y
59. f y y 2, 0 ≤ y ≤ 3
n
Area
x
4
6
8
10
372
Chapter 4
Integration
62. hy y3 1, 1 ≤ y ≤ 2 Note: y Sn 1
n
y
5
h1 nn
n
i
4
1
3
i1
n
i1
i
1
n
3
2
1
1
n
1
x
1 n
i3
3i2 3i
2 3 2 n i1
n
n
n
2
1
3 19
1 4
2
4
63. f x x2 3, 0 ≤ x ≤ 2, n 4
Let ci 64. f x x2 4x, 0 ≤ x ≤ 4, n 4
xi xi1
.
2
Let ci 1
3
5
7
1
x , c1 , c2 , c3 , c4 2
4
4
4
4
n
4
f c x c
2
i
i
i1
3
i1
12
n
69
8
4
14 2 94 6 254 10 494 14
66. f x sin x, 0 ≤ x ≤
,n4
xi xi1
.
2
16
4ci1
i1
53
65. f x tan x, 0 ≤ x ≤
n
2
i
i
i1
x 4
f c x c
Area
1 1
9
25
49
3 3 3 3
2 16
16
16
16
Let ci xi xi1
.
2
1
3
5
7
x 1, c1 , c2 , c3 , c4 2
2
2
2
, c1 Let ci 32
f ci x i1
67. f x 10
n 12 1 n 12n 1 3n 1
n24
2
n2
2n
n →
8
Area lim Sn 2 Area
6
1
1 n2n 12
3 nn 12n 1 3 nn 1
2n 3
2
n
n
4
n
6
n
2
2
Area
4
, c2 3
5
7
,c ,c 32 3
32 4
32
i1
i1
0.345
Exact value is 16 3.
16
, c2 3
5
7
,c ,c 16 3
16 4
16
4
i
x on 0, 4.
Approximate area
, c1 n
Area
3
5
7
tan
tan
tan
tan 16 32
32
32
32
n
8
f c x sin c 8 4
i
,n4
xi xi1
.
2
x tan c 16
2
4
8
12
16
20
5.3838
5.3523
5.3439
5.3403
5.3384
8
i
i1
sin 16 sin 316 sin 516 sin 716 1.006
Section 4.2
68. f x Area
373
8
on 2, 6.
x2 1
n
Approximate area
69. f x tan
4
8
12
16
20
2.3397
2.3755
2.3824
2.3848
2.3860
4
8
12
16
20
2.2223
2.2387
2.2418
2.2430
2.2435
4
8
12
16
20
1.1041
1.1053
1.1055
1.1056
1.1056
8x on 1, 3.
n
Approximate area
70. f x cos x on 0, 2.
n
Approximate area
71.
72.
y
y
4
4
3
3
2
2
1
1
1
(b) A
2
3
x
4
6 square units
73. We can use the line y x bounded by x a and x b.
The sum of the areas of these inscribed rectangles is the
lower sum.
1
(a) A
2
3
x
4
3 square units
The sum of the areas of these circumscribed rectangles is the
upper sum.
y
y
x
a
b
x
a
b
We can see that the rectangles do not contain all of the area in
the first graph and the rectangles in the second graph cover
more than the area of the region.
The exact value of the area lies between these two sums.
74. See the definition of area. Page 265.
374
Chapter 4
75. (a)
Integration
y
(b)
y
8
8
6
6
4
4
2
2
x
1
3
2
x
4
1
Lower sum:
s4 0 4 513 6 1513 46
3 15.333
(c)
y
2
3
4
Upper sum:
1
2
11
326
S4 4 53 6 65 2115 15 21.733
(d) In each case, x 4n. The lower sum uses left endpoints, i 14n. The upper sum uses right endpoints,
i4n. The Midpoint Rule uses midpoints, i 12 4n.
8
6
4
2
x
1
3
2
4
Midpoint Rule:
M4 223 445 557 629 6112
315 19.403
(e)
n
4
8
20
100
200
sn
15.333
17.368
18.459
18.995
19.06
Sn
21.733
20.568
19.739
19.251
19.188
Mn
19.403
19.201
19.137
19.125
19.125
(f) sn increases because the lower sum approaches the exact value as n increases. Sn decreases because the upper sum
approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the
exact value, whereas the upper sum is always larger.
2 y
76. f x sin x, 0,
1.0
Let A1 area bounded by f x sin x, the x-axis, x 0 and x 2. Let A2 area of the
rectangle bounded by y 1, y 0, x 0, and x 2. Thus, A2 21 1.570796.
In this program, the computer is generating N2 pairs of random points in the rectangle whose
area is represented by A2. It is keeping track of how many of these points, N1, lie in the region
whose area is represented by A1. Since the points are randomly generated, we assume that
A1 N1
N1
⇒ A1 A.
A2 N2
N2 2
( π2 , 1)
f (x) = sin(x)
0.75
0.5
0.25
π
π
4
2
The larger N2 is the better the approximation to A1.
77. True. (Theorem 4.2 (2))
78. True. (Theorem 4.3)
79. Suppose there are n rows and n 1 columns in the figure. The stars on the left total 1 2 . . . n, as do the stars
on the right. There are nn 1 stars in total, hence
21 2 . . . n nn 1
1 2 . . . n 12nn 1.
x
Section 4.2
80. (a)
2
n
375
r
h
r
(b) sin
Area
h
θ
r
h r sin
1
1
1
A bh rr sin r2 sin
2
2
2
(c) An n
12r
2
Let x 2
sin
2
n
r2n sin 2n
2
r2
sin22 n n
n. As n → , x → 0.
lim An lim
n→ x→0
81. (a) y 4.09
r2
sinx x r 21 r2
105x3 0.016x2 2.67x 452.9
(b)
500
(c) Using the integration capability of a graphing utility,
you obtain
76,897.5 ft2.
A
0
350
0
For n even,
82. For n odd,
n 1,
1 row,
1 block
n 2, 1 row,
2 block
n 3,
2 rows,
4 blocks
n 4, 2 rows,
6 blocks
n 5,
3 rows,
9 blocks
n 6, 3 rows,
12 blocks
n,
n1
rows,
2
n 2 1 blocks,
2
n
83. (a)
n
2i nn 1
(b)
i1
i
3
i1
The formula is true for n 1: 2 11 1 2.
Assume that the formula is true for n k:
k
2i kk 1.
n2 2n
blocks,
4
n2n 12
4
The formula is true for n 1 because
121 12 4
13 1.
4
4
Assume that the formula is true for n k:
i1
Then we have
n
rows,
2
n,
k1
k
i1
i1
2i 2i 2k 1
kk 1 2k 1
k 1k 2
which shows that the formula is true for n k 1.
k
i
3
i1
k2k 12
.
4
k1
Then we have
i
3
i1
k
i
3
k 13
i1
k2k 12
k 13
4
k 12 2
k 4k 1
4
k 12
k 22
4
which shows that the formula is true for n k 1.
376
Chapter 4
Integration
84. Assume that the dartboard has corners at ± 1, ± 1.
y
A point x, y in the square is closer to the center than the top edge if
(x, 1)
1
x2 y2 ≤ 1 y
(x, y)
x 2 y 2 ≤ 1 2y y 2
−1
1
1 x2.
2
y ≤
x
1
(0, 0)
−1
By symmetry, a point x, y in the square is closer to the center than the
right edge if
1
1 y 2.
2
x ≤
1
1
In the first quadrant, the parabolas y 21 x2 and x 21 y 2 intersect at 2 1, 2 1. There are 8 equal regions
that make up the total region, as indicated in the figure.
y
1
(
2 − 1,
2 − 1(
(1, 1)
−1
1
x
−1
Area of shaded region S 21
Probability Section 4.3
12 1 x xdx 2 3 2 65
2
0
42 5
22 5
8S
2
Area square
3
6
3
3
Riemann Sums and Definite Integrals
1. f x x, y 0, x 0, x 3, ci 3i2
n2
y
3
3i2 3i 12
3
xi 2 22i 1
n
n2
n
n
lim
n → i1
3i
n
3
2i 1
n2
33
2
n
f ci xi lim
2
2
n → i1
lim
n3
n →
n
2i
1
i
i1
n →
n 13n2n 1 lim 33
n →
2
23 0 23 3.464
33
n1
2n2
x
...
3
n2
33 nn 12n 1 nn 1
2
n3
6
2
lim
y=
2
3(2)2 . . . 3(n − 1)2 3
n2
n2
x
Section 4.3
3
x, y 0, x 0, x 1, ci 2. f x n n
i
i 3 3i 2 3i 1
3
n3
n
f c x lim
i3
n3
i3
i 13 3i 2 3i 1
n3
n3
n3
xi n→ i1
Riemann Sums and Definite Integrals
lim
i
3
n→ i1
1 n
3i 3 3i 2 i
n→ n4 i1
lim
1
nn 1
n2n 12
nn 12n 1
3
3
4
n→ n
4
6
2
lim
lim
1 3n4 6n3 3n2 2n3 3n2 n n2 n
n4
4
2
2
lim
1
1
1 3n4 n3 n2
3
3
lim
n→ 4
n4 4
2
4
2n 4n2
4
n→ n→ 3. y 6 on 4, 10 .
n
n
4
i
i1
i1
10 4 6
, → 0 as n → n
n
Note: x f c x f
i
n
n 36
1
6
6
1 n
6
36 36n 36
n
n
ni1
n
i1
i1 n
6i
n
10
6 dx lim 36 36
n→ 4
4. y x on 2, 3 .
n
Note: x n
f c x f
i
2 i
i1
i1
10 3
2
n→ n
n
i1
i
1 i1
2 2i
n
1
1
x3 dx lim
5
25 n
10 2
i
n
n i1
n→ 1 1 2
, → 0 as n → n
n
n
2
2i
1 n
n
i1
3
n
2
6i 12i 2 8i 3
1 2 3
n
n
n
n
i1
24 n 2 16 n 3
12 n
i 3
i 4
i
n2 i1
n i1
n i1
2 6 1 n
5
5i
2 n
n
i1
25 nn 1
25
1
5 25
10 1
n2
2
2
n
2 2n
Note: x f c x f
i
5i
n
5 25
5
2 2n
2
x dx lim
5. y x3 on 1, 1 .
3 2 5
, → 0 as n → n
n
2
0
n
1
3
1
2
1
2
4 2 2 4 1 2 n
n n
n n
n
2n
377
378
Chapter 4
Integration
31 2
, → 0 as n → n
n
6. y 3x 2 on 1, 3 . Note: x n
n
f c x f
i
i
i1
i1
n
2
n
2i
n
1
2i
3 1 n
2
2
n
i1
6 n
4i 4i2
1 2
ni1
n
n
6
4 nn 1
4 nn 12n 1
n
2
n
n
2
n
6
6 12
3
3x 2 dx lim
n →
1
n1
n 12n 1
4
n
n2
n 1 4n 12n 1
n
n2
6 12
6 12 8 26
7. y x2 1 on 1, 2 .
Note: x n
n
f c x f
i
i
i1
i1
2
1
2
x2 1 dx lim
n→ 1
n
1
n
i1
i
n
f ci xi i1
n
f
2
1
i
n
2
1n
1
1 n n
n
1n
1
1 3i
n
2 1 3
; → 0 as n → n
n
3
n
2
3 n
3i
3 1 ni1
n
3 n
6i 9i2
3 1 2 2
n i1
n
n
3
18 nn 1 27 nn 12n 1
3n 2
2n
n
n
2
n
6
2
n →
i2
2
i1
3x 2 2dx lim
2i
10
10
3
1
3
2n 6n2
3
i1
15 1
1 n 2
1
1
3
10
1
3
1
2 n
i 3
i 2 1
2 2 2
n i1
n i1
n
6
n n
3
2n 6n2
8. y 3x 2 2 on 1, 2 . Note: x n
21 1
, → 0 as n → n
n
27n 1 27 n 12n 1
n
2
n2
15 27
n 1 27 n 12n 1
n
2
n2
15 27 27 15
n
9. lim
→0 i1
5
3ci 10 xi on the interval 1, 5 .
1
3x 10 dx
n
10. lim
→0 i1
4
6ci4 ci2 xi on the interval 0, 4 .
0
6x4 x2 dx
Section 4.3
3
n
11. lim
→0 i1
ci2 4 xi n
x2 4 dx
12. lim
0
2
14.
3 dx
2
4 x2 dx
18.
3
1
3
dx
x2
4 x dx
16.
sin x dx
20.
4
19.
2
x2 dx
0
1
dx
1 x 1
2
2
21.
15.
1
2
4
4 2x dx
0
0
3
xi ci2
on the interval 1, 3 .
5
17.
→0 i1
on the interval 0, 3 .
13.
Riemann Sums and Definite Integrals
4
0
tan x dx
0
2
y 3 dy
22.
0
y 22 dy
23. Rectangle
0
A bh 34
3
A
4 dx 12
0
y
5
3
Rectangle
2
1
x
1
25. Triangle
24. Rectangle
a
A
a
4 dx 8a
4
x dx 8
A
0
y
4
5
1
1
A bh 42
2
2
4
A
3
26. Triangle
1
1
A bh 44
2
2
A bh 24a
2
0
x
dx 4
2
y
y
5
3
4
3
2
Triangle
Triangle
Rectangle
2
1
2
1
x
1
−a
x
a
x
2
27. Trapezoid
b1 b2
59
h
2
A
2
2
2x 5 dx 14
0
4
−1
29. Triangle
1
1
A bh 88 32
2
2
A
3
4
28. Triangle
2
A
2
1
1
A bh 21
2
2
1
8
8 xdx 32
A
0
1
1 x dx 1
y
y
y
10
9
1
Triangle
8
6
6
Trapezoid
x
1
4
3
2
x
1
2
x
3
2
4
6
8
10
1
379
380
Chapter 4
Integration
30. Triangle
1
1
A bh 2aa
2
2
a
A
a
32. Semicircle
31. Semicircle
1
A r2
2
1
1
A r 2 32
2
2
3
a x dx a2
A
3
9 x2 dx y
r
9
2
A
r 2 x 2 dx r
y
y
4
a
1 2
r
2
Semicircle
Semicircle
r
Triangle
−a
x
x
a
4
2
2
−r
4
x
r
−r
4
In Exercises 33–40,
2
x dx 4x dx 4
2
33.
dx 2
2
34.
4
36.
2
4
2
4
15 dx 15
2
4
x 8 dx 4
x dx 46 24
2
x3 dx 0
2
4
2
x dx 6
2
39.
2
2
4
37.
4
x dx 6,
4
4
35.
4
x3 dx 60,
4
4
x dx 8
2
dx 6 82 10
38.
1 3
1
x 3x 2 dx 2
2
4
4
x3 dx 3
2
4
x dx 2
2
4
x3 4dx 2
2
dx 152 30
2
4
x3 dx 4
2
4
dx
2
40.
7
0
0
0
(b)
5
f x dx 2
2
42. (a)
5
(b)
(c)
f x dx 310 30
f x dx 2
gx dx
g x f x dx 2
2gx dx 2
6
gx dx 2 10 12
2
f x dx gx dx 22 4
f x dx 310 30
1
0
1
1
f x dx f x dx 0 5 5
0
0
f x dx 1
f x dx 5 5 10
1
1
(d)
1
1
(c)
f x dx 51 5
3
0
f x dx
2
6
3f x dx 3
6
1
2
6
5f x dx 5
1
(b)
6
2
44. (a)
2
6
2
6
f x dx 0
0
6
10 2 8
6
2
6
2
(d)
f x dx 1 1
3
3
f x gx dx (d)
0
6
(c)
6
f x dx 6
5
3f x dx 3
0
(b)
f x dx 4 1 3
3
3
5
43. (a)
0
3
f x dx 0
5
(d)
x3 dx
2
6
f x dx 6
0
5
(c)
3
f x dx 10
4
x dx 2
3
f x dx 0
5
5
f x dx 6
f x dx 10 3 13
4
dx 2
62 26 60 36
7
f x dx 4
6 2x x3dx 6
1
60 36 22 16
2
41. (a)
dx 60 42 68
2
3f x dx 3
1
f x dx 30 0
1
3f x dx 3
0
f x dx 35 15
Section 4.3
45. Lower estimate: 24 12 4 20 362 48
Riemann Sums and Definite Integrals
381
46. (a) 6 8 30(2) 64 left endpoint estimate
Upper estimate: 32 24 12 4 202 88
(b) 8 30 802 236 right endpoint estimate
(c) 0 18 502 136 midpoint estimate
If f is increasing, then (a) is below the actual value and (b)
is above.
47. (a) Quarter circle below x-axis: 14 r 2 14 22 (b) Triangle: 12 bh 12 42 4
(c) Triangle Semicircle below x-axis: 12 21 12 22 1 2
(d) Sum of parts (b) and (c): 4 1 2 3 2
(e) Sum of absolute values of (b) and (c): 4 1 2 5 2
(f) Answer to (d) plus 210 20: 3 2 20 23 2
1
48. (a)
1
f x dx 0
f xdx 0
1
2
y
2
(3, 2) (4, 2)
4
(b)
(11, 1)
3f (x) dx 32 6
1
3
(c)
0
1 1
1
1
f (x) dx 22 2 22 5
2 2
2
2
11
(d)
x
−2
7
f (x) dx 5
−1
2
4
8
12
(0, 1)
(8, 2)
−2
1 1
1
42 3
2 2
2
11
(e)
1
1
f (x) dx 2 2 2 4 2
2
2
0
10
(f)
f (x) dx 2 4 2
4
5
49. (a)
0
0
5
(c)
8
0
f x dx f x dx 2
4,x,
3
5
2 dx 4 10 14
(b)
2
0
5
f x 2 dx f x dx 4
Let u x 2.
0
5
5
5
50. f x 5
f x 2 dx f x dx 24 8
f even
(d)
5
0
f x dx 0
f odd
y
x < 4
x ≥ 4
(8, 8)
8
1
f x dx 44 44 44 40
2
4
x
4
51. The left endpoint approximation
will be greater than the actual area: >
8
52. The right endpoint approximation
will be less than the actual area: <
53. f x 1
x4
is not integrable on the interval
3, 5 because f has a
discontinuity at x 4.
382
Chapter 4
Integration
54. f x x x is integrable on 1, 1 , but is not continuous on 1, 1 .
There is discontinuity at x 0. To see that
y
2
1
1
x
dx
x
1
55.
56.
y
4
3
3
2
2
1
2
3
x
x
4
1
4
3
2
1
1
2
x
3
2
2
1 2 3 4 5 6 7 8 9
1
0
(c) Area 27.
1
2 sin x dx 12 1
2
3
x3 x dx
0
n
4
8
12
16
20
Ln
3.6830
3.9956
4.0707
4.1016
4.1177
Mn
4.3082
4.2076
4.1838
4.1740
4.1690
Rn
3.6830
3.9956
4.0707
4.1016
4.1177
4
8
12
16
20
Ln
7.9224
7.0855
6.8062
6.6662
6.5822
Mn
6.2485
6.2470
6.2460
6.2457
6.2455
Rn
4.5474
5.3980
5.6812
5.8225
5.9072
3
60.
0
5
dx
x2 1
n
1
9
8
7
6
5
4
3
2
1
2
1
3
4
y
58.
y
1
2
1
2
(b) A 43 square units
(a) A 5 square units
59.
−2
1
1
(d)
1
y
4
57.
x
−2
is integrable, sketch a graph of the region bounded by f x x x and the x-axis for
1 ≤ x ≤ 1. You see that the integral equals 0.
x
2
Section 4.3
Riemann Sums and Definite Integrals
383
2
61.
sin2 x dx
0
n
4
8
12
16
20
Ln
0.5890
0.6872
0.7199
0.7363
0.7461
Mn
0.7854
0.7854
0.7854
0.7854
0.7854
Rn
0.9817
0.8836
0.8508
0.8345
0.8247
4
8
12
16
20
Ln
2.8186
2.9985
3.0434
3.0631
3.0740
Mn
3.1784
3.1277
3.1185
3.1152
3.1138
Rn
3.1361
3.1573
3.1493
3.1425
3.1375
3
62.
x sin x dx
0
n
63. True
64. False
1
1
xx dx 0
65. True
1
x dx
0
x dx
0
68. True. The limits of integration are
the same.
67. False
66. True
2
x dx 2
0
69. f x x2 3x, 0, 8
x0 0, x1 1, x2 3, x3 7, x4 8
x1 1, x2 2, x3 4, x4 1
c1 1, c2 2, c3 5, c4 8
4
f c x f 1 x
i
1
f 2 x2 f 5 x3 f 8 x4
i1
41 102 404 881 272
70. f x sin x, 0, 2
x0 0, x1 x1 c1 , x , x3 , x4 2
4 2
3
2
, x2 , x3 , x4 4
12
3
2
3
, c , c3 ,c 6 2
3
3 4
2
2
3
x1 f
x2 f
x3 f
x4
6
3
3
2
4
f c x f
i
i
i1
1
2
3
4
2
3
12
2
2
1 0.708
3
384
Chapter 4
71. x Integration
ba
ba
, ci a ix a i
n
n
b
x dx lim
n
f c x
i
0 i1
a
n→
n
i1
ba
n
lim
n→
a i1
ba
n
b n a an lim
ab a b n a
n→ 2
i
2 n
i1
nn 1
2
n1
2
b a2
2
ba
2
b a a ba
n
2
ab a n
b n a
lim
n→ 72. x ba
n
a i
lim
b aa b b 2 a 2
2
2
ba
ba
, ci a ix a i
n
n
b
x 2 dx lim
n
f c x
0 i1
a
i
lim
a i
lim
n
n→ i1
n→ lim
n→ ba
n
n
na
2
2
a2 i1
a b a lim
n→
ba
n
ba
n
2
ba
n
2aib a
ba
i2
n
n
2
2ab a nn 1
ba
n
2
n
2
nn 12n 1
6
ab a2n 1
b a3 n 12n 1
n
6
n2
1
a2b a ab a2 b a3
3
1
b3 a3
3
1,
73. f x 0,
x is rational
x is irrational
is not integrable on the interval 0, 1 . As → 0,
f ci 1 or f ci 0 in each subinterval since there
are an infinite number of both rational and irrational
numbers in any interval, no matter how small.
0, x 0
74. f x 1 ,
x 0 < x ≤ 1
The limit
y
1
2
f(x) = x
1
n
lim
f c x
0 i1
i
i
x
1
does not exist. This does not
contradict Theorem 4.4 because
f is not continuous on 0, 1 .
2
Section 4.4
75. The function f is nonnegative between x 1 and x 1.
The Fundamental Theorem of Calculus
385
2
76. To find 0 x dx, use a geometric approach.
y
y
2
3
f(x) = 1 − x 2
2
1
x
−2
1
2
−1
x
−1
1
3
−1
−2
Hence,
2
Thus,
b
2
1 x 2 dx
a
x dx 12 1 1.
0
is a maximum for a 1 and b 1.
77. Let f x x2, 0 ≤ x ≤ 1, and xi 1 n. The appropriate Riemann Sum is
n
i1
i1
lim
n→ ni 1n n1
2
n
f ci xi n
3
i 2.
i1
1 2
1
1 22 32 . . . n2 lim 3
n→ n
n3
lim
n→ Section 4.4
1. f x 0
n2n 1n 1
6
1
2n2 3n 1
1
1
1
lim
n→ 3
6n2
2n 6n2
3
The Fundamental Theorem of Calculus
4
x2 1
2. f x cos x
5
4
dx is positive.
x2 1
−5
2
cos x dx 0
−2
−2
5
3. f x x x2 1
4. f x x 2 x
2
5
2
2
x x2 1 dx 0
−5
5
2
x 2 x dx is negative.
−2
−5
2x dx x2
x 2 dx 0
0
0
7.
1
5
8.
6.
2 2x
0
x2
1
0
1
t3 2t
3
12 2 52
752 20 6 8 392
t2 2 dt 7
1
1
5
2
13 2 31 2 103
3 dv 3v
2
3
3v 4 v v2 4v
2
2
1
9.
7
101
2
−5
1
1
5.
0
0
5
2
37 32 15
386
Chapter 4
Integration
27 3
10.
5x 2
4x
2
3x 2 5x 4dx x3 1
3
45
5
12 1 4
2
2
1
38
2t 12 dt t3 9t dt 1
11.
1
0
1
4
3
v1 3 dv 1
1
34v 3
4 3
3
34t
t 2 dt 2 t t dt t1 3 t2 3 dt 1
x x2
dx 3
2
2 x
2t
1
1
8
3
3
t5
5
0
1
3
43t
3 2
1
2
0
2
t5
5
1
2
82 2
1 1 2
1
3 2 3
18
t15t20 6t
2
2
0
0
2 2
16 2
20 12 15
15
27
34 53 20
1
x2 3 x5 3 dx
8
1 3 53 3 8
x x
2 5
8
1
3
8
1
80 24 15x
x5
3
8
3
3 2x dx 0
8
2x
1
0
3
8
2t1 2 t3 2 dt 3 2
2x 3 dx 23 4 32
4 4 4 1
2
1 x2 2 3
x
3 2
3
0
4
2
22x1
x x1 2 dx 4 3
23
2
34 2 34 2 4
1
34t
4u1
4
0
1
0
3 2
2
1
23.
23u
3 4 3 3 0
x1 2 dx 0
22.
4 3
3
1
0
21.
3
4
8
2
2
20.
12 1 2 21 2
1
1
x x
dx 3
3
0
1
u1 2 2u1 2 du 1
19.
1
4
2
dx x
1
3
2
8
18.
3
1
4
21
3
3
0
1
2
3
1
14 29 14 29 0
2
2
2t2 t
3
2 2 3 1 2
u2
du u
1
17.
1
u u1 du u2 u1
2
16.
1
1
15.
9
t2
2
4
3
3
1 dx x
x2
x
1
14.
14t
2
13.
4
0
1
12.
3t
4t2 4t 1 dt 3 2
0
1
32
4569
39 144 80
80
80
2x 3 dx split up the integral at the zero x 3 2
3x x2
x2 3x
3
3 2
3
2
92 49 0 9 9 94 29 292 49 29
Section 4.4
4
24.
1
3
6 x dx
x2
2
3
x2 92 21 24 8 18 29
2 3
1
6x 0
2
x 2 4 dx 8
4
27.
x3
3
x3
x2 4x 3 dx 3
x3 2x
x3 2x
13 2 3 9 18 9 13 2 3 643 32 12 9 18 9
4 4
4
0 04
3
3 3
3
3x
2
d 2t cos t dt t 2 sin t
4
3
x x2 dx 3 x x dx 1
0
2
4
3
3 1
2
0
3
3
2
0
x3 2x
3x
3
3
2
1
3x
4
3
2
6
4
29.
6
0 6
sec2 x dx tan x
6
4
2
2 1 2 1 2 2
4
1 2
1 2
1
1
6
34. A 2
3x1 2 x3 2 dx 2x3 2 x5
5
0
3
2
0
1
1 x 4 dx x x5
5
1
x x
10 2x
5
3
0
12 3
5
3
3
2 3
3
3
3
42 42 0
2
x2 x3 3
0
3
2
4
0
1
1 0 1 2 0
4 sec tan d 4 sec 35. A split up the integral at the
zeros x 1, 3
x2 4x 3 dx
3
0
4
1
2 csc2 x dx 2x cot x
33. A x2 4x 3 dx 2
2
1 sin2 d cos2 3
3
8
8
9 12 8
3
3
4
32.
3 4x
0
2
31.
0
0
30.
2
1 sin x dx x cos x
4
28.
1
0
x 2 4 dx
23
3
x2 4x 3 dx 0
3
2
4x 26.
4
3
4 x 2 dx 0
9 13
2
2
4 16 18 3
3 x 3 dx
3
4
x dx 1
387
4
3 x 3 dx 1
25.
3
3 1x 31 dx The Fundamental Theorem of Calculus
1
1
8
5
388
Chapter 4
1
1
dx x2
x
x sin x dx 2
36. A 1
38. A Integration
2
1
1
1
2
2
1
0
2 cos x
x2
0
37. A 2
2
cos x dx sin x
0
1
0
2
2 4
2
2
2
39. Since y ≥ 0 on 0, 2 ,
2
A
2
3x2 1 dx x3 x
0
8 2 10.
0
40. Since y ≥ 0 on 0. 8 ,
8
8
3
1 x1 3 dx x x 4
4
0
Area 3
0
3
8 16 20.
4
41. Since y ≥ 0 on 0, 2 ,
42. Since y ≥ 0 on 0, 3 ,
2
A
x3 x dx 0
2
43.
0
x 2 x dx x2 2
x4 x2
4
2
4x3
3
2 2
0
2
0
3
4 2 6.
A
0
2
8 2
3
3
44.
1
9
9
dx 2
x3
2x
3
1
32x
2
9
2
c3
34 2
c2 c
3
c3 34 2
1
c2 c1
3
9
2
c
64 2
c 1 3
9
1.6510
2
3
2
64 2
3
c1±
64 2
3
c 1 ±
2
c 0.4380 or c 1.7908
4
45.
4
4
2 sec2 x dx 2 tan x
21 21 4
4
4 4 4
f c
2 sec2 c 8
sec2 c 4
sec c ±
3
46.
3
cos x dx sin x
3
3
3 3 f c
cos c 3
3 3
2
c ± 0.5971
2
c ± arcsec
2
± arccos
± 0.4817
2
x3
3
1 9
4
2 2
f c3 1 4
68 2
f c2 0 3
3x x2 dx 3
3
0
9
.
2
Section 4.4
47.
2
1
2 2
2
Average value 2
1
1
4x x3
4
3
4 x2 dx 2
1
4
The Fundamental Theorem of Calculus
8 3 8 3 3
8
8
8
(−
2
3
, 8
3
3
(
(
5
2
3
3
, 8
3
(
8
3
−3
4 x2 48.
1
31
3
1
3
8
8
2 3
when x2 4 or x ±
± 1.155.
3
3
3
4x 2 1
dx 2
x2
3
1 x2dx 2 x 1
1
x
−1
3
49.
1
1
16
2 3
3
3
1
0
Average value sin x 4x2 1 16
⇒ x±
x2
3
1
2 0
2
cos x dx 0
Average value cos x sin x dx 0
16
Average value 3
50.
389
1
cos x
0
2
2
(0.690, π2 (
2
2
sin x
2
0
2
2
(2.451, π2 (
−
2
3
2
−1
x 0.690, 2.451
3
2
51. The distance traveled is 0 vt dt. The area under
the curve from 0 ≤ t ≤ 8 is approximately
18 squares 30 540 ft.
8
1.5
(0.881, π2 (
2
0
2.71
−0.5
x 0.881
52. The distance traveled is 0 vt dt. The area under
the curve from 0 ≤ t ≤ 5 is approximately
29 squares 5 145 ft.
5
53. If f is continuous on a, b and F x f x on a, b ,
b
then
f x dx Fb Fa.
a
54. (a)
7
7
f x dx Sum of the areas
(b) Average value 1
f x dx
71
1
8 4
6 3
(c) A 8 62 20
A1 A2 A3 A4
1
1
1
3 1 1 2 2 1 31
2
2
2
Average value 8
20 10
6
3
y
7
y
6
5
4
A1
4
3
A2
A3
3
A4
2
2
1
1
x
1
2
3
4
5
6
7
x
1
2
3
4
5
6
7
0
6
2
55.
f x dx area of region A 1.5
56.
6
f x dx area or region B 2
0
3.5 1.5 5.0
2
f x dx 0
f x dx
390
Chapter 4
6
57.
2
0
f x dx 1.5 5.0 6.5
6
2 f x dx 0
2
58.
2
6
59.
6
f x dx f x dx 0
Integration
2
2f x dx 2
0
6
2 dx 0
60. Average value f x dx
f x dx 21.5 3.0
0
0
1
6
6
0
1
f x dx 3.5 0.5833
6
12 3.5 15.5
61. (a) F x k sec2 x
(b)
F 0 k 500
1
30
3
500 sec2 x dx 0
F x 500 sec2 x
1500
3
1500
tan x
0
3 0
826.99 newtons
827 newtons
62.
63.
kR r dr 0.1729t 0.1522t2 0.0374t3 dt 1
R0
1
50
64. (a)
R
2
2
0
3 R
k 2
r
Rr
R
3
0
2
2kR
3
5
0
1
0.08645t2 0.05073t3 0.00935t4
5
(b)
1
0
0
0.5318 liter
10
24
0
−1
24
0
The area above the x-axis equals the area below the
x-axis. Thus, the average value is zero.
65. (a) v 8.61
(b)
The average value of S appears to be g.
104t3 0.0782t2 0.208t 0.0952
90
−10
70
−10
60
(c)
5
vt dt 0
8.61
104t 4
4
66. (a) histogram
0.0782t3 0.208t 2
0.0952t
3
2
60
0
2476 meters
N
18
16
14
12
10
8
6
4
2
t
1 2 3 4 5 6 7 8 9
(b) 6 7 9 12 15 14 11 7 2 60 8360 4980 customers
—CONTINUED—
Section 4.4
The Fundamental Theorem of Calculus
66. —CONTINUED—
(c) Using a graphing utility, you obtain
Nt 0.084175t3 0.63492t2 0.79052 4.10317.
(d)
16
−2
10
−2
9
(e)
Nt dt 85.162
0
The estimated number of customers is 85.16260 5110.
7
(f) Between 3 P.M. and 7 P.M., the number of customers is approximately
Nt dt 60 50.2860 3017.
3
Hence, 3017 240 12.6 per minute.
x
67. Fx t 5 dt 0
2 5t
t2
F2 4
52 8
2
F5 25
25
55 2
2
F8 64
58 8
2
x
68. Fx x
0
x2
5x
2
x
t3 2t 2 dt 4 t
4 x
x4
x 2 2x 4
4
2
t4
2
x4
2
2t
2
2x 4 4 4
Note: F2 t
2
F2 4 4 4 4 0
3
2
F5 625
25 10 4 167.25
4
F8 84
64 16 4 1068
4
x
69. Fx 10
2 dv 1 v
x
10v2 dv 1
10
1
10 10 1 x
x
10
v
x
1
2t 2 dt 0
x
70. Fx 2
3 dt 2 t
F2 1 1
0
4 4
x
2t3 dt 2
F2 10
12 5
F5 1
1
21
0.21
25 4
100
F5 10
45 8
F8 1
1
15
64 4
64
F8 10
78 354
1
t2
x
2
1
1
x2 4
391
392
Chapter 4
Integration
x
71. Fx cos d sin 1
x
sin x sin 1
1
F2 sin 2 sin 1 0.0678
F5 sin 5 sin 1 1.8004
F8 sin 8 sin 1 0.1479
x
72. Fx x
sin d cos 0
cos x cos 0 1 cos x
0
F2 1 cos 2 1.4161
F5 1 cos 5 0.7163
F8 1 cos 8 1.1455
x
x
73. gx 74. gx f t dt
0
f tdt
0
(a) g0 f t dt 0
2
2
g2 f t dt 4 2 1 7
g4 f t dt 7 2 9
g6 f t dt 9 1 8
f t dt 8 2 4 2
0
8
0
8
g8 f t dt 1244 8
0
6
0
6
g6 f t dt 1224 4
0
4
0
4
g4 f t dt 0
0
0
g2 0
0
(a) g0 g8 f t dt 8 3 5
f t dt 2 6 4
0
0
(b) g increasing on 0, 4 and decreasing on 4, 8
(b) g decreasing on 0, 4 and increasing on 4, 8
(c) g is a maximum of 9 at x 4.
(c) g is a minimum of 8 at x 4.
(d)
y
(d)
y
10
4
8
2
6
−2
4
x
2
−2
4
8
10
−4
2
−6
x
2
4
6
x
75. (a)
t 2 dt 0
(b)
77. (a)
1
x2 2x
2
x
tt2 1 dt t3 t dt 0
14t
4
1
t2
2
x
0
1
1
x2
x4 x2 x2 2
4
2
4
d 1 4 1 2
x x x3 x xx2 1
dx 4
2
x
3
t dt 8
(b)
0
0
(b)
x
t2
2t
2
d 1 2
x 2x x 2
dx 2
x
76. (a)
−8
8
d 3 4
x
dx 4
3
34t x
4 3
8
3
x 4
4
3
3
16 x 4
4
12 x1 3 3 x
x
3
12
78. (a)
t dt 4
(b)
d 2 3
x
dx 3
2
23t x
3 2
4
2
x3
3
16
x1
3
2
2
x
16 2 3
x
3
3
2
8
Section 4.4
x
79. (a)
x 4
(b)
x
x 4
tan x 1
80. (a)
(b)
sec t tan t dt sec t
2
82. F x t 2 2t dt
F x x2 2x
x
4
F x d
sec x 2 sec x tan x
dx
1
83. F x F x x2
x 1
t 4 1 dt
x4 1
x
86. Fx t cos t dt
sec3 t dt
0
F x sec3 x
F x x cos x
1
2
0
F x 4 x
x
t2
dt
2
t
1
1
x
85. F x t dt
sec x 2
3
x
x
84. F x x
3
d
tan x 1 sec2 x
dx
81. F x x
sec2 t dt tan t
The Fundamental Theorem of Calculus
x2
87. Fx 4t 1 dt
Alternate solution:
x
x2
2t2 t
x2
Fx 4t 1 dt
x
x
0
2x 22 x 2 2x2 x
8x 10
x2
4t 1 dt x
0
x
F x 8
4t 1 dt
x2
4t 1 dt 0
4t 1 dt
0
F x 4x 1 4x 2 1 8
x
88. Fx t 3 dt x
t4 4 x
x
0
Alternate solution:
x
Fx F x 0
t3 dt
x
0
x
x
t3 dt 0
x
t3 dt
x
t3 dt 0
t3 dt
0
F x x31 x3 0
sin x
89. Fx t dt 0
3t 2
3 2
sin x
0
2
sin x3
3
2
Alternate solution:
2
x
t3 dt 2
t2
2
x2
2
1
2t2
t dt
0
F x 90. Fx sin x
Fx F x sin x1 2 cos x cos x sin x
x2
2
sin x
d
sin x dx
sin xcos x
1 1
⇒ F x 2x5
2x4
8
Alternate solution: F x x 232x 2x5
x3
91. Fx x
sin t 2 dt
0
F x sinx32
92. Fx 2
sin 2 d
0
3x2 3x2 sin x 6
F x sin x 22 2x 2x sin x 4
393
394
Chapter 4
Integration
x
93. gx y
f t dt
0
2
1
1
g0 0, g1 , g2 1, g3 , g4 0
2
2
f
1
g
x
1
g has a relative maximum at x 2.
2
3
4
−1
−2
94. (a)
x
1
2
3
4
5
6
7
8
9
10
gx
1
2
0
2
4
6
3
0
3
6
(b)
y
6
4
2
(c) Minimum of g at 6, 6.
x
−2
(d) Minimum at 10, 6. Relative maximum at 2, 2.
(e) On 6, 10 , g increases at a rate of
2
4
6
10
12
−4
−6
12
3.
4
(f) Zeros of g: x 3, x 8
45t x
95. (a) Cx 5000 25 3
0
x
5000 25 3
5 4
4
5000 25 (b)
96. (a) gt 4 t1 4 dt
lim gt 4
t→ 0
12 5
x
5
4
t2
1000125 12x
Horizontal asymptote: y 4
5 4
x
(b) Ax C1 1000125 121 $137,000
4
1
C5 1000125 1255 4 $214,721
4t C10 1000125 12105 4 $338,394
4
t
4
dt
t2
x
1
4x 4x2 8x 4 4x 12
x
x
lim Ax lim 4x x→ 4
8
x
x→ 4
8 08
x
The graph of Ax does not have a horizontal asymptote.
97. xt t 3 6t 2 9t 2
x t 3t 2 12t 9
3t 2 4t 3
3t 3t 1
5
Total distance x t dt
0
5
3 t 3t 1 dt
0
1
3
t 2 4t 3 dt 3
3
0
4 4 20
28 units
1
5
t 2 4t 3 dt 3
3
t 2 4t 3 dt
Section 4.4
The Fundamental Theorem of Calculus
4
98. xt t 1t 32 t 3 7t 2 15t 9
x t dt
99. Total distance 1
x t 14t 15
3t2
4
Using a graphing utility,
1
4
5
Total distance x t dt 27.37 units.
0
vt dt
1
2t1
dt
t
1
4
2
1
22 1 2 units
100. P 2
2
sin d 0
2
cos 2
0
2
2
0 1 63.7%
102. True
101. True
103. The function f x x2 is not continuous on 1, 1 .
1
0
1
x2 dx 1
1
x2 dx 104. Let Ft be an antiderivative of f t. Then,
vx
x2 dx
0
Each of these integrals is infinite. f x has a nonremovable discontinuity at x 0.
x2
ux
d
dx
vx
ux
vx
f t dt Ft
f t dt ux
Fvx Fux
d
Fvx Fux
dx
F vxv x F uxu x
f vxv x f uxu x.
1 x
105. f x 0
1
dt t2 1
x
1
2 1 dt
t
0
By the Second Fundamental Theorem of Calculus, we have
1
1
1
2
1 x2 1 x2
x 1
f x 1
1
0.
1 x2 x2 1
Since f x 0, f x must be constant.
x
106. Gx s
f t dt ds
s
0
0
0
(a) G0 0
f t dt ds 0
f x 0
Gx Fs ds
0
0
f 0 f t dt.
x
f t dt
0
(d) G 0 0
(b) Let Fs s
0
0
x
(c) G x x
s
s
s
f t dt 0
G x Fx x
x
f t dt
0
0
G 0 0
0
f t dt 0
395
396
Chapter 4
Section 4.5
Integration
Integration by Substitution
f gxgx dx
1.
2.
3.
4.
5.
6.
7.
u gx
du gx dx
5x2 1210x dx
5x2 1
10x dx
x2x3 1 dx
x3 1
3x2 dx
x2 1
2x dx
sec 2x tan 2x dx
2x
2 dx
tan2 x sec2 x dx
tan x
sec2 x dx
cos x
dx
sin2 x
sin x
cos x dx
x
x2 1
1 2x42 dx Check:
8.
x2 94
C
4
4x2 93
d x2 94
C 2x x2 932x
dx
4
4
9 x21 22x dx d 2
9 x23
dx 3
2
9 x23 2
C 9 x23 2 C
3 2
3
2
C 2
3
3
29 x21 22x 9 x22x
3
1 2x21 34x dx 1 2x24 3 C
4
d 3
1 2x24
dx 4
x3x4 32 dx Check:
12.
Check:
11.
x2 932x dx Check:
10.
1 2x5
C
5
d 1 2x5
C 21 2x4
dx
5
Check:
9.
dx
C 3
4
4
31 2x21 34x 1 2x21 34x
1
1 x4 33
x4 33
x4 324x3 dx C
C
4
4
3
12
3x4 32 3
d x4 33
C 4x x4 32x3
dx
12
12
x2x3 54 dx Check:
3
1
1 x3 55
x3 55
C
C
x3 543x2 dx 3
3
5
15
5x3 543x2
d x3 55
C x3 54x2
dx
15
15
Section 4.5
13.
16.
18.
2
C C
1
6
1
C t4 53
6
C 15
8
4
3 1 x2
31 x21 32x 5x1 x21 3 5x
2
C 2
9
C
1 x24 3
15
C 1 x24 3 C
4 3
8
1
1 u3 23
u3 21 23u2 du 3
3
3 2
d 2u3 23
du
9
2
3
3
2
2 t4 51 24t3 t4 51 2 t3
5
5
1 x21 32x dx 2
2
t2 23 2
C
3
3 2t2 21 22t
t2 21 2t
3
C 2
d
15
1 x24
dx
8
u2u3 2 du 2
d 1 4
t 53
dt 6
1 4
1 t4 5 3
t 51 24t3 dt 4
4
3 2
5x1 x21 3 dx 2
C
2u3 23
9
2
C
3
2 u3 21 23u2 u3 21 2 u2
x
1
1 1 x22
1
1 x232x dx C
C
2 3 dx 1 x 2
2
2
41 x22
d
1
x
1
C 21 x232x dx 41 x22
4
1 x23
x3
1
1
1
dx 1 x424x3 dx 1 x41 C C
1 x42
4
4
41 x4
d
1
1
x3
C 1 x424x3 dx 41 x 4
4
1 x42
1
1 1 x31
1
x2
dx 1 x323x2 dx C
C
3 2
1 x 3
3
1
31 x3
Check:
22.
1 2
1 t2 23
t 21 22t dt 2
2
3 2
Check:
21.
t3t4 5 dt Check:
20.
1
1 4x2 34
4x2 34
C
4x2 338x dx C
8
8
4
32
d t2 23
dt
3
Check:
19.
tt2 2 dt Check:
d 4x2 34
44x2 338x
C x4x2 33
dx
32
32
Check:
17.
x4x2 33 dx Check:
5x3 143x2
d x3 15
C x2x3 14
dx
15
15
Check:
15.
1
1 x3 15
x3 15
x3 143x2 dx C
C
3
3
5
15
x2x3 14 dx Check:
14.
Integration by Substitution
d
1
1
x2
C 11 x323x2 3
dx 31 x 3
1 x32
1
1 16 x31
1
x2
dx 16 x323x2 dx C
C
3
2
16 x 3
3
1
316 x3
Check:
d
1
1
x2
C 116 x323x2 dx 316 x3
3
16 x32
397
398
23.
Chapter 4
x
x3
1
t
1
3
1
dt t2
x2 1
dx 3x2
31.
1
t
3
C
3
2
C
x3
1 x4
1
1 1 t
dt t2
4
1
1
C 4 1 4
t
1 x4
4
C
1
1
1
2 1
t2
t
t
3
1
x3 1 x1
x3
1
3x4 1
C
C
x2 x2 dx C 9
3
9 1
3
9x
9x
1
2x
1
1
1 x1 2
dx x 1 2 dx C 2x C
2
2 1 2
2x
d
1
1
2x C 2x1 22 dx
2
2x
dx 1 1 2
1 x1 2
x
dx C x C
2
2 1 2
1
d
x C dx
2x
x2 3x 7
dx x
Check:
1
1
2
1 x41 24x3 2
Check:
30.
C 1 x2 C
1
1 2x1 2
1
dx 2x1 2 2 dx C 2x C
2
2
1 2
2x
4
Check:
29.
2
d 1 3 1 1
1
1
x x C x2 x2 x2 dx 3
9
9
3x2
Alternate Solution:
28.
d
1 1 t
dt
4
Check:
27.
d 1 x4
1
C dx
2
2
Check:
26.
1
1 1 x41
1 x41 24x3 dx 4
4
1 2
dx 1 x4
1
1 1 x21
1 x21 22x dx 2
2
1 2
d
1
x
1 x21 2 C 1 x21 22x 1 x2
dx
2
Check:
25.
dx 1 x2
Check:
24.
Integration
d 2 5
x
dx 5
t 2t2
t
dt 2
2
2
x3 2 3x1 2 7x1 2 dx x5 2 2x3 2 14x1 2 C xx2 5x 35 C
5
5
2x3 2 14x1
2
C x2 3x 7
x
2
4
2
t1 2 2t3 2 dt t3 2 t 5 2 C t 3 25 6t C
3
5
15
Check:
d 23
t
dt 3
2
dt t
Check:
d 14
2
t t2 C t 3 2t t 2 t dt 4
t
t2 t 2
4
t5 2 C t1
5
2
2t3
2
1
t3 2t dt t 4 t2 C
4
t 2t2
t
Section 4.5
32.
1
t3
2 dt 3
4t
Check:
33.
1 3 1 2
1 t4
1 t1
1
1
t t
dt C t4 C
3
4
3 4
4 1
12
4t
d 1 4
1
1
1
t C t3 2
dt 12
4t
3
4t
9 yy dy Check:
34.
9y1 2 y3 2 dy 9
2 3
y
3
2
2
y5
5
d 2 32
d
2
y 15 y C 6y3 2 y5
dy 5
dy
5
2
2
35. y d
2
d 4 y2
14 y3 2 C 2 4y2 y7
dy 7
dy
7
4x 4x
16 x2
dx
2
C y3 215 y C
5
C 9y1
2
2 y8 y3 2 dy 2 8y y5 2 dy 2 4y2 y7
7
Check:
2
2
x2
16 x21
2
2
1 2
2
y3
4 y2
14 y3 2 C
7
36. y C
x1
dx
x2 2x 32
9 yy
2
C 16 y 2 y5 2 2 y8 y3 2
2x2 416 x2 C
37. y 2
C
4 x dx 2 16 x21 22x dx
4
Integration by Substitution
10x2
1 x3
dx
10
1 x31 23x2 dx
3
10 1 x31
3
1 2
20
1 x3 C
3
38. y 2
C
x4
x2 8x 1
dx
1
x2 2x 322x 2 dx
2
1
x2 8x 11 22x 8 dx
2
1 x2 2x 31
C
2
1
1 x2 8x 11
2
1 2
2
x2
y
(b)
3
x
2
C
dy
x4 x2, 2, 2
dx
y
−2
2
x2 8x 1 C
1
C
2x 3
39. (a)
x4 x2 dx 1
2
1
4 x21 22x dx
2
1
2
34 x23 2 C 34 x23 2 C
−1
1
2, 2: 2 4 223 2 C ⇒ C 2
3
1
y 4 x23
3
2
2
2
−2
2
−1
399
400
Chapter 4
40. (a)
Integration
(b)
y
dy
x 2x3 12, 1, 0
dx
2
y
x
−2
2
−2
x2x3 12dx 1
x3 123x2 dx u x3 1
3
1 x3 13
1
C x3 13 C
3
3
9
0C
1
y x3 13
9
41. (a)
42. (a)
y
y
4
3
x
−4
4
x
−3
3
−4
(b)
−3
dy
x cos x2, 0, 1
dx
y
x cos x2 dx (b)
1
cosx22x dx
2
dy
2 sec2x tan2x, 0, 1
dx
y
1
sinx2 C
2
2 sec2x tan2x dx u 2x
sec2x C
1
0, 1: 1 sin0 C ⇒ C 1
2
1
y sinx2 1
2
1 sec0 C ⇒ C 0
y sec2x
5
−6
6
−3
43.
45.
47.
49.
sin x dx cos x C
sin 2x dx 1
1
sin 2x2x dx cos 2x C
2
2
1
1
1
1
1
cos d cos 2 d sin C
2
sin 2x cos 2x dx 46.
48.
4x3 sin x4 dx cos 6x dx 1
2
sin x44x3 dx cos x4 C
1
1
cos 6x6 dx sin 6x C
6
6
x sin x2 dx 1
1
1 sin 2x2
C sin2 2x C
sin 2x2 cos 2x dx 2
2
2
4
sin 2x cos 2x dx sin 2x cos 2x dx 44.
1
1
sin x22x dx cos x2 C
2
2
OR
1
1 cos 2x2
1
cos 2x2 sin 2x dx C1 cos2 2x C1
2
2
2
4
2 sin 2x cos 2x dx 1
2
1
sin 4x dx cos 4x C2
8
OR
Section 4.5
50.
51.
tan5 x
1
C tan5 x C
5
5
53.
sec1 x tan1 x dx tan4 x sec2 x dx 55.
tan x sec2 x dx 2
tan x3 2
C tan x3 2 C
3 2
3
1
1
1
1
cot x2
C
C tan2 x C sec2 x 1 C sec2 x C1
2
2 cot2 x
2
2
2
sin x
cos x2
1
1
dx cos x3sin x dx C
C sec2 x C
3
cos x
2
2 cos2 x
2
57. f x cos
56.
x
x
dx 2 sin C
2
2
58. f x csc2 x 1 dx cot x x C
Since f 0 3 2 sin 0 C, C 3. Thus,
f x 2 sin
59. f x sin 4x,
f x f
52.
cot2 x dx csc2
x
x
dx 2 csc2
2
2
1
x
dx 2 cot
C
2
2
sec x tan x dx sec x C
Since f 1 3 1 sec 3 C, C 1. Thus
f x sec x 1.
x
3.
2
3
,
4 4
60. f x sec 22x,
1
cos 4x C
4
1
cos 4
4
4
4
,2
2
1
f x tan2x C
2
C
3
4
f
1
tan 2
2
2
2
1
3
1 C 4
4
1
0 C 2
2
C 1
C2
1
f x cos 4x 1
4
61. f x 2x4x2 102, 2, 10
f x C2
1
tan2x 2
2
62. f x 2x8 x 2, 2, 7
f x 4x2 103
22x2 53
C
C
12
3
f x 28 x232
C
3
f 2 28 53
C 18 C 10 ⇒ C 8
3
f 2 243 2
16
5
C
C7 ⇒ C
3
3
3
f x 28 x23 2 5
3
3
2
f x 2x2 53 8
3
401
sec1 x tan1 x 1 dx sec1 x C
csc2 x
dx cot x3csc2 x dx
cot3 x
54.
Integration by Substitution
402
Chapter 4
Integration
1
1
64. u 2x 1, x u 1, dx du
2
2
63. u x 2, x u 2, dx du
xx 2 dx u 2u du
x2x 1 dx u3 2 2u1 2 du
2
u5
5
2
4
u3 2 C
3
1
1
u 1u du
2
2
1
3
u
4
2
du
1 2
u
1 2 52 2 3
u u
4 5
3
2
C
2u3 2
3u 10 C
15
u3 2
3u 5 C
30
2
x 23
15
1
2x 13
30
2
x 23 23x 4 C
15
1
2x 13 26x 2 C
30
1
2x 13 23x 1 C
15
2
3x 2 10 C
65. u 1 x, x 1 u, dx du
x21 x dx 1 u2u du
u1 2 2u3
2 3
u
3
2
4
u5
5
2
u5 2 du
2
2
u7
7
2
C
2u3 2
35 42u 15u2 C
105
2
1 x3
105
2
1 x3 215x2 12x 8 C
105
35 421 x 151 x2 C
2
66. u 2 x, x 2 u, dx du
x 12 x dx 3 uu du
3u
2u3
1 2
2
du
3 2
u
2
u5
5
2
C
2u3 2
5 u C
5
2
2 x3
5
2
5 2 x C
2
2 x3 2x 3 C
5
2
32x 1 5 C
Section 4.5
Integration by Substitution
403
1
1
67. u 2x 1, x u 1, dx du
2
2
1 2u 1 2 1 1
du
2
u
x2 1
dx 2x 1
1 1
u
8
u2 2u 1 4 du
2
1
u3
8
2
2u1 2 3u1 2 du
1 2 5
u
8 5
2
4
u3 2 6u1
3
u1 2 2
3u 10u 45 C
60
2x 1
60
2
C
32x 12 102x 1 45 C
1
2x 112x2 8x 52 C
60
1
2x 13x2 2x 13 C
15
68. Let u x 4, x u 4, du dx.
2x 1
x 4
dx 2u 4 1
du
u
2u1 2 7u1 2 du
4
u3
3
69. u x 1, x u 1, dx du
x
dx x 1 x 1
u 1
du
u u
1 u1 2 du
2
u1 22u 21 C
3
u 2u1 2 C
2
x 4 2x 4 21 C
3
2
x 42x 13 C
3
u 2u C
x 1 2x 1 C
x 2x 1 1 C
x 2x 1 C1
where C1 1 C.
70. u t 4, t u 4, dt du
3 t 4 dt t
u 4u1 3 du
u4 3 4u1 3 du
3
u7
7
3
3u4
3
C
3u4 3
u 7 C
7
3
t 44
7
3
t 4 7 C
3
t 44 3t 3 C
7
u u 1
14u1 2 C
2
du
u 1 u 1
404
Chapter 4
Integration
71. Let u x2 1, du 2x dx.
1
1
1
2
xx2 13 dx 1
1
x2 132x dx 18x
2
14
1
1
0
72. Let u x3 8, du 3x2 dx.
4
2
x2x3 82 dx 1
3
4
2
x3 823x2 dx 13 x
3
83
3
4
2
1
64 83 8 83 41,472
9
73. Let u x3 1, du 3x2 dx.
2
1
1
3
2x2x3 1 dx 2 2
x3 11 23x2 dx
1
3
2 x3 13
3 2
2 2
4 3
x 13
9
1
2
2
1
8
4
27 22 12 2
9
9
74. Let u 1 x2, du 2x dx.
1
x1 x2 dx 0
1
1
2
1
1 x21 22x dx 1 x23
3
0
1
0
2
0
1 1
3 3
75. Let u 2x 1, du 2 dx.
4
0
1
1
dx 2
2x 1
4
2x 11 22 dx 2x 1
0
4
9 1 2
0
76. Let u 1 2x2, du 4x dx.
2
0
1
x
dx 4
1 2x2
77. Let u 1 x, du 9
1
2
1 2x21 24x dx 0
21 2x 2
1
2
0
3 1
1
2 2
1
dx.
2x
1
dx 2
2
x 1 x 9
1
1 x 2
1
2x
dx 2
1 x
9
1
1
1
1
2
2
78. Let u 4 x2, du 2x dx.
2
3 4 x2 dx x
0
1
2
2
4 x21 32x dx 0
84 x 3
2
2 4 3
0
3
84
8
3
33
44 3 6 4 3.619
2
79. u 2 x, x 2 u, dx du
When x 1, u 1. When x 2, u 0.
2
1
x 12 x dx 0
1
0
2 u 1 u du 1
u3 2 u1 2 du 25u
5 2
2
u3
3
0
2
1
25 3 15
2
4
Section 4.5
Integration by Substitution
1
80. Let u 2x 1, du 2 dx, x u 1.
2
When x 1, u 1. When x 5, u 9.
5
1
x
dx 2x 1
9
1
1 2 3
u
4 3
1
4
16
3
x cos x dx 3
2
2u1
9
u1 2 u1 2 du
1
9
2
1
2
2
27 23 2
3
3
2
3
2
x dx sin x
3
2
3
2
83.
cos
0
82.
1 2u 1 1
1
du u
2
4
2
81.
2
0
2
3 3
33
2 2
4
2
x2 sin x
3
5 2 2 3
2
2 3
1 8
18
2
72
2
dy
18x22x3 12, 0, 4
dx
y 3 2x3 12 6x2 dx
y3
84.
u 2x3 1
y 48 3x 53 dx
2x3 13
C 2x3 13 C
3
48
4 13 C ⇒ C 3
y 2x3 13 3
85.
dy
2x
, 5, 4
dx 2x2 1
y
1
2x2 11 2 4x dx u 2x2 1
2
y
1 2x2 11 2
C 2x2 1 C
2
12
4 49 C 7 C ⇒ C 3
y 2x2 1 3
dy
48
, 1, 3
dx 3x 53
86.
1
3x 53 3 dx
3
163x 52
C
2
8
C
3x 52
3
8
8
C C ⇒ C1
31 52
4
y
8
1
3x 52
dy
9x2
4x , 0, 2
3
dx
3x 13 2
y
4x 3x3 13 2 9x2 dx
2x 2 2x2 20
3x3 11
1 2
2
3x3 1
2
C
C
2
C ⇒ C4
1
y 2x2 2
3x3 1
4
405
406
Chapter 4
Integration
87. u x 1, x u 1, dx du
When x 0, u 1. When x 7, u 8.
7
Area 8
3
x
x 1 dx 0
3
u 1
u du
1
8
u4 3 u1 3 du 1
37u
7 3
3
u4
4
8
3
1
384
3 3
1209
12 7
7 4
28
88. u x 2, x u 2, dx du
When x 2, u 0. When x 6, u 8.
6
Area 89. A 2
sin x cos 2x dx cos x 0
2 3
sec2
2
x
dx 2
2
2 3
4
93.
4
2
csc 2x cot 2x dx 12
x
2
sec2
92. Let u 2x, du 2 dx.
Area 1
2
0
12 7
u
7
3
3u4
8
3
0
4
12
7
x3x 2 dx 7.581
95.
0
xx 3 dx 28.8 3
15
3
1
97.
cos
0
d 7.377
6
5
1
0
2x 12 dx 2
−1
1
1
4
1
2x 12 2 dx 2x 13 C1 x3 2x2 x C1
2
6
3
6
1
They differ by a constant: C2 C1 .
6
2
4
4
4x2 4x 1 dx x3 2x2 x C2
3
sin 2x dx 1.0
0
0
−1
2x 12 dx 2
98.
5
50
0
0
x2x 1 dx 67.505
8
0
2
0
5
4752
35
1
2
5
2 3 1
2
2
−1
99.
2 3
20
−1
2
1
csc 2x cot 2x2 dx csc 2x
2
12
94.
10 3
4
3
96.
0
1
x
dx 2 tan
2
2
3
1
sin 2x
2
10u
1
cos 2x
2
4
10
x
dx 3.333 3
2x 1
0
u7 3 4u4 3 4u1 3 du 0
2 sin x sin 2x dx 2 cos x 91. Area 8
3
u 22 u du 0
0
90. A 8
3
x2 x 2 dx −1
144
5
Section 4.5
100.
sin x cos x dx sin x1 cos x dx sin2 x
C1
2
sin x cos x dx cos x1sin x dx Integration by Substitution
cos2 x
C2
2
1 sin2 x
sin2 x 1
cos2 x
C2 C2 C2
2
2
2
2
1
They differ by a constant: C2 C1 .
2
102. f x xx2 13 is odd.
101. f x x2x2 1 is even.
2
2
2
x2x2 1 dx 2
x4 x2 dx 2
0
5
32
2
x5 x3
5
3
2
2
2
0
8
272
3
15
103. f x sin2 x cos x is even.
2
sin2 x cos x dx 2
xx2 13 dx 0
104. f x sin x cos x is odd.
2
2
sin2 xcos x dx
2
0
sin x cos x dx 0
2
sin3 x
3
2
0
2
105.
x2 dx 0
x3 3 2
0
(a)
2
0
8
; the function x2 is an even function.
3
2
x2 dx 0
0
4
4
2
2
cos x dx 2
2
(d)
x2 dx 4
cos x dx 2
4
(c)
2
x2 dx 2
0
8
3
(d)
2
x2 dx 0
16
3
2
3x2 dx 3
x2 dx 8
0
sin x dx 0 since sin x is symmetric to the origin.
4
(b)
2
0
106. (a)
(b)
2
x2 dx 2
8
3
x2 dx 2
(c)
2
3
4
cos x dx 2 sin x
0
2
cos x dx 2 sin x
0
0
2
0
2 since cos x is symmetric to the y-axis.
2
sin x cos x dx 0 since sinx cosx sin x cos x and hence, is symmetric to the origin.
2
x3 6x2 2x 3 dx sin 3x cos 3x dx 4
107.
4
4
108.
4
4
x3 2x dx sin 3x dx 4
4
6x2 3 dx 0 2
0
cos 3x dx 0 2
0
6x2 3 dx 2 2x3 3x
cos 3x dx 23 sin 3x
0
0
4
0
232
407
408
Chapter 4
Integration
109. If u 5 x2, then du 2x dx and
x5 x23 dx 1
1
5 x232x dx u3 du.
2
2
2
110. f x xx2 12 is odd. Hence,
2
xx2 12 dx 0.
dQ
k100 t2
dt
111.
Qt k
k100 t2 dt 100 t3 C
3
Q100 C 0
k
Qt 100 t3
3
k
Q0 1003 2,000,000 ⇒ k 6
3
Thus, Qt 2100 t3. When t 50, Q50 $250,000.
112.
k
dV
dt
t 12
Vt 113. R 3.121 2.399 sin0.524t 1.377
(a)
k
k
C
dt t 12
t1
8
V0 k C 500,000
0
1
V1 k C 400,000
2
12
0
Relative minimum: 6.4, 0.7 or June
Solving this system yields k 200,000 and
C 300,000. Thus,
Relative maximum: 0.4, 5.5 or January
12
200,000
300,000.
t1
(b)
When t 4, V4 $340,000.
(c)
Vt 114.
1
ba
(a)
(b)
(c)
b
74.50 43.75 sin
a
t
262.5
1
74.50t cos
3
6
t
1
262.5
t
dt 74.50t cos
6
ba
6
262.5
t
1
74.50t cos
3
6
3
0
1
t
262.5
74.50t cos
12
6
115. (a)
6
3
b
a
1
262.5
447 223.5 102.352 thousand units
3
12
0
1
262.5 262.5
894 74.5 thousand units
12
Maximum flow: R 61.713 at t 9.36.
18.861, 61.178 is a relative maximum.
24
0
9
1
Rt dt 13 4.33 inches
3
24
12
1
262.5
223.5 102.352 thousand units
3
0
(b) Volume 1
3
70
0
Rt dt 37.47 inches
0
Rt dt 1272 (5 thousands of gallons)
Section 4.5
116.
1
ba
(a)
(b)
b
1
1
1
cos60 t sin120t
b a 30
120 2 sin60 t cos120 t dt a
1 60
1
1
1
cos60 t sin120 t
1 60 0 30
120
0
1 30
0
1
240
0
1
1
1
cos60 t sin120 t
1 30 0 30
120
b
a
1.273 amps
4
302 120
1 240
1
1
1
cos60 t sin120 t
1 240 0 30
120
1
1
0 30
30
60
1
1
30
2
5 22 1.382 amps
(c)
Integration by Substitution
1
1
30
30
30
0 amps
117. u 1 x, x 1 u, dx du
When x a, u 1 a. When x b, u 1 b.
b
Pa, b a
15
4
15
15
x1 x dx 4
4
1b
1 uu du
1a
1b
u3 2 u1 2 du 1a
(a) P0.50, 0.75 1 x
2
3 2
3x 2
1 x3 2
3x 2
2
(b) P0, b 15 2 5
u
4 5
b
0
0.75
0.50
2
1b
2
u3
3
2
1a
15 2u3 2
3u 5
4 15
1b
1a
1 x3 2
3x 2
2
0.353 35.3%
1 b3 2
3b 2 1 0.5
2
1 b3 23b 2 1
b 0.586 58.6%
118. u 1 x, x 1 u, dx du
When x a, u 1 a. When x b, u 1 b.
b
Pa, b a
1155 3
1155
x 1 x3 2 dx 32
32
1155
32
1b
1 u3 u3 2 du
1a
1b
u9 2 3u7 2 3u5 2 u3 2 du 1a
2u5 2
1b
1155
105u3 385u2 495u 231
32 1155
(a) P0, 0.25 16 105u
385u2 495u 231
(b) P0.5, 1 16 105u
385u2 495u 231
u5
u5
2
3
2
3
20
119. (a) C 0.1
12 sin
8
18
(b) C 0.1
10
12 sin
1155 2 112 2 9
u
u
32 11
3
1a
0.025 2.5%
0.736 73.6%
0.75
1
0
0.5
20
8
Savings 9.17 3.14 $6.03.
1b
2
1a
1b
1a
14.4
1 1 $9.17
t 8
14.4
t 8
6 dt 0.6t
cos
12
12
6
2
u7 2 u5
7
5
385u2 495u 231
3
t 8
t 8
14.4
dt cos
12
12
16 105u
u5 2
2
18
3
14.4
2
10
14.4
10.8 3
2
6 $3.14
b
a
409
410
Chapter 4
120.
1
365
Integration
365
100,000 1 sin
0
121. (a)
2t 60
100,000
365
2t 60
dt t
cos
365
365
2
365
0
100,000 lbs.
(b) g is nonnegative because the graph of f is positive at the
beginning, and generally has more positive sections than
negative ones.
4
g
9.4
0
365
f
−4
(c) The points on g that correspond to the extrema of f are
points of inflection of g.
(e) The graph of h is that of g shifted 2 units downward.
(d) No, some zeros of f, like x 2, do not correspond to
an extrema of g. The graph of g continues to increase
after x 2 because f remains above the x-axis.
4
9.4
0
−4
t
gt f x dx
0
2
t
f x dx f x dx 2 ht.
2
0
123. (a) Let u 1 x, du dx, x 1 u
122. Let f x sin x, 0 ≤ x ≤ 1.
x
Let
x 0 ⇒ u 1, x 1 ⇒ u 0
1
and use righthand endpoints
n
1
i
ci , i 1, 2, . . . , n.
n
0
n
sini n
lim
lim
f ci x
n→ i1
x→0 i1
n
1
1
x 51 x2 dx
0
sin x dx
1
cos x
u51 u2 du
0
(b) Let u 1 x, du dx, x 1 u
0
1 u2u5du
1
1
n
0
x21 x5 dx x 0 ⇒ u 1, x 1 ⇒ u 0
1
0
1
2
1
1 1 0
x a1 xb dx 0
1 uaubdu
1
1
ub1 ua du
0
1
x b1 xa dx
0
124. (a) sin x cos
Let u x and cos x sin
x
2
2
sin2x dx 0
2
cos2
0
2
0
x dx 2
cos2u du x as in part (a):
2
sinn x dx 0
cos2udu
2
cos2x dx
2
0
0
2
0
2
x, du dx, x u:
2
2
2
(b) Let u 2
0
cosn
x dx 2
cosn u du 0
2
cosnudu
2
0
cosn x dx
Section 4.5
125. False
126. False
127. True
10
10
10
1
1
2x 12 2 dx 2x 13 C
2
6
2x 12 dx ax3 bx2 cx d dx 10
xx2 12 dx 10
ax3 cx dx Odd
10
b
a
129. True
a
Even
cos b cos a cosb 2 cos a b2
sin x dx
a
4 sin x cos x dx 2 sin 2x dx cos 2x C
130. False
1
1 sin 2x3
1
sin 2x22 cos 2x dx C sin3 2x C
2
2
3
6
sin2 2x cos 2x dx 131. Let u cx, du c dx:
b
c
cb
f cx dx c
a
du
c
f u
ca
cb
f u du
ca
cb
f x dx
ca
132. (a)
d
sin u u cos u C cos u cos u u sin u u sin u
du
Thus,
u sin u du sin u u cos u C.
(b) Let u x, u2 x, 2 u du dx.
0
2
sinx dx sin u2u du
0
2
u sin u du
0
2 cos
2
2 sin u u cos u
(part (a))
0
bx2 d dx
0
b
sin x dx cos x
411
1
1
x2 12x dx x2 12 C
2
4
10
bx2 d dx 0 2
128. True
Integration by Substitution
412
Chapter 4
Integration
133. Because f is odd, f x f x. Then
a
0
a
f x dx a
a
f x dx 0
a
f x dx
a
f x dx 0
f x dx.
0
Let x u, dx du in the first integral.
When x 0, u 0. When x a, u a.
1
a
a
f x dx 0
a
a
f udu f x dx
0
a
f u du 0
f x dx 0
0
134. Let u x h, then du dx. When x a, u a h. When x b, u b h. Thus,
b
bh
f x h dx a
bh
f u du ah
f x dx.
ah
1
135. Let f x a0 a1x a2 x2 . . . an x n.
1
f x dx a0 x a1
0
a0 136. 2
x2
x3
x n1
a2 . . . an
2
3
n1
an
a1 a2 . . .
0
2
3
n1
0
1
f x dx 1 2
2
1
2
0
f xx dx 2 2
2
0
1
(Given)
f xx2 dx 2
0
By the Mean Value Theorem for Integrals, there exists c
in 0, 1 such that
Adding,
1
1
f x dx f c1 0
2 f x 2 x f x x2f x dx 0
0
0
1
0 f c.
f x x2 dx 0.
0
Thus the equation has at least one real zero.
Section 4.6
Numerical Integration
2
1. Exact:
x2 dx 0
2
Trapezoidal:
x2 dx 0
2
Simpson’s:
x2 dx 0
1
2. Exact:
0
1
Trapezoidal:
0
1
Simpson’s:
Since x2 ≥ 0, f 0. Hence, there are no such
functions.
0
3x 1
2
3
0
8
2.6667
3
2
2
1
1
02
4
2
1
1
04
6
2
212 2
3
2
2
212 4
3
2
2
x2
x3
1 dx x
2
6
1
0
11
2.7500
4
8
2.6667
3
22 22 7
1.1667
6
x2
1
1 42
1 22
3 42
12
75
1 dx 1 2
1 2
1 2
1 1 1.1719
2
8
2
2
2
2
64
x2
1
1 42
1 22
3 42
12
1 dx 14
1 2
1 4
1 1
2
12
2
2
2
2
67 1.1667
Section 4.6
2
3. Exact:
x3 dx 4
x3 dx 1
1
02
4
2
0
2
Trapezoidal:
0
2
Simpson’s:
x3 dx 0
2
4. Exact:
x3 dx 0
2
x3 dx 0
2
x3 dx 0
2
4
0
3
8
Trapezoidal:
3
0
8
Simpson’s:
3
0
34x 8
4 3
0
x dx 5
22
16
x dx 4
9
3 2
4
2
2
4
6
2
4
6
2
3
3
2
4
2
2
18 5
24
24
3
4 x2 dx 4x 1
3
4 x2 dx 1
3
Simpson’s:
3
24
4.0000
6
23 2
2
4
7
2
4
7
4
2
1
0.5090
4
1
0.5004
4
2
4
2
3
4
3
3
4
3
4
213 2
3
213 4
3
5
4
3
5
4
3
6
4
2
2
6
4
2
3
4
3
7
4
7
4
8 4.0625
3
8 4.0000
16 38
12.6667
3
3
37
2
8
21
2
4
47
2
8
26
2
4
57
2
8
31
2
4
67
3
8
12.6640
9
Trapezoidal:
3
2
17
4.2500
4
12.0000
23x 4
8. Exact:
3
x dx 9
Simpson’s:
213 4
23 1
x dx 0 4 2 3 2 4 3 3 2 3 4 4 3 5 2 3 6 4 3 7 2 11.8632
3
4
Trapezoidal:
3
1
x dx 0 2 2 3 2 2 3 3 2 3 4 2 3 5 2 3 6 2 3 7 2 11.7296
2
9
7. Exact:
2
1
1
04
12
4
0
3
2
4.0000
1
1
02
8
4
x dx 213 2
0.5000
1
14x 8
6. Exact:
2
2
Simpson’s:
3
1
4
1
dx 14
x2
12
5
1
Trapezoidal:
1
1
04
6
2
2
5. Exact:
3
1
4
1
dx 1 2
x2
8
5
1
Simpson’s:
2
413
4.0000
0
1
1
dx x2
x
1
Trapezoidal:
2
x4
Numerical Integration
1
4 x2 dx 37
8
x3
3
3
1
21 4
26 4
57
8
31 4
2
11
0.6667
3
3
3
1
3
32 4
4
2
47
8
20 24 2
5
2
2
5 0.7500
1
9
25
34 4
04 4
5 0.6667
6
4
4
67
3 12.6667
8
414
Chapter 4
Integration
2
9. Exact:
1
1
1
dx x 12
x1
2
Trapezoidal:
2
1 1 32
8
32
1
0.1676
8 4 81 25 121 9
x x2 1 dx 0
2
x x2 1 dx 0
2
x x2 1 dx 0
2
11. Trapezoidal:
0
2
Simpson’s:
0
8
64
1
1 1 64
0.1667
12 4 81 25 121 9
2
Simpson’s:
Trapezoidal:
1 1 1
0.1667
3 2 6
1 1
1
1
1
1
1
dx 2
4
4
x 12
12 4
5 4 12
3 2 12
7 4 12
9
1
1
10. Exact:
2
1 1
1
1
1
1
1
dx 2
2
2
x 12
8 4
5 4 12
3 2 12
7 4 12
9
1
Simpson’s:
1 2
x 13
3
2
2
0
1
53
3
2
1 3.393
1 22 1 21 12 1 2
3
2
3 22 1 2 22 1 3.457
1 22 1 21 12 1 4
3
2
3 22 1 2 22 1 3.392
1
1
02
4
2
1
1
04
6
2
1
1 x3 dx 1 2 1 1 8 2 2 2 1 27 8 3 3.283
4
1
1 x3 dx 1 4 1 1 8 2 2 4 1 27 8 3 3.240
6
Graphing utility: 3.241
2
12. Trapezoidal:
0
2
Simpson’s:
0
1
2
1 1 23
1
2
1 13
1
1
1.397
3
1 3 23
1
2
1 1 23
1
4
1 13
1
1
1.405
3
1 3 23
1
1
dx 1 2
4
1 x3
1
1
dx 1 4
6
1 x3
Graphing utility: 1.402
1
13.
1
x 1 x dx 0
x1 x dx
0
x1 x dx 1
02
8
x1 x dx 1
04
12
1
Trapezoidal:
0
1
Simpson’s:
0
1
1
1
2
4
4
1
1
1
2
4
4
1
1
1
2
2
2
1
1
1
4
2
2
3
3
1
4
4
3
3
1
4
4
0.342
0.372
Graphing utility: 0.393
14. Trapezoidal:
2
Simpson’s:
2
x sin x dx 24
x sin x dx Graphing utility: 1.458
16
1 2
2
4
2
5
5
sin
2
8
8
5
5
sin
2
8
8
3
3
sin
2
4
4
3
3
sin
4
4
4
7
7
sin
0 1.430
8
8
7
7
sin
0 1.458
8
8
Section 4.6
15. Trapezoidal:
2
2
4
2
2
4
2
2
cos 0 2 cos
8
cosx2 dx 0
2 cos
2
2
2
2 cos
2
2
2
Numerical Integration
2 cos
3 2
4
2
4 cos
3 2
4
2
415
cos
2
2
cos
2
2
0.957
Simpson’s:
2
2
cos 0 4 cos
12
cosx2 dx 0
0.978
Graphing utility: 0.977
16. Trapezoidal:
4
4
4
2
4
4
2
4
tan 0 2 tan
8
tanx2 dx 0
2 tan
4
2
2
2 tan
4
2
2
2 tan
3 4
4
2
4 tan
3 4
4
2
tan
4
2
tan
4
2
0.271
Simpson’s:
4
4
tan 0 4 tan
12
tanx2 dx 0
0.257
Graphing utility: 0.256
1.1
17. Trapezoidal:
sin x2 dx 1
1.1
Simpson’s:
sin x2 dx 1
1
sin1 2 sin1.0252 2 sin1.052 2 sin1.0752 sin1.12 0.089
80
1
sin1 4 sin1.0252 2 sin1.052 4 sin1.0752 sin1.12 0.089
120
Graphing utility: 0.089
2
18. Trapezoidal:
1 cos2 x dx 2 2 1 cos2 8 2 1 cos2 4 2 1 cos23 8 1 1.910
16
1 cos2 x dx 2 4 1 cos2 8 2 1 cos2 4 4 1 cos23 8 1 1.910
24
0
2
Simpson’s:
0
Graphing utility: 1.910
4
19. Trapezoidal:
x tan x dx 0
4
Simpson’s:
x tan x dx 0
2
2
3
3
02
tan
2
tan
2
tan
0.194
32
16
16
16
16
16
16
4
2
2
3
3
04
tan
2
tan
4
tan
0.186
48
16
16
16
16
16
16
4
Graphing utility: 0.186
20. Trapezoidal:
0
Simpson’s:
0
sin x
2 sin 4 2 sin 2 2 sin3 4
1
dx 0 1.836
x
8
4
2
3 4
sin x
4 sin 4 2 sin 2 4 sin3 4
dx 1
0 1.852
x
12
4
2
3 4
Graphing utility: 1.852
416
21.
Chapter 4
Integration
22. Trapezoidal: Linear polynomials
y
Simpson’s: Quadratic polynomials
y = f ( x)
x
a
b
The Trapezoidal Rule overestimates the area if the graph
of the integrand is concave up.
23.
f x x3
24.
f x 2x 3
fx 3x2
fx 2
f x 6x
f x 0
f x 6
f 4x
The error is 0 for both rules.
0
(a) Trapezoidal: Error ≤
2 03
12 0.5 since
1242
f x is maximum in 0, 2 when x 2.
(b) Simpson’s: Error ≤
2 05
0 0 since
18044
f 4x 0.
25.
f x 1
x1
26.
f x x 12
fx 2x 13
1
fx x 12
f x 6x 14
f x 24x 15
f x 2
x 13
f 4x 120x 16
f x 6
x 14
(a) Trapezoidal: Error ≤
f 4x 24
x 15
since f x is a maximum of 6 at x 2.
1
1 02
(a) Trapezoidal: Error ≤
0.01 since
2 1242
96
f x is maximum in 0, 1 when x 0.
(b) Simpson’s: Error ≤
4 23
1
6 ,
1242
4
1 05
1
0.0005
24 18044
1920
since f 4x is maximum in 0, 1 when x 0.
(b) Simpson’s: Error ≤
4 25
1
120 18044
12
since f 4x is a maximum of 120 at x 2.
Section 4.6
27.
f x cos x
28.
fx cosx
f x cos x
f x 2 sinx
f x sin x
f x 3 cosx
f 4x cos x
f 4x 4 sinx
03
3
0.1615
1 1242
192
(b) Simpson’s: Error ≤
1 03 2
2
0.0514,
2
124 192
since f x ≤ 2 on 0, 1.
05
5
0.006641
1 18044
46,080
(b) Simpson’s: Error ≤
because f 4x is at most 1 on 0, .
29. f x (a) Trapezoidal: Error ≤
because f x is at most 1 on 0, .
417
f x sinx
fx sin x
(a) Trapezoidal: Error ≤
Numerical Integration
1 05 4
4
0.0021,
18044
46,080
since f 4 x ≤ 4 on 0, 1.
2
in 1, 3.
x3
(a) f x is maximum when x 1 and f 1 2.
23
Trapezoidal: Error ≤
2 < 0.00001, n2 > 133,333.33, n > 365.15; let n 366.
12n2
24
f 4x 5 in 1, 3.
x
(b) f 4x is maximum when x 1 and f 41 24.
25
Simpson’s: Error ≤
24 < 0.00001, n4 > 426,666.67, n > 25.56; let n 26.
180n4
30. f x 2
in 0, 1.
1 x3
(a) f x is maximum when x 0 and f 0 2.
Trapezoidal: Error ≤
f 4x 1
2 < 0.00001, n2 > 16,666.67, n > 129.10; let n 130.
12n2
24
in 0, 1
1 x5
(b) f 4x is maximum when x 0 and f 40 24.
Simpson’s: Error ≤
1
24 < 0.00001, n4 > 13,333.33, n > 10.75; let n 12. (In Simpson’s Rule n must be even.)
180n4
418
31.
Chapter 4
Integration
f x x 21 2,
1
fx x 21
2
0 ≤ x ≤ 2
f 4x 15
x 27
16
2
(a) Maximum of f x 1
4x 23
2
2
0.0884.
16
is
2
f x 15 7
x
8
2
f 4x 105 9
x
16
2
n ≥ 76.8. Let n 77.
(b) Maximum of f 4 x 15
16x 27
2
15 2
0.0829.
256
1
2
n2 ≥
3
4x 5
2
is
3
on 1, 3.
4
10 5
n ≥ 223.6. Let n 224
(b) Maximum of f 4 x is
Simpson’s: Error ≤
25 15 2
≤ 0.00001
180n4 256
105
105
is
on 1, 3.
16x9 2
16
25 105
≤ 0.00001
180n4 16
7
6
n4 ≥
3215 2 5
2 5
10 10
180256
96
n4 ≥
3 13 3
≤ 0.00001
Trapezoidal: Error ≤
12n2 4
8 2 5
2 5
10 10
1216
24
n2 ≥
Simpson’s: Error ≤
2
(a) Maximum of f x 2 03 2
≤ 0.00001
Trapezoidal: Error ≤
12n2
16
1 ≤ x ≤ 3
1 3
x
2
3
f x x5
4
2
2
f x x1 2,
fx 2
1
f x x 23
4
3
f x x 25
8
32.
10 5
n ≥ 18.5. Let n 20 (even)
n ≥ 6.2. Let n 8 (even).
33.
f x cosx, 0 ≤ x ≤ 1
34.
fx sinx
f x sin x
f x 3 sinx
f x cos x
f 4x 4 cosx
(a) Maximum of f x 2cosx is 2.
Trapezoidal: Error ≤
1 03 2
≤ 0.00001
12n2
2
12
n2 ≥
10 5
f 4 x
Simpson’s: Error ≤
n4 ≥
x is
4cos
4.
1
4 ≤ 0.00001
180n4
4
180
f 4x sin x
All derivatives are bounded by 1.
(a) Trapezoidal: Error ≤
n2 ≥
n ≥ 286.8. Let n 287.
(b) Maximum of
2
fx cos x
f x 2 cosx
f x sin x, 0 ≤ x ≤
10 5
n ≥ 15.3. Let n 16.
23
1 ≤ 0.00001
12n2
3
96
10 5
n ≥ 179.7. Let n 180.
(b) Simpson’s: Error ≤
n4 ≥
25
1 ≤ 0.00001
180n4
5
5760
10 5
n ≥ 8.5. Let n 10 (even).
Section 4.6
35. f x Numerical Integration
419
1x
(a) f x 1
41 x3
2
in 0, 2.
1
f x is maximum when x 0 and f 0 4.
Trapezoidal: Error ≤
(b) f 4x 8 1
12n2 4
< 0.00001, n2 > 16,666.67, n > 129.10; let n 130.
15
in 0, 2
161 x7 2
15
f 4x is maximum when x 0 and f 40 16.
32 15
180n4 16
Simpson’s: Error ≤
36. f x x 12
(a) f x < 0.00001, n4 > 16,666.67, n > 11.36; let n 12.
3
2
9x 14
3
in 0, 2.
2
f x is maximum when x 0 and f 0 9.
Trapezoidal: Error ≤
(b) f 4x 8 2
12n4 9
< 0.00001, n2 > 14,814.81, n > 121.72; let n 122.
56
in 0, 2.
81x 110 3
56
f 4x is maximum when x 0 and f 40 81.
Simpson’s: Error ≤
32 56
180n4 81
< 0.00001, n4 > 12,290.81, n > 10.53; let n 12. (In Simpson’s Rule n must be even.)
37. f x tanx2
(a) f x 2 sec2x21 4x2 tanx2 in 0, 1.
f x is maximum when x 1 and f 1 49.5305.
Trapezoidal: Error ≤
1 03
49.5305 < 0.00001, n2 > 412,754.17, n > 642.46; let n 643.
12n2
(b) f 4x 8 sec2x212x2 3 32x4 tanx2 36x2 tan2x2 48x4 tan3x2 in 0, 1
f 4x is maximum when x 1 and f 41 9184.4734.
Simpson’s: Error ≤
1 05
9184.4734 < 0.00001, n4 > 5,102,485.22, n > 47.53; let n 48.
180n4
38. f x sinx2
(a) f x 22x2 sinx2 cosx2 in 0, 1.
f x is maximum when x 1 and f 1 2.2853.
1 03
2.2853 < 0.00001, n2 > 19,044.17, n > 138.00; let n 139.
12n2
(b) f 4x 16x4 12 sinx2 48x2 cosx2 in 0, 1
Trapezoidal: Error ≤
f 4x is maximum when x 0.852 and f 40.852 28.4285.
Simpson’s: Error ≤
1 05
28.4285 < 0.00001, n4 > 15,793.61, n > 11.21; let n 12.
180n4
420
Chapter 4
Integration
39. (a) b a 4 0 4,
n4
40. n 8, b a 8 0 8
4
0
8
4
f x dx 3 27 29 27 0
8
(a)
f x dx 0
210 29 26 0
1
49
49 24.5
2
2
1
88 44
2
4
(b)
0
8
0 21.5 23 25.5 29
16
4
f x dx 3 47 29 47 0
12
8
(b)
77
25.67
3
0
f x dx 8
0 41.5 23 45.5 29
24
410 29 46 0
134
1
134 3
3
41. The program will vary depending upon the computer or programmable calculator that you use.
42. f x 2 3x2 on 0, 4.
Ln
Mn
Rn
Tn
Sn
4
12.7771
15.3965
18.4340
15.6055
15.4845
8
14.0868
15.4480
16.9152
15.5010
15.4662
10
14.3569
15.4544
16.6197
15.4883
15.4658
12
14.5386
15.4578
16.4242
15.4814
15.4657
16
14.7674
15.4613
16.1816
15.4745
15.4657
20
14.9056
15.4628
16.0370
15.4713
15.4657
n
43. f x 1 x2 on 0, 1.
Ln
Mn
Rn
Tn
Sn
4
0.8739
0.7960
0.6239
0.7489
0.7709
8
0.8350
0.7892
0.7100
0.7725
0.7803
10
0.8261
0.7881
0.7261
0.7761
0.7818
12
0.8200
0.7875
0.7367
0.7783
0.7826
16
0.8121
0.7867
0.7496
0.7808
0.7836
20
0.8071
0.7864
0.7571
0.7821
0.7841
n
44. f x sin x on 0, 4.
Ln
Mn
Rn
Tn
Sn
4
2.8163
3.5456
3.7256
3.2709
3.3996
8
3.1809
3.5053
3.6356
3.4083
3.4541
10
3.2478
3.4990
3.6115
3.4296
3.4624
12
3.2909
3.4952
3.5940
3.4425
3.4674
16
3.3431
3.4910
3.5704
3.4568
3.4730
20
3.3734
3.4888
3.5552
3.4643
3.4759
n
Section 4.6
Numerical Integration
421
2
45. A x cos x dx
0
Simpson’s Rule: n 14
2
x cos x dx 0
84
cos
2
28
28
0 cos 0 4
cos
4
14
14
3 . . .
3
cos
28
28
cos
2
2
0.701
y
1
1
2
π
π
4
2
x
46. Simpson’s Rule: n 8
2
8 3
2 2
3
sin d 3
6
17.476
1
0
1
2 2
sin 0 4
3
1
2 2 sin
2
3
16
1
2 2 ...
sin
3
8
1
2 2
sin
3
2
5
47. W 100x 125 x 3 dx
0
Simpson’s Rule: n 12
5
100x 125 x3 dx 0
5
5
0 400
312
12
125 5
12
3
15
12
125 15
12
3
400
200
10
12
125 10
12
3
. . . 0 10,233.58 ft lb
48. (a) Trapezoidal:
2
f x dx 0
2
4.32 24.36 24.58 25.79 26.14 27.25 27.64 28.08 8.14 12.518
28
Simpson’s:
2
f x dx 0
2
4.32 44.36 24.58 45.79 26.14 47.25 27.64 48.08 8.14 12.592
38
(b) Using a graphing utility,
y 1.3727x3 4.0092x2 0.6202x 4.2844.
2
Integrating,
y dx 12.53.
0
1 2
49.
0
6
1 x2
dx
Simpson’s Rule, n 6
1
0
2
6 46.0209 26.0851 46.1968 26.3640 46.6002 6.9282
36
1
113.098 3.1416
36
422
Chapter 4
Integration
50. Simpson’s Rule: n 6
1
4
0
1
4
4
2
4
2
4
1
dx 1
1 x2
36
1 1 62 1 2 62 1 3 62 1 462 1 5 62 2
3.14159
51. Area 1000
125 2125 2120 2112 290 290 295 288 275 235 89,250 sq m
210
52. Area 120
75 281 284 276 267 268 269 272 268 256 242 223 0
212
7435 sq m
53. Let f x Ax3 Bx2 Cx D. Then f 4x 0.
Simpson’s: Error ≤
b a5
0 0
180n4
Therefore, Simpson’s Rule is exact when approximating the integral of a cubic polynomial.
1
Example:
x3 dx 0
1
1
04
6
2
3
1 1
4
This is the exact value of the integral.
t
54.
sinx dx 2, n 10
0
By trial and error, we obtain t 2.477.
55. The quadratic polynomial
px x x2x x3
x x1x x3
x x1x x2
y y y
x1 x2x1 x3 1 x2 x1x2 x3 2 x3 x1x3 x2 3
passes through the three points.
Review Exercises for Chapter 4
1.
2.
y
y
f
f′
f
x
x
f
3.
2
1
2x2 x 1 dx x3 x2 x C
3
2
4.
2
2
2 x2 3
dx 3
x1 3dx 3
C
3
3 2 3
3x
3
3
3 3
x2 3 C
3x2 3 C
Review Exercises for Chapter 4
5.
x3 1
dx x2
x
1
1
1
dx x2 C
x2
2
x
6.
x3 2x2 1
dx x2
423
x 2 x2 dx
1
1
x2 2x C
2
x
7.
4x 3 sin x dx 2x2 3 cos x C
9. fx 2x, 1, 1
f x 8.
5 cos x 2 sec2 x dx 5 sin x 2 tan x C
10. f x 6x 1
fx 2x dx x2 C
When x 1:
6x 1 dx 3x 12 C1
Since the slope of the tangent line at 2, 1 is 3, it follows
that f2 3 C1 3 when C1 0.
y 1 C 1
fx 3x 12
C2
y 2 x2
f x 3x 12 dx x 13 C2
f 2 1 C2 1 when C2 0.
f x x 13
11. (a)
12. (a)
y
2
y
6
x
−2
6
(6, 2)
x
7
−6
(b)
−2
dy
2x 4, 4, 2
dx
y
2x 4 dx x 2 4x C
2 16 16 C ⇒ C 2
y
x2
4x 2
1
−4
8
−7
(b)
dy 1 2
x 2x, 6, 2
dx 2
y
1 2
1
x 2x dx x 3 x2 C
2
6
1
2 63 6 2 C ⇒ C 2
6
1
y x3 x2 2
6
424
13.
Chapter 4
Integration
at a
vt 14. 45 mph 66 ft sec
30 mph 44 ft sec
a dt at C1
at a
v0 0 C1 0 when C1 0.
vt at 66 since v0 66 ft sec.
a
st t 2 66t since s0 0.
2
Solving the system
vt at
st a
at dt t2 C2
2
vt at 66 44
s0 0 C2 0 when C2 0.
a
s(t t2 66t 264
2
a
st t 2
2
we obtain t 24 5 and a 55 12. We now solve
55 12t 66 0 and get t 72 5. Thus,
a
s30 302 3600 or
2
23600
8 ft sec2.
a
302
s
2
66
475.2 264 211.2 ft.
16. at 9.8 m sec2
15. at 32
vt 32t 96
vt 9.8t v0 9.8t 40
st st 4.9t2 40t s0 0
16t2
96t
(a) vt 32t 96 0 when t 3 sec.
(a) vt 9.8t 40 0 when t (b) s3 144 288 144 ft
(c) vt 32t 96 (c) vt 9.8t 40 20 when t i1
n
20.
i1
10
21.
i1
20
23.
3
n
i1
n
2
3 11
n
n
2
i 2 1 2 2 2 3 2 . . . 12 2
21
22
23
212
i1 2i
12
1
1
1
1
...
4i
4
1
4
2
4
8
i1
n
18.
3 21
n
n
2
...
3 n1
n
n
2
3i 2 i 12
n 12
4
9
3 2
6 2
. . . 3n 2 n
n
n
n
3i 3
1011
165
2
i 12 i1
20
i2 2i 1
i1
2021
202141
2
20
6
2
2870 420 20 3310
20
2.04 sec.
9.8
(d) s2.04 61.2 m
8
19.
40
4.08 sec.
9.8
(b) s4.08 81.63 m
96
3
when t sec.
2
2
3
9
3
16
96
108 ft
2
4
2
(d) s
72
475.2 ft.
5
Stopping distance from 30 mph to rest is
v30 830 240 ft sec
17.
55 12 72
72
5
2
5
20
4i 1 4
22.
i1
12
24.
ii 2 1 i1
2021
20 820
2
12
i 3 i
i1
122132 1213
4
2
6084 78 6006
Review Exercises for Chapter 4
10
25. (a)
2i 1
i1
n
i3
(b)
i1
10
(c)
4i 2
i1
26. x1 2, x2 1, x3 5, x4 3, x5 7
(a)
1
16
1 5
x 2 1 5 3 7 5 i1 i 5
5
5
(b)
i1
5
(c)
1
1
1 1 1
37
1 xi 2
5 3 7 210
2xi xi2 22 22 21 12 25 52 23 32 27 72 56
i1
5
(d)
xi xi1 1 2 5 1 3 5 7 3 5
i2
27. y 10
1
, x , n 4
x2 1
2
Sn S4 1 10
10
10
10
2 1
1 22 1 12 1 3 22 1
13.0385
s n s4 10
10
10
10
1
2 1 22 1 1 1 3 22 1 22 1
9.0385
9.0385 < Area of Region < 13.0385
1
28. y 9 x2, x 1, n 4
4
9 41 4
1
1
1
9 9 9 16 9 25
4
4
4
S4 1
22.5
9 49
1
s4 1
1
1
9 16 9 25 9 9
4
4
14.5
4
29. y 6 x, x , right endpoints
n
Area lim
n→
lim
n→
n
6
4
i1
n
6
i1
4i 4
n n
4
4 nn 1
6n n
n
2
lim
24 8
n→
8
f ci x
lim
n→
y
n1
24 8 16
n
2
x
−2
2
−2
4
6
8
425
426
Chapter 4
Integration
30. y x2 3, x Area lim
n→
lim
n→
2
n
y
right endpoints
12
10
n
f ci x
8
i1
n
i1
6
2i
n
2
4
2
n
3
2
x
lim
2 4 nn 12n 1
3n
n n2
6
lim
3
n→
n→
4 n 12n 1
26
8
6 6
n2
3
3
31. y 5 x2, x Area lim
n→
lim
n→
n
3
n
y
6
4
f ci x
3
i1
n
i1
2
5 2 3i
n
2
1
3
n
−4 −3
1
2
3
4
−2
lim
3
12 nn 1
9 nn 12n 1
2
n
n
n
2
n
6
lim
3 18 n n 1 92 n 1n2n 1
n→
x
−1
3 n
12i 9i2
2
lim
1
n→
n i1
n
n
n→
2
1
2 n 4i2
lim
3
n→
n i1 n2
2
3 18 9 12
1
2
32. y x3, x 4
n
Area lim
n→
n
y
20
f ci x
15
i1
10
n
1
2i
lim
2
n→
4
n
i1
3
2
n
1 n
24i 24i2 8i3
lim
8
2 3
n→
2n i1
n
n
n
lim
4 n
3i 3i2
i3
1 2 3
n i1
n
n
n
lim
4
3 nn 1
3 nn 12n 1
1 n2n 12
n
2
3
n
n
2
n
6
n
4
n→
n→
4 6 4 1 15
5
x
1
2
3
4
Review Exercises for Chapter 4
33. x 5y y2, 2 ≤ y ≤ 5, y n
Area lim
n→
5 2 n
3i
i1
n
3
n
2
y
6
3i
n
2
3
n
4
9i2
2
3
i
3
15i
4 12 2
10 n i1
n
n
n
lim
n→
1
x
1
3 n
3i 9i2
lim
6 2
n→
n i1
n
n
b
4
b
2b
m
4
4
n
(b) Sn f
i1
n1
i0
n→
b
n1
bi
m
n
i0
mb2
5mb2
b
1 2 3 4 4
16
8
b
4b
m
4
4
→0 i1
b
b
m
n
n
b
b
m
n
n
2 n
i
y
y = mx
i1
mb2 nn 1
mb2n 1
2
n
2
2n
x=b
mb2 n 1n
mb2n 1
i 2
n
2
2n
i0
2 n1
b
0
1
mb2
2
2ci 3 xi n
2x 3 dx
36. lim
3
3ci9 ci 2 xi →0 i1
4
3x9 x2 dx
1
3
0
37.
6
mb2
3mb2
b
1 2 3 4
16
8
b
3b
m
4
4
6
n
5
mb2n 1
mb2n 1 1 2 1
1
lim
mb bmb baseheight
n→
2n
2n
2
2
2
2
0
b
n
12mx
mx dx b
2b
m
4
4
n
b
mbi
n
n
i1
bi
n
bi
f
n
(c) Area lim
35. lim
b
3b
m
4
4
b
b
m
4
4
s m0
4
9
27
9 2
2
18 (d)
3
n→
sn 2
9 nn 12n 1
3
3 nn 1
2
6n n
n
2
n
6
lim
34. (a) S m
2
3x 6 dx
38.
3
y
39.
9 x2dx
40.
y
12
6
9
5
6
3
Triangle
2
3
1
x
−3
3
6
x
9
−4 −3 −2 −1
−3
5
0
5 x 5 dx (triangle)
427
1
2
3
4
−2
5
0
5 5 x dx 5
0
x dx 25
2
4
4
16 x2 dx (semicircle)
1
42 8
2
x
428
Chapter 4
Integration
6
41. (a)
6
f x gx dx 2
2
6
(b)
2
6
f x dx 2
f x dx 510 50
2
6
3
f x dx 0
0
3
(b)
gx dx 210 33 11
2
6
5f x dx 5
2
6
f x dx 3
2
6
42. (a)
6
2f x 3gx dx 2
2
(d)
gx dx 10 3 7
2
6
(c)
gx dx 10 3 13
2
6
f x gx dx 6
f x dx 6
f x dx 43.
3
2 x dx 2x 4t 3 2t dt t 4 t 2
4
f x dx 4 1 3
0
x2
2
4
0
8
6
f x dx 6
f x dx 1 1
3
4
(c)
f x dx 0
4
6
(d)
6
10 f x dx 10
3
t2 2 dt x4 2x2 5 dx 1
44.
1
t3 2t
2
9
x5
4
2
48.
1
1
1
dx x2 x3
4
50.
25x
1
1
0
2
9
5 2
4
2
9 5 4 5 52243 32 422
5
5
1
1
x2 x3 dx 2
x
2x
1
2
2x3
5x
3
2
sec2 t dt tan t
4
1
sin d cos
0
45.
52
15
x3 2 dx 4
3 4
49.
1
14
3
32 16
32 16
10 10
5
3
5
3
9
xx dx 1
5 47.
1
3
2
46.
f x dx 101 10
3
3 4
0
4
4
2
2
2
1
1 1
1
1
1 2 8
2
8
11
2
2
2 2
2
3
1 1 2
51.
3
2x 1 dx x2 x
1
y
6
5
4
3
2
1
−2 −1
−1
x
1
2
3
4
5
6
1
6
16
16
2
Review Exercises for Chapter 4
2
52.
x 4 dx 0
2 4x
x2
2
0
4
10
53.
3 9x
x3
x2 9 dx 3
y
3
64
36 9 27
3
7
4
6
64 54 10
3
3
3
5
4
y
2
1
8
7
x
−2 −1
2
1
3
4
5
6
5
4
3
2
1
x
1
2
54.
1
x2 x 2 dx x3 x2
2x
3
2
4
55.
6
7
x x3 dx 0
8
1 1
2
24 3
3 2
5
1
2
1
2
8
x2 x4
4 1
2
y
1
10 7 9
3
6 2
x
1
y
1
3
2
x
−2
1
3
−1
−2
1
56.
x1 x dx x1
2
1
x3 2 dx
57. Area 0
sin x dx
0
23x
2 2
4
3 5 15
3 2
2
x5
5
y
1
2
cos x
0
x
1
58. Area 2
x cos x dx
0
2 sin x
x2
2
8
1
2
0
0
cos1 1
1 cos1
0.460
1
1
0
1 1 1
2 4 4
429
430
Chapter 4
9
59. Area 4
x
1
Integration
dx 1 2
4x1
2 9
1
83 1 16
60. Area 3
y
sec2 x dx
5
0
y
tan x
10
3
0
4
3
3
2
8
6
π
6
4
2
−2
61.
x
2
−2
4
9
1
94
6
1
x
4
8
10
15 2x
dx 9
4
2
2
3 2 Average value
5
5
y
2
2
1
5 x
1
x x
2
25
x
4
62.
1
20
25 2
,
4 5
5
2
2
x3 dx 0
x4
8
2
0
y
4
6
8
10
63. Fx x21 x3
2
8
x3 2
6
3 2
x 4
2
( 3 2 , 2)
x
1
64. Fx 67.
68.
1
x2
65. Fx x2 3x 2
x2 13 dx x
1
x
2
dx x6 3x4 3x2 1 dx x2 2 x2 dx 69. u x3 3, du 3x2 dx
x2
dx 3
x3
2
x3 31 2 x2 dx 66. Fx csc2 x
x7 3 5
x x3 x C
7
5
x3
1
2x C
3
x
1
2
x3 31 2 3x2 dx x3 31
3
3
2
C
70. u x3 3, du 3x2 dx
x2x3 3 dx 1
2
x3 31 2 3x2 dx x3 33
3
9
2
C
71. u 1 3x2, du 6x dx
x1 3x24 dx 1
1
1
1 3x246x dx 1 3x25 C 3x2 15 C
6
30
30
π
3
x
Review Exercises for Chapter 4
431
72. u x2 6x 5, du 2x 6 dx
x3
2x 6
1
1 2
1
dx x 6x 51 C C
x2 6x 52 2 x2 6x 52
2
2x2 6x 5
73.
75.
sin
1 cos
76.
sin x
77.
79.
80.
csc2
81.
xx2 4 dx x2x3 13 dx sin3 x cos x dx cos x
dx 1 4
sin x C
4
sin x1 2 cos x dx 2sin x1 2 C 2sin x C
tann1 x
C, n
n1
1
2
x
1
dx 6
3x2 8
6
3
sec 2x tan 2x dx 1
1
sec 2x tan 2x2 dx sec 2x C
2
2
1
1
1 sec x2 sec x tan x dx 1 sec x3 C
3
2
1
1
3
1
dx 1 x
0
3
78.
0
84.
1
1
d cot 4csc2 d cot 5 C
5
1
83.
1
1
sin 3x26x dx cos 3x2 C
6
6
1 cos 1 2 sin d 21 cos 1 2 C 21 cos C
1 sec x2 sec x tan x dx 1
x sin 3x2 dx tann x sec2 x dx cot4
d 2
82.
74.
x2 42x dx 1 x2 42
2
2
1
x3 133x2 dx 0
3
6
x2 81 22x dx 3
1
1
x3 14
12
1 x1 2 dx 21 x1
0
2
3
2
0
13x
2
1
9
0 9 4
4
1
0
1
5
16 1 12
4
422
81
6
2
3
1
27 1
3
85. u 1 y, y 1 u, dy du
When y 0, u 1. When y 1, u 0.
1
2
0
y 11 y dy 2
0
1 u 1u du
1
0
2
u3 2 2u1 2 du 2
1
5u
2
5 2
4
u3
3
0
2
1
28
15
86. u x 1, x u 1, dx du
When x 1, u 0. When x 0, u 1.
0
2
1
1
x2x 1 dx 2
u 12u du
0
1
2
0
u5 2 2u3 2 u1 2 du 2
27u
7 2
4
u5
5
2
2
u3
3
1
2
0
32
105
432
Chapter 4
87.
Integration
x
dx 2
2
cos
0
89. (a)
cos
0
x 1
x
dx 2 sin
2 2
2
0
2
4
88.
sin 2x dx 0 since sin 2x is an odd function.
4
y
2
x
−3
3
dy
x9 x2,
dx
(b)
y
0, 4
3
9 x 2 1 2
1 9 x23
x dx 2
3 2
2
1
C 9 x23
3
2
C
−6
6
1
1
4 9 03 2 C 27 C ⇒ C 5
3
3
1
y 9 x23 2 5
3
90. (a)
y
−5
(b)
dy
1
x sinx2,
dx
2
3
y
0, 0
1
1
x sinx2 dx 2
4
sinx22x dx
u x2
x
1
cosx2 C
4
−3
1
cosx2 C
4
xx 11 3 dx. Let u x 1, du dx.
1
8
A
y
1
1
cosx2 4
4
92.
cos x sin2x dx sin x 0
8
u 1u1 3 du 0
1
1
cos 0 C ⇒ C 4
4
2
9
91.
0
u 4 3 u1 3 du
1
0
3u7
7
3
3u 4
4
1
cos2x
2
0
3
468
3
128 16 7
4
7
C
15,000
M
t1
p ds t
15,000
M
t1
1.20 0.04s ds
t
(a) 2000 corresponds to t 10.
C
15,000
M
(b) 2005 corresponds to t 15.
11
1.20 0.04t dt
C
10
15,000
1.20t 0.02t2
M
11
10
24,300
M
15,000
1.20t 0.02t2
M
0
1
1
0
2
2
2
3 8
93. p 1.20 0.04t
2
16
15
27,300
M
Problem Solving for Chapter 4
2
94.
1.75 sin
0
t
t
2
dt 1.75 cos
2
2
2
433
2
7
1.751 1 2.2282 liters
Increase is
7
5.1 1.9
0.6048 liters.
2
1
2
Simpson’s Rule n 4:
1
1
1
1
2
2
2
1
dx 0.257
1 x3
8 1 13 1 1.253 1 1.53 1 1.753 1 23
95. Trapezoidal Rule n 4:
1
1
1
4
2
4
1
dx 0.254
1 x3
12 1 13 1 1.253 1 1.53 1 1.753 1 23
Graphing utility: 0.254
1
96. Trapezoidal Rule n 4:
0
1
Simpson’s Rule n 4:
0
1
21432
21232
23432
1
x32
0
0.172
2 dx 2
2
3x
8
3 14
3 12
3 342 2
1
41432
21232
43432
1
x32
0
0.166
2 dx 2
2
3x
12
3 14
3 12
3 342 2
Graphing utility: 0.166
97. Trapezoidal Rule n 4:
2
x cos x dx 0.637
0
Simpson’s Rule n 4: 0.685
Graphing Utility: 0.704
98. Trapezoidal Rule n 4:
1 sin2 x dx 3.820
0
Simpson’s Rule n 4: 3.820
Graphing utility: 3.820
Problem Solving for Chapter 4
1
1. (a) L1 1
1
dt 0
t
x
(c) Lx 1 2.718
1
1
1
dt for x 2.718
t
1
dt 0.999896
t
(Note: The exact value of x is e, the base of the
natural logarithm function.)
1
by the Second Fundamental Theorem of Calculus.
x
(b) Lx L1 1
x1
(d) We first show that
1
To see this, let u x1
Then
1
1
dt t
Now, Lx1x2 1
dt t
1
1x1 t
1
x1 du ux
1x1 1
x1x2
1
dt t
x2
1
1
du u
1
1x1 u
x2
1
du 1x1 u
x1
dt.
1
t
and du dt.
x1
x1
1
1
1
1
1
x2
1
Lx1 Lx2 .
1
1
du 1x1 u
1
dt.
1x1 t
du using u 1
du
u
1
du
u
1
t
x1
434
Chapter 4
Integration
x
2. (a) Fx sin t2 dt
2
x
Fx
(b) Gx 0
1.0
1.5
1.9
2.0
2.1
2.5
3.0
4.0
5.0
0.8048
0.4945
0.0265
0.0611
0
0.0867
0.3743
0.0312
0.0576
0.2769
1
x2
x
Gx
x
sin t2 dt
2
1.9
1.95
1.99
2.01
2.1
0.6106
0.6873
0.7436
0.7697
0.8671
lim Gx 0.75
x→2
Fx F2
x2
(c) F2 lim
x→2
1
x→2 x 2
lim
x
sin t2 dt
2
lim Gx
x→2
Since Fx sin x2, F2 sin 4 lim Gx. Note: sin 4 0.7568
x→2
3. y x 4 4x 3 4x 2,
ci 0, 2 ,
2i
n
2
(a) x , f x x 4 4x 3 4x 2
n
n
A lim
n→ i1
n
lim
n→ i1
n
(b)
i1
4
2i
n
f ci x
4
2i
n
4
4
3
2i
n
4
9n
8n 16n 15n
n
8n 1n 15n
3
2
3
2
3
n→
n
n→ i1
lim
n
n→ i1
12
2i
n
2
ci n 1 2 16
n 15
2i
,
n
x 2
n
f ci x
2
5
8n 12 8n 12n 1 2
n
3n
n
3
(a) A lim
i1
2n
n 1 2
n
3
0, 2 ,
2n
2
2
n 1
n
8n 1n 15n
1
4. y x 5 2x 3,
2
n
2i
n
2i
n
4
3
(c) A lim
(b)
3
2i
n
1 2i
n
2
5
2i
n
n→
2
3
2i
n
2n
3
2n
4n 1 2n3n 2n 1 4n n 1 2n
8n 1 5n3n 2n 1
2
2
3
2
2
4
8n 1 5n3n 2n 1 403
(c) A lim
2
2
2
4
Problem Solving for Chapter 4
t2
dt
2
x
5. Sx sin
0
(b)
y
(a)
435
y
1.00
2
0.75
1
0.50
x
1
3
0.25
−1
x
−2
1
−0.25
2
The zeros of y sin
3
2
5
6
72 23
x2
correspond to the relative
2
extrema of Sx.
(c) Sx sin
x2
x2
0 ⇒
n ⇒ x2 2n ⇒ x 2n, n integer.
2
2
Relative maximum at x 2 1.4142 and x 6 2.4495
Relative minimum at x 2 and x 8 2.8284
(d) S x cos
x2
x2 n ⇒ x2 1 2n ⇒ x 1 2n, n integer
x 0 ⇒
2
2
2
Points of inflection at x 1, 3, 5, and 7.
1
6. (a)
1
cos x dx cos 1
1
1
3
1
cos x dx sin x
1
cos
1
3
2 cos
1
3
1.6758
2 sin1 1.6829
Error: 1.6829 1.6758 0.0071
1
(b)
1 1
1
1
1
3
dx x2
1 13 1 13 2
(Note: exact answer is 2 1.5708)
(c) Let px ax3 bx2 cx d.
1
1
p 3
7. (a) Area ax4
4
px dx 1
3
p
1
3
bx3 cx2
dx
3
2
3
9 x2 dx
2b
2d
3
b
b
2b
d d 2d
3
3
3
(c) Let the parabola be given by y b2 a2x2, a, b > 0.
0
ba
Area 2
3
b2 a2x2 dx
2
2
(b) Base 6, height 9, Area bh 69 36
3
3
b2
2 b2x a2
2 b2
x3 ba
3
0
2
b
a b
a
3 a
3
y
b3 1 b3
4 b3
2
a
3 a
3 a
10
8
7
6
5
4
3
2
1
Base x
1 2
4 5
y
0
0
2 27 9 36
−2 −1
1
3
9 x2 dx 2
x3
2 9x 3
−4
1
−
b
a
2b
, height b2
a
Archimedes’ Formula: Area 2 2b 2
4 b3
b 3 a
3 a
b
a
x
436
Chapter 4
Integration
8. Let d be the distance traversed and a be the uniform
acceleration. We can assume that v0 0 and s0 0.
Then
y
9. (a)
5
4
3
2
1
at a
(6, 2)
(8, 3)
f
(0, 0)
x
2
−1
−2
−3
−4
−5
vt at
1
s t at2.
2
st d when t 2d
2ad.
a
The lowest speed is v 0.
1
2ad 0 2
ad
.
2
The time necessary to traverse the distance d at the mean
speed is
d
t
ad2
(2, − 2)
(b)
2da.
The highest speed is v a
The mean speed is
4 5 6 7 8 9
x
0
1
Fx
0
1
2
2
2
which is the same as the time calculated above.
7
2
4
4
5
7
2
6
7
8
2
1
4
3
0 ≤ x < 2
x,
x 4, 2 ≤ x < 6
(c) f x 1
x 1, 6 ≤ x ≤ 8
2
0 ≤ x < 2
x22,
2
Fx f t dt x 2 4x 4, 2 ≤ x < 6
0
14x2 x 5, 6 ≤ x ≤ 8
x
2d
a
3
Fx f x. F is decreasing on 0, 4 and increasing
on 4, 8. Therefore, the minimum is 4 at x 4, and
the maximum is 3 at x 8.
1,
1,
(d) F x fx 1,
2
0 < x < 2
2 < x < 6
6 < x < 8
x 2 is a point of inflection, whereas x 6 is not.
10. (a)
v
100
80
60
40
20
t
0.2
0.4
0.6
0.8
1.0
(b) v is increasing (positive acceleration) on 0, 0.4 and 0.7, 1.0.
(c) Average acceleration v0.4 v0 60 0
150 mihr2
0.4 0
0.4
(d) This integral is the total distance traveled in miles.
1
vt dt 0
1
385
38.5 miles
0 220 260 240 240 65 10
10
(e) One approximation is
a0.8 v0.9 v0.8 50 40
100 mihr2
0.9 0.8
0.1
(other answers possible)
Problem Solving for Chapter 4
x
11.
x
f tx t dt 0
d
dx
Thus,
0
x
x f t dt x
t f t dt x
0
x
x
0
0
t f t dt
0
x
f tx t dt x f x f t dt x
f t dt x f x 0
f t dt
0
Differentiating the other integral,
d
dx
x
x
0
x
f v dv dt 0
f v dv.
0
Thus, the two original integrals have equal derivatives,
x
x
f tx t dt 0
t
0
f v dv dt C.
0
Letting x 0, we see that C 0.
12. Consider Fx f x
⇒ Fx 2f x fx. Thus,
2
b
b
f x fx dx a
a
1
Fx dx
2
Sn b
1
Fb Fa
2
1
f b2 f a2 .
2
1
x5 dx 0
x6
6
1
0
1
0
2
. The corresponding
3
Riemann Sum using right-hand endpoints is
a
14. Consider
2
3
x dx x32
0
12 Fx
1
13. Consider
Thus, lim
n→ 1
n
1
n
2 . . .
n
1
1 2 . . . n .
n32
1 2 . . . n
n32
1
.
6
The corresponding Riemann Sum using right endpoints is
Sn 1
n
1
n
5
2
n
5
n
. . .
n
5
1 5
1 25 . . . n5 .
n6
Thus, lim Sn lim
n→ n→ 15 25 . . . n5 1
.
n6
6
b
15. By Theorem 4.8, 0 < f x ≤ M ⇒
b
f x dx ≤
a
b
f x dx.
m dx ≤
a
M dx Mb a.
a
b
Similarly, m ≤ f x ⇒ mb a a
b
Thus, mb a ≤
f x dx ≤ Mb a. On the interval 0, 1 , 1 ≤ 1 x4 ≤ 2 and b a 1.
a
1
Thus, 1 ≤
0
1
1 x4 dx ≤ 2.
Note:
0
1 x4 dx 1.0894
nn 2
.
3
437
438
Chapter 4
Integration
b
16. (a) Let A 0
f x
dx.
f x f b x
0
Let u b x, du dx.
0
A
b
b
0
b
0
Then,
3
0
x
3
dx 2
x
3 x
f b u
du
f b u f u
f b x
dx
f b x f x
b
0
f x
dx f x f b x
b
0
f b x
dx
f b x f x
b
1 dx b.
0
b
Thus, A .
2
17. (a) 1 i3 1 3i 3i2 i3 ⇒ 1 i3 i3 3i2 3i 1
3i2 3i 1 i 13 i3
(b)
n
n
3i2 3i 1 i1
i 13 i3
i1
23 13 33 23 . . . n 13 n3
n 13 1
n
Hence, n 13 3i2 3i 1 1.
i1
(c) n 13 1 n
n
3i2 3i 1 i1
3i2 i1
3nn 1
n
2
3nn 1
n
⇒
3i2 n3 3n2 3n 2
i1
n
n
⇒
2n3 6n2 6n 3n2 3n 2n
2
2n3 3n2 n
2
nn 12n 1
2
i2 nn 12n 1
6
i1
18. Since f x ≤ f x ≤ f x ,
b
f x dx ≤
a
b
a
f x dx ≤
b
a
b
f x dx ⇒
a
40
f 0 4f 1 2f 2 4f 3 f 4
34
1
1
1
4 42 21 4
5.417
3
2
4
f x dx ≤
19. (a) R < I < T < L
(b) S4 sin x
1
dx sin1 x sin x
2
(c) b 3, f x x
f b u
du
f b u f u
2A 1
(b) b 1 ⇒
b
a
f x dx.
Problem Solving for Chapter 4
x
20. Six 0
sin t
dt
t
(a)
2
−12
12
−2
(b) Si x sin x
x
Six 0 for x 2n
For positive x, x 2n 1
For negative x, x 2n
Maxima at , 3, 5, . . . and 2, 4, 6, . . .
(c) Si x x cos x sin x
0
x2
x cos x sin x for x 4.4934
Si4.4934 1.6556
(d) Horizontal asymptotes at y ±
lim Six x→ 2
lim Six x→
2
2
439
Problem Solving for Chapter 4
2
94.
1.75 sin
0
t
t
2
dt 1.75 cos
2
2
2
433
2
7
1.751 1 2.2282 liters
Increase is
7
5.1 1.9
0.6048 liters.
2
1
2
Simpson’s Rule n 4:
1
1
1
1
2
2
2
1
dx 0.257
1 x3
8 1 13 1 1.253 1 1.53 1 1.753 1 23
95. Trapezoidal Rule n 4:
1
1
1
4
2
4
1
dx 0.254
1 x3
12 1 13 1 1.253 1 1.53 1 1.753 1 23
Graphing utility: 0.254
1
96. Trapezoidal Rule n 4:
0
1
Simpson’s Rule n 4:
0
1
21432
21232
23432
1
x32
0
0.172
2 dx 2
2
3x
8
3 14
3 12
3 342 2
1
41432
21232
43432
1
x32
0
0.166
2 dx 2
2
3x
12
3 14
3 12
3 342 2
Graphing utility: 0.166
97. Trapezoidal Rule n 4:
2
x cos x dx 0.637
0
Simpson’s Rule n 4: 0.685
Graphing Utility: 0.704
98. Trapezoidal Rule n 4:
1 sin2 x dx 3.820
0
Simpson’s Rule n 4: 3.820
Graphing utility: 3.820
Problem Solving for Chapter 4
1
1. (a) L1 1
1
dt 0
t
x
(c) Lx 1 2.718
1
1
1
dt for x 2.718
t
1
dt 0.999896
t
(Note: The exact value of x is e, the base of the
natural logarithm function.)
1
by the Second Fundamental Theorem of Calculus.
x
(b) Lx L1 1
x1
(d) We first show that
1
To see this, let u x1
Then
1
1
dt t
Now, Lx1x2 1
dt t
1
1x1 t
1
x1 du ux
1x1 1
x1x2
1
dt t
x2
1
1
du u
1
1x1 u
x2
1
du 1x1 u
x1
dt.
1
t
and du dt.
x1
x1
1
1
1
1
1
x2
1
Lx1 Lx2 .
1
1
du 1x1 u
1
dt.
1x1 t
du using u 1
du
u
1
du
u
1
t
x1
434
Chapter 4
Integration
x
2. (a) Fx sin t2 dt
2
x
Fx
(b) Gx 0
1.0
1.5
1.9
2.0
2.1
2.5
3.0
4.0
5.0
0.8048
0.4945
0.0265
0.0611
0
0.0867
0.3743
0.0312
0.0576
0.2769
1
x2
x
Gx
x
sin t2 dt
2
1.9
1.95
1.99
2.01
2.1
0.6106
0.6873
0.7436
0.7697
0.8671
lim Gx 0.75
x→2
Fx F2
x2
(c) F2 lim
x→2
1
x→2 x 2
lim
x
sin t2 dt
2
lim Gx
x→2
Since Fx sin x2, F2 sin 4 lim Gx. Note: sin 4 0.7568
x→2
3. y x 4 4x 3 4x 2,
ci 0, 2 ,
2i
n
2
(a) x , f x x 4 4x 3 4x 2
n
n
A lim
n→ i1
n
lim
n→ i1
n
(b)
i1
4
2i
n
f ci x
4
2i
n
4
4
3
2i
n
4
9n
8n 16n 15n
n
8n 1n 15n
3
2
3
2
3
n→
n
n→ i1
lim
n
n→ i1
12
2i
n
2
ci n 1 2 16
n 15
2i
,
n
x 2
n
f ci x
2
5
8n 12 8n 12n 1 2
n
3n
n
3
(a) A lim
i1
2n
n 1 2
n
3
0, 2 ,
2n
2
2
n 1
n
8n 1n 15n
1
4. y x 5 2x 3,
2
n
2i
n
2i
n
4
3
(c) A lim
(b)
3
2i
n
1 2i
n
2
5
2i
n
n→
2
3
2i
n
2n
3
2n
4n 1 2n3n 2n 1 4n n 1 2n
8n 1 5n3n 2n 1
2
2
3
2
2
4
8n 1 5n3n 2n 1 403
(c) A lim
2
2
2
4
Problem Solving for Chapter 4
t2
dt
2
x
5. Sx sin
0
(b)
y
(a)
435
y
1.00
2
0.75
1
0.50
x
1
3
0.25
−1
x
−2
1
−0.25
2
The zeros of y sin
3
2
5
6
72 23
x2
correspond to the relative
2
extrema of Sx.
(c) Sx sin
x2
x2
0 ⇒
n ⇒ x2 2n ⇒ x 2n, n integer.
2
2
Relative maximum at x 2 1.4142 and x 6 2.4495
Relative minimum at x 2 and x 8 2.8284
(d) S x cos
x2
x2 n ⇒ x2 1 2n ⇒ x 1 2n, n integer
x 0 ⇒
2
2
2
Points of inflection at x 1, 3, 5, and 7.
1
6. (a)
1
cos x dx cos 1
1
1
3
1
cos x dx sin x
1
cos
1
3
2 cos
1
3
1.6758
2 sin1 1.6829
Error: 1.6829 1.6758 0.0071
1
(b)
1 1
1
1
1
3
dx x2
1 13 1 13 2
(Note: exact answer is 2 1.5708)
(c) Let px ax3 bx2 cx d.
1
1
p 3
7. (a) Area ax4
4
px dx 1
3
p
1
3
bx3 cx2
dx
3
2
3
9 x2 dx
2b
2d
3
b
b
2b
d d 2d
3
3
3
(c) Let the parabola be given by y b2 a2x2, a, b > 0.
0
ba
Area 2
3
b2 a2x2 dx
2
2
(b) Base 6, height 9, Area bh 69 36
3
3
b2
2 b2x a2
2 b2
x3 ba
3
0
2
b
a b
a
3 a
3
y
b3 1 b3
4 b3
2
a
3 a
3 a
10
8
7
6
5
4
3
2
1
Base x
1 2
4 5
y
0
0
2 27 9 36
−2 −1
1
3
9 x2 dx 2
x3
2 9x 3
−4
1
−
b
a
2b
, height b2
a
Archimedes’ Formula: Area 2 2b 2
4 b3
b 3 a
3 a
b
a
x
436
Chapter 4
Integration
8. Let d be the distance traversed and a be the uniform
acceleration. We can assume that v0 0 and s0 0.
Then
y
9. (a)
5
4
3
2
1
at a
(6, 2)
(8, 3)
f
(0, 0)
x
2
−1
−2
−3
−4
−5
vt at
1
s t at2.
2
st d when t 2d
2ad.
a
The lowest speed is v 0.
1
2ad 0 2
ad
.
2
The time necessary to traverse the distance d at the mean
speed is
d
t
ad2
(2, − 2)
(b)
2da.
The highest speed is v a
The mean speed is
4 5 6 7 8 9
x
0
1
Fx
0
1
2
2
2
which is the same as the time calculated above.
7
2
4
4
5
7
2
6
7
8
2
1
4
3
0 ≤ x < 2
x,
x 4, 2 ≤ x < 6
(c) f x 1
x 1, 6 ≤ x ≤ 8
2
0 ≤ x < 2
x22,
2
Fx f t dt x 2 4x 4, 2 ≤ x < 6
0
14x2 x 5, 6 ≤ x ≤ 8
x
2d
a
3
Fx f x. F is decreasing on 0, 4 and increasing
on 4, 8. Therefore, the minimum is 4 at x 4, and
the maximum is 3 at x 8.
1,
1,
(d) F x fx 1,
2
0 < x < 2
2 < x < 6
6 < x < 8
x 2 is a point of inflection, whereas x 6 is not.
10. (a)
v
100
80
60
40
20
t
0.2
0.4
0.6
0.8
1.0
(b) v is increasing (positive acceleration) on 0, 0.4 and 0.7, 1.0.
(c) Average acceleration v0.4 v0 60 0
150 mihr2
0.4 0
0.4
(d) This integral is the total distance traveled in miles.
1
vt dt 0
1
385
38.5 miles
0 220 260 240 240 65 10
10
(e) One approximation is
a0.8 v0.9 v0.8 50 40
100 mihr2
0.9 0.8
0.1
(other answers possible)
Problem Solving for Chapter 4
x
11.
x
f tx t dt 0
d
dx
Thus,
0
x
x f t dt x
t f t dt x
0
x
x
0
0
t f t dt
0
x
f tx t dt x f x f t dt x
f t dt x f x 0
f t dt
0
Differentiating the other integral,
d
dx
x
x
0
x
f v dv dt 0
f v dv.
0
Thus, the two original integrals have equal derivatives,
x
x
f tx t dt 0
t
0
f v dv dt C.
0
Letting x 0, we see that C 0.
12. Consider Fx f x
⇒ Fx 2f x fx. Thus,
2
b
b
f x fx dx a
a
1
Fx dx
2
Sn b
1
Fb Fa
2
1
f b2 f a2 .
2
1
x5 dx 0
x6
6
1
0
1
0
2
. The corresponding
3
Riemann Sum using right-hand endpoints is
a
14. Consider
2
3
x dx x32
0
12 Fx
1
13. Consider
Thus, lim
n→ 1
n
1
n
2 . . .
n
1
1 2 . . . n .
n32
1 2 . . . n
n32
1
.
6
The corresponding Riemann Sum using right endpoints is
Sn 1
n
1
n
5
2
n
5
n
. . .
n
5
1 5
1 25 . . . n5 .
n6
Thus, lim Sn lim
n→ n→ 15 25 . . . n5 1
.
n6
6
b
15. By Theorem 4.8, 0 < f x ≤ M ⇒
b
f x dx ≤
a
b
f x dx.
m dx ≤
a
M dx Mb a.
a
b
Similarly, m ≤ f x ⇒ mb a a
b
Thus, mb a ≤
f x dx ≤ Mb a. On the interval 0, 1 , 1 ≤ 1 x4 ≤ 2 and b a 1.
a
1
Thus, 1 ≤
0
1
1 x4 dx ≤ 2.
Note:
0
1 x4 dx 1.0894
nn 2
.
3
437
438
Chapter 4
Integration
b
16. (a) Let A 0
f x
dx.
f x f b x
0
Let u b x, du dx.
0
A
b
b
0
b
0
Then,
3
0
x
3
dx 2
x
3 x
f b u
du
f b u f u
f b x
dx
f b x f x
b
0
f x
dx f x f b x
b
0
f b x
dx
f b x f x
b
1 dx b.
0
b
Thus, A .
2
17. (a) 1 i3 1 3i 3i2 i3 ⇒ 1 i3 i3 3i2 3i 1
3i2 3i 1 i 13 i3
(b)
n
n
3i2 3i 1 i1
i 13 i3
i1
23 13 33 23 . . . n 13 n3
n 13 1
n
Hence, n 13 3i2 3i 1 1.
i1
(c) n 13 1 n
n
3i2 3i 1 i1
3i2 i1
3nn 1
n
2
3nn 1
n
⇒
3i2 n3 3n2 3n 2
i1
n
n
⇒
2n3 6n2 6n 3n2 3n 2n
2
2n3 3n2 n
2
nn 12n 1
2
i2 nn 12n 1
6
i1
18. Since f x ≤ f x ≤ f x ,
b
f x dx ≤
a
b
a
f x dx ≤
b
a
b
f x dx ⇒
a
40
f 0 4f 1 2f 2 4f 3 f 4
34
1
1
1
4 42 21 4
5.417
3
2
4
f x dx ≤
19. (a) R < I < T < L
(b) S4 sin x
1
dx sin1 x sin x
2
(c) b 3, f x x
f b u
du
f b u f u
2A 1
(b) b 1 ⇒
b
a
f x dx.
Problem Solving for Chapter 4
x
20. Six 0
sin t
dt
t
(a)
2
−12
12
−2
(b) Si x sin x
x
Six 0 for x 2n
For positive x, x 2n 1
For negative x, x 2n
Maxima at , 3, 5, . . . and 2, 4, 6, . . .
(c) Si x x cos x sin x
0
x2
x cos x sin x for x 4.4934
Si4.4934 1.6556
(d) Horizontal asymptotes at y ±
lim Six x→ 2
lim Six x→
2
2
439