Problemas pares

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.