C H A P T E R Integration
4
Section 4.1
Antiderivatives and Indefinite Integration . . . . . . . . . 450
Section 4.2
Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
Section 4.3
Riemann Sums and Definite Integrals . . . . . . . . . . . 462
Section 4.4
The Fundamental Theorem of Calculus . . . . . . . . . . 466
Section 4.5
Integration by Substitution . . . . . . . . . . . . . . . . . 472
Section 4.6
Numerical Integration
Review Exercises
. . . . . . . . . . . . . . . . . . . 479
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
C H A P T E R Integration Section 4.1
4
Antiderivatives and Indefinite Integration
Solutions to Even-Numbered Exercises
2.
1 d 4 1 x C 4x3 2 dx x x
4.
d 2 x2 3 d 2 3 2 x 2x 1 2 C C dx dx 3 3 x
x1 2 x 3 2 6.
dr d
8.
r C Check:
2x 2 1 C 2 c 2 x
Check:
d 1 C 2x 3 dx x 2
Given
Rewrite
Integrate
Simplify
10.
1 dx x2
x 1 C 1
1 C x
12.
x x2 3 dx
x3 3x dx
x4 x2 3 C 4 2
1 4 3 2 x x C 4 2
14.
1 x 1 C 9 1
1 C 9x
16.
1 dx 3x 2
1 2 x dx 9
5 x dx 5x
x2 C 2
Check:
20.
18.
x4 2x 2 2x C 4
4x3 6x 2 1 dx x 4 2x 3 x C d 4 x 2x 3 x C 4x 3 6x 2 1 dx
d x4 2x 2 2x C x3 4x 2 dx 4
x
dx x 2 x 1
1 2
1 x3 2 1 x1 2 2 x 1 2 dx C x3 2 x1 2 C 2 3 2 2 1 2 3
d 2 3 2 1 1 x x1 2 C x1 2 x 1 2 x dx 3 2 2 x
Check:
x 3 4x 2 dx
Check:
450
x2 d 5x C 5 x dx 2
Check:
22.
x 2 dx
x2 1 x3 2
dy 2x 3 dx y
d C d
Section 4.1
24.
4 x3 1 dx
Check:
28.
4 x3 4 1 dx x7 4 x C 7
x 2 2x 3 3x 4 dx
x 1 2x 2 3x 3 C 1 2 3
1 1 1 2 3 C x x x
36.
40.
x 4 dx
d 1 1 C 4 dx 3x3 x
2t2 1 2 dt
30.
4t4 4t2 1 dt
4 4 t5 t3 t C 5 3 Check:
d 45 43 t t t C 4t 4 4t2 1 dt 5 3 2t2 1 2
x 2 2x 3 x4
1 3 t2 3t3 dt t3 t4 C 3 4
3 dt 3t C
34.
d 13 34 t t C t2 3t3 1 3t t2 dt 3 4
Check:
1 t 2 sin t dt t3 cos t C 3
1 2 sec2 d 3 tan C 3
sec y tan y sec y dy
Check:
sec y tan y sec2 y dy
42.
d 3t C 3 dt
38.
d 13 t cos t C t2 sin t dt 3
d 1 3 tan C 2 sec2 d 3
cos x dx 1 cos2 x
sec y tan y C
d Check: sec y tan y C sec y tan y sec2 y dy
Check:
sec y tan y sec y
cos x dx sin2 x
46. f x x
y 6
f x C=3
x2 C 2 y
C=0 C = −2 4
6
d 1 csc x C csc x cot x dx sin x
y
2
x
f ( x) = 2
2
f′ f (x) = − 1x + 1
f′
4
−4
−4
x
−2
4 −2
−4
−2
cos x sin x
1 x2
6
−2
cos x 1 cos2 x
2 f ( x) = x + 2
8 8
x dx cos sin x
1 f x C x
x
−2
1 sin x
48. f x
4 2
csc x cot x dx csc x C
44. f x x
451
x 3 1 C 3 C 3 3x
1 3t t2 dt
Check:
d 1 1 1 2 3 C x 2 2x 3 3x 4 dx x x x
Check:
1 dx x4
Check:
32.
26.
d 4 7 4 4 x3 1 x x C x3 4 1 dx 7
x 2 2x 3 dx x4
Check:
Antiderivatives and Indefinite Integration
x 2
4
−4
f (x) = − 1x
452
50.
Chapter 4
Integration
dy 2 x 1 2x 2, 3, 2 dx y
52.
2 x 1 dx x2 2x C
2 3 2 2 3 C ⇒ C 1 y x2 2x 1
54. (a)
y
(b)
5
4
3
1 C ⇒ C 2 1
y
1 2, x > 0 x
x 2 dx
1 C x
5
(−1, 3)
y
x3 x C 3
3
1 3 1 C 3
−4
4
−5
1 3 1 C 3
−5
C y
56. g x 6x 2, g 0 1 g x
y
dy x2 1, 1, 3 dx
x
−4
1 dy 2 x 2, 1,3 dx x
6x 2 dx 2x3 C
7 3 7 x3 x 3 3 58. f s 6s 8s 3, f 2 3 f s
6s 8s 3 ds 3s 2 2s 4 C
g 0 1 2 0 3 C ⇒ C 1
f 2 3 3 2 2 2 2 4 C 12 32 C ⇒ C 23
g x 2x3 1
f s 3s 2 2s 4 23 62. f x sin x
60. f x x2 f 0 6
f 0 1
f 0 3 f x
f 0 6
1 x 2 dx x3 C1 3
f x
sin x dx cos x C1
f 0 0 C1 6 ⇒ C1 6
f 0 1 C1 1 ⇒ C1 2
1 f x x3 6 3
f x cos x 2
f x
1 3 1 x 6 dx x 4 6x C2 3 12
f 0 0 0 C2 3 ⇒ C2 3 f x
1 4 x 6x 3 12
f x
cos x 2 dx sin x 2x C2
f 0 0 0 C2 6 ⇒ C2 6 f x sin x 2x 6
Section 4.1
64.
dP k t, 0 ≤ t ≤ 10 dt P t
2 kt1 2 dt kt3 2 C 3
Antiderivatives and Indefinite Integration
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 , .
P 0 0 C 500 ⇒ C 500
y 3
2 P 1 k 500 600 ⇒ k 150 3 2 P t 150 t3 2 500 100t3 2 500 3
f′
2
−3
x
−2
1
70. v0 16 ft sec
f 0 s0
(a)
s t 16t2 16t 64 0 16 t2 t 4 0
32 dt 32t C1
t
f 0 0 C1 v0 ⇒ C1 v0 f t 32t v0 f t s t
1 ± 17 2
Choosing the positive value,
32t v0 dt 16t2 v0t C2
f 0 0 0 C2 s0 ⇒ C2 s0
3
−3
s0 64 ft
2
−2
f
f 0 v0
f t v t
f ′′
1
P 7 100 7 3 2 500 2352 bacteria 68. f t a t 32 ft sec2
453
t
1 17 2.562 seconds. 2 v t s t 32t 16
(b)
f t 16t 2 v0t s0 v
1 2 17 32 1 2 17 16
16 17 65.970 ft sec 72. From Exercise 71, f t 4.9t2 1600. (Using the canyon floor as position 0.) f t 0 4.9t2 1600 4.9t2 1600 t2
1600 ⇒ t 326.53 18.1 sec 4.9
74. From Exercise 71, f t 4.9t2 v0t 2. If f t 200 4.9t2 v0t 2, then v t 9.8t v0 0 for this t value. Hence, t v0 9.8 and we solve 4.9
9.8 v0
2
v0
9.8 2 200 v0
v2 4.9 v02 0 198 2 9.8 9.8 4.9 v02 9.8 v02 9.8 2 198 4.9 v02 9.8 2 198 v02 3880.8 ⇒ v0 62.3 m sec.
454
76.
Chapter 4
v dv GM
Integration
1 dy y2
1 2 GM v C 2 y
15t 9
(b) v t > 0 when 0 < t <
5 and 3 < t < 5. 3
7 (c) a t 6t 14 0 when t . 3
1 GM C v02 2 R 1 2 GM 1 2 GM v v0 2 y 2 R
v
73 3 73 5 73 3 2 32 34
2GM 2GM v02 y R
v2 v02 2GM
80. (a) a t cos t
f t
7t2
a t v t 6t 14
1 2 GM v C 2 0 R
v t
t3
(a) v t x t 3t2 14t 15 3t 5 t 3
When y R, v v0.
v2
0 ≤ t ≤ 5
78. x t t 1 t 3 2
1y R1
a t dt
cos t dt sin t C1 sin t since v0 0
v t dt
sin t dt cos t C2
f 0 3 cos 0 C2 1 C2 ⇒ C2 4 f t cos t 4 (b) v t 0 sin t for t k , k 0, 1, 2, . . . v 0 45 mph 66 ft sec
(a) 16.5t 66 44
30 mph 44 ft sec
t
15 mph 22 ft sec a t a
s
v t at 66
(b) 16.5t 66 22 t
ec
132 when a a t 16.5 v t 16.5t 66
s t 8.25t 2 66t
mp
h=
66
30
33 16.5. 2
44
ft/s
h=
(c)
66 66 a
mp
2
44 16.5 117.33 ft ft/s ec mp 0m h= ph 22 f t/se c
s
66 at 66 0 when t . a
44 2.667 16.5
0
15
v t 0 after car moves 132 ft.
66 a 66 s a 2 a
22 1.333 16.5
22 16.5 73.33 ft
a s t t2 66t Let s 0 0. 2
45
82.
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.
Section 4.1
Antiderivatives and Indefinite Integration
30
84. No, car 2 will be ahead of car 1. If v1 t and v2 t are the respective velocities, then
0
86. (a) v 0.6139t3 5.525t2 0.0492t 65.9881
30
v2 t dt >
455
v1 t dt.
0
0.6139t4 5.525t3 0.0492t2 65.9881t 4 3 2 Note: Assume s 0 0 is initial position
(b) s t
v t dt
s 6 196.1 feet
88. Let the aircrafts be located 10 and 17 miles away from the airport, as indicated in the figure. vA t kA t 150
vB kB t 250
1 sA t kA t2 150t 10 2
Airport
A
B
0
10
17
1 sB kB t2 250t 17 2
(a) When aircraft A lands at time tA you have vA tA kA tA 150 100 â&#x2021;&#x2019; kA
50 tA
1 sA tA kA t2A 150tA 10 0 2
1 50 2 t 150tA 10 2 tA A 125tA 10 tA
kA
10 . 125
50 125 625 2 t 150t 10 50 625 â&#x2021;&#x2019; SA t S1 t tA 10 2
Similarly, when aircraft B lands at time tB you have vB tB kB tB 250 115 â&#x2021;&#x2019; kB
135 tB
1 sB tB kB t2B 250tB 17 0 2
1 135 2 t 250tB 17 2 tB B 365 t 17 2 B tB
kB (b)
34 . 365
135 365 49,275 49,275 2 135 â&#x2021;&#x2019; SB t S2 t t 250t 17 tB 34 34 68 (c) d sB t sA t
20
20
Yes, d < 3 for t > 0.0505. sB
d 3
sA 0.1
0
0
0
90. True
0.1
0
92. True
456
Chapter 4
Integration
94. False. f has an infinite number of antiderivatives, each differing by a constant.
96.
d s x 2 c x 2 2s x s x 2c x c x dx 2s x c x 2c x s x 0 Thus, s x c x 2 k for some constant k. Since, 2
s 0 0 and c 0 1, k 1. Therefore,
s x 2 c x 2 1. [Note that s x sin x and c x cos x satisfy these properties.]
Section 4.2
Area 5
6
2.
k k 2 3 1 4 2 5 3 6 4 50
4.
4
i 1
2
1
1
1
47
j 3
k 3
6.
1
j 3 4 5 60
i 1 3 0 8 1 27 4 64 9 125 238
i 1
10.
2
2 n 2i 1 1 n i 1 n
18.
i
10
15
2i 3 2 i 3 15
i 1
12.
j 1
15
16.
j 1 4
4
5 1 i i 1
2
i 1
i 1
10
2
1
i 1
10
i
3
i 1
10
i
22. sum seq x 15
i
10 11 102 11 2 3080 4 2
3
2i
15 2 15 1 2 15 15 1 2 4 2 15 2 16 2 15 16 14,160 4
2 1 55 3 2 6
2 1 1 9 4.5 3 2 3 2
26. S 5 2 1 s 2 1
14 2 14 12 2 14 34 2 14
i 1
3 2x, x, 1, 15, 1 14,160 (TI-82)
i 1
s 4 4 2 0 1 10
28. S 8
1 2
14
54 2 14 32 2 14 74 2 14
2 2
5 6 7 1 1 2 3 16 1 2 6.038 4 2 2 2 2 2 2
1 s 8 0 2 4
10
24. S 5 5 4 2 1 16
i 1 n
1 n 1 n i 0
1
i 1
2
14.
10 11 21 10 375 6
15 216 45 195
i i
2
i 1
2
20.
10
i
1
>
15
8.
14 2 14 12 2 14 . . . 74 2 14 5.685
14
2
Section 4.2
30. S 5 1 s 5
15 1 15 15 1 25 15 1 35 15 1 45 15 2
2
n→
2
24 21 16 9 1 1
0.859 5 5 5 5 5
1 15 15 1 25 15 1 35 15 1 45 15 0 0.659 2
2
2
64n n n 1 6 2n 1 646 lim 2n
3
32. lim
2
3
n→
2
3n2 n 64 64 2 n3 6 3
n1 n n 2 1 21 lim n n n 21 1 21 2
34. lim n→
2
4j 3 1 n 1 4n n 1 2n 5 3n S n 2 4j 3 2 2 n j 1 n 2 n j 1 n n
36.
2
n→
S 10
25 2.5 10
S 100 2.05 S 1000 2.005 S 10,000 2.0005 4 n 3 4 n2 n 1 2 n n 1 2n 1 4i2 i 1 4 i i2 4 4 n n n 4 6 i 1 i 1 n
38.
4 n3 2n2 n 2n2 3n 1 n3 4 6
1 3n3 6n2 3n 4n2 6n 2 3n3
1 3n3 2n2 3n 2 S n 3n3
S 10 1.056 S 100 1.006566 S 1000 1.00066567 S 10,000 1.000066657
40. lim
n n
42. lim
1 n n
n
2i
2
n→ i 1
n
n→ i 1
2i
2
lim
n→
2
4 n 4 n n 1 4 1 i lim 2 1 2 lim n→ n n→ 2 n2i 1 2 n
lim
n→
2 lim
n n
2
i 1
i 4 i
n
n
i 1
i 1
4n
2
2 3 n n 1 4 n n 1 2n 1 n 4n n3 2 6
n→
2 n→ n3 n→
2 n n 2i 2 n3i 1
lim lim
1 2 n2 34 n2 3n2
2 1 2
2
4 26 3 3
Area
457
458
Chapter 4
1 n n 2 lim n n 2i n
3
2i
44. lim n→
Integration
2
n
1
n→
i 1
3
4
i 1
1 n 3 n 6n2i 12ni 2 8i 3 n→ n4 i 1
2 lim 2 lim
1 4 n n 1 n n 1 2n 1 n2 n 1 2 n 6n2 12n 8 n4 2 6 4
2 lim
1 3 n3 4 n6 n2 2 n4 n2
n→
2 lim 10
46. (a)
2
n→
n→
2
4 13 2 20 n n
(b) x
y
3 1 2 n n
4
Endpoints: 3
1 < 1
2 1
1 <1 1
x 2
4
2 4 2n < 1 <...< 1 3 n n n
2n < 1 2 2n < . . . < 1 n 1 2n < 1 n 2n
(c) Since y x is increasing, f mi f xi 1 on xi 1, xi . s n
n
f x
i 1
x
i 1
f 1 i 1 n n 1 i 1 n n n
2
n
2
i 1
2
2
i 1
(d) f Mi f xi on [xi 1, xi
f x x f 1 i n n 1 i n n n
S n
n
i 1
(e)
2
2
x
i 1
2
2
i 1
5
10
50
100
s n
3.6
3.8
3.96
3.98
S n
4.4
4.2
4.04
4.02
1 i 1 n n n
2
(f) lim n→
n
i
2
i 1
lim
n→
lim
n→
1 i n n n
lim
n→ i 1
2
2
lim
n→
lim
n→
2n n n2 n n 2 1 n
2
2n 2 4 2 lim 4 4 n→ n n n
2 2 n n 1 n n n 2
2
2 n 1 2 lim 4 4 n→ n n
Section 4.2
Note: x 5 n 2 3n
48. y 3x 4 on 2, 5 .
3i 4 n
n
3i f 2 S n n i 1
n
3 2
i 1
8
n→
6
i 12
4
i 1
2
x 4
6
8
10
12
27 39 2 2
50. y x 2 1 on 0, 3 . Note: x
n 3 n l 1
3i S n f n l 1
2 n
3 3 18 3 n n
Area lim S n 6
n
12 10
459
y
3 n
27 1 27 n 1 n 6 1 n2 2 2 n
6
Area
3i n
2
3 0 3 n n
y 12
3 n
1
10 8 6
27 n 2 3 n l 1 n 3 l 1 n l 1
27 n n 1 2n 1 3 9 2n 3n 1 n 3 n3 6 n 2 n2
4 2 x
2
1
3
2
9 Area lim S n 2 3 12 n → 2 52. y 1 x2 on 1, 1 . Find area of region over the interval 0, 1 .
n
i s n f n i 1
Note: x n1 3
n 1 i 1 n n i 1
2
1 n
2
n n 1 2n 1 1 3 1 1 n 2 i 1 1 2 2 n3 i 1 6n3 6 n n
1
y
x
−2
2 −1
1 2 1 Area lim s n 1 n→ 2 3 3 Area
4 3
Note: x 1 n 0 1n
54. y 2x x3 on 0, 1 .
y
Since y both increases and decreases on 0, 1 , T n is neither an upper nor lower sum.
2.0 1.5
T n
f n n 2 n n n n
i
n
1
i 1
i
i
3
1
1.0
i 1
0.5
2 1 n n 1 1 n n 1 i 4 i3 4 n2 i 1 n i 1 n2 n 4 n
1
2 1 1 1 n 4 4n 4n2
Area lim T n 1 n→
n
1 3 4 4
2
2
x 0.5
1.0
2.0
460
Chapter 4
Integration
Note: x 0 n 1 n1
56. y x2 x3 on 1, 0 .
n
i s n f 1 n i 1
n
2
i 1
2
n 1 n i 1
5i 4i 2 i3 2 3 n n n
i 1 n
i 1 n
n→
1 n
2 1
5 n 4 n 2 1 n 3 1 2 2 i 3 i 4 i n n i 1 n i 1 n i 1
−2
60. f y 4y y 2, 1 ≤ y ≤ 2. Note: y
2
S n
i 1
n
1 2i 2 n i 1 2
2 1 n n 1 n 1 n 2 n n 2 n
2 2 i 1 n n i 1 n
Area lim S n 2 1 3 n →
5
n
1
i
2
1 n i i 4 1 1 n i 1 n n
1 n 4i 2i i2 4 1 2 n i 1 n n n
1 n 2i i2 3 2 n i 1 n n
1 2 n n 1 1 n n 1 2n 1 3n 2 n n 2 n 6
3
3 2 1 4
n 1 n 1 2n 1 n 6
Area lim S n 3 1 n →
x 3
5 y 5
3 2 1 x 1
62. h y y3 1, 1 ≤ y ≤ 2 Note: y S n
1 n
i
2
3
4
5
h 1 n n n
1
y
i 1
1 n n
i
3
5
n
1
i 1
1
1 n i3 3i2 3i 2 3 2 n i 1 n n n
4
n 1 2 1 n 1 2n 1 3 n 1 n24 2 n2 2n
Area lim S n 2 n →
3
1 1 n2 n 1 2 3 n n 1 2n 1 3 n n 1 2n 3 2 n n 4 n 6 n 2 2
f 1 n n
4
2
2 1 1 n n
y
1
2
i 1 n
1 −1
g 2 n n 2i
x
−1
5 4 1 7 2 3 4 12
1 4 2 2 58. g y y, 2 ≤ y ≤ 4. Note: y 2 n n n
3
3
1 5 4 2 2 1 1 5 2 2n 3 n 3n3 4 2n 4n2
Area lim s n 2
S n
2
y
1 3 19 1 4 2 4
2 1
x 2
4
6
8
10
1 11 3 3
Section 4.2 66. f x sin x, 0 ≤ x ≤
64. f x x2 4x, 0 ≤ x ≤ 4, n 4 Let ci
xi xi 1 . 2
Let ci
1 3 5 7 x 1, c1 , c2 , c3 , c4 2 2 2 2 Area
n
4
c
f ci x
2 i
i 1
4ci 1
Area
4 2 4 6 4 1
9
25
494 14
53 68. f x
Approximate area
4
8
12
16
20
2.3397
2.3755
2.3824
2.3848
2.3860
n
4
i
4
8
12
16
20
Approximate area
1.1041
1.1053
1.1055
1.1056
1.1056
72. See the Definition of Area. Page 259. 3 x, 0 ≤ x ≤ 8 74. f x
n
10
20
50
100
200
s n
10.998
11.519
11.816
11.910
11.956
S n
12.598
12.319
12.136
12.070
12.036
M n
12.040
12.016
12.005
12.002
12.001
(Note: exact answer is 12.) 76.
y
78. True.
4 3 2 1 1
2
3
4
a. A 3 square units
x
i
i 1
3 5 7 sin sin sin sin
1.006 8 16 16 16 16
70. f x cos x on 0, 2 . n
f c x sin c 8
8 on 2, 6 . x2 1
n
,n 4 2
3 5 7 ,c ,c ,c ,c 8 1 16 2 16 3 16 4 16
i 1
10
461
xi xi 1 . 2
x
i 1
Area
(Theorem 4.3)
462
Chapter 4
80. (a)
Integration
2 n
r
h (b) sin r
h
θ r
h r sin 1 1 1 A bh r r sin r2 sin 2 2 2 (c) An n
12r
2
sin
2 r 2n 2 sin 2 n sin r2 n 2 n 2 n
Let x 2 n. As n → , x → 0. lim An lim r 2
n→
x→0
sinx x r 1 r 2
2
n
82. (a)
n
2i n n 1
i
3
(b)
i 1
i 1
The formula is true for n 1: 2 1 1 1 2
n2 n 1 2 4
The formula is true for n 1 because 12 1 1 2 4 13 1 4 4
Assume that the formula is true for n k: k
2i k k 1 .
Assume that the formula is true for n k:
i 1
k 1
Then we have
k
k
2i 2i 2 k 1
i 1
i
3
i 1
i 1
k k 1 2 k 1
k2 k 1 2 4 k 1
Then we have
k 1 k 2
i
3
i 1
Which shows that the formula is true for n k 1.
k
i
3
k 1 3
i 1
k2 k 1 2 k 1 3 4
k 1 2 2 k 4 k 1
4
k 1 2 k 2 2 4
which shows that the formula is true for n k 1.
Section 4.3
Riemann Sums and Definite Integrals
3 2. f x x, y 0, x 0, x 1, ci
xi n
lim
n→ i 1
i3 n3
i3 i 1 3 3i 2 3i 1 3 n n3 n3
f ci xi lim
i 3i n n
3
3
n→ i 1
3
2
3i 1 n3
lim
1 n 3i 3 3i 2 i n4 i 1
lim
1 n2 n 1 2 n n 1 2n 1 n n 1 3 3 4 n 4 6 2
n→
n→
1 3n4 6n3 3n2 2n3 3n2 n n2 n n→ n4 4 2 2
lim
1 3n4 n3 n2 3 3 1 1 lim n→ n4 n→ 4 4 2 4 2n 4n2 4
lim
Section 4.3
Riemann Sums and Definite Integrals
Note: x 3 n 2 n5, â&#x2020;&#x2019; 0 as n â&#x2020;&#x2019;
4. y x on 2, 3 .
f c x f 2 n n 2 n n 10 n i n
n
i
5i
i 1
i 1
3
2
x dx lim
nâ&#x2020;&#x2019;
f ci xi
i 1
n
25 2
i 1
2
25 52 2n 25 3 1 2 , â&#x2020;&#x2019; 0 as n â&#x2020;&#x2019; n n
f 1 n n n
2i
2
i 1
n
3 1
i 1
2i n
2n 2
6 n 4i 4i2 1 2 ni 1 n n
6 4 n n 1 4 n n 1 2n 1 2 n n n 2 n 6
6 12
5
25 25n n n 2 1 10 252 1 n1 25 2n
6. y 3x 2 on 1, 3 . Note: x
5i
i 1
10
n
n
5
i
3
3x 2 dx lim
n â&#x2020;&#x2019;
1
n 1 n 1 2n 1 4 n n2
6 12 nn 1 4 n 1n 2n 1 2
6 12 8 26
8. y 3x 2 2 on 1, 2 . Note: x
2 1 3 ; â&#x2020;&#x2019; 0 as n â&#x2020;&#x2019; n n
f c x f 1 n n n
n
i
3i
i 1
i 1
2
1
2
3 n 3i 3 1 ni 1 n
3 n 6i 9i2 3 1 2 2 n i 1 n n
3 18 n n 1 27 n n 1 2n 1 3n 2 2n n n 2 n 6
2
15
3
i
3x 2 2 dx lim
n â&#x2020;&#x2019;
27 n 1 27 n 1 2n 1 n 2 n2
15 27
n 1 27 n 1 2n 1 n 2 n2
15 27 27 15 n
10. lim
â&#x2020;&#x2019;0
4
6ci 4 ci 2 xi
i 1
6x 4 x 2 dx
â&#x2020;&#x2019;0
0
0
i 1
i
1
2
16.
0
x2 dx
18.
1 dx 1 x 1 2
4
1
2
2
4 2x dx
3
2 i
on the interval 1, 3 .
on the interval 0, 4 .
14.
3 3 c x x dx n
12. lim
20.
0
tan x dx
463
464
Chapter 4
Integration
2
22.
y 2 2 dy
26. Triangle
24. Rectangle
0
1 1 A bh 4 2 2 2
A bh 2 4 a
a
A
a
4 dx 8a
4
A
0
y
x dx 4 2
y 5 3 2
3
Triangle Rectangle
2
1
1
x 1
â&#x2C6;&#x2019;a
a
A
8
8 x dx 32
a
0
3
4
32. Semicircle
1 1 A bh 2a a 2 2
1 1 A bh 8 8 32 2 2
2
â&#x2C6;&#x2019;1
30. Triangle
28. Triangle
A
x
a
1 A r2 2
r
a x dx a2
A
r 2 x 2 dx
r
y
y
Semicircle
r
y a
1 2 r 2
Triangle
10 8
â&#x2C6;&#x2019;a
x
a
â&#x2C6;&#x2019;r
x
r
6 4
â&#x2C6;&#x2019;r
2 x 2
4
6
8
10
4
In Exercises 34â&#x20AC;&#x201C; 40,
2
x3 dx 0
x3 4 dx
4
x3 dx 60,
x dx 6,
2
4
dx 2.
2
2
34.
2
15 dx 15
6 2x x3 dx 6
4
2
4
38.
4
36.
4
2
2
4
4
x3 dx 4
dx 15 2 30
2
dx 60 4 2 68
2
40.
4
2
4
dx 2
2
4
x dx
2
x3 dx
2
6 2 2 6 60 36
6
42. (a)
3
f x dx
0
0
3
(b)
f x dx
6
f x dx 4 1 3
44. (a)
f x dx 1 1
(b)
6
(d)
3
(c)
1
3
f x dx 5 1 5
(d)
0
f x dx 0 5 5
0
f x dx 5 5 10
1
1
6
5f x dx 5
1
1
1
f x dx 0
1
f x dx
0
f x dx
0
3
3
f x dx
1
6
1
1
3
3
(c)
0
6
f x dx
3f x dx 3
1
f x dx 3 0 0
1
3f x dx 3
0
f x dx 3 5 15
Section 4.3
5
46. (a)
0
5
f x dx
0
5
(c)
5
f x 2 dx
3
2 dx 4 10 14
(b)
5
f even
(d)
5
0
f x dx 4
Let u x 2.
0
f x dx 0
f odd
50. The average of Exercise 39 and Exercise 40 consists of a trapezoidal approximation, and is greater than the exact area: >
52. 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
54.
y
4
1
f x 2 dx
5
f x dx 2 4 8
48. The right endpoint approximation will be less than the actual area: <
1
465
5
2
0
5
f x dx 2
Riemann Sums and Definite Integrals
3
x dx x
2 1
is integrable, sketch a graph of the region bounded by f x x x and the x-axis for 1 â&#x2030;¤ x â&#x2030;¤ 1. You see that the integral equals 0.
x
1 4
1 2
3 4
1
4
b. A 3 square units
y
2 1 x
â&#x2C6;&#x2019;2
1
2
â&#x2C6;&#x2019;2
3
y
56.
58.
0
9 8 7 6 5 4 3 2 1
n
1 2 3 4 5 6 7 8 9
x
5 dx x2 1 4
8
12
16
20
L n
7.9224
7.0855
6.8062
6.6662
6.5822
M n
6.2485
6.2470
7.2460
6.2457
6.2455
R n
4.5474
5.3980
5.6812
5.8225
5.9072
c. Area 27.
3
60.
x sin x dx
0
4
8
12
16
20
L n
2.8186
2.9985
3.0434
3.0631
3.0740
M n
3.1784
3.1277
3.1185
3.1152
3.1138
R n
3.1361
3.1573
3.1493
3.1425
3.1375
n
62. False
1
0
x x dx
x dx 1
0
64. True
1
0
x dx
66. False
4
2
x dx 6
466
Chapter 4
Integration
68. f x sin x, 0, 2
x0 0, x1
, x , x3 , x4 2 4 2 3
2 , x2 , x3 , x4 4 12 3
x1
2 3 , c , c3 ,c 6 2 3 3 4 2
c1
f c x f 6 x 4
i
i
3 x
f
1
2
i 1
f
23 x
3
f
32 x
4
12 4 23 12 23 23 1 0.708
2
70. To find 0 x dx, use a geometric approach. y
3 2 1 x
â&#x2C6;&#x2019;1
1
2
3
â&#x2C6;&#x2019;1
Thus,
2
x dx 1 2 1 1.
0
Section 4.4
The Fundamental Theorem of Calculus
2. f x cos x
4. f x x 2 x
2
2
cos x dx 0
0
2
0
x 2 x dx is negative.
7
7
3 dv 3v
2
2
3 7 3 2 15
5
8.
752 20 6 8 392
3 3v 4 dv v2 4v 2 2
2
5x 2 4x 2
3x 2 5x 4 dx x3
1
5
27
3
10.
3 1
45 5 12 1 4 2 2
38
1
12.
1
t3 9t dt
1
14.
2
4t 1
4
1
9 t2 2
1
1
u u1 du u2 u1 2
2
2
â&#x2C6;&#x2019;2
2
â&#x2C6;&#x2019;5
â&#x2C6;&#x2019;2
6.
5
14 29 14 29 0
12 1 2 21 2
Section 4.4
3
16.
3
v1 3 dv
8
18.
1
34v
3 3
2
20.
1
8
1
2t1 2 t3 2 dt
1 x x2 dx 3 2 2 x
1
1 3 5 3 3 8 3 x x 2 5 8
3t 4
1
8
4
0
4
1
x80 24 15x 5 3
34. P
2
2
0
1
36. A
6 x dx
x2 2
40. A
0
1 32 4569 39 144 80 80 80
3 x 3 dx
92 21 24 8 18 29
1
6x
4 3
9 13 2 2
1
3
x2 4x 3 dx
0
4
x2 4x 3 dx
1
split up the integral at the zeros x 1, 3
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 0 4 3 3 3
3
d
4
2
x sin x dx
4
2
2
2
2 cos
sin d
0
1
x2 cos x 2
0
3
3
2
1
3x
4 3
4
0
x3 2x
3x
4
2
2 1 2 1 2 2
4
1
2
1 2
2 2 0 1 63.7%
0
1
3
2
4
0
1
3x
2
1 1 x 4 dx x x5 5 1
2 2 16 2 20 12 15 15
3
2 3
8
2t cos t dt t 2 sin t
2
4
x dx
2 csc2 x dx 2x cot x
2
0
1 sin2 d cos2
2
2
0
x2
4
2
x2 4x 3 dx
0
t t
15 20 6t
3
4 16 18
2 t5 2 5
3 2
8 2 2
4
1
32.
1
3 x 3 dx
3
30.
1
3
3 1x 31 dx
28.
8
x2 3 x5 3 dx
8
1
26.
8
1
4
24.
2 2x
x 1 2 dx 2 2 x1 2
0
8
2
2 t t dt
0
22.
2 dx 2 x
467
3 3 3 3 4 0 3 4 4
4 3
The Fundamental Theorem of Calculus
0
2
8 5
38. A
1
2 2 4 2 2 2
1 1 dx x2 x
2 1
1 1 1 2 2
468
Chapter 4
Integration
42. Since y â&#x2030;Ľ 0 on 0. 8 ,
8
8
3 1 x1 3 dx x x 4 3 4 0
Area
0
3 8 16 20 4
3
44. Since y â&#x2030;Ľ 0 on 0, 3 ,
46.
1
3
A
0
3 x3 3x x dx x2 2 3 2
3 0
9 . 2
9 9 dx 2 x3 2x
3 1
1 9 4 2 2
f c 3 1 4 9 2 c3 c3 c
3
48.
3
cos x dx sin x
3
3
3
50.
1 3 1
3
1
9 2
92 1.6510 3
4 x 2 1 dx 2 x2
3 3 3
1 x 2 dx 2 x
1
f c
cos c
3
2 3
1 16 3 3
3 3 2
c Âą 0.5971
52.
1 2 0
2
cos x dx
0
Average value
2
cos x
2
2
2 sin x
0
2
1.5
(0.881, Ď&#x20AC;2 ( 0
2.71
â&#x2C6;&#x2019;0.5
x 0.881 54. (a)
7
7
f x dx Sum of the areas
(b) Average value
1
7 1
1
A1 A2 A3 A4 1 1 1 3 1 1 2 2 1 3 1 2 2 2
f x dx
(c) A 8 6 2 20 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 x 1
2
3
4
5
6
7
2
3
4
5
6
7
8 4 6 3
1 x
3
1
Section 4.4
6
56.
6
f x dx area or region B
2
0
2
2
f x dx
The Fundamental Theorem of Calculus
f x dx
58.
2
2f x dx 2
0
0
469
f x dx 2 1.5 3.0
0
3.5 1.5 5.0 60. Average value
1 6
6
0
1 f x dx 3.5 0.5833 6
62.
1 R 0
R
k R r dr 2
2
0
3 R
k 2 r Rr R 3
0
2
2kR 3
64. P 5 t 30 (a)
t
1
2
3
4
5
6
P
155
157.071
158.660
160
161.180
162.247
954.158 1 159.026 Average profit 155 157.071 158.660 160 161.180 162.247 6 6 (b)
1 6
6.5
0.5
5 t 30 dt
2 1 5 t3 2 30t 6 3
6.5
5.0
954.061 159.010 6
(c) The definite integral yields a better approximation. 66. (a) R 2.33t 4 14.67t3 3.67t2 70.67t
4
(c)
R t dt
0
(b)
100
2.33t 5
5
14.67t4 3.67t3 70.67t2 4 3 2
181.957
â&#x2C6;&#x2019;1
5 â&#x2C6;&#x2019; 20
68. (a) histogram
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 83 60 4980 customers (c) Using a graphing utility, you obtain N t 0.084175t3 0.63492t2 0.79052 4.10317. (d)
16
â&#x2C6;&#x2019;2
10 â&#x2C6;&#x2019;2
9
(e)
N t dt 85.162
0
The estimated number of customers is 85.162 60 5110.
7
(f) Between 3 P.M. and 7 P.M., the number of customers is approximately
3
Hence, 3017 240 12.6 per minute.
N t dt 60 50.28 60 3017.
4 0
470
Chapter 4
Integration
x
70. F x
x
t3 2t 2 dt
4 t
4 x
x4 x 2 2x 4 4
2
t4
x4
F 2 4 4 4 4 0
2t
2
2
84 64 16 4 1068 4 2 3 dt 2 t
x
2t 3 dt
2
1 t2
t3 2t 2 dt 0
2
F 8
x
2
625 25 10 4 167.25 4
2x 4 4 4
Note: F 2
F 5
72. F x
2
x
2
1 1 x2 4
1 1 F 2 0 4 4 F 5
1 1 21 0.21 25 4 100
F 8
1 1 15 64 4 64
x
74. F x
x
sin d cos
0
0
cos x cos 0 1 cos x
F 2 1 cos 2 1.4161 F 5 1 cos 5 0.7163 F 8 1 cos 8 1.1455
x
76. (a)
0
(b)
14t
1 t2 2
4
x 0
1 1 x2 x4 x2 x2 2 4 2 4
d 1 4 1 2 x x x3 x x x2 1 dx 4 2
t dt
4
(b)
t3 t dt
0
x
78. (a)
x
t t2 1 dt
3t 2
x
3 2
4
80. (a)
3
d 2 3 2 16 x x1 2 x dx 3 3
(b)
x
82. F x F x
t2 dt 2 t 1 1
x2 x 1 2
84. F x
x
2 16 2 3 2 x 8 x3 2 3 3 3
x 4 t
1
4 x F x
x
sec t tan t dt sec t
sec x 2
3
d sec x 2 sec x tan x dx
x
dt
86. F x
sec3 t dt
0
F x sec3 x
Section 4.4
x
88. F x
t 3 dt
x
4 t4
x x
0
The Fundamental Theorem of Calculus
471
Alternate solution
x
F x
F x 0
t3 dt
x 0
x
x
t3 dt
0
x
t3 dt
x
t3 dt
0
t3 dt
0
F x x 3 1 x3 0
x
90. F x
2
t 3 dt
2
t 2
2
x2 2
1 2t2
x2 2
1 1 â&#x2021;&#x2019; F x 2x 5 2x4 8
Alternate solution: F x x 2 3 2x 2x 5
x
92. F x
2
sin 2 d
0
F x sin x 2 2 2x 2x sin x 4 94. (a)
x
1
2
3
4
5
6
7
8
9
10
g x
1
2
0
2
4
6
3
0
3
6
(b)
y 6 4 2
(c) Minimum of g at 6, 6 .
x
(d) Minimum at 10, 6 . Relative maximum at 2, 2 . (e) On 6, 10 g increases at a rate of
2
â&#x2C6;&#x2019;2
4
6
10
4
4 dt t2
12
â&#x2C6;&#x2019;4 â&#x2C6;&#x2019;6
12 3. 4
(f) Zeros of g: x 3, x 8.
96. (a) g t 4
4 t2
x
(b) A x
1
lim g t 4
4t
tâ&#x2020;&#x2019;
Horizontal asymptote: y 4
4 t
x 1
4x
4 8 x
4x2 8x 4 4 x 1 2 x x
lim A x lim 4x
xâ&#x2020;&#x2019;
xâ&#x2020;&#x2019;
4 8 0 8 x
The graph of A x does not have a horizontal asymptote. 98. True
100. Let F t be an antiderivative of f t . Then,
v x
u x
d dx
v x
u x
v x
f t dt F t
f t dt
u x
F v x F u x
d F v x F u x dx
F v x v x F u x u x f v x v x f u x u x .
472
Chapter 4
Integration
x
102. G x
s
f t dt ds
s
0
0
0
(a) G 0
s
s
0
(b) Let F s s
f t dt ds 0
s
x
x
(c) G x x
f x
G x
f t dt
0
f 0
F s ds
0
0
(d) G 0 0
f t dt.
0
0
G x F x x
f t dt 0
0
x
f t dt
0
0
G 0 0
0
104. x t t 1 t 3 2 t 3 7t 2 15t 9 x t 3t2 14t 15 Using a graphing utility,
5
Total distance
Section 4.5
2. 4. 6.
8.
f g x g x dx
u g x
du g x dx
x2 x3 1 dx
x3 1
3x2 dx
sec 2x tan 2x dx
2x
2 dx
cos x dx sin2 x
sin x
cos x dx
x2 9 3 2x dx
d x2 9 4 4 x2 9 3 C 2x x2 9 3 2x dx 4 4
3 1 2x2 1 3 4x dx 1 2x2 4 3 C 4
x2 x3 5 4 dx
4
3 1 2x2 1 3 4x 1 2x2 1 3 4x
1 1 x3 5 5 x3 5 5 x3 5 4 3x2 dx C C 3 3 5 15
d x3 5 5 5 x3 5 4 3x2 C x3 5 4x2 dx 15 15
x 4x2 3 3 dx
Check:
d 3 3 1 2x2 4 3 C dx 4 4
Check:
14.
x2 9 4 C 4
Check:
12.
Integration by Substitution
Check:
10.
x t dt 27.37 units
0
1 1 4x2 3 4 4x2 3 4 4x2 3 3 8x dx C C 8 8 4 32
d 4x2 3 4 4 4x2 3 3 8x C x 4x2 3 3 dx 32 32
f t dt 0
Section 4.5
16.
t3 t4 5 dt
Check:
18.
x2
1
3
2 u3 2 1 2 3u2 u3 2 1 2 u2
d 1 1 x3 4 2 3 4 C 1 x 4x dx 4 1 x 4 1 x4 2
1 x2 1 d C 1 16 x3 2 3x2 3 dx 3 16 x 3 16 x3 2
2 x
1 dx 3x 2
x3 1 x4
1 x3 1 x 1 x3 1 3x4 1 x2 x 2 dx C C C 9 3 9 1 3 9x 9x
1 1 2 1 x1 2 x dx C x C 2 2 1 2
d x C 1 dx 2 x
dt
2 4 2 t1 2 2t3 2 dt t3 2 t 5 2 C t 3 2 5 6t C 3 5 15
d 2 3 2 4 5 2 t 2t2 t t C t1 2 2t3 2 dt 3 5 t
t3 1 2 dt 3 4t
Check:
1 2
1 x4 1 2 4x3
d 1 3 1 1 1 1 x x C x2 x 2 x2 dx 3 9 9 3x 2
t 2t2 t
d 1 x4 1 C dx 2 2
dx
Check:
32.
Check:
30.
1 x4 x3 1 1 1 x4 1 2 dx 1 x4 1 2 4x3 dx C C 4 1 x 4 4 1 2 2
d 2 u3 2 3 2 2 C du 9 9
Check:
28.
2 u3 2 3 2 1 1 u3 2 3 2 C C u3 2 1 2 3u2 du 3 3 3 2 9
1 1 16 x3 1 1 x2 dx 16 x3 2 3x2 dx C C 16 x3 2 3 3 1 3 16 x3
Check:
26.
3
2 t4 5 1 2 4t3 t4 5 1 2 t3
Check:
24.
1 1 1 x3 dx 1 x4 2 4x3 dx 1 x4 1 C C 1 x4 2 4 4 4 1 x4
Check:
22.
u2 u3 2 du
Check:
20.
1 4 1 t4 5 3 2 1 t 5 1 2 4t3 dt C t4 5 3 2 C 4 4 3 2 6
1 d 1 4 t 5 3 2 C dt 6 6
Integration by Substitution
1 3 1 2 1 t4 1 t 1 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
473
474
34.
Chapter 4
Integration
2 4 y2 2 y 8 y3 2 dy 2 8y y5 2 dy 2 4y2 y7 2 C 14 y3 2 C 7 7
Check:
36. y
d 2 d 4 y2 14 y3 2 C 2 4y2 y7 2 C 16 y 2 y5 2 2 y 8 y3 2 dy 7 dy 7
10x2 dx 1 x3
38. y
x 4
x2 8x 1
dx
10 1 x3 1 2 3x2 dx 3
1 x2 8x 1 1 2 2x 8 dx 2
10 1 x3 1 2 C 3 1 2
1 x2 8x 1 1 2 C 2 1 2
20 1 x3 C 3
40. (a)
x2 8x 1 C
(b)
y
dy x cos x2, 0, 1 dx
3
y x
â&#x2C6;&#x2019;5
5
â&#x2C6;&#x2019;2 â&#x2C6;&#x2019;3
x cos x2 dx
1 sin x2 C 2
0, 1 : 1 y
42.
46.
1 1 sin x2 2x dx cos x2 C 2 2
48.
50.
52.
54.
4x3 sin x4 dx
x sin x2 dx
sin x4 4x3 dx cos x4 C
44.
1 cos x2 2x dx 2
1 sin 0 C â&#x2021;&#x2019; C 1 2 1 sin x2 1 2
cos 6x dx
1 1 cos 6x 6 dx sin 6x C 6 6
sec 1 x tan 1 x dx sec 1 x tan 1 x 1 dx sec 1 x C
tan x sec2 x dx
tan x 3 2 2 C tan x 3 2 C 3 2 3
1 1 sin x cos x 2 dx cos x 3 sin x dx C C sec2 x C 3 cos x 2 2 cos2 x 2
csc2
2x dx 2 csc 2x 12 dx 2 cot 2x C 2
56. f x
sec x tan x dx sec x C
Since f 1 3 1 sec 3 C, C 1. Thus f x sec x 1.
Section 4.5 1 1 58. u 2x 1, x u 1 , dx du 2 2
x 2x 1 dx
Integration by Substitution
60. u 2 x, x 2 u, dx du
x 1 2 x dx 3 u u du
1 1 u 1 u du 2 2
3u
1 3 2 1 2 u u du 4
1 2x 1 3 2 3 2x 1 5 C 30
1 2x 1 3 2 6x 2 C 30
1 2x 1 3 2 3x 1 C 15 64. u t 4, t u 4, dt du
3 t 4 dt t
2u1 2 7u 1 2 du
2 x 4 2 x 4 21 C 3
3 t 4 4 3 t 4 7 C 7
2 x 4 2x 13 C 3
3 t 4 4 3 t 3 C 7
1 3
4
2
1
x 1 x2 dx
0
x3 8 2 3x2 dx
1 x3 8 3 3
3
4 2
1
1 2
1 1 x2 1 2 2x dx 1 x2 3 2 3 0
1 0
0
1 1 3 3
70. Let u 1 2x2, du 4x dx.
0
u4 3 4u1 3 du
2 u1 2 2u 21 C 3
68. Let u 1 x2, du 2x dx.
2
u 4 u1 3 du
3 u7 3 3u4 3 C 7
1 64 8 3 8 8 3 41,472 9
4 u3 2 14u1 2 C 3
x2 x3 8 2 dx
2u3 2 5 u C 5
2 2 x 3 2 x 3 C 5
2 u 4 1 du u
4
2
2 2 x 3 2 5 2 x C 5
66. Let u x3 8, du 3x2 dx.
du
3 2
u
62. Let u x 4, x u 4, du dx. 2x 1 dx x 4
1 2
2 2u3 2 u5 2 C 5
1 2 5 2 2 3 2 u u C 4 5 3
u3 2 3u 5 C 30
1 x dx 4 1 2x2
2
0
475
1 2x2 1 2 4x dx
12 1 2x
2
2
0
3 1 1 2 2
3u4 3 u 7 C 7
476
Chapter 4
Integration
72. Let u 4 x2, du 2x dx.
2 3 x 4 x2 dx
0
1 2
2
8 4 x
2
3
4 x2 1 3 2x dx
2 4 3
0
0
3 33 84 3 44 3 6 4 3.619 8 2
1 74. 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
3
9
u1 2 u 1 2 du
1
2 2 27 2 3 2 3 3
9 1
16 3
2
76.
1 2 3 2 u 2u1 2 4 3
1 4
1 2 u 1 1 1 du u 2 4
x cos x dx
2
x2 sin x 2
3
8
2
18 23 572 2
1
2
2 3 2
78. u x 2, x u 2, dx du When x 2, u 0. When x 6, u 8.
6
Area
80. A
2
8
3 x 2 dx x2
0
82. Let u 2x, du 2 dx. Area
4
csc 2x cot 2x dx
12
1 2
0
10 3
4
4
1 csc 2x cot 2x 2 dx csc 2x 2 12
12
86.
0
x2 x 1 dx 67.505
1
20
2
88.
2
90.
5
1 0
sin x cos x dx
sin x 1 cos x dx
sin2 x C1 2
sin x cos x dx cos x 1 sin x dx
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
0
sin 2x dx 1.0
2
2
0
0
8
0
50
0
12 7 3 u 3u4 3 7
1 2
5
x3 x 2 dx 7.581
2
103 u
1 sin 2x 2
2
84.
u7 3 4u4 3 4u1 3 du
0
sin x cos 2x dx cos x
0
8
3 u du u 2 2
â&#x2C6;&#x2019;1
4752 35
Section 4.5 92. f x sin2 x cos x is even.
2
sin2 x cos x dx
2
Integration by Substitution
94. f x sin x cos x is odd.
2
2
sin2 x cos x dx
2
0
sin x cos x dx 0
2
sin3 x 3
2
0
4
96. (a)
sin x dx 0 since sin x is symmetric to the origin.
4
4
(b)
2
cos x dx 2
2
4
cos x dx 2 sin x
0
2
(d)
4
cos x dx 2
4
(c)
2 3
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 sin x cos x sin x cos x and hence, is symmetric to the origin.
2
98.
sin 3x cos 3x dx
sin 3x dx
100. If u 5 x2, then du 2x dx and
x 5 x2 3 dx
dQ k 100 t 2 dt
102.
Q t
cos 3x dx 0 2
cos 3x dx
0
23 sin 3x
0
0
1 1 5 x2 3 2x dx u3 du. 2 2
104. R 3.121 2.399 sin 0.524t 1.377
k k 100 t 2 dt 100 t 3 C 3
(a)
8
Q 100 C 0 0
k Q t 100 t 3 3
12 0
Relative minimum: 6.4, 0.7 or June
k Q 0 100 3 2,000,000 â&#x2021;&#x2019; k 6 3 Thus, Q t 2 100 t When t 50, Q 50 $250,000. 3.
Relative maximum: 0.4, 5.5 or January
12
(b)
R t dt 37.47 inches
0
(c)
1 3
12
9
1 R t dt 13 4.33 inches 3
24
106. (a)
(b) Volume
70
0
0
24 0
Maximum flow: R 61.713 at t 9.36. 18.861, 61.178 is a relative maximum.
R t dt 1272 (5 thousand of gallons)
477
478
Chapter 4
108. (a)
Integration
4
(b) g is nonnegative because the graph of f is positive at the beginning, and generally has more positive sections than negative ones.
g 9.4
0
f â&#x2C6;&#x2019;4
(c) The points on g that correspond to the extrema of f are points of inflection of g.
(e)
(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
â&#x2C6;&#x2019;4
The graph of h is that of g shifted 2 units downward.
t
g t
f x dx
0
2
0
110. False
t
f x dx
f x dx 2 h t .
2
1 1 x2 1 2x dx x2 1 2 C 2 4
x x2 1 2 dx
112. True
b
b
sin x dx cos x
a
114. False
a
cos b cos a cos b 2 cos a
a
0
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
a
0
f u du
a
f u du
0
sin x dx
a
116. Because f is odd, f x f x . Then a
b 2
1 1 sin 2x 3 1 C sin3 2x C sin 2x 2 2 cos 2x dx 2 2 3 6
sin2 2x cos 2x dx
f x dx
0
a
0
f x dx 0
Section 4.6
Section 4.6
0
1
Trapezoidal:
0
1
Simpsonâ&#x20AC;&#x2122;s:
0
2
4. Exact:
3 x dx
8 3 x dx
0 8 3 x dx
0
2
34x
8
4 3
0
4 x2 dx 4 x2 dx
x x2 1 dx
2
x x2 1 dx
0 2
x x2 1 dx
0
2
0 2
Simpsonâ&#x20AC;&#x2122;s:
0
67 1.1667
2 2
46
2
2
46
2 2
47
4
2
47
2
1 0.5090 4
1 0.5004 4
x3 3
3 1
3
11 2 0.6667 3 3
2 0 2 4 52 5 0.7500
2
2
1 9 25 3 4 4 0 4 4 5 0.6667 6 4 4
0
12. Trapezoidal:
1 3 3 2 4 4 2
2
Simpsonâ&#x20AC;&#x2122;s:
12.0000
1
Trapezoidal:
2
4 x2 dx 4x
3
1 3 2 4 3 3 2 3 4 4 3 5 2 3 6 4 3 7 2 11.8632
0 4 2 3
1
10. Exact:
1 3 2 2 3 3 2 3 4 2 3 5 2 3 6 2 3 7 2 11.7296
0 2 2 2
3
Simpsonâ&#x20AC;&#x2122;s:
1
Trapezoidal:
0.5000
1
3
8. Exact:
7 1.1667 6
0
Simpsonâ&#x20AC;&#x2122;s:
1 x2 1 4 2 1 2 2 3 4 2 12 1 dx 1 4 1 2 1 4 1 1 2 12 2 2 2 2
8
Trapezoidal:
0
1 1 4 dx 1 4 x2 12 5
1
1
1 75 x2 1 4 2 1 2 2 3 4 2 12 1 dx 1 2 1 2 1 2 1 1 1.1719 2 8 2 2 2 2 64
2
1 1 4 dx 1 2 x2 8 5
1
6. Exact:
2
Simpsonâ&#x20AC;&#x2122;s:
x2 x3 1 dx x 2 6
1 1 dx x2 x
1
Trapezoidal:
479
Numerical Integration 1
2. Exact:
Numerical Integration
1 2 x 1 3 2 3
2 0
1 53 2 1 3.393 3
1 3 1 1 2 2 1 2 1 12 1 2 3 2 2 1 2 22 1 3.457 0 2 4 2 2 1 1 3 1 2 2 1 2 1 12 1 4 3 2 2 1 2 22 1 3.392 0 4 6 2 2
1 1 1 1 1 1 dx 1 2 2 2 1.397 3 3 3 3 4 3 1 x 1 1 2 1 1 1 3 2 1 1 x3
Graphing utility: 1.402
dx
1 1 1 1 1 1 4 2 4 1.405 6 3 1 1 2 3 1 13 1 3 2 3
480
Chapter 4
Integration
14. Trapezoidal:
2 1 2 58 sin 58 2 34 sin 34 2 78 sin 78 0 1.430 5 5 3 3 7 7 x sin x dx 4 sin 2 sin 4 sin 0 1.458 24 2 8 8 4 4 8 8
x sin x dx
2
Simpson’s:
16
2
Graphing utility: 1.458
4
16. Trapezoidal:
tan x2 dx
0
4
8
tan 0 2 tan
4
tan 0 4 tan
4
4
2
2
2 tan
4
2 tan
4
2
2
2
4
2 tan
3
4 tan
3
4
2
2
4
tan
2
0.271
4
Simpson’s:
tan x2 dx
0
4
12
4
2
4
4
4
tan
2
0.257 Graphing utility: 0.256
2
18. Trapezoidal:
1 cos2 x dx
0
2
Simpson’s:
1 cos2 x dx
0
2 2 1 cos2 8 2 1 cos2 4 2 1 cos2 3 8 1 1.910 16
2 4 1 cos2 8 2 1 cos2 4 4 1 cos2 3 8 1 1.910 24
Graphing utility: 1.910
20. Trapezoidal:
0
Simpson’s:
0
2 sin 4 2 sin 2 2 sin 3 4 sin x dx 1 0 1.836 x 8 4 2 3 4
4 sin 4 2 sin 2 4 sin 3 4 sin x dx 1 0 1.852 x 12 4 2 3 4
Graphing utility: 1.852 22. Trapezoidal: Linear polynomials Simpson’s: Quadratic polynomials
24.
f x
1 x 1
f x
1 x 1 2
f x
2 x 1 3
f x
6 x 1 4
f 4 x
24 x 1 5
(a) Trapezoidal: Error ≤
1 0 2 1 0.01 since 2 12 42 96
f x is maximum in 0, 1 when x 0. (b) Simpson’s: Error ≤
1 1 0 5 0.0005 24 180 44 1920
since f 4 x is maximum in 0, 1 when x 0.
Section 4.6 26. f x
Numerical Integration
481
2 in 0, 1 . 1 x 3
(a) f x is maximum when x 0 and f 0 2. 1 2 < 0.00001, n2 > 16,666.67, n > 129.10; let n 130. 12n2
Trapezoidal: Error ≤ f 4 x
24 in 0, 1 1 x 5
(b) f 4 x is maximum when x 0 and f 4 0 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
28. f x x 1 2 3 (a) f x
2 in 0, 2 . 9 x 1 4 3 2
f x is maximum when x 0 and f 0 9.
< 0.00001, n
8 2 12n4 9
Trapezoidal: Error ≤ (b) f 4 x
> 14,814.81, n > 121.72; let n 122.
2
56 in 0, 2 81 x 1 10 3 56
f 4 x is maximum when x 0 and f 4 0 81.
< 0.00001, n
32 56 Simpson’s: Error ≤ 180n4 81 be even.)
4
> 12,290.81, n > 10.53; let n 12. (In Simpson’s Rule n must
30. f x sin x2 (a) f x 2 2x2 sin x2 cos x2 in 0, 1 .
f x is maximum when x 1 and f 1 2.2853. 1 0 3 2.2853 < 0.00001, n2 > 19,044.17, n > 138.00; let n 139. 12n2 (b) f 4 x 16x4 12 sin x2 48x2 cos x2 in 0, 1 Trapezoidal: Error ≤
f 4 x is maximum when x 0.852 and f 4 0.852 28.4285. Simpson’s: Error ≤
1 0 5 28.4285 < 0.00001, n4 > 15,793.61, n > 11.21; let n 12. 180n4
32. The program will vary depending upon the computer or programmable calculator that you use. 34. f x 1 x2 on 0, 1 . L n
M n
R n
T n
S n
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
482
Chapter 4
Integration
sin x on 1, 2 . x
36. f x
L n
M n
R n
T n
S n
4
0.7070
0.6597
0.6103
0.6586
0.6593
8
0.6833
0.6594
0.6350
0.6592
0.6593
10
0.6786
0.6594
0.6399
0.6592
0.6593
12
0.6754
0.6594
0.6431
0.6593
0.6593
16
0.6714
0.6594
0.6472
0.6593
0.6593
20
0.6690
0.6593
0.6496
0.6593
0.6593
n
38. Simpson’s Rule: n 8
2
8 3
1 32 sin d 2
0
3
1 32 sin 0 4 1 32 sin 16 2 1 32 sin 8 . . . 1 32 sin 2
2
2
2
2
6 17.476
40. (a) Trapezoidal:
2
f x dx
0
2
4.32 2 4.36 2 4.58 2 5.79 2 6.14 2 7.25 2 7.64 2 8.08 8.14 12.518 2 8
Simpson’s:
2
f x dx
0
2
4.32 4 4.36 2 4.58 4 5.79 2 6.14 4 7.25 2 7.64 4 8.08 8.14 12.592 3 8
(b) Using a graphing utility, y 1.3727x3 4.0092x2 0.6202x 4.2844
2
Integrating,
y dx 12.53
0
42. Simpson’s Rule: n 6
1
4
0
4 4 2 4 2 4 1 1 dx 1 1 x2 3 6 1 1 6 2 1 2 6 2 1 3 6 2 1 4 6 2 1 5 6 2 2
3.14159 44. Area
120
75 2 81 2 84 2 76 2 67 2 68 2 69 2 72 2 68 2 56 2 42 2 23 0 2 12
7435 sq m 46. The quadratic polynomial p x
x x2 x x3 x x1 x x3 x x1 x x2 y y y x1 x2 x1 x3 1 x2 x1 x2 x3 2 x3 x1 x3 x2 3
passes through the three points.