The midpoint Riemann sum will be larger than ∫ a () f x dx .

21. The midpoint Riemann sum will be larger than ∫ a () f x dx .

If f is concave down, then f () c < 0 for any c ∈ () ab , . Thus, the error E n =

f ′′ () c < . Since the error 0

24 n 2 is negative, then the Riemann sum must be greater than the integral.

22. b The Trapezoidal Rule approximation will be smaller than ∫ f x dx

a () .

If f is concave down, then f ′′ () c < 0 for any c ∈ () ab , . Thus, the error E =−

2 f ′′ () c > . Since the 12 0 n

error is positive, then the Trapezoidal Rule approximation must be less than the integral.

[ a k 1 – (– ) k a + 1 = 1 a k + ∫ 1 ⎢ ] [ – a k + 1 ]0 – = a

A corresponding argument works for all n.

24. a. T ≈ 48.9414; fx ′ () = 4 x 3 [4(3) – 4(1) ](0.25) 3 3 2

12 The correct value is 48.4 .

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[cos – cos 0] π π ()

The correct value is 2.

25. The integrand is increasing and concave down. By problems 19-22, LRS < TRAP < MRS < RRS.

26. The integrand is increasing and concave up. By problems 19-22, LRS < MRS < TRAP < RRS

27. A ≈ [75 2 71 2 60 2 45 2 45 2 52 2 57 2 60 59] +⋅+⋅ +⋅ +⋅ +⋅ +⋅ +⋅ + = 4570 ft 2

28. A ≈ [23 4 24 2 23 4 21 2 18 4 15 2 12 4 11 2 10 4 8 0] +⋅ +⋅ +⋅+⋅+⋅+⋅+⋅+⋅+⋅+ = 465 ft 2

V =⋅≈ A 6 2790 ft 3

29. A ≈ [0 4 7 2 12 4 18 2 20 4 20 2 17 4 10 0] +⋅+⋅+⋅+⋅ +⋅ +⋅+⋅+ = 2120 ft 2

4 mi/h = 21,120 ft/h

(2120)(21,120)(24) = 1,074,585,600 ft 3

30. Using a right-Riemann sum,

4. False:

fx () = x 2 + 2 x + 1 and

Distance = ∫ v t dt () ≈ ∑ vt () Δ t

gx () = x + 7 x − 5 are a

i = 1 counterexample.

The two sides will in general differ by 852

60 5. False:

a constant term.

At any given height, speed on the downward trip is the negative of

31. Using a right-Riemann sum,

speed on the upward.

Water Usage = ∫ 0 () F t dt

148) n − 1 12(71 68 a n − 2 + a n + a n − 1 ∑ i i = 1

∑ (2 i −= 1) 2 ∑∑ i − 1

4.7 Chapter Review

Concepts Test

∑ ( a i 2 + 1) = a ∑ 2 i + 2 ∑ a i + ∑ 1

1. True:

Theorem 4.3.D

9. True:

2. True:

Obtained by integrating both sides of

the Product Rule

10. False:

f must also be continuous except at a

3. True:

If Fx () = () f x dx f x , () is a

finite number of points on [a, b ]. ∫

derivative of F(x).

11. True:

The area of a vertical line segment is 0.

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12. False:

1 x dx is a counterexample.

28. False:

fx () = x ∫ 3 − 1 is a counterexample.

13. False:

A counterexample is

29. False:

fx () = x is a counterexample.

fx () =⎨

⎧ 0, x ≠ 0

⎩ 1, x = 0 30. True:

All rectangles have height 4,

1 2 regardless of x i .

with ∫ ⎡ ⎣ fx ⎤ dx = − 1 () ⎦ 0 .

If b fx () is continuous, then

31. True:

Fb () − Fa () = ∫ ′

a () F x dx

[ ( )] fx 2 ≥ 0 , and if [ ( )] fx 2 is greater

a () G x dx = Gb () − Ga () than 0 on [a, b], the integral will be also.

32. False:

= − f x dx

a () f x dx 2 ∫ 0 () because f

is even.

15. True:

sin x + cos x has period 2 π , so

33. False:

zt () = t 2 is a counterexample.

x +π ∫ 2

(sin x + cos ) x x dx b

34. False:

() f x dx = Fb () − F (0)

is independent of x .

Odd-exponent terms cancel x → a x → a themselves out over the interval, since

lim [ fx () + gx () ] they are odd.

= lim fx () + lim ( ) gx when all the

x → a x a 36. False:

= 0, b = 1, f(x) = –1, g(x) = 0 is a

counterexample.

limits exist.

37. False:

a = 0, b = 1, f(x) = –1, g(x) = 0 is a

17. True:

sin 13 x is an odd function.

counterexample.

18. True:

Theorem 4.2.B

The statement is not true if c > d.

≤ a 1 + a 2 + a 3 ++ a n because any negative values of a i make the

20. False:

0 left side smaller than the right side.

2 dt ⎥ =

39. True:

Note that − fx () ≤ fx () ≤ fx ()

21. True:

Both sides equal 4.

and use Theorem 4.3.B.

22. True:

Both sides equal 4.

40. True:

Definition of Definite Integral

23. True:

If f is odd, then the accumulation

41. True:

Definition of Definite Integral

function x Fx () = ∫

0 () f t dt is even,

and so is () + C for any C.

42. False: Consider

fx () = x 2 is a counterexample.

43. True.

Right Riemann sum always bigger.

44. True. Midpoint of x coordinate is midpoint

25. False:

fx () = x 2 is a counterexample.

of y coordinate.

45. False.

Trapeziod rule overestimates integral.

26. False:

fx () = x 2 is a counterexample.

46. True.

Parabolic Rule gives exact value for

27. False:

fx () = x 2 , v(x) = 2x + 1 is a

quadratic and cubic functions.

counterexample.

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Sample Test Problems

12. u = 2 y 3 + 3 y 2 + 6, y du = (6 y 2 + 6 y + 6) dy

1 5 ( y 2 ++ y 1

1. x − x + 2 x

dy = ∫ u du

− 9 cos1 i = 1 2 4

7. u = tan(3 x 2

u 3 + C 15. ∫ 0 (2 − x + 1) dx

⎤ = 1 3 π+ 2 = ∫ ( x +− 54 x + 1 ) dx

1 25 u − 1/ 2

9 du = ⎡ 1/ 2 ⎤ 4 = 2 ⎣ u ⎦ 1 9 1 5 12 5

16. ∫ 3 x 2 x 3 − dx = ⎡ 4 ( x 3 − 4) 3/2 ⎤

3 17. ∫ 2 ⎜ 5 − 2 ⎟ dx = ⎢ 5 x ⎣ + x ⎥ =

⎣ 9 y 3 y ⎦ ⎥ 2 27 n

18. (3 i − 3 i − ∑ 1 )

11. ∫ ( x + 1) sin ( x + 2 x + 3 ) dx

=−+ (3 1) (3 2 −+ 3) (3 3 − 3) 2 ++ (3 n − 3 n − 1 ) = 1

sin ∫ 2 ( x + 2 x + 32 ) ( x + 2 ) dx

2 ∫ sin u du

19. ∑ (6 i 2 − 8) i = 6 ∑ 2 i − 8 ∑ i

=− cos ( x + 2 x ++ 3 C

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20. a. ++=

e. ∫ f ( − x dx ) =−

0 ∫ () f x dx =− 2 0

21. a. ∑

b. ∑ nx 4 1 1

∫ x − 1 dx = (1)(1) +

∑ ⎢− 16 ⎜⎟⎜⎟ ⎥ x dx =++= 1236

lim

⎝⎠⎝⎠ n ⎥ ⎦ n ⎧ ⎪ n ⎡ 48 27

= lim

∑ ⎢ n 3 ⎥ ⎬ c.

∫ ( x − x ) dx = ∫ x dx − ∫ x dx

n →∞ n ∑ 3 ∑

27 ⎡ nn ( + 1)(2 n + 1) ⎫

⎥ ⎬ 25. a. ∫ − 2 () f x dx = 2 ∫ f x dx =−=− 0 () 2( 4) 8 n →∞

lim 48 ⎨ 3 ⎢

= lim 48 ⎨ − ⎢ 2 ++ 2 ⎥ ⎬ b. Since fx () ≤ 0 , fx () =− fx () and

n →∞ ⎩

∫ − () f x dx =− 2 ∫ − 2 () f x dx

() f x dx 2 ∫ 0 () ∫ 8

c. ∫ () g x dx = 0

b. ∫ () f x dx =− ∫ () f x dx =− 1 0 4

d. ∫ − 2 [ fx () + f ( − x dx ) ]

3 ∫ 0 () 3(2) 6 = 2 ∫ () f x dx + 2 f x dx

d. ∫ 0 [ 2()3() gx − f x dx ]

2 ∫ gx ()3 − ∫ () f x dx 0 0

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2 1 x e. ∫ 2()3()

20 ∫ f z dz x fx x

g x dx + 3 ∫ () f x dx

0 0 x d. Gx ′ () = () f t dt

f. gx ∫ () g x dx =− ∫ () g x dx =− 5 e. Gx () = du = [ ( )] gu ()

gx ()

− 2 0 ∫ 0 du

26. ∫ 3 + − 100 ( x sin 5 x dx ) = 0 Gx ′ () = ggxgx ′ ( ( )) ( ) ′

1 2 2 f. Gx () = ∫ f () − t dt = fu ( )( − du )

27. ∫ x dx = c −+

30. a. ∫ 0 x dx = ⎡ x 3/2 ⎤ =

c =− 7 ≈− 2.65 4 2 4 16

28. a. Gx ′ () =

x 2 + 1 b. 32 ∫ 3 1 x dx = ⎡⎤

31. fx () = ∫ x dt = ∫ dt − ∫ dt

32. Left Riemann Sum: ∫

4 dx ≈ 0.125[ ( fx 0 ) + fx () 1 +…+ fx ( 7 )] 0.2319 ≈ 1 1 + x

Right Riemann Sum: ∫

4 dx ≈ 0.125[ ( ) fx 1 + fx ( 2 ) +…+ fx ( )] 0.1767 ≈ 1 1 8 + x

Midpoint Riemann Sum: ∫

dx ≈ 0.125[ ( fx 0.5 ) + fx ( 1.5 ) +…+ fx ( 7.5 )] 0.2026 ≈

dx ≈

[( fx 0 )2() + fx 1 +…+ 2( fx 7 ) + fx ( )] 0.2043

4 (5 c − 3) (4)(2 ) (5)(2 ) 3 ( − )

f ''( ) c =

Remark: A plot of '' f shows that in fact f '' () c < 1.5 , so E n < 0.002 .

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dx ≈

[( fx 0 )4()2( + fx 1 + fx 2 ) +…+ 4( fx 7 ) + fx ( )] 1.1050 8 ≈

0 12 + x

f () 4 () c =

( 12 + c )

E n =−

⋅ f () c ≤

c 4 (5 c − 3) (4)(2 ) (5)(2 ) 3 ( + )

≈ 138,333 so n > 138,333 ≈ 371.9 Round up to n = 372 .

Remark: A plot of '' f shows that in fact f '' () c < 1.5 which leads to n = 36 .

36. f () () c =

( 12 + c )

4 4 384 n ⋅ > ≈ 21,845,333 , so n ≈ 68.4 . Round up to n = 69 .

37. The integrand is decreasing and concave up. Therefore, we get: Midpoint Rule, Trapezoidal rule, Left Riemann Sum

Review and Preview Problems

6. ( x +− h x 2 ) + ( ( x + h ) 2 − x 2 )

2 ⎜⎟ ⎝⎠ 2 2 4 4 = h 2 + ( 2 xh + h 2 )

2. x − x 2 7. V = ( π ⋅ 2 2 )0.4 1.6 = π