False; A counterexample is fx () = 0 for all x,

58. False; A counterexample is fx () = 0 for all x,

except f () 1 = 1 . Thus, ∫ 0 () f x dx = 0 , but f is

not identically zero. ⎧ t

⎪ 5 du ,

0 ≤≤ t 100

62. a. st () = ⎨ ∫ 0 5 du + ∫ 100 ⎜ 6 − ⎟ du

0 5 du + ∫ 100 ⎜ 6 − ⎟ du ⎩ + ⎪ ⎝ 100 ⎠ ∫ 700 () − 1 du , t > 700

⎥ −− ( t 700 ) t > 700

2400 − t ⎪⎩ > , t 700

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2. ∫ x dx 4 =

t > 600 . So, t = 600 is the point at which

the object is farthest to the right of the origin.

At t = 600 , st () = 1750 .

3. (3 ∫ 2 x − 2 x + 3) dx = ⎡ x 3 − x 2 + x ⎤

c. st () == 0 2400 − tt ; = 2400

63. − fx () ≤ fx () ≤ fx () , so

b b 4. ∫ 1 (4 x 7) dx ⎣ x 7 x ⎦ 1

∫ a − () f x dx ≤ ∫ a () f x dx ⇒ = (16 + 14) – (1 + 7) = 22

a () f x dx ≥− ∫ a () f x dx 4 1 4 ⎡ 1 ⎤ ⎛ 1 ⎞

5. ∫ 2 dw =− ⎢ ⎥ =− ⎜ ⎟ −−= ( 1)

and combining this with

a () f x dx ≥ ∫ a () f x dx , 3

we can conclude that

b 1 ⎜ ⎟ b −−= t 3 ⎣ t 2 ⎦ 1 ⎝ 9 ⎠ 9

6. ∫ dt =− ⎢ ⎥ =− ( 1)

∫ a () f x dx ≤ ∫ a () f x dx

∫ t dt =⎢ t ⎥ = ⎜ ⋅−= 8 0

Boundedness Property. If x <, a

⎛ 3 ⎞⎛ 3 ⎞ 45 () f x dx =−

′ () f x dx ≥− Mx ( − a ) by

8. ∫ w dw =⎢ w

∫ a 1 ⎟⎜ ⎣ ⎟ 4 ⎦ ⎥ 1 ⎝ 4 ⎠⎝ 4 ⎠ 4

the Boundedness Property. Thus

′ () f x dx ≤ Mx − a .

9. ∫ ⎜ y +

y 3 ⎟ ⎟ dy = ⎢ − 2 ⎥ x x ⎝ ⎠ ⎣ ⎢ 3 2 y ⎦ ⎥

From Problem 63,

′ () f x dx ≥

′ () f x dx .

a ∫ a ⎛ 81 ⎞⎛ 64 1 ⎞ 1783

∫ ′ () f x dx = fx () − a fa () ≥ fx () − fa () ⎝ 38 ⎠⎝ 3 32 ⎠ 96

Therefore, fx () − fa () ≤ Mx − or a 4 4 4 s − 8 4 2 − 2 ⎡ s 3 8 ⎤

10. ∫ 2 ds = ∫ ( s − 8 s ) ds = +

4.4 Concepts Review

11. ∫ cos x dx = [ sin x ] 0 =1–0=1

1. antiderivative; F(b) – F(a)

12. ∫ 2sin t dt =− [ 2 cos t ] π /6 =+ 0 3 = 3

2. F(b) – F(a)

3. F ()() d − F c

13. (2 x 4 − 3 x 2 ⎡ 2 ∫ ⎤ + 5) dx = 5 ⎢ x − x 3 + 5 x

4. u du ∫

Problem Set 4.4

14. ∫ ( x 4/3 − 2 x 1/ 3 ) dx = x 7/3 − x 4/3 ⎢ ⎥

1. ∫ x dx = ⎢ ⎥ =−=

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25. u = x 2 + du 4, 2 = x dx

⋅ 1 u 2 du = u 3/2 += 2 (3 x + 2) ∫ 3/2 C + C 1 1

3 9 9 ∫ sin( ) u ⋅ du =− cos u + C

16. u = 2x – 4, du = 2 dx

cos( x 2 ++ 4) C

1 1 1 ∫ cos u ⋅ du = sin u += C sin( x 3 ++ 5) C ∫ cos( ) u ⋅ du = sin u += C sin(3 x ++ 2) C 3 3 3 3 3 3

∫ sin u ⋅ du =− cos u + C 2 2

∫ sin u du =− cos u +=− C cos x 2 ++ 4 C

cos(2 x −+ 4) C

∫ sin u ⋅ 1 =− du 1 cos u + C 6 6 3 3 3 ∫ cos u ⋅ du = sin u += C sin 32 z ++ 3 C

=− cos(6 x −+ 7) C

cos u ⋅ du = sin u += C sin( π− v 7) + C 1 ∫ 1 cos u ⋅ du = sin u + C

u ⋅ du = u 3/2 += C ( x 2 + 4) 3/2 + C 30. u = (7 x 7 +π ) , 441 (7 9 du = x 6 x 7 +π ) 8 dx ∫

∫ sin u ⋅ du =− cos u + C

cos(7 x 1 7 1 =− +π+ ) 9 C

u ∫ 9 ⋅ du = u 10 += C ( x 3 + 5) 10 + C 441

31. u = sin( x 2 + 4), 2 cos( du = x x 2 + 4) dx

23. u = x 2 + du 3, 2 = x dx 1 1

∫ u ⋅ du = u 3/2 + − C 12 / 7 1 7

∫ u ⋅ du =− u − 5/7 + C 2 3 2 10 1 3/2

= ⎡ sin( x 2 + 4) ⎤ + C

32. u = cos(3 x 7 + 9)

24. u = 3 v 2 +π , du = 23 v dv du =− 21 x 6 sin(3 x 7 + 9) dx

u 7/8 1 ∫ 4 ⋅ du = u 15 / 8 + C 3 ⎛ 1 ⎞

∫ u ⋅− ⎜ ⎟ du =− u + C ⎝ 21 ⎠ 28

3 v +π ) + C =− ⎡ x 7

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33. u = cos( x 3 + 5), 3 du =− x 2 sin( x 3 + 5) dx 41. u =+ t 72, 4 2 du = t dt

u ⋅− ⎜ ⎟ du =− u 10 + C ∫ 72 + t 2 (8 ) t dt = 2 + t 2 ⋅ () 4 ∫ t dt 72

∫ u du =⎢ u 25 ⎣ 3 ⎥ ⎦

34. u = tan( x 3 + 1) , du =− − 3 x 4 2 x sec ( − 3 + 1) dx = (125)

18 ⎣ tan( x + 1) ⎦

∫ ( x + 1) (2 ) x dx = u du

3 2 ∫ 0 cos 2 x sin x dx =− cos 2 x − sin 36. x dx u = x + du 1, 3 = x dx ∫ 0 ( )

02 ⎡ u 3 ⎤

∫ x + 1 (3 x ) dx = ∫ udu =⎢ u ⎥ =− ∫ u du − =− 1 0 1 ⎢ ⎣ ⎥ 3 ⎦

44. u = sin 3 , 3cos 3 x du = x dx

sin 3 3cos 3 2 x x dx =

∫ y − 1 dy =

2 ∫ udu u

= ∫ ( x 2 + x 2 ) 2( 2 x + 1) dx

∫ 3 x + 1 dx = ∫ 3 x +⋅ 13 dx = ∫ u du 3 5 3 3 5 3 16 1 32 ⎡ u ⎤ 9

= ∫ 0 u du = ⎢ ⎥ =

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47. u = sin , cos θ du = θθ d 56. u =π sin , cos θ du =π θθ d

1 1 ∫ cos u du = [ sin u

] −π = 0 ∫ u du = = −= 0 −π

57. u = cos( x 2 ), 2 sin( du =− x x 2 ) dx

48. u = cos , sin θ du =− θθ d cos1

− ∫ u − 3 du = ⎡ u − 2 ⎤ = −= 1 ⎜ − ⎟ 2 ∫ u du =−

cos u du = [ sin u

] 3 = (0 sin( 3)) −

3 58. u = sin( x 3 ), 3 du = x 2 cos( x dx 3 )

= sin 3

1 sin( π 3 / 8) 2 1 3 sin( π 3 / 8)

∫ 3 u du ⎡⎤ u

2 π ∫ sin u du =− cos u =− ( 1 1) −− 0 = [ 2 ] π 0 2 π 9

59. a. Between 0 and 3, fx

( 1 1) −− b. Since f is an antiderivative of f ' [ , ]

cos u =−

∫ 0 f '( ) x dx = f (3) − f (0)

52. u = 2 x 5 , du = 10 x dx 4 3 c. f ''( ) x dx = f '(3) − f '(0)

cos u du =

sin u ] =−−=−< 10 10

(sin(2 ) 0) = sin(2 5 π )

d. Since f is concave down at 0, f ''(0) < 0 .

∫ f '''( ) x dx = f ''(3) − f ''(0)

u = x du 2, 2 = dx

=− 0 (negative number) > 0

cos u du + ∫

sin u du

60. a. On [] 0, 4 , () fx > . Thus, 0 ∫ () f x dx > 0 .

= [ sin u

− [ cos u ] 0

1 1 b. Since f is an antiderivative of ' f ,

'( ) f x dx 0 = f (4) − f (0)

54. u = x du 3, 3;5, 5 = dx v = x dv = dx =−=−< 12 10

cos u du +

∫ sin v dv −π 5/2

c. ∫ f ''( ) x dx = f '(4) − f '(0)

sin u ] −π 3/2 − [ cos v ] 5/2

d. ∫ f '''( ) x dx = f ''(4) − f ''(0)

55. u = cos , sin x du =− x dx = ( negative )( − positive ) < 0

0 − 0 ∫ sin

u du = [ cos u

1 ] 1 =− 1 cos1

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61. Vt

Vt ′

20 − t dt = 20 t − t 2 + C b b () n ∫ () ∫ ( ) 67. a. ∫ x dx n = B n ; ∫ n n y dy = A