Calculus of a single variable early transcendental functions 6th edition larson solutions manual dow by philip.vanhorn565

C H A P T E R 5

Integration Section5.1 Antiderivativesand IndefiniteIntegration.........................................428 Section5.2 Area 436 Section5.3 RiemannSums and DefiniteIntegrals 451 Section5.4 TheFundamentalTheoremof Calculus............................................460 Section5.5 Integrationby Substitution.................................................................473 Section5.6 NumericalIntegration.........................................................................487 Section5.7 The NaturalLogarithmicFunction: Integration 496 Section5.8 InverseTrigonometricFunctions:Integration 506 Section5.9 HyperbolicFunctions.........................................................................516 Review Exercises 527 ProblemSolving 535

C H A P T E R 5 Integration Section 5.1

Antiderivatives

1. d ⎛ 2 + C⎞ = d (2x 3 + C) = 6x 4 = 6 4. dy = 5 dx⎜ x3 ⎟ dx x 4 dt ⎝ ⎠ 2. d ⎛2x 4 1 + C⎞ = d ⎛2x 4 1 x 1 + C⎞ y = 5t + C d dx⎜ 2x ⎟ dx⎜ 2 ⎟ Check: [5t + C] = 5 ⎝ ⎠ ⎝ ⎠ dt = 8x3 + 1 x 2 = 8x3 + 1 2 2x2 5. dy dx = x3 2 3. dy dt = 9t2 y = 2 x5 2 + C 5 y = 3t3 + C Check: d ⎡2 x5 2 + C⎤ = x3 2 Check: d ⎡3t3 + C⎤ = 9t2 dx ⎢5 ⎥ dt ⎣ ⎦ 6. dy dx ⎣ ⎦ = 2x 3 2x 2 1 y = + C = + C 2 x 2 Check: d ⎡ 1 + C⎤ = 2x 3 dx ⎢ x 2 ⎥ ⎣ ⎦ Given Rewrite Integrate Simplify 7. ∫ 3 x dx ∫ x1 3 dx x 4 3 + C 4 3 3 x 4 3 + C 4 8. 1 dx 4x2 1 x dx 4 1 x 1 + C 4 1 1 + C 4x 9. 1 dx x x ∫ x 3 2 dx x 1 2 + C 1 2 2 + C x 10. 1 dx (3x)2 1 x dx 9 1⎛ x 1 ⎞ 9⎜ 1⎟ + C 1 + C 9x 11. ∫(x + 7) dx = x 2 + 7x + C 14. ⎛ x + 1 ⎞ dx = ⎛ x1 2 + 1 x 1 2 ⎞ dx 2 ∫ ⎜ 2 x ⎟ ∫ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ Check: d ⎡ x 2 ⎢ ⎤ + 7x + C⎥ = x + 7 x3 2 = + 1⎛ x1 2⎞ ⎜ ⎟ + C

and Indefinite Integration

428 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. dx ⎣ 2 ⎦ 3 2 2⎜1 2⎟ 12. ∫(8x3 9x2 + 4) dx = 2x4 3x3 + 4x + C = 2 x3 2 + x 1 2 + C 3 Check: d (2x4 3x3 + 4x + C) = 8x3 9x2 + 4 dx Check: d ⎛ 2 x 3 2 + x 1 2 + C⎞ = x 1 2 + 1 x 1 2 dx⎜3 ⎟ 2 13. ∫(x3 2 + 2x + 1) dx = 2 x5 2 + x2 + x + C ⎝ ⎠ = x + 1 5 2 x Check: d ⎛ 2 x 5 2 + x 2 + x + C⎞ = x 3 2 + 2x + 1 dx⎜5 ⎟ ⎝ ⎠

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ∫ ∫ x 3 ∫ ∫( 2 4) x +6 ⎝ ⎛ ∫ 4 ∫ 6 2 Section 5.1 Antiderivatives and Indefinite Integration 429 15. 5 3 3 x2 dx = x2 3 dx = + C = x5 3 + C 5 3 5 20. x 4 3x 2 +5 dx = 1 3x + 5x dx x4 1 3 Check: d ⎛ 3 x 5 3 + C⎞ = x 2 3 = 3 x 2 = x 3x + 5x + C dx⎜5 ⎟ 1 3 ⎝ ⎠ = x + 3 5 + C 16. ∫( 4 x 3 + 1) dx = ∫(x 3 4 + 1) dx = 4 x7 4 + x + C 7 Check: x 3x3 Check: d ⎛ 4 x 7 4 + x + C⎞ = x 3 4 + 1 = 4 x 3 + 1 d ⎡ x + 3 5 + C⎤ = d ⎡ x + 3x 1 5 x 3 + C⎤ dx⎜7 ⎟ ⎢ 3 ⎥ ⎢ ⎥ ⎝ ⎠ dx ⎣ x 3x ⎦ dx ⎣ 3 ⎦ = 1 3x 2 + 5x 4 1 x 4 1 17. ∫ x 5 dx = x 5 dx = + C = + C 4 4x = 1 3 + 5 x 2 x 4 Check: d ⎛ 1 + C⎞ = d ⎛ 1 x 4 + C⎞ x4 3x2 + 5 dx⎜4x 4 ⎟ dx⎜ 4 ⎟ = ⎝ ⎠ ⎝ ⎠ x 4 1( 4x 5) 1 3 = = 4 3x 6 x5 1 21. ∫(x + 1)(3x 2) dx = ∫(3x2 + x 2) dx = x3 + 1 x2 2x + C 18. ∫ x7 dx = 3x 7 dx = + C = + C 6 2x d ⎛ x3 + 1 x2 2 x + C⎞ = x 2 + x Check: d ⎛ 1 + C⎞ = d ⎛ 1 x 6 + C⎞ Check: ⎜ 2 dx 2 ⎟ 3 2 dx⎜ 2x6 ⎟ dx⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = (x + 1)(3x 2) = ⎛ 1⎞( 6)x 7 = 3 ⎜ 2⎟ x7 19. ⎝ ⎠ ∫ dx = ∫(x1 2 + 6x 1 2) dx 22. ∫(4t2 + 3) dt = ∫(16t4 + 24t2 + 9) dt 16t5 3 x x3 2 x1 2 = + 8t 5 + 9t + C = + 6 + C d ⎛16t5 3 ⎞ 4 2 3 2 1 2 = 2 x3 2 + 12x1 2 + C Check: dt ⎜ 5 + 8t + 9t + C⎟ = 16t ⎠ + 24t + 9 2 3 = 2 x 1 2(x + 18) + C 3 23. = (4t2 + 3) ∫(5 cos x + 4sin x) dx = 5sin x 4 cos x + C Check: d ⎛ 2 x 3 2 + 12x 1 2 + C⎞ Check: dx⎜3 ⎟ ⎝ ⎠ = 2⎛3 x1 2 ⎞ + 12 1 x 1 2 ⎞ d (5 sin x 4 cos x + C) = 5 cos x + 4 sin x dx 3⎜2 ⎟ ⎜2 ⎟ ⎝ ⎠ ⎝ ⎠ ∫ θ + sec θ dθ = 1 + tan θ + C

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. = x1 2 + 6x 1 2 = x + 6 x 24. ( 2 2 ) θ 3 3 Check: d ⎛1 3 + tan θ + C⎞ = θ 2 + sec2 θ ⎜ θ ⎟ dθ⎝3 ⎠ 25. 26. ∫(2 sin x 5ex ) dx = 2 cos x 5ex + C Check: d ( 2 cos x 5e x + C) = 2 sin x 5e x dx ∫ sec y(tan y sec y) dy = ∫(sec y tan y sec2 y) dy = sec y tan y + C Check: d (sec y tan y + C) = sec y tan y sec 2 y dy = sec y(tan y sec y)

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ∫ 4 ∫ 3 x ⎝ 430 Chapter 5 Integration 27. ∫(tan2 y + 1) dy = ∫ sec 2 y dy = tan y + C 35. f ′(x) = 6x, f (0) = 8 Check: d (tan y + C) = sec 2 y = tan2 y + 1 ( ) ∫ f x = 6x dx = 3x2 + C dy ( ) = = ( )2 + ⇒ = f 0 8 3 0 C C 8 28. ∫(4x csc 2 x) dx = 2x 2 + cot x + C f (x) = 3x 2 + 8 29. Check: d (2x2 + cot x + C) = 4x csc2 x dx x (2x 4x ) dx = x2 + C ln 4 36. f ′(s) = 10s 12s3 , f (3) = 2 f (s) = ∫(10s 12s3) ds = 5s2 3s4 + C f (3) = 2 = 5(3)2 3(3)4 + C = 45 243 + C ⇒ C = 200 f (s) = 5s 2 3s 4 + 200 d ⎛ 4x ⎞ Check: x2 + C = 2x 4x 37. f ′′(x) = 2 dx ⎜ ln 4 ⎟ 30. ⎝ ⎠ x (cos x + 3x ) dx = sin x + + C ln 3 Check: d ⎛ sin x + 3 + C ⎞ = cos x + 3x f ′(2) = 5 f (2) = 10 f ′(x) = ∫ 2 dx = 2x + C1 f ′(2) = 4 + C1 = 5 ⇒ C1 = 1 dx ⎜ ln 3 ⎟ f (x) = 2x + 1 ⎛ 5⎞ ⎝ ⎠ x2 ′ f (x) = ∫(2x + 1) dx = x 2 + x + C2 ∫ ⎝ x ⎠ 2 f (2) = 6 + C = 10 ⇒ C = 4 31. ⎜x ⎟ dx = 5 ln x + C 2 2 Check: d ⎛ x2 dx ⎜ 2 ⎞ 5 5 ln x + C⎟ = x ⎠ x 38. f (x) = x 2 + x + 4 f ′′(x) = x 2 32. ⎛ 4 + sec2 x ⎞ dx = 4 ln x + tan x + C f ′(0) = 8 ∫ ⎜ x ⎟ ⎝ ⎠ Check: d (4 ln x + tan x + C) = 4 + sec 2 x f (0) = 4 f ′(x) = x2 dx = 1 x3 + C dx x ∫ 3 1 33. f ′(x) = 4 f (x) = 4x + C Answers will vary y f(x) = 4x + 2 5 f ′ f ′(0) = 0 + C1 = 8 ⇒ C1 = 8 f ′(x) = 1 x3 + 8 3 f (x) = ∫ ⎛1 x3 + 8⎞ dx = 1 x4 + 8x + C 3 ⎜3 ⎟ 12 2 2 f(x) = 4x x ⎝ ⎠ f (0) = 0 + 0 + C2 = 4 ⇒ C2 = 4

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 34. f ′(x) = x f (x) = x + C 2 3 2 1 1 2 3 39. f (x) = 1 x 4 + 8x + 4 12 f ′′(x) = x 3 2 f ′(4) = 2 f (0) = 0 y f(x) x2 + 2 f ′ x = x dx = 2x + C = 2 + C 2 8 f(x) = x2 2 ( ) ∫ 3 2 2 1 2 1 1 x 6 f ′(4) = + C1 = 2 ⇒ C1 = 3 2 4 f ′ f ′(x) = 2 + 3 x x 4 2 2 4 f (x) = ∫( 2x 1 2 + 3) dx = 4x1 2 + 3x + C2 Answers will vary f (0) = 0 + 0 + C2 = 0 ⇒ C2 = 0 f (x) = 4x 1 2 + 3x = 4 x + 3x

5.1 Antiderivatives and Indefinite Integration

40. f ′′(x) = sin x f ′(0) = 1 (b) dy = x 2 1, ( 1, 3) dx x 3 f (0) = 6 f ′(x) = ∫ sin x dx = cos x + C1 f ′(0) = 1 + C1 = 1 ⇒ C1 = 2 f ′(x) = cos x + 2 f (x) = ∫( cos x + 2) dx = sin x + 2x + C2 y = x + C 3 ( 1)3 3 = ( 1) + C 3 C = 7 3 x3 7 5 (−1, 3) 4 4 f (0) = 0 + 0 + C2 = 6 ⇒ C2 = 6 y = x + 3 3 5 41. f (x) = sin x + 2x + 6 f ′′(x) = e x f ′(0) = 2 f (0) = 5 44. (a) Answers will vary Sample answer: y (1, 3) f ′(x) = ∫ e x dx = e x + C1 1 7 f ′(0) = 2 = e 0 + C1 ⇒ C1 = 1 f ′(x) = ex + 1 f (x) = ∫(ex + 1) dx = ex + x + C2 (b) dy 1 , x dx x2 > 0, (1, 3) f (0) = 5 = e0 + 0 + C2 ⇒ C2 = 4 1 x 1 1 y = ∫ dx = ∫ x 2 dx = + C = + C f (x) = e x + x + 4 2 x 3 = 1 + C ⇒ C = 2 1 1 x 42. f ′′(x) = x2 f ′(1) = 4 f (1) = 3 2 2 y = 1 + 2 x 5 f ′(x) = ∫ dx = 2x 2 dx = + C x2 1 8 f ′(1) = 4 = 2 + C1 ⇒ C1 = 6 1 f ′(x) 2 + 6 = x⎛ 2 ⎞ 45. (a) 9 f (x) = ∫ ⎜ + 6⎟ dx = 2 ln x + 6x + C2 ⎝ x ⎠ 3 3 f (1) = 3 = 6 + C2 ⇒ C2 f (x) = 2 ln x + 6x 3 = 3 9 dy 43. (a) Answers will vary Sample answer. y (b) = 2x, ( 2, 2) dx 5 x 4 4

Section

431

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (c) y = ∫ 2 x d x = x 2 + C 2 = ( 2) 2 + C = 4 + C ⇒ C = 6 y = x 2 6 1 2 5 15 15 8

They are the same In both cases you are finding a function

)such that

49. Because f ′′is negative on ( ∞ , 0), f ′ is decreasing on ( ∞ , 0) Because f ′′is positive on (0, ∞), f ′ is

on (0, ∞) f ′has a relative minimumat

0) Because f ′is positive on ( ∞ , ∞), f is increasing on ( ∞ , ∞)

f (0) = 4. Graph of f ′is given.

(a)

(b) No. The slopes of the tangent lines are greater than

2 on [0, 2] Therefore, f must increase more than

(d) f is a maximumat x = 3.5 because and the First Derivative Test.

f is concave upward when f ′is increasing on (

is concave downward on (1,

Points of inflection at x = 1, 5.

bacteria

s(0) = 6 = C2 s(t) = 16t2 + 60t + 6, Position function

The ball reaches its maximum height when v(t) = 32t + 60 = 0 32t = 60 t = 15 seconds.

(a) 20 50.

46.

f ′(4) ≈ 1 0

0 6

(b) dy = 2 dx x, (4, 12) (c) No, f (5) < f (4) because f is decreasing on [4, 5] 4 y = 2x1 2 dx = x3 2 + C 3 12 = 4(4)3 2 + C = 4(8)+ C = 32 + C ⇒ C = 4

4 units on [0, 4].

f ′

3.5

≈ 0 3 3 3 3 y = 4 x3 2 + 4 3 3

∞

(c) 20 0 6 0 51. (a) (b) h(t) = ∫(1.5t + 5) dt = 0.75t2 + 5t + C h(0) = 0 + 0 + C = 12 ⇒ C = 12 h(t) = 0.75t2 + 5t + 12 h(6) = 0 75(6)2 + 5(6) + 12 = 69 cm 47.

F′(x) = f (x) 52. dP = k t , 0 ≤ t ≤ 10 dt 48. f (x) = tan2 x ⇒ f ′(x) = 2 tan x ⋅ sec 2 x P(t) = ∫ kt1 2 dt = 2kt3 2 + C g(x) = sec 2 x ⇒ g ′(x) = 2sec x ⋅ sec x tan x = f ′(x) 3 P(0) = 0 + C = 500 ⇒ C = 500 The derivatives are the same, so f and g

a constant In fact, tan2 x + 1 = sec2 x P(1) = 2k + 500 = 600

150

(

)

(e)

, 1)and (5, ∞) f

(

differ by

⇒ k

P(t) = 2(150)t3 2 + 500 = 100t3 2 + 500 3

P(7) = 100(7)3 2 + 500 ≈ 2352

2 v

1 3 f ′ 2 1 3 2 2 f 3 f ″ x 1 2 3 v

increasing

(0,

53. a(t) = 32 ft sec

(t) = ∫ 32 dt = 32t + C

(0) = 60 = C1 s(t) = ∫( 32t + 60) dt = 16

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (15) (15)2 (15) s 8 = 16 8 + 60 8 + 6 = 62.25 feet

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 Section 5.1 Antiderivatives and Indefinite Integration 433 54. a(t) = 32 ft sec2 57. From Exercise 56, f (t) = 4.9t2 + v0t + 2.If v(t) = ∫ 32 dt = 32t + C1 f (t) = 200 = 4.9t2 + v0t + 2, v(0) = 0 + C1 = V0 ⇒ C1 = V0 Then v(t) = 9.8t + v0 = 0 s ′(t) = 32t + V0 s(t) = ∫ ( 32t + V0)dt = 16t2 + V0t + C2 for this t value. So, t = v0 9.8 and you solve ⎛ v ⎞ ⎛ v ⎞ 4.9 0 + v 0 + 2 = 200 ⎜ ⎟ 0⎜ ⎟ s(0) = 0 + 0 + C2 = S0 ⇒ C2 = S0 ⎝9.8⎠ ⎝9.8⎠ 2 ⎛ 2 ⎞ s(t) = 16t2 + V0t + S0 4.9v0 + ⎜ v0 ⎟ = 198 (9.8)2 v ⎝9.8⎠ s ′(t) = 32t + v0 maximumheight. s ⎛ v0 ⎞ = 16⎜ v0 ⎟ = 0 when t = 0 = time to reach 32 2 v + v ⎜ 0 ⎟ = 550 4.9v0 2 + 9.8v0 2 4.9v0 2 v0 = (9.8)2198 = (9.8)2198 = 3880.8 ⎛ ⎞ ⎛ ⎞ 2 ⎜32⎟ 32 0 32 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ v0 ≈ 62.3 m sec v 2 v 2 0 + 0 = 550 64 32 58. From Exercise 56, f (t) = 4.9t2 + 1800.(Using the v0 2 v0 = 35,200 ≈ 187.617 ft sec canyon floor as position 0 ) f (t) = 0 = 4.9t2 + 1800 55. v0 = 16 ft sec 4.9t2 = 1800 s0 (a) = 64 ft s(t) = 16t2 + 16t + 64 = 0 t2 = 1800 ⇒ t ≈ 9.2 sec 4.9 16(t2 t 4) = 0 t = 1 ± 17 2 59. a = 1.6 v(t) = ∫ 1.6 dt = 1 6t + v0 stone was dropped, v0 = 0. = 1.6t, because the Choosing the positive value, s(t) = ∫( 1.6t) dt = 0 8t2 + s0 2 t = 1 + 17 2 ≈ 2.562 seconds. s(20) = 0 ⇒ 0.8(20) + s0 s0 = 0 = 320 (b) v(t) = s ′(t) = 32t + 16 ⎛ ⎞ So, the height of the cliff is 320 meters. v(t) = 1.6t ⎛1 + 17 ⎞ 1 + 17 v⎜ ⎟ = 32⎜ ⎟ + 16 v(20) = 32 m sec ⎝ 2 ⎠ ⎝ 2 ⎠ 56. a(t) = 9.8 = 16 17 ≈ 65.970 ft sec 60. 1 ∫ v dv = GM ∫ y2 dy

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 0 0 ⎝ ⎠ v(t) = ∫ 9.8 dt = 9.8t + C1 1 v 2 2 = GM + C y v(0) = v0 = C1 ⇒ v(t) = 9 8t + v0 When y = R, v = v0 f (t) = ∫( 9.8t + v0) dt = 4.9t2 + v0t + C2 1 v0 2 = GM + C f (0) = s0 = C2 ⇒ f (t) = 4 9t2 + v0t + s0 2 R C = 1 v 2 GM So, f (t) = 4.9t2 + 10t + 2. 1 v2 2 = GM R + 1 v 2 GM v(t) = −9.8t + 10 = 0 (Maximum height when v = 0) 2 y 2 0 R 9.8t = 10 v 2 = 2GM + v 2 2GM t = 10 9.8 y R ⎛1 1 ⎞ f ⎛ 10 ⎞ ≈ 7.1 m ⎜9.8⎟ v 2 = v0 2 + 2GM⎜ ⎟ ⎝ y R ⎠

It takes 1 333 seconds to reduce the speed from 45 mi h to 30 mi h, 1.333 seconds to reduce the speed

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. = 3 . ⎜ ⎟ 16.5 2 3 434 Chapter 5 Integration 61. x(t) = t3 6t2 + 9t 2, 0 ≤ t ≤ 5 66. v(0) = 45 mi h = 66 ft sec (a) (b) v(t) = x ′(t) = 3t2 12t + 9 = 3(t2 4t + 3) = 3(t 1)(t 3) a(t) = v ′(t) = 6t 12 = 6(t 2) v(t) > 0 when 0 < t < 1or 3 < t < 5. 30 mi h = 44 ft sec 15 mi h = 22 ft sec a(t) = a v(t) = at + 66 (c) a(t) = 6(t 2) = 0 when t = 2. s(t) a t2 + 66t (Let s(0) = 0.) 2 v(2) = 3(1)( 1) = 3 v(t) = 0 after car moves132 ft. 62. x(t) = (t 1)(t 3)2 0 ≤ t ≤ 5 at + 66 = 0 when t = 66 a (a) = t3 7t2 + 15t 9 v(t) = x ′(t) = 3t2 14t + 15 = (3t 5)(t 3) s ⎛66⎞ a⎛ 66⎞ 66⎞ 66⎜ ⎜ ⎟ = ⎜ ⎟ + ⎛ ⎟ (b) a(t) = v ′(t) = 6t 14 v(t) > 0 when 0 < t < 5 and 3 < t < 5. ⎝ a ⎠ 2⎝ a ⎠ ⎝ a ⎠ = 132 when a = 33 = 16.5. 2 (c) a(t) = 6t 14 = 0 when t = 7 a(t) = 16.5 (7) ((7) )(7 ) ( 2) 4 v t = 16.5t + 66 v 3 = 3 3 5 3 3 = 2 3 = 3 () s(t) = 8.25t2 + 66t 63. v(t) = 1 = t 1 2 t t > 0 (a) 16.5t + 66 = 44 x(t) = ∫ v(t)dt = 2t1 2 + C x(1) = 4 = 2(1) + C ⇒ C = 2 Position function: x(t) = 2t1 2 + 2 Acceleration function: a(t) = v ′(t) = 1 t 3 2 = 1 2 2t3 2 (b) t = 22 ≈ 1.333 16.5 s ⎛ 22 ⎞ ≈ 73.33 ft ⎝16.5⎠ 16.5t + 66 = 22 t = 44 ≈ 2.667 64. (a) a(t) = cos t v(t) = ∫ a(t) dt = ∫ cos t dt = sin t + C1 = sin t (because v0 = 0) (c) 16.5 s( 44 ) ≈ 117.33 ft f (t) = ∫ v(t) dt = ∫ sin t dt = cos t + C2 0 132 f (0) = 3 = cos(0) + C2 f (t) = cos t + 4 = 1 + C2 ⇒ C2 = 4 73.33 117.33 feet feet

v(t)

0 = sin t for t = kπ, k = 0, 1, 2,

(b)

65. (a) v(0) = 25 km h = 25 1000

250 m

3600 36 from 30 mi h to 15 mi

3600

v(t)

sec

h, and 1.333 seconds to v(13) = 80

h = 80 1000 = 800 m sec

36 a(t) = a (constant acceleration)

= at + C

reduce the speed from 15 mi h to 0 mi h Each time, less distance is needed to reach the next speed reduction. v

⇒

(13) = 800 = 13a + 250 550 36 36 36 = 13a a = 550 = 275 ≈ 1.175 m sec2 (b) 468 234 s(t) = a t + 250 t 2 36 2 (s(0) = 0) s(13) = 275 (13) + 250(13) ≈ 189.58 m 234 2 36

)

250

v(t) = at + 250 v

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ⎪ ⎪ ⎨2 2 ⎪ ⎢⎣ ⎦ ⎣ ⎦ ⎥ 2 2 y 2 Section 5.1 Antiderivatives and Indefinite Integration 435 67. Truck: v(t) = 30 ⎧ 1, 0 ≤ x < 2 ′ ⎪ s(t) = 30t (Let s(0) = 0) Automobile: a(t) = 6 v(t) = 6t (Let v(0) = 0) s(t) = 3t2 (Let s(0) = 0) 72. f (x) = ⎨2, 2 < x < 3 ⎩0, 3 < x ≤ 4 ⎧ x + C1, 0 ≤ x < 2 f (x) = ⎪ x + C , 2 < x < 3 ⎩C3, 3 < x ≤ 4 At the point where the automobileovertakes the truck: 30t = 3t2 f (0) = 1 ⇒ C1 = 1 0 = 3t2 30t 0 = 3t(t 10) when t = 10 sec f continuous at x = 2 ⇒ 2 + 1 = 4 + C2 ⇒ C2 = 5 (a) s(10) = 3(10)2 = 300 ft f continuous at x = 3 ⇒ 6 5 = C3 = 1 ⎧ x + 1, 0 ≤ x < 2 ⎪ (b) v(10) = 6(10) = 60 ft sec ≈ 41 mi h f (x) = ⎨2x 5, 2 ≤ x < 3 ⎩1, 3 ≤ x ≤ 4 68. a(t) = k v(t) = kt s(t) = k t2 because v(0) = s(0) = 0. 2 1 x 1 2 3 4 At the time of lift-off, kt = 160 and (k 2)t2 = 0 7. Because (k 2)t2 = 0 7, t = 1.4 k 73. d ⎡⎡s(x)⎤2 + ⎡c(x)⎤2⎤ = 2s(x)s ′(x) + 2c(x)c ′(x) dx ⎣ ⎦ ⎛ 1.4⎞ v = k 1.4 = 160 = 2s x c x 2c x s x = 0 ⎜ ⎟ ( ) ( ) ( ) ( ) ⎝ k ⎠ k 1602 So, ⎡s(x)⎤2 + ⎡c(x)⎤2 = k for some constant k 1.4k = 1602 ⇒ k = 1.4 ⎣ ⎦ ⎣ ⎦ Because, s(0) = 0and c(0) = 1, k = 1. ≈ 18,285.714 mi h2 ≈ 7.45 ft sec2 69. False f has an

of antiderivatives,

a constant. Therefore, ⎡ ⎣s(x)⎦ ⎤ + ⎣ ⎡c(x)⎦ ⎤ = 1. [Note that s(x) = sin x and c(x) = cos properties.] x satisfy these 70. True 74. d (ln Cx ) = d (ln C + ln x ) = 0 + 1 = 1 71. f ′′(x) = 2x dx dx x x f ′(x) = x2 + C 75. d (ln x + C) = 1 + 0 = 1 f ′(2) = 0 ⇒ 4 + C = 0 ⇒ C = 4 x 3 f (x) = 4x + C1 3 dx x x

infinite number

each differing by

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ( ) ( ) 8 16 f 2 = 0 ⇒ 8 + C1 = 0 ⇒ C1 = 3 3 x3 16 f x = 4x + 3 3

436 Chapter 5 Integration

76. f (x + y) =

g(x + y) =

f ′(0) = 0

f (x) f ( y) g(x)g(y)

f (x)g( y) + g(x) f ( y)

[Note: f (x) = cos x and g(x) = sin x satisfy these conditions]

f ′(x + y) =

g ′(x + y) =

f (x) f ′( y) g(x)g ′(y) (Differentiatewith respect to y)

f (x)g ′(y) + g(x) f ′(y) (Differentiatewith respect to y)

So, 2 f (x) f ′(x) = 2 f (x)g(x)g ′(0) 2

k =0

All

be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ⎣ ⎦⎣ ⎦

Rights Reserved May not

g(x)g ′(0) = g(x)g ′(0) g ′(x) = f (x)g ′(0) + g(x) f ′(0) = f (x)g ′(0)

Letting y = 0, f ′(x) = f (x) f ′(

)

g(x)g ′(x) = 2g(x) f (x)g ′(0) Adding, 2 f (x) f ′(x) + 2g(x)g ′(x) = 0. Integrating, f (x)2 + g(x)2 = C Clearly C ≠ 0, for if C = 0,then f (x)2 = g(x)2 ⇒ f (x) = g(x) = 0, which contradicts that f, g are nonconstant Now, C = f (x + y)2 + g(x + y)2 = ( f (x) f ( y) g(x)g( y))2 + ( f (x)g(y) + g(x) f ( y))2 = f (x)2 f (y)2 + g(x)2 g( y)2 + f (x)2 g( y)2 + g(x)2 f ( y)2 = ⎡ f (x)2 + g(x)2⎤⎡ f ( y)2 + g( y)2⎤ = C2 So, C = 1and you have Section 5.2 Area f (x)2 + g(x)2 = 1. 6 6 6 1. ∑(3i + 2) = 3∑i + ∑2 = 3(1 + 2 + 3 + 4 + 5 + 6) + 12 = 75 i =1 i =1 i =1 9 2. ∑(k2 + 1) = (32 + 1) + (42 + 1) + + (92 + 1) = 287 k =3 4 1 1 1 1 1 158 3. ∑ k2 + 1 = 1 + 2 + 5 + 10 + 17 = 85 6 4. ∑ 3 = 3 + 3 + 3 = 37 j =4 j 4 5 6 20 4 5. ∑c = c + c + c + c = 4c k =1 4 6. ⎡(i 1)2 + (i + 1)3⎤ = (0 + 8) + (1 + 27) + (4 + 64) + (9 + 125) = 238 ∑⎣ ⎦ i =1 11 1 6 ⎡ ⎛ j ⎞ ⎤ 7. ∑ 9. ∑ ⎢7⎜6⎟ + 5⎥ i =1 5i j =1 ⎣ ⎝ ⎠ ⎦

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 1 + i 4 14 9 4 ⎡ ⎛ ⎞ ⎤ 8. ∑ i =1 10. ∑ j =1 ⎢1⎢ ⎣ j ⎜ ⎟ ⎥ ⎝ ⎠ ⎥ ⎦

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. n n n n n 4 n n Section 5.2 Area 437 11. 2 ⎡⎛2i ⎞3 ⎛ 2i ⎞⎤ 17. 20 ∑(i 1)2 19 = ∑ i2 ⎡19(20)(39)⎤ = ⎢ ⎥ = 2470 ∑ ⎢⎜ ⎟ ⎜ ⎟⎥ i =1 ⎢ ⎣⎝ ⎠ ⎝ ⎠⎥ ⎦ n ⎡ ⎛ ⎞2⎤ i =1 10 2 i =1 ⎣ 10 10 2 6 ⎦ ⎡10(11)(21)⎤ 12. 3∑⎢2⎜1 + 3i ⎟ ⎥ 18. ∑(i 1) = ∑ i ∑1 = ⎢ ⎥ 10 = 375 n i =1 ⎝ n ⎠ 12 i =1 15 i =1 15 2 3 i =1 ⎣ 6 ⎦ 15 15 2 13. ∑7 = 7(12) = 84 i =1 19. ∑ i(i 1) i =1 = ∑ i i =1 2 2∑ i i =1 2 + ∑ i i =1 30 15 (16) 15(16)(31) 15(16) = 2 + 14. ∑( 18) = ( 18)(30) = 540 i =1 4 6 2 = 14,400 2480 + 120 = 12,040 24 24 ⎡24(25)⎤ 15. ∑4i = 4∑i = 4⎢ ⎥ = 1200 25 25 25 3 3 i =1 i =1 ⎣ 2 ⎦ 20. ∑(i i =1 2i) = ∑ i i =1 2∑ i i =1 2 2 16. ( i 4) = 5 i 4(16) = 5 ⎡16(17)⎤ 64 = 616 ( )( ) = ( ) 2 16 16 ∑ 5 ∑ ⎢ 2 ⎥ 25 26 25 26 4 2 i=1 i=1 ⎣ ⎦ = 105,625 650 = 104,975 21. ∑ 2i 1 = 1 ∑(2i + 1) = 1 ⎢2 n(n 1) + n⎥ = n 2 = 1 + 2 = S(n) n + n 2 n 2 ⎡ + ⎤ + n 2 2 n n i =1 n S(10) = 12 10 i =1 ⎣ ⎦ = 1.2 S(100) = 1.02 S(1000) = 1.002 S(10,000) = 1.0002 n 7 j + 4 1 n 22. ∑ 2 j =1 = 2 ∑(7 j + 4) j =1 1 ⎡ = 2 ⎢7⎣ n(n + 1) 2 ⎤ + 4n⎥ ⎦ 7n 2 + 7n = 2n2 + 4n n2 = 7n + 15 2n = S(n) S(10) = 17 = 4.25 S(100) = 3.575 S(1000) = 3.5075

S(1000) = 1.999998

S(10,000) = 1.99999998

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. n S(10,000) = 3.50075 23. n 6k(k 1) ∑ = 6 ∑(k2 k) = 6 ⎡n(n + 1)(2n + 1) ⎢ n(n + 1)⎤ ⎥ k =1 n3 n3 k =1 2 n3 ⎣ 6 2 ⎦ = 6 ⎡2n +3n +1 3n 3⎤ = 1 ⎡2n2 2⎤ = 2 2 = S(n) n2 ⎢ 6 ⎥ n2 ⎣ ⎦ n2 ⎣ ⎦

S(10) = 1.98

S(100) = 1.9998

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ⎡ ⎤ ⎣ ⎦ 438 Chapter 5 Integration n 3 n 24. ∑ 2i 3i i =1 n4 = 1 ∑(2i3 3i) n4 i =1 1 ⎡ = ⎢2 n 2(n + 1)2 n(n + 1)⎤ 3 ⎥ n 4 4 2 2 3 2 (n +1) = 2n 2 S(10) = 0.5885 3(n +1) = 2n 3 n +2n 2n 3 = S(n) 2n3 S(100) = 0.5098985 S(1000

S(10,000

25. y y 10 10 8 8 6 6 4 4 2 2 x x 1 2 1 2 ∆x = 2 0 = 1 4 2 Left endpoints: Area ≈ 1[5 + 6 + 7 + 8] = 26 = 13 2 2 Right endpoints: Area ≈ 1[6 + 7 + 8 + 9] = 30 = 15 2 2 13 < Area < 15 26. y y 10 10 8 8 6 6 4 4 2 2 x 1 2 3 4 x 1 2 3 4 ∆x = 4 2 = 1 6 3 Left endpoints: Area ≈ 1 7 + 20 + 19 + 6 + 17 + 16 = 37 ≈ 12.333 3⎢ 3 3 3 3 ⎥ 3 ⎣ ⎦ Right endpoints: Area ≈ 1⎡20 + 19 + 6 + 17 + 16 + 15⎤ = 35 ≈ 11.667 35 < Area < 37 3 3 3⎢ 3 3 3 3 3 ⎥ 3

)

0.5009989985

) = 0.50009999

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 2 2 2 ⎦ ⎦ Section 5.2 Area 439 27. y y 50 50 40 40 30 30 20 20 10 10 x 1 2 3 4 5 x 1 2 3 4 5 ∆x = 5 2 = 1 6 2 Left endpoints: Area ≈ Right endpoints: Area ≈ 55 < Area < 74.5 1[5 + 9 + 14 + 20 + 27 + 35] = 55 1[9 + 14 + 20 + 27 + 35 + 44] = 149 = 74.5 28. y y 10 10 8 8 6 6 4 4 2 2 x 1 2 3 x 1 2 3 ∆x = 3 1 = 1 8 4 Left endpoints: 1 41 ⎣ 13 65 97 29 137⎦ 155 Area ≈ ⎡ + + 4 16 4 + 16 + 5 + 16 + + ⎤ = 4 16 16 = 9.6875 Right endpoint: Area ≈ 1 ⎡41 + 13 + 65 + 5 + 97 + 29 + 137 + 10⎤ = 11.6875 4⎣16 4 16 16 4 16 ⎦ 9.6875 < Area < 11.6875 29. y y 1 1 π π π π 4 2 4 2 π 0 ∆x = 2 = π 4 8 Left endpoints: Area≈ π⎢cos(0) + cos⎜π⎟ + cos⎜π⎟ + cos⎜3π⎟⎤ ≈ 1.1835 ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 8 ⎣ ⎝ 8 ⎠ ⎝ 4 ⎠ ⎝ 8 ⎠⎥ Right endpoints: Area≈ π⎢cos⎜π⎟ + cos⎜π⎟ + cos⎜3π⎟ + cos⎜π⎟⎤ ≈ 0.7908 ⎡ 8 ⎣ 0.7908 < Area < 1.1835 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝ 8 ⎠ ⎝ 4 ⎠ ⎝ 8 ⎠ ⎝ 2 ⎠⎥

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1 π 2 ⎛ ⎛ 2 2 ⎛ 5 5 5 5 5 5 5 5 440 Chapter 5 Integration 30. y y 1 x x π 2 ∆x = π 0 = π 6 6 Left endpoints: Area ≈ π sin 0 + sin π + sin π + sin π + sin 2π + sin 5π ≈ 1.9541 ⎡ ⎤ 6⎢ 6 3 2 3 6 ⎥ ⎣ ⎦ Right endpoints: Area≈ π sin π + sin π + sin π + sin 2π + sin 5π + sin π ≈ 1.9541 ⎡ ⎤ 6⎢ 6 3 2 3 6 ⎥ ⎣ ⎦ By symmetry, the answers are the same The exact area (2) is larger. 31. S = ⎡3 + 4 + 9 + 5⎤(1) = 33 = 16.5 ⎣ 2 ⎦ 2 s = ⎡1 + 3 + 4 + 9⎤(1) = 25 = 12.5 ⎣ 2⎦ 2 32. S = [5 + 5 + 4 + 2](1) = 16 s = [4 + 4 + 2 + 0](1) = 10 33. S(4) = 1⎛ 1⎞ + 1⎛ 1⎞ + 3⎛ 1⎞ + 1 1⎞ = 1 + 2 + 3 +2 ≈ 0.768 4⎜4⎟ 2⎜4⎟ 4⎜4⎟ ⎜4⎟ 8 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ s(4) = 0⎜1⎞ + 1⎛ 1⎞ + 1⎛ 1⎞ + 3⎛ 1⎞ = 1 + 2 + 3 ≈ 0.518 ⎝4⎟ 4⎜4⎟ 2⎜4⎟ 4⎜4⎟ 8 34. 35. ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ S(4) = 4(e 0 + e 0.5 + e 1 + e 1.5 )1 ≈ 4.395 s(4) = 4(e 0.5 + e 1 + e 1.5 + e 2)1 ≈ 2.666 S(5) = 1⎜1⎞ + 1 ⎛1⎞ + 1 ⎛1⎞ + 1 ⎛1⎞ + 1 ⎛ 1⎞ = 1 + 1 + 1 + 1 + 1 ≈ 0.746 5⎟ 6 5⎜5⎟ 7 5⎜5⎟ 8 5⎜5⎟ 9 5⎜5⎟ 5 6 7 8 9 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ s(5) = 1 ⎛1⎞ + 1 ⎛1⎞ + 1 ⎛1⎞ + 1 ⎛ 1⎞ + 1⎛1⎞ = 1 + 1 + 1 + 1 + 1 ≈ 0.646 6 5⎜5⎟ 7 5⎜5⎟ 8 5⎜5⎟ 9 5⎜5⎟ 2⎜5⎟ 6 7 8 9 10 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 2 2 2 36. S(5) = 1⎜1⎟ + 1 1 1 ⎜ ⎟ ⎜ ⎟ + 2 1 1 ⎜ ⎟ ⎜ ⎟ + 3 1 1 ⎜ ⎟ ⎜ ⎟ + 4 1 1 ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ 1⎡ 24 21 16 9⎤ = ⎢1 + + + + ⎥ ≈ 0.859 5 5 5 5 5 2 2 2 2 s(5) = 1 ⎛1⎞ ⎛1⎞ + 1 ⎛ 2⎞ ⎛ 1⎞ + 1 ⎛ 3⎞ ⎛1⎞ + 1 ⎛ 4⎞ ⎛1⎞ + 0 ≈ 0.659 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ n ⎛ 24i ⎞ 24 n 24⎛ n(n+1)⎞ ⎡ ⎛ n 2 +n ⎞⎤ ⎛ 1⎞

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 37. lim ∑⎜ ⎟ = lim ∑ i = lim ⎜ ⎟ = lim⎢12⎜ ⎟⎥ = 12 lim⎜1 + ⎟ = 12 n→∞ ⎝ n 2 ⎠ n→∞ n 2 n→∞ n 2 2 n→∞ n 2 n→∞⎝ n ⎠ i =1 i =1 ⎝ ⎠ ⎣ ⎝ ⎠⎦

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ∑ ∑ n ⎢ ⎥ ∑ 3 Section 5.2 Area 441 38. n ⎛ 3i ⎞⎛ 3⎞ lim ⎜ ⎟⎜ ⎟ = lim 9 n i = lim 9 ⎡n(n+1)⎤ ⎢ ⎥ = lim 9⎛ n +1⎞ 9 ⎜ ⎟ = n →∞ i =1⎝ n ⎠⎝ n ⎠ n →∞ n2 i =1 n→∞ n2⎣ 2 ⎦ n→∞ 2⎝ n ⎠ 2 39. lim∑ 1 (i 1)2 = lim 1 n 1 ∑ i2 = lim 1 ⎡(n 1)(n)(2n 1)⎤ n→∞ i =1 n 3 n→∞n 3 =1 n→∞ n 3⎣ 6 ⎦ i 1⎡2n3 3n2 + n⎤ ⎡1⎛2 (3 n) + (1 n2)⎞⎤ 1 = lim ⎢ 6 n ⎥ = lim⎢ ⎜ 6 ⎟⎥ = 1 3 n→∞ ⎣ ⎦ n→∞⎢ ⎣ ⎜ ⎟⎥ ⎝ ⎠⎦ n ⎛ ⎞2⎛ ⎞ n 40. lim∑⎜1 + 2i ⎟ ⎜2⎟ = lim 2 ∑(n + 2i)2 n→∞ i =1 ⎝ n ⎠ ⎝ n ⎠ n→∞ n 3 i =1 n n n = lim 2 ⎡ n2 + 4n i + 4 i2⎤ n→∞ n3 ⎢∑ ∑ ∑ ⎥ ⎣i =1 i =1 i =1 ⎦ = lim 2 ⎢n3 + (4n)⎜ n(n+1)⎞ + 4(n)(n +1)(2n +1)⎥ ⎡ ⎛ ⎤ n→∞ n3 2 ⎟ 6 ⎝ ⎠ ⎡ 2 4 2 2 ⎤ ⎛ 4⎞ 26 = 2 lim ⎢1 + 2 + + + + ⎥ = 2⎜1 + 2 + ⎟ = n→∞ n 3 n 3n 2 3 3 ⎣ ⎦ ⎝ ⎠ n ⎛ i ⎞⎛ 2⎞ 1⎡ n 1 n ⎤ 1 ⎡ 1⎛ n(n+1)⎞⎤ ⎡ n 2 +n⎤ ⎛ 1⎞ 41. lim ∑⎜1 + ⎟⎜ ⎟ = 2 lim ⎢∑1 + ∑ i⎥ = 2 lim ⎢n + ⎜ ⎟⎥ = 2lim⎢1 + ⎥ = 2⎜1 + ⎟ = 3 n→∞ i =1 ⎝ n ⎠⎝ n⎠ n→∞ n⎣i =1 ni =1 ⎦ n→∞ n ⎢ ⎣ n⎝ 2 ⎠ n→∞⎣ 2n2 ⎦ ⎝ 2⎠ n 3 n 3 42. lim∑⎛2 + 3i ⎞ ⎛ 3⎞ = lim 3∑⎡2n +3i⎤ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ n →∞ i =1 ⎝ n ⎠ ⎝ n ⎠ n →∞ n i =1 ⎣ n ⎦ n = lim 3 (8n3 + 36n2i + 54ni2 + 27i3) n →∞ n 4 i =1 3 ⎛ n(n + 1) n(n + 1)(2n + 1) n2(n + 1)2 ⎞ = lim ⎜8n 4 + 36n 2 + 54n + 27 ⎟ n →∞ n4 ⎜ 2 6 4 ⎟ ⎝ ⎠ ⎛ (n + 1) 9(n + 1)(2n + 1) 27 (n + 1)2 ⎞ = lim 3⎜8 + 18 + + ⎟ n →∞ ⎜ n n2 4 n2 ⎟ ⎝ ⎠ ⎛ 27⎞ = 3⎜8 + 18 + 18 + ⎟ = 609 = 152.25 ⎝ 4 ⎠ 4 43. (a) y 3 2

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ( ) 1 (b) ∆x = x 1 3 2 0 = 2 n n⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞ Endpoints: 0 < 1⎜ ⎟ < 2⎜ ⎟ < < n 1 ⎜ ⎟ < n⎜ ⎟ = 2 ⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ ⎝ n⎠ (c) Because y = x is increasing, f (mi ) = f (xi 1) on [xi 1, xi ] n n ⎛2i 2⎞⎛2⎞ n ⎡ ⎛2⎞⎤⎛2⎞ s(n) = ∑ f (xi 1)∆x = ∑ f ⎜ ⎟⎜ ⎟ = ∑⎢(i 1)⎜ ⎟⎥⎜ ⎟ i =1 i =1 ⎝ n ⎠⎝ n⎠ i =1 ⎣ ⎝ n⎠⎦⎝ n ⎠

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. x 5 10 50 100 s(n) 1.6 1.8 1.96 1.98 S(n) 2.4 2.2 2.04 2.02 x 5 10 50 100 s(n) 3.6 3.8 3.96 3.98 S(n) 4.4 4.2 4.04 4.02 ( ) 1 442 Chapter 5 Integration (d) f (M i ) = f (xi ) on [xi 1, xi ] n n ⎛ ⎞ n ⎡⎛ ⎞⎤⎛ ⎞ S(n) = f (x )∆x = f 2i 2 = i 2 2 (e) ∑ i i =1 ∑ i =1 ⎜ ⎟ ⎝ n ⎠n ∑⎢ ⎜ ⎟⎥⎜ ⎟ i =1 ⎣ ⎝ n⎠⎦⎝ n ⎠ (f ) lim ⎡( 1)⎛2⎞⎤⎛2⎞ lim 4 ( 1) lim 4 ⎡n(n+1) ⎤ ⎡2(n +1) 4⎤ lim 2 n n n→∞ ∑ ⎢ i ⎜ ⎟⎥⎜ ⎟ = ⎝ n ⎠ ⎝ n ⎠ n→∞ n2∑ i = n→∞ n2 ⎢ 2 n⎥ = ⎢ n→∞ n n ⎥ = i =1 ⎣ ⎦ i =1 ⎣ ⎦ ⎣ ⎦ n ⎡ ⎛ 2⎞⎤⎛2⎞ 4 n ⎛ 4 ⎞n(n+1) 2(n +1) lim ∑⎢i⎜ ⎟⎥⎜ ⎟ = lim ∑ i = lim ⎜ ⎟ = lim = 2 n→∞ i =1 ⎣ ⎝ n⎠⎦⎝ n ⎠ n→∞ n 2 i =1 n→∞ ⎝ n2 ⎠ 2 n→∞ n 44. (a) y 4 3 2 1 x 2 4 (b) ∆ x = 3 1 = 2 n n Endpoints: 1 < 1 + 2 < 1 + 4 < " < 1 + 2n = 3 n n n ⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞ ⎛2⎞ 1 < 1 + 1⎜ ⎟ < 1 + 2⎜ ⎟ < " < 1 + n 1 ⎜ ⎟ < 1 + n⎜ ⎟ ⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ (c) Because y = x is increasing, f (mi ) = f (xi 1) on [xi 1, xi ] n n ⎡ ⎛2⎞⎤⎛2⎞ n ⎡ ⎛2⎞⎤⎛2⎞ s(n) = ∑ f (xi 1)∆x = ∑ f ⎢1 + (i 1)⎜ ⎟⎥⎜ ⎟ = ∑ ⎢1 + (i 1)⎜ ⎟⎥⎜ ⎟ (d) i =1 f (M i ) = i =1 ⎣ f (xi ) on [xi 1, xi ] ⎝ n ⎠⎦⎝ n ⎠ i =1 ⎣ ⎝ n ⎠⎦⎝ n ⎠ S(n) = n n f (x )∆x = f ⎡ + i⎛ 2⎞⎤⎛2⎞ = n ⎡ 1 + i ⎛ 2⎞⎤⎛2⎞ (e) ∑ i i =1 ∑ ⎢ i =1 ⎣ ⎜ ⎟⎥⎜ ⎟ ⎝ n⎠⎦⎝ n ⎠ ∑ ⎢ i =1 ⎣ ⎜ ⎟⎥⎜ ⎟ ⎝ n⎠⎦⎝ n ⎠

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (f ) lim ∑⎢1 + ( 1)⎜2⎟⎥⎜2⎟ = 2 2 n(n 1) lim⎜ ⎟⎢n + ⎜ n⎟⎥ = lim 2 + 2n 2 4 = lim 4 2 = 4 n ⎡ ⎛ ⎞⎤⎛ ⎞ ⎛ ⎞⎡ ⎛ + ⎞⎤ ⎡ + ⎤ ⎡ ⎤ i ⎣ ⎝ ⎠⎦⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥ ⎢ ⎥ n n n n n n n n →∞ i =1 n →∞ ⎣ ⎝ 2 ⎠⎦ n→∞⎣ ⎦ n →∞⎣ ⎦ n ⎡ ⎛ 2⎞⎤⎛2⎞ 2⎡ ⎛ 2⎞n(n+1)⎤ ⎡ 2(n +1)⎤ ⎡ 2⎤ lim ∑⎢1 + i⎜ ⎟⎥⎜ ⎟ = lim ⎢n + ⎜ ⎟ ⎥ = lim⎢2 + ⎥ = lim ⎢4 + ⎥ = 4 n →∞ i =1 ⎣ ⎝ n ⎠⎦⎝ n ⎠ n →∞ n⎣ ⎝ n ⎠ 2 ⎦ n→∞⎣ n ⎦ n→∞ ⎣ n⎦

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 3 1 1 2 3 ⎜ ⎜ ⎢ = f 2 + n n ⎛ 3⎞ n n n Section 5.2 Area 443 45. y = 4x + 5 on [0, 1] ⎛Note: ∆x = 1⎞ n ⎟ 47. y = x2 + 2 on [0, 1] ⎛Note: ∆x = 1⎞ n ⎟ ⎝ ⎠ ⎝ ⎠ n ⎛ ⎞⎛ ⎞ n ⎛ ⎞ ⎤⎛ ⎞ n ⎛ i ⎞⎛1⎞ s(n) = ∑ f ⎜ i ⎟⎜ 1⎟ = ∑ ⎡ 4⎜ i ⎟ + 5⎥⎜1⎟ S(n) = ∑ f ⎜ ⎟⎜ ⎟ i =1 ⎝ n ⎠⎝ n ⎠ i =1 ⎣ n ⎝ n ⎠ ⎦⎝ n ⎠ i =1 n ⎝ n ⎠⎝ n ⎠ 2 ⎛ ⎞ ⎤⎛ ⎞ 4 ∑ i + 5 ⎡ i 1 = + 2⎥ n 2 ∑⎢⎜ ⎟ ⎜ ⎟ i =1 4 n(n+1) i =1 ⎢ ⎣⎝ n ⎠ n ⎥ ⎦⎝ n ⎠ = + ⎡ 1 = ⎤ i + 2 5 n2 2 ⎢ n 3 ∑ 2⎥ ⎣ i =1 ⎦ ⎛ 1⎞ = 2⎜1 + ⎟ + 5 n(n+1)(2n +1) 1 3 1 ⎝ n ⎠ = + 2 = ⎛2 + + ⎞ + 2 6n 3 6⎜ n n 2 ⎟ Area = lim s(n) = 3 n →∞ Area = ⎝ ⎠ lim S(n) = 7 y n→∞ 3 5 y 4 3 2 1 2 1 x 1 2 3 x 46. y = 3x 2 on [2, 5] ⎛Note: ∆x = 5 2 = 3⎞ ⎜ n n ⎟ ⎛ 2 0 2⎞ ⎝ ⎠ 48. y = 3x 2 + 1on [0, 2] ⎜Note: ∆x = = ⎟ n S(n) = ∑ ⎛ 3i ⎞⎛ 3⎞ ⎜ ⎟⎜ ⎟ ⎝ n n ⎡ ⎛ ⎞2 n n ⎠ ⎤⎛ ⎞ i =1 ⎝ n ⎠⎝ n ⎠ S(n) = f ⎛2i ⎞⎛2⎞ = ⎢3 2i + 1⎥ 2 ∑ ⎜ ⎟⎜ ⎟ ∑ ⎜ ⎟ ⎜ ⎟ n ⎡ ⎛ ⎞ ⎤⎛ ⎞ i =1 ⎝ n ⎠⎝ n ⎠ i =1 ⎝ n ⎠ ⎥ ⎦⎝ n ⎠ = ∑ ⎢3⎜2 + 3i ⎟ 2⎥⎜3⎟ i =1 ⎣ ⎝ n ⎠ ⎦⎝ n ⎠ = 24∑i2 + 2∑1 2 = 18 + 3⎜ ⎟ ∑ i 6 3 i =1 i =1 ⎝ n ⎠ i =1 24⎛ n(n+1)(2n +1)⎞ 2 = 3 ⎜ ⎟ + (n) 27⎛(n +1)n ⎞ 27⎛ 1⎞ n ⎝ 6 ⎠ n = 12 + ⎜ n2 ⎟ = 12 + 2 2 ⎜1 + n ⎟ 4(n +1)(2n +1) ⎝ ⎠ ⎝ ⎠ = + 2 2 Area = lim S(n) = 12 + 27 = 51 n

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. n→∞ 2 2 y Area = y lim S(n) = 8 + 2 = 10 n →∞ 15 3 10 2 5 1 2 3 x 4 5 6 x 1 2 3

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ⎜ 2 ⎢4 ⎥ 3 ∑ ⎜ ⎟ ⎜ ⎟ n 444 Chapter 5 Integration 49. y = 25 x2 on [1, 4] ⎛Note: ∆x = 3⎞ n ⎟ 50. y = 4 x2 on [ 2, 2] Find area of region over the ⎝ ⎠ interval [0, 2] ⎛Note: ∆x = 2⎞ ⎜ ⎟ n ⎛ ⎞⎛ ⎞ n ⎡ ⎛ ⎞ ⎤⎛ ⎞ ⎝ n ⎠ s(n) = ∑ f ⎜1 + 3i ⎟⎜3⎟ = ∑⎢25 ⎜1 + 3i ⎟ ⎥⎜3⎟ i =1 ⎝ n ⎠⎝ n ⎠ i =1 ⎝ n ⎠ ⎥ ⎦⎝ n ⎠ ( ) n ⎛ 2i ⎞⎛ 2⎞ s n = ∑ f ⎜ ⎟⎜ ⎟ 3 n ⎡ 9i2 6i⎤ i =1 ⎝ n ⎠⎝ n ⎠ = n∑⎢24 n 2 n ⎥ n ⎡ 2⎤ i =1 ⎣ ⎦ = ⎛ 2i ⎞ ⎛ 2⎞ 3⎡ 9 n(n+1)(2n +1) 6 n(n+1)⎤ i =1 ⎝ n ⎠ ⎥ ⎦⎝ n ⎠ = ⎢24n ⎥ n ⎣ n2 6 n 2 ⎦ = 8 8 ∑ i2 = 72 9 (n + 1)(2n + 1) 9(n + 1) 2n 2 n n3 i =1 =8n(= n + 1)(2n +1) = = 4⎛ + 3 + 1 ⎞ Area = lim s(n) = 72 9 9 = 54 8 6n3 8 3⎜2 n n2 ⎟ n→∞ y 1 Area = ⎝ ⎠ lim s(n) = 8 8 = 16 20 15 10 5 x 1 1 2 3 4 5 5 2 n→∞ 3 3 Area = 32 3 y 3 2 1 x 1 1 51. y = 27 x3 on [1, 3] ⎛Note: ∆x = 3 1 = 2⎞ ⎜ n n ⎟ ⎝ ⎠ n ⎛ ⎞⎛ ⎞ n ⎡ ⎛ ⎞ ⎤⎛ ⎞ s(n) = ∑ f ⎜1 + 2i ⎟⎜2⎟ = ∑⎢27 ⎜1 + 2i ⎟ ⎥⎜2⎟ i =1 ⎝ 2 n ⎡ n ⎠⎝ n ⎠ 8i3 i =1 12i2 ⎝ 6i⎤ n ⎠ ⎥ ⎦⎝ n ⎠ y = n∑⎢26 n3 n 2 n ⎥ 30 i =1 ⎣ ⎦ 24 2⎡ 8 n2(n + 1)2 12 n(n + 1)(2n + 1) 6 n(n + 1)⎤ 18 = ⎢26n n⎢ ⎣ n3 4 n 2 6 ⎥ 12 n 2 ⎥ ⎦ 6 x = 52 4 (n + 1)2 4 (n + 1)(2n + 1) 6n +1 2 1 6 1 2 4 5 n 2 n 2 n Area = lim s(n) = 52 4 8 6 = 34 n→∞ 52. y = 2x x3 on [0, 1] ⎛Note: ∆x = 1 0 = 1⎞ ⎜ n n ⎟

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ⎝ ⎠ Because y both increases and decreases on [0, 1], T(n) is neither an upper nor lower sum. T(n) = ∑ f ⎜ i ⎟⎜1⎟ = ∑⎢2⎜ i ⎟ ⎜ i ⎟ ⎥⎜1⎟ y 2.0 1.5 n 2 i =1 n 4 i=1 n2 n4 4 n 4 4n 4n2 1.0 0.5 Area = lim T(n) = 1 1 = 3 0.5 1.0 x 2.0

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 ∑ ∑ ∑ Section 5.2 Area 445 53. y = x2 x3 on [ 1, 1] ⎛ 1 ( 1) 2⎞ Note: ∆x = = ⎜ n n ⎟ ⎝ ⎠ Because y both increases and decreases on [ 1, 1], T(n) is neither an upper nor a lower sum. T(n) = n ⎛ 2i ⎞⎛ 2⎞ n f 1 + = ⎡⎛ 1 + 2i ⎞2 ⎛ 2i ⎞3⎤⎛2⎞ 1 + ∑ ⎜ ⎟⎜ ⎟ ∑⎢⎜ ⎟ ⎜ ⎟ ⎥⎜ ⎟ i =1 ⎝ n ⎠⎝ n ⎠ i =1 ⎢ ⎣⎝ n ⎠ ⎝ n ⎠ ⎥ ⎦⎝ n ⎠ n ⎡⎛ 4i 4i2 ⎞ ⎛ 6i 12i2 8i3 ⎞⎤⎛2⎞ = ∑⎢⎜1 n + n2 ⎟ ⎜ 1 + n n2 + n3 ⎟⎥⎜ n ⎟ i =1 ⎣⎝ ⎠ ⎝ ⎠⎦⎝ ⎠ n ⎡ 10i 16i2 8i3 ⎤⎛2⎞ 4 n 20 n 32 n 16 n = ∑⎢2 n + n 2 3 ⎥⎜ ⎟ = n ⎝ n ⎠ n ∑1 n 2 ∑ i + i2 i3 n3 n4 i =1 ⎣ ⎦ i =1 i =1 i =1 2 i =1 2 = 4(n) 20 n(n+1) + 32 n(n+1)(2n +1) 16 n (n +1) n n 2 2 n 3 6 n 4 4 1⎞ 16⎛ 3 1 ⎞ 2 1 ⎞ = 4 10⎜1 + + 2 + + 4⎜1 + + ⎛ ⎝ n ⎟ 3 ⎜ n n2 ⎟ ⎛ ⎝ n n2 ⎟ ⎠ ⎝ ⎠ ⎠ Area = lim T(n) = 4 10 + 32 4 = 2 n→∞ 3 3 y 2 1 x 1 1 54. y = 2x3 x2 on [1, 2] ⎛Note: ∆x = 2 1 = 1⎞ ⎜ n n ⎟ ⎝ ⎠ n ⎛ ⎞⎛ ⎞ n ⎡ ⎛ ⎞3 ⎛ ⎞ ⎤⎛ ⎞ s(n) = ∑ f ⎜1 + i 1 i ⎟⎜ ⎟ = ⎢2⎜1 + ⎟ ⎜1 + i 1 ⎟ ⎥⎜ ⎟ i =1 ⎝ n ⎠⎝ n ⎠ i =1 ⎝ n ⎠ ⎝ n ⎠ ⎥ ⎦⎝ n⎠ n ⎛2i3 5i2 4i ⎞⎛1⎞ = ∑⎜ + + + 1⎟⎜ ⎟ n 3 n 2 n ⎝ n ⎠ i =1 ⎝ ⎠ 2 n2(n + 1)2 5 n(n + 1)(2n + 1) 4 n(n + 1) = ⋅ + ⋅ + ⋅ + 1 n 4 4 n 3 6 n 2 2 Area = lim sn n →∞

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. = 1 + 5 + 2 + 1 = 31 2 3 6 y 12 10 8 6 4 2 −3 −2 −1 x 1 2 3

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. n ⎜ g 2 + 2 + = ∑ ∑ 2 = 5 446 Chapter 5 Integration 55. f (y) = 4y, 0 ≤ y ≤ 2 ⎛Note: ∆y = 2 0 = 2⎞ 57. f (y) = y2 , 0 ≤ y ≤ 5 ⎛Note: ∆y = 5 0 = 5⎞ ⎜ n n ⎟ ⎜ n n ⎟ ⎝ ⎠ ⎝ ⎠ n n ⎛5i ⎞⎛5⎞ S(n) = ∑ f (mi )∆y S(n) = ∑ f ⎜ n ⎟⎜ n ⎟ i =1 i =1 ⎝ ⎠⎝ ⎠ n ⎛ 2i ⎞⎛ 2⎞ n ⎛ 5i ⎞2⎛ 5⎞ = ∑ f ⎜ ⎟⎜ ⎟ = ∑⎜ ⎟ ⎜ ⎟ i =1 n ⎝ n ⎠⎝ n ⎠ ⎛ ⎞⎛ ⎞ i =1 ⎝ n ⎠ ⎝ n ⎠ 125 n = ∑4⎜2i ⎟⎜2⎟ = ∑ i2 i =1 ⎝ n ⎠⎝ n ⎠ n3 i =1 16 n 125 n(n + 1)(2n + 1) = 2 ∑ i i =1 = n3 6 ⎛16⎞ n(n+1) 8(n +1) 8 125⎛2n2 + 3n + 1⎞ 125 125 125 = ⎜ ⎟ = = 8 + = ⎜ ⎟ = + + ⎝ n 2 ⎠ 2 n n n2 ⎝ 6 ⎠ 3 2n 6n2 Area = lim S(n) = lim⎛8 + 8⎞ = 8 Area lim S(n) = lim ⎛125 + 125 + 125⎞ = 125 n→∞ n→∞ n ⎟ ⎜ 2 ⎟ ⎝ ⎠ n→∞ y y n→∞ ⎝ 3 2n 6n ⎠ 3 4 3 2 1 x 2 4 6 8 1 6 4 2 x 5 5 10 15 20 25 2 4 6 56. g(y) = 1 y, 2 ≤ y ≤ 4. ⎛Note: ∆y = 4 2 = 2⎞ ⎛ ⎞ 2 ⎜ n n ⎟ 58. f (y) = 4y y2 , 1 ≤ y ≤ 2. Note: ∆y = 2 1 = 1 ⎝ ⎠ ⎜ n n ⎟ S(n) = n ⎛ 2i ⎞⎛ 2⎞ ⎜ ⎟⎜ ⎟ S(n) = n ⎛ f 1 + ⎝ ⎠ i ⎞⎛ 1⎞ i =1 ⎝ n ⎠⎝ n ⎠ ∑ ⎜ n ⎟⎜ n ⎟ n n ⎛ ⎞ i =1 ⎝ ⎠⎝ ⎠ 1⎛ 2i ⎞⎛ 2⎞ = ⎜ ⎟⎜ ⎟ 2∑⎜1 + i ⎟ 1 n ⎡ ⎛ i ⎞ ⎛ i ⎞ ⎤ i =1 2⎝ n ⎠⎝ n ⎠ n i=1 ⎝ n ⎠ = ∑⎢4⎜1 + ⎟ ⎜1 + ⎟ ⎥ 2⎡ 1 n(n + 1)⎤ n + 1 n i =1 ⎝ n ⎠ ⎝ n ⎠ = ⎢n + ⎥ = 2 + 1 n ⎛ 4i 2i i2 ⎞ n⎣ n 2 ⎦ n = n ∑⎜4 + n 1 n n 2 ⎟ Area = lim S(n) = 2 + 1 = 3 i =1 ⎝ ⎠ n→∞ 1 n ⎛ 2i i2 ⎞ y n ∑⎜3 + n n 2 ⎟

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. i =1 ⎝ ⎠ 1 ⎡ 2 n(n+1) 1 n(n+1)(2n +1)⎤ 4 = ⎢3n + ⎥ 3 2 n + 1 1 n (n + 1)(2n + 1) 6 x 1 2 3 4 5 Area = lim S(n) = 3 + 1 1 = 11 n→∞ 3 3 y 5 3 2 1 x 1 2 3 4 5

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 3 ⎜ 3 n Section 5.2 Area 447 59. g(y) = 4y2 y3 , 1 ≤ y ≤ 3. ⎛Note: ∆y = 3 1 = 2⎞ ⎜ n n ⎟ ⎝ ⎠ n ⎛ ⎞⎛ ⎞ S(n) = ∑ g⎜1 + 2i ⎟⎜ 2⎟ i =1 ⎝ n ⎡ ⎛ n ⎠⎝ n ⎠ 2i ⎞ ⎛ 2i ⎞ ⎤ 2 = ∑⎢4⎜1 + ⎟ ⎜1 + ⎟ ⎥ i =1 ⎝ n ⎠ ⎝ n ⎠ ⎥ ⎦ n 2 n ⎡ 4i 4i2 ⎤ ⎡ 6i 12i2 8i3 ⎤ = n ∑4⎢1 + n + n 2 ⎥ ⎢1 + n + n 2 + n 3 ⎥ i =1 ⎣ ⎦ ⎣ ⎦ 2 n ⎡ 10i 4i2 8i3 ⎤ = n ∑⎢3 + n + n 2 n3 ⎥ i =1 ⎣ ⎦ 2⎡ 10 n(n + 1) 4 n(n + 1)(2n + 1) 8 n2(n + 1)2 ⎤ = ⎢3n + + ⎥ n n 2 n 2 6 n 2 4 Area = lim S(n) = 6 + 10 + 8 4 = 44 n→∞ 3 3 y 10 8 6 2 x 4 2 60. 2 4 h(y) = y3 + 1, 1 ≤ y ≤ 2 ⎛Note: ∆y = 1⎞ n ⎟ ⎝ ⎠ n ⎛ ⎞⎛ ⎞ S(n) = ∑ h⎜1 + i ⎟⎜ 1⎟ i =1 ⎝ n ⎡⎛ n ⎠⎝ n ⎠ i ⎞ ⎤ 1 = ∑⎢⎜1 + ⎟ i =1 ⎢ ⎣⎝ ⎠ + 1⎥ ⎥ ⎦ n y = ∑⎜2 + + + ⎟ 5 i =1 ⎝ ⎠ 4 1⎡ 1 n2(n + 1)2 3 n(n + 1)(2n + 1) 3 3n(n + 1)⎤ 3 = ⎢2n + + n n 3 4 n 2 6 + ⎥ n 2n 2 1 = 2 + + + x 2 4 6 8 10 n24 2 n2 2n Area = lim S(n) = 2 + 1 + 1 + 3 = 19 n→∞ 4 2 4

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 4 61. f (x) = x 2 + 3, 0 ≤ x ≤ 2, n = 4 x + x Let ci = i i 1 2 ∆x = 1 , c = 1 , c = 3 , c = 5 , c = 7 2 1 4 2 4 3 4 4 4 Area ≈ n f (c )∆x = ⎡c2 + 3⎤⎛1⎞ = 1⎡⎛ 1 + 3⎞ + ⎛ 9 + 3⎞ + ⎛ 25 + 3⎞ + ⎛ 49 + 3⎞⎤ = 69 ∑ i ∑⎣ i ⎦⎜ ⎟ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ i =1 i =1 ⎝2⎠ 2⎣⎝16 ⎠ ⎝16 ⎠ ⎝16 ⎠ ⎝16 ⎠⎦ 8

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 448 Chapter 5 Integration 62. f (x) = x 2 + 4x, 0 ≤ x ≤ 4, n = 4 x + x Let ci = i i 1 2 1 3 5 7 ∆x = 1, c1 = n , c2 2 = , c3 = 2 4 , c4 = 2 2 Area≈ f (c )∆x = ⎡c 2 + 4c ⎤(1) = ⎡⎛1 + 2⎞ + ⎛ 9 + 6⎞ + ⎛ 25 + 10⎞ + ⎛ 49 + 14⎞⎤ = 53 ∑ i ∑⎣ i i ⎦ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ i =1 i =1 ⎣⎝4 ⎠ ⎝4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠⎦ 63. f (x) = tan x, 0 ≤ x ≤ π , n = 4 4 x + x Let ci = i i 1 2 ∆x = π , c = π , c = 3π , c = 5π , c = 7π 16 1 n 32 2 32 3 4 32 4 32 Area≈ f (c )∆x = (tan c )⎛π ⎞ = π⎛ tan π + tan 3π + tan 5π + tan 7π⎞ ≈ 0.345 ∑ i ∑ i ⎜ ⎟ ⎜ ⎟ i =1 i =1 ⎝16⎠ 16⎝ 32 32 32 32 ⎠ 64. f (x) = cos x, 0 ≤ x ≤ π , n = 4 2 Let ci = xi + xi +=1 2 ∆x = π , c = π , c = 3π , c = 5π , c = 7π 8 1 16 n 2 16 3 4 16 4 16 Area ≈ f (c )∆x = cos(ci)⎛π⎞ = π⎛ cos π + cos 3π + cos 5π + cos 7π⎞ ≈ 1.006 ∑ i ∑ ⎜ ⎟ ⎜ ⎟ i =1 i =1 ⎝ 8 ⎠ 8⎝ 16 16 16 16 ⎠ 65. f (x) = ln x, 1 ≤ x ≤ 5, n = 4 x + x Let ci = i i 1 , ∆ x = 1 2 c = 3 , c = 5 , c = 7 , c = 9 1 2 2 n 2 3 2 4 2 4 Area ≈ ∑ f (ci )∆x = ∑⎡ ⎣ln(ci )⎤ ⎦(1) ≈ 0.40547 + 0.91629 + 1.25276 + 1.50408 ≈ 4.0786 i =1 i =1 66. f (x) = xe x , 0 ≤ x ≤ 2, n = 4

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 4 3 2 1 1 2 3 4 4 3 2 1 1 2 3 4 ⎦ Let c = xi + xi 1 , ∆ x = 1 i 2 2 c = 1 , c = 3 , c = 5 , c = 7 1 4 2 n 4 3 4 4 4 4 Area ≈ ∑ f (c )∆ x = ∑ ⎡c eci ⎤⎛1⎞ ≈ [0.32101 + 1.58775 + 4.36293 + 10.07055]⎛1⎞ ≈ (16.34224)⎛1⎞ ≈ 8.1711 i i =1 ⎣ i i =1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝2⎠ 67. y 68. y x x (b) A ≈ 6 square units (a) A ≈ 3 square units

69. You 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.

The sum of the areas of these circumscribed rectangles is the upper sum.

(d) In each case, ∆x = 4 n The lower sum uses left end-points, (i 1)(4 n) The upper sum uses right endpoints, (i)(4 n) The Midpoint Rule uses midpoints, (i 1)(4 n)

(e)

You 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.

70. See the definition of area, page 296.

the lower sum is always smaller than the exact

(f ) s(n) increases because the lower sum approaches the exact value as n increases. S(n)decreases because the upper sum approaches the exact value as n increases. Because of the shape of the graph, 71. (a)

72. (a) Left endpoint of first subinterval is 1.

4 Left endpoint of last subinterval is 4

(b) Right endpoint of first subinterval is 1 +

(d) The heights would be equal to that constant.

73. True. (Theorem5.2 (2))

74. True. (Theorem5.3)

75. Suppose there are n rows and n + 1columns in the figure. The stars on the left total 1 + 2 + " + n, as do the stars on the right. There are n(n + 1) stars in total, so

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. N 4 8 20 100 200 s(n) 15.333 17.368 18.459 18.995 19.06 S(n) 21.733 20.568 19.739 19.251 19.188 M(n) 19.403 19.201 19.137 19.125 19.125 2 Section 5.2 Area 449

y (c) y 8 6 4 2 x 1 2 3 4 Midpoint Rule: x ( ) 2 4 5 2 6112 a b M 4 = 2 3 + 4 5 + 5 7 + 6 9 = 315 ≈ 19.403

x a b

8 6

4 4 2

= 15

5 x 4 4 1 2 3 4 2 2 Lower sum: ( ) 1 1 46

s 4 (b) y 8 6 = 0 + 4 + 53 + 6 = 153 = 3 ≈ 15.333

4 2 x 1 2 3 4

Upper sum: ( ) 1 2 11 326 2[1 + 2 + " + n] = n(n + 1) 1 S 4 = 4 + 5 3 + 6 + 6 5 = 21 15 = 15 ≈ 21.733 1 + 2 + " + n = 2(n)(n + 1).

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ⎜ ⎟ ⎜ ⎟ θ 450 Chapter 5 Integration 76. (a) θ = 2π n r 77. For n odd, n = 1, 1 row, 1 block (b) sin θ = h h r r n = 3, 2 rows, 4 blocks n = 5, 3 rows, 9 blocks h = r sin θ n, n +1 rows, ⎛ n +1⎞2 blocks, A = 1bh = 1 r(r sin θ) = 1 r2 sin θ 2 ⎜ 2 ⎟ 2 2 2 ⎝ ⎠ For n even, (c) A = n ⎛1 r2 sin 2π⎞ n = 2, 1 row, 2 block n ⎜2 n ⎟ ⎝ ⎠ n = 4, 2 rows, 6 blocks r 2 n 2π = sin = πr2⎛ sin(2π n)⎞ n = 6, 3 rows, 12 blocks 2 n ⎝ 2π n ⎠ n n2 +2n Let x = 2π n As n → ∞ , x → 0 n, rows, 2 blocks, 4 lim An = lim πr2⎛ sin x ⎞ = πr2(1) = π r2 n→∞ x→0 ⎝ x ⎠ 78. (a) n ∑ 2i = n(n + 1) i =1 The formula is true for n = 1: 2 = 1(1 + 1) = 2. Assume that the formula is true for n = k: k ∑2i = k(k + 1) i =1 k +1 k Then you have ∑2i = ∑2i + 2(k + 1) = k(k + 1) + 2(k + 1) = (k + 1)(k + 2) i =1 i =1 which shows that the formula is true for n = k + 1. (b) n ∑ i3 = i =1 n 2(n + 1)2 4 12(1 + 1)2 4 The formula is true for n = 1because 13 = = = 1. 4 4 Assume that the formula is true for n = k: k ∑ i3 = i =1 k2(k + 1)2 4 k +1 k 2 2 2 2 Then you have ∑ i3 = ∑ i3 + (k + 1)3 = k (k +1) + (k + 1)3 = (k +1) ⎡k2 + 4(k + 1)⎤ = (k +1) (k + 2)2 i =1 i =1 4 4 ⎣ ⎦ 4 which

the formula is

for n = k + 1.

shows that

true

Section 5.3 Riemann Sums and Definite Integrals 451

79. Assume that the dartboard has corners at (±1, ±1).

point (x, y) in the square is closer to the center than the top edge if y

In the first quadrant, the parabolas y

regions that make up the total region, as indicated in the figure.

Section 5.3 Riemann Sums and Definite Integrals

x 2 + y 2 ≤ 1 y 1 (x, 1) y ≤ 1(1 x2) (x, y) x 1 (0, 0) 1 x ≤ 2 1 y 1 2)

= 1(1 x2 ) and x = 1(1 y2) intersect at ( 2 1, 2 1) There are 8 equal

y 1 x Area of shaded region S = 2 1 ⎡1(1 x2) x ⎤dx = 2 2 5 ∫0 ⎢2 ⎥ 3 6 ⎣ ⎦ Probability = 8S = 2 ⎡2 2 5⎤ = 4 2 5 Area square ⎢ 3 6 ⎥ 3 3 ⎣ ⎦

1. f (x) = x, y = 0, x = 0, x = 3, ci 2 2 3i2 = n2 ∆x = 3i 3(i 1) = 3 (2i 1) i n2 n2 n2 n n 3i2 3 lim ∑ f (ci )∆xi n→∞ i =1 = lim∑ n→∞ i =1 n 2 n 2 (2i 1) n = lim 3 3 ∑(2i2 i) n→∞ n 3 i =1 = lim 3 3⎡ ⎢2 n(n + 1)(2n + 1) n(n + 1)⎤ ⎥ n→∞ n3 ⎣ 6 2 ⎦ ⎡(n + 1)(2n + 1) n + 1⎤ = lim3 3⎢ 3n2 2n 2 ⎥ n→∞ ⎣ ⎦ = 3 3⎡2 0 = 2 3 ≈ 3.464 ⎢3 ⎥ ⎣ ⎦

© 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ∑( ) ⎢ ⎢ f 2 + ⎜ ⎟ ∑ 452 Chapter 5 Integration 2. f (x) = 3 x, y = 0, x = 0, i3 x = 1, ci = n 3 ∆xi i3 (i 1)3 = n3 n3 3i2 3i + 1 = n3 n n i3 ⎡3i2 3i + 1⎤ lim∑ f (ci )∆xi = lim ∑ 3 3 ⎢ 3 ⎥ n→∞ i =1 n→∞ i=1 n ⎣ n ⎦ n = lim 1 3i3 3i2 + i n→∞ n4 i =1 1 ⎡ ⎛ n2(n +1)2 ⎞ ⎛ n(n +1)(2n +1)⎞ n(n +1)⎤ = lim ⎢3⎜ ⎟ 3⎜ ⎟ + ⎥ n→∞ n4 ⎢ ⎜ 4 ⎟ ⎝ 6 ⎠ 2 ⎥ ⎣ ⎝ ⎠ 1 ⎡3n4 + 6n3 + 3n2 ⎦ 2n3 + 3n2 + n n2 + n⎤ = lim n4 4 2 + 2 ⎥ n→∞ ⎣ ⎦ 1 ⎡3n4 n3 n2 ⎤ ⎡3 1 1 ⎤ 3 = lim n4 4 + 2 4 ⎥ = lim⎢4 + 2n 4n2 ⎥ = 4 n→∞ 3. y = 8 on [2, 6] ⎛Note: ⎣ ∆x = 6 2 = ⎦ n→∞⎣ ⎦ 4 , ∆ → 0 as n → ∞ ⎞ ⎜ n n ⎟ ⎝ ⎠ n n n ⎛ ⎞ n n f (c )∆x = f ⎛2 + 4i ⎞⎛ 4⎞ = 8 4 = 32 = 1 32 = 1(32n) = 32 ∑ i i ∑ ⎜ ⎟⎜ ⎟ ∑ ⎜ ⎟ ∑ ∑ i =1 6 i =1 ⎝ n ⎠⎝ n ⎠ i =1 ⎝ n ⎠ i =1 n n i=1 n ∫2 8 dx = lim 32 = 32 n→∞ 4. y = x on [ 2, 3] ⎛ 3 ( 2) 5 Note: ∆x = = , ⎞ ∆ → 0 as n → ∞ ⎜ n n ⎟ ⎝ ⎠ n n ∑ f (ci ) ∆xi ⎛ 5i ⎞⎛ 5⎞ = ⎜ ⎟⎜ ⎟ i =1 i =1 ⎝ n ⎛ n ⎠⎝ n ⎠ 5i ⎞⎛ 5⎞ 25 n ⎛ 25⎞ n(n+1) 25⎛ 1⎞ 5 25 = ∑⎜ 2 + ⎟⎜ ⎟ = 10 + ∑ i = 10 + ⎜ ⎟ = 10 + ⎜1 + ⎟ = + i =1 ⎝ n ⎠⎝ n ⎠ n 2 i=1 ⎝ n 2 ⎠ 2 2 ⎝ n ⎠ 2 2n 3 ∫ 2 x dx = lim ⎛ 5 + 25⎞ = 5 n→∞ ⎝2 2n ⎠ 2 5. y = x3 on [ 1, 1] ⎛ Note:∆x = 1 ( 1) = 2 , ⎞ ∆ → 0 as n → ∞ ⎜ n n ⎟ ⎝ ⎠

Calculus of a single variable early transcendental functions 6th edition larson solutions manual dow