EVEN04 - CHAPTER Integration Section 4.1 Section 4.2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER Integration Section 4.1 Section 4.2 Section 4.3 Section 4.4 Section 4.5 Section 4.6 4 Antiderivatives and Indefinite Integration . . . . . . . . . 450 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456 Riemann Sums and Definite Integrals . . . . . . . . . . . 462 The Fundamental Theorem of Calculus . . . . . . . . . . 466 Integration by Substitution . . . . . . . . . . . . . . . . . 472 Numerical Integration . . . . . . . . . . . . . . . . . . . 479 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 CHAPTER Integration Section 4.1 4 Antiderivatives and Indefinite Integration Solutions to Even-Numbered Exercises 2. d4 x dx 1 x C 4x3 1 x2 4. d 2 x2 3 dx 3x C d 23 x dx 3 x1 2 2 2x 32 12 C x x2 1 x3 2 6. dr d r d Check: d C C 8. dy dx y 2x 2x 2 3 2 C 1 x2 C c 2x 3 Check: Rewrite x 2 1 d dx x 2 Given 10. 1 dx x2 Integrate x 1 Simplify 1 x 14 x 4 1 9x C dx 1 x4 4 C 12. x x2 3 dx x3 3x dx 3 x2 2 C 32 x 2 C 14. 1 dx 3x 2 1 x 2dx 9 x2 2 x2 2 C 1x1 91 C C 16. 5 Check: x dx d 5x dx 5x C 5 x 18. 4x3 Check: 6x 2 d4 x dx 1 dx 2x 3 x4 x C 2x 3 x 4x 3 C 6x 2 1 20. x3 Check: 4x d x4 dx 4 1 2 dx 2x 2 x4 4 2x 2x 2 C 2x x3 C 4x 2 22. x Check: 2x d 23 x dx 3 dx 2 x1 x1 2 2 1 x 2 x1 12 dx 1 x 2 x3 2 32 12 1 x1 2 212 x 1 2x C 23 x 3 2 x1 2 C C 2 450 S ection 4.1 Antiderivatives and Indefinite Integration 451 24. 4 x3 1 dx d 47 x dx 7 3 4 x3 x 4 1 dx C x3 4 47 x 7 1 4 x 4 C 1 26. 1 dx x4 Check: d dx x 4 dx x 3 3 C C 1 x4 1 3x3 C Check: x3 1 3x3 28. x2 2x x4 dx x x 1 2 2x 2x 1 x2 2 3 3x 3x 3 4 dx C 30. 2t2 1 2 dt 4t4 45 t 5 4t2 43 t 3 t C t 1 dt C 4t 4 2t2 4t2 1 2 1 1 x Check: d dx 1 x 1 x2 1 x3 2 1 x3 C x x2 3 C 2 Check: 2x 2x x4 34 t 4 1 3 d 45 t dt 5 43 t 3 1 3x 4 3 32. 1 Check: 3t t2 dt d 13 t dt 3 t2 34 t 4 13 t 3 cos t 3t3 dt C t2 13 t 3 3t3 C 3t t2 34. 3 dt Check: 3t d 3t dt C C 3 36. t2 Check: sin t dt d 13 t dt 3 cos t C C t2 sin t 38. 2 sec2 d1 d3 d 3 1 3 tan 3 tan C 2 C sec2 Check: 40. sec y tan y sec y dy sec y tan y sec y tan y sec2 y dy C sec2 y 42. 1 cos x dx cos2 x cos x dx sin2 x csc x cot x dx 1 sin x cos x dx sin x C cos x sin x csc x 1 sin x d sec y Check: dy tan y C sec y tan y sec y tan y Check: sec y d dx csc x C csc x cot x cos x cos2 x 1 44. f x y 6 4 x 46. f x fx C=3 C=0 x x2 2 y 48. f x C f ( x) = x2 + 2 1 x2 1 x y fx 2 f ( x) = 2 x 2 C 2 −2 −2 −4 C = −2 x 4 6 8 8 6 4 f′ f ( x) = − 1 + 1 x f′ −4 −2 −2 4 x f ( x) = − 1 x −4 −2 x 2 4 −4 452 Chapter 4 Integration 50. dy dx y 2 y 2x 2x 3 x2 2 1 2x 1 dx 2, 3, 2 x2 2x C 1 52. dy dx y 3 y 1 1 1 x 1 x2 x 2 x dx 2, 1,3 1 x 2 C 23 2x 1 C⇒C C⇒C 2, x > 0 54. (a) 5 y (b) dy dx y x2 x3 3 1 3 1 3 7 3 x3 3 1, x 3 1, 3 (−1, 3) 5 C 1 C −4 4 −4 x 4 3 3 C y −5 −5 1 C x 7 3 58. f s 6s 6s 3 3s 2 sin x 1 6 sin x dx 1 C1 2 2 dx sin x 6 2x C2 cos x 1 ⇒ C1 C1 2 32 8s 3, f 2 8s 3 ds 2 56. g x gx g0 gx 60. f x f0 f0 fx f0 fx fx f0 fx 6x 2, g 0 6x 2 dx 1 2x3 x2 6 3 x 2 dx 0 13 x 3 13 x 3 0 14 x 12 0 C1 6 13 x 3 20 1 2x3 3 1 C C⇒C 1 3 3s 2 4 fs f2 fs 62. f x f0 f0 2s 4 C 12 C 32 C⇒C 23 22 23 2s 4 C1 6 fx f0 fx 6 ⇒ C1 cos x cos x 0 0 sin x 6 dx C2 6x 3 14 x 12 3 ⇒ C2 6x 3 C2 fx f0 fx C2 2x 6 ⇒ C2 6 S ection 4.1 Antiderivatives and Indefinite Integration 453 64. dP dt Pt P0 P1 Pt P7 k t, 0 ≤ t ≤ 10 kt1 2 dt 0 2 k 3 C 500 2 23 kt 3 2 C 500 , 0 , f is decreasing on 66. Since f is negative 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 , . y 3 500 ⇒ C 600 ⇒ k 500 500 150 2 f′ 2 1 f ′′ x 1 2 3 2 150 t3 3 100 7 100t3 500 −3 −2 −2 32 2352 bacteria f −3 68. f t f0 f0 ft f0 ft ft f0 ft at v0 s0 vt 0 C1 32t st 0 0 16t 2 32 ft sec2 70. v0 s0 (a) 16 ft sec 64 ft st 16 t2 t 4 t 0 1± 2 17 16t2 16t 64 0 32 dt v0 ⇒ C1 v0 32t C2 v0t 32t v0 C1 Choosing the positive value, v0 dt s0 ⇒ C2 s0 v 1 2 17 32 1 2 17 16 16t2 s0 v0t C2 (b) t 1 2 vt st 17 2.562 seconds. 32t 16 16 17 65.970 ft sec 72. From Exercise 71, f t 4.9t2 canyon floor as position 0.) ft 4.9t2 t2 0 1600 1600 ⇒t 4.9 4.9t2 1600 1600. (Using the 74. From Exercise 71, f t ft then vt 9.8t v0 0 200 4.9t2 4.9t2 v0t v0t 2, 2. If 326.53 18.1 sec for this t value. Hence, t 4.9 v0 9.8 2 v0 9.8 and we solve 2 200 v0 4.9 v02 9.8 2 v0 9.8 v02 9.8 198 9.8 2 198 9.8 2 198 3880.8 ⇒ v0 62.3 m sec. 4.9 v02 9.8 v02 4.9 v02 v02 454 Chapter 4 Integration 76. v dv 12 v 2 When y 12 v 20 C 12 v 2 v2 v2 GM GM y R, v GM R 12 v 20 GM y 2GM y v02 1 dy y2 C v0. C GM R 12 v 20 v02 2GM 1 y GM R 2GM R 1 R 78. x t (a) v t at t t3 1t 7t2 xt vt 3 2 0≤t≤5 9 14t 14 5 and 3 < t < 5. 3 7 . 3 2 2 3 4 3 15 3t 5t 3 15t 3t2 6t (b) v t > 0 when 0 < t < (c) a t v 7 3 6t 3 14 7 3 0 when t 5 7 3 3 80. (a) a t vt ft f0 ft (b) v t 82. v0 30 mph 15 mph at vt st vt at 66 cos t a t dt v t dt 3 cos t 0 cos 0 4 k ,k 0, 1, 2, . . . (a) 16.5t 66 t s 66 66t Let s 0 0. 44 16.5 (b) 16.5t 22 16.5 66 t s (c) 66 h= mp cos t dt sin t dt C2 sin t cos t 1 C1 C2 sin t since v0 0 C2 ⇒ C2 4 sin t for t 45 mph 44 ft sec 22 ft sec a at a2 t 2 66 ft sec 44 22 16.5 73.33 ft 22 44 16.5 2.667 1.333 0 after car moves 132 ft. 0 when t a 66 2a 2 66 . a 66 66 a 33 2 16.5. 117.33 ft ft/s ec mp 0m h= ph 22 f t/se c mp 30 66 s a ft/s ec h= 44 132 when a at vt st 16.5 16.5t 8.25t 2 66 45 0 73.33 feet 117.33 feet 66 t 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. 15 132 S ection 4.1 Antiderivatives and Indefinite Integration 30 30 455 84. No, car 2 will be ahead of car 1. If v1 t and v2 t are the respective velocities, then 0 v2 t dt > 0 v1 t dt. 86. (a) v 0.6139t3 5.525t2 0.0492t 65.9881 (b) s t 0.6139t4 5.525t3 0.0492t2 4 3 2 Note: Assume s 0 0 is initial position v t dt 196.1 feet 65.9881t s6 88. Let the aircrafts be located 10 and 17 miles away from the airport, as indicated in the figure. vA t sA t kA t 12 kt 2A 150 150t 10 vB sB kB t 12 kt 2B 250 250t 17 A irport 0 A 10 B 17 (a) When aircraft A lands at time tA you have vA tA sA tA kA tA 12 kt 2AA 150 150tA 100 ⇒ kA 10 150tA 125tA tA 50 tA 125 10 0 10 10 10 . 125 S1 t 625 2 t 2 150t 10 50 tA 1 50 2 t 2 tA A kA 50 625 ⇒ SA t Similarly, when aircraft B lands at time tB you have vB tB sB tB kB tB 12 kt 2BB 250 250tB 115 ⇒ kB 17 250tB 365 t 2B tB 135 tB 365 34 0 17 17 34 . 365 S2 t 49,275 2 t 68 (c) d sB sA 0 0 0.1 0 0 135 tB 1 135 2 t 2 tB B kB (b) 20 135 49,275 ⇒ SB t 34 250t sB t 17 sA t 20 Yes, d < 3 for t > 0.0505. d 3 0.1 90. True 92. True 456 Chapter 4 Integration d sx dx 94. False. f has an infinite number of antiderivatives, each differing by a constant. 96. 2 cx 2 2s x s x 2s x c x 0 2c x c x 2c x s x Thus, s x s0 Therefore, sx 2 2 cx 2 k for some constant k. Since, 1, k 1. 0 and c 0 cx 2 1. cos x satisfy these [Note that s x properties.] sin x and c x Section 4.2 6 Area 5 2. k 3 4 kk 2 31 42 53 64 50 4. j 1 3j 1 3 1 4 1 5 47 60 6. i 1 i 1 2 i 1 3 0 8 1 27 4 64 9 125 238 15 8. i 1 5 1 i 15 4 10. j 1 1 j 4 2 12. 2n 1 ni 1 10 2i n 10 2 1 14. 1n ni 1 1 0 i n 2 15 10 16. i 1 2i 3 2 i i 1 3 15 45 195 18. i 1 i2 1 i i2 1 i 1 1 10 375 2 15 16 2 10 10 11 21 6 10 10 20. i 1 i i2 1 i i3 1 i 2 1 i 10 11 2 3080 22. sum seq x 15 3 2i 2x, x, 1, 15, 1 15 15 2 14,160 (TI-82) 2 > 102 11 4 i3 i 1 15 4 16 2 1 2 15 16 15 15 2 1 2 4 24. S s 5 4 5 4 4 2 21 01 16 10 s 2 1 26. S 5 2 1 2 3 2 3 1 2 1 2 1 3 14,160 55 6 9 2 4.5 28. S 8 1 4 2 1 4 1 2 5 4 2 1 4 1 4 5 2 1 2 3 4 3 2 6 2 2 1 4 2 1 4 1 4 7 2 ... 1 7 4 2 2 1 4 2 1 4 2 2 1 4 2 1 1 4 2 1 16 4 s8 0 2 1 2 1 4 2 2 1 4 3 2 2 6.038 7 4 1 4 2 5.685 S ection 4.2 1 5 24 5 1 64 n n n3 1 5 2 Area 457 30. S 5 1 1 1 5 21 5 1 5 1 2 1 5 16 5 1 2 5 1 9 5 2 2 5 2 1 5 1 3 5 2 1 5 1 4 5 2 1 5 1 1 5 s5 0.859 1 3n2 n3 1 2 2n n 5 n 3 5 2 1 5 2n3 1 5 64 3 1 4 5 2 1 5 0 0.659 32. lim n→ 1 2n 6 64 lim 6 n→ 64 2 6 34. lim n→ n 1 nn 1 n2 2 3 n2 S 10 1n 4j n2 j 1 25 10 2.05 2.005 2.0005 4n3 i n4 i 1 2.5 1 n2 n lim 2 n→ n2 1 4n n 1 n2 2 1 1 2 36. j 1 4j 3 3n Sn S 100 S 1000 S 10,000 n 38. i 4i2 i 1 n4 1 i2 4 n2 n 1 n4 4 4 n3 n3 1 3n3 3n3 1 3n3 3n3 2n2 4 2 nn n 2n2 1 2n 6 3n 6 6n Sn 1 1 6n2 2n2 3n 3n 4n2 2 2 S 10 S 100 S 1000 S 10,000 n 1.056 1.006566 1.00066567 1.000066657 2i n 2 n 2i n 2 40. lim n→ i 1 n→ lim 4n i n2i 1 n→ lim 4 nn 1 n2 2 n→ lim 4 1 2 1 n 2 n 42. lim n→ i 1 1 2 n n→ lim 2n n n3i 1 2 n3 n 2i 2 n n n→ lim lim n2 i 1 4n i 1 i 4 i i2 1 n→ 23 n n3 1 2 2 4 3 4n 2 n nn 2 4 3 1 2 n 4n n 2 3n2 1 2n 6 1 2 lim n→ 21 26 3 458 Chapter 4 n 3 Integration 1n n n4i 1 1n 3 n n4 i 1 14 n n4 1 3 13 n 44. lim n→ i 1 1 2i n 2 n 2 lim n→ 2i 3 2 lim n→ 6n2i 6n2 3 n 4 n2 nn 2 4 12ni 2 1 6 n 20 2 n2 8i 3 12n 2 nn 4 n 1 2n 6 2 n2 1 8 n2 n 4 1 2 2 lim n→ 2 lim n→ 2 lim 10 n→ 46. (a) 4 y (b) x 3 n 1 2 n Endpoints: 3 2 1 x 2 4 1<1 1 <1 1 2 <1 n 2 4 ... < <1 n 2 <...< 1 n 2n n n 3 1 2 n <1 n 2 n 2 <1 n (c) Since y sn x is increasing, f mi n f xi 1 on xi 1, xi . f xi i n 1 1 x i 1, f1 i 1 1 xi n 2 n 2 n n 1 i 1 i 1 2 n 2 n (d) f Mi Sn i f xi on [xi n f xi 1 x i 1 f1 50 3.96 4.04 2 n 2 n i 2 n 100 3.98 4.02 2 n n 1 i 1 i 2 n 2 n (e) x sn Sn n 5 3.6 4.4 10 3.8 4.2 (f) lim n→ i 1 1 i 1 n→ lim lim 2 n 2 2 n n 2 n 2n 2 nn 1 n 2 2 n 4 n n lim 4 2 n 4 n→ n n→ n→ lim i 1 1 i 2 n 2 n n→ lim 2 nn 1 n 2 2n n 1 n→ n→ lim lim 4 2 n 4 S ection 4.2 5 n 2 3 n 12 10 8 Area 459 48. y 3x Sn 4 on 2, 5 . n Note: x 3i n 3i n 3 n 4 3 n 6 39 2 3 y f2 i n 1 32 i 1 18 27 1 2 3 3 n 1 n 2n 6 i i 1 12 4 2 x 4 6 8 10 12 6 Area 27 n 1 n n2 2 6 27 2 n→ lim S n 50. y Sn x2 n l 1 on 0, 3 . Note: x 3i f n 1 3 n n l 1 0 n 3 n 3 n 12 10 8 6 y 3i n 2 1 27 n 2 l n3 l 1 27 n n n3 Area n→ 3n 1 nl 1 1 2n 6 9 2 2 1 3 n n 3 12 9 2n 2 2 4 2 3n n2 1 x 3 1 2 3 lim S n 52. y 1 sn x2 on n i 1, 1 . Find area of region over the interval 0, 1 . 1 n n Note: x 1 n 3 2 y i f n 1 1 i 1 i n 2 1 n 1 1 1 2 6 3 n 1 n2 1 1 Area 2 Area 1n2 i n3 i 1 1 1 1 3 nn 2 3 1 2n 6n3 −2 −1 x 2 n→ lim s n 4 3 1 n 0 1 n 2.0 1.5 54. y 2x x3 on 0, 1 . Note: x y Since y both increases and decreases on 0, 1 , T n is neither an upper nor lower sum. n Tn i 1 f 2 n2 i 1 n i n i 1 1 n 1 n4 i 1 4 1 n n 2 i 1 i n 1 n2 i n 3 1 n 1nn 1 n4 4 2 2 1.0 0.5 x 0.5 1.0 2.0 i3 1 nn 1 4n2 3 4 1 n 2 4n 1 4 Area n→ lim T n 460 Chapter 4 Integration 0 n 1 i 1 56. y x2 sn x3 on n 1, 0 . 1 2 5i n 5 2n 2 5 2 i n Note: x 1 n 4i 2 n2 4 3 4 3 2 n 1 4 4 n n 1 i n 2 1 4 2 1 n 3 y f i n i 1 1 1 5n i n2 i 1 1 2n 1 4n2 i n 3 1 n 1n3 i n4 i 1 2 1 x 1 −1 2 i3 n3 1 n 2 3n3 7 12 2 4n2 i n3 i 1 −2 −1 2 Area 5 2 n→ lim s n 58. g y Sn 1 y, 2 ≤ y ≤ 4. Note: y 2 n 2 n 60. f y Sn 4y n y 2, 1 ≤ y ≤ 2. Note: y f1 i n 1 n i n 4i n 2i n 1 i2 n2 1 nn n2 1 2n 6 1 1 3 1 11 3 1 2i n i n 2 2 n 1 1 n g2 i n i 1 2i n 2i n 2 n 2 n 2 ni 2 1 3 n i 1 1 2 12 1 1 i n 1 n 1n 41 ni 1 1n 4 ni 1 1n 3 ni 1 1 3n n 3 Area y 5 2 n n Area y 5 4 3 2 1 1nn 1 n 2 2 n i2 n2 n→ lim S n 2nn 1 n 2 1 n n 1 2n 6 1 n x 1 2 3 4 5 n→ lim S n 3 3 2 1 x 1 2 3 4 5 62. h y Sn y3 n 1, 1 ≤ y ≤ 2 Note: y h1 i n i n 3 1 n y 5 4 3 i n 1 1 n 1 3i2 n2 2 1 i 1 1 n 3i n 3 nn n2 1 2n n2 1 3 2 1 19 4 1 2n 6 1 3nn 1 n 2 1n 2 ni 1 1 2n n 2 Area n i3 n3 2 1 x 2 4 6 8 10 1 n2 n 1 n3 4 1 n24 2 1n 2 2 1 4 3n 1 2n n→ lim S n S ection 4.2 64. f x Let ci x n Area 461 x2 xi 4x, 0 ≤ x ≤ 4, n xi 2 1 4 66. f x Let ci sin x, 0 ≤ x ≤ xi xi 2 , c1 1 2 ,n 4 . 1 ,c 22 4 . , c2 4 1, c1 f ci i 1 3 ,c 23 ci2 i 1 5 ,c 24 4ci 1 25 4 10 7 2 Area x n 8 16 x 3 ,c 16 3 sin ci 8 5 16 5 ,c 16 4 7 16 Area x 2 9 4 f ci i 1 i 1 1 4 53 68. f x n 8 x2 1 6 49 4 14 8 sin 16 sin 3 16 sin sin 7 16 1.006 on 2, 6 . 4 2.3397 8 2.3755 12 2.3824 16 2.3848 20 2.3860 Approximate area 70. f x n cos x on 0, 2 . 4 1.1041 8 1.1053 12 1.1055 16 1.1056 20 1.1056 Approximate area 72. See the Definition of Area. Page 259. 74. f x n sn Sn Mn 3 x, 0 ≤ x ≤ 8 10 10.998 12.598 12.040 20 11.519 12.319 12.016 50 11.816 12.136 12.005 100 11.910 12.070 12.002 200 11.956 12.036 12.001 (Note: exact answer is 12.) 76. y 78. True. (Theorem 4.3) 4 3 2 1 1 2 3 4 x a. A 3 square units 462 Chapter 4 2 n h r h A (c) An Let x n→ n Integration 80. (a) (b) sin r θ r h r sin 1 bh 2 n 1 r r sin 2 12 r sin 2 r 2n 2 sin 2 n , x → 0. sin x x r2 1 r2 n 2 12 r sin 2 n 2 n. As n → x→0 r2 sin 2 n 2n lim An lim r2 82. (a) i 2i 1 nn 1 1: 2 11 k: 1 2 (b) i i3 1 n2 n 4 1 2 The formula is true for n Assume that the formula is true for n k 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: 2i i 1 kk 1. k 1 k k Then we have i 1 2i i 2i 1 2k 1 1k 1 i 1 i3 1 k2 k 4 k 1 1 2 kk k 2k 2 k Then we have i 1 i3 i i3 1 k 1 4 1 2 2 1 3 Which shows that the formula is true for n k 1. k2 k k 4 k 4 1 k k2 4k 2 2 1 3 1 2 k which shows that the formula is true for n k 1. Section 4.3 2. f x 3 Riemann Sums and Definite Integrals 0, x xi 0, x i3 n3 n→ x, y 1, ci i n3 n 3 i 1 i3 n3 1 3 3i 2 3i n3 3i n3 i 3 3n2 n2 4 1 1 n n→ lim i 1 f ci xi lim i 3 3i 2 n3 3i 2 2 1n lim 4 3i 3 n→ ni1 n→ lim lim 1 n2 n 1 43 n 4 1 3n4 n4 1 3n4 n4 4 6n3 4 n3 2 nn 2n3 1 2n 6 3n2 2 3 4 1 2n n 1 n2 nn 2 n 2 3 4 1 n→ n→ lim n→ lim 1 4n2 S ection 4.3 3 n 5i n 5 n n Riemann Sums and Definite Integrals 463 4. y x on n 2, 3 . n Note: x f i 1 2 5 , n 2 → 0 as n → 5i n 25 1 2 5 n 1 n 10 5 2 25 n i n2 i 1 25 2n f ci i 1 xi 2 i 1 10 3 25 n n 1 n2 2 5 2 25 2n 3 n 2 n 2 10 x dx 2 n→ lim 5 2 1 2 , n 6. y n 3x 2 on 1, 3 . Note: x n → 0 as n → f ci xi i 1 i n 1 f1 31 i 1 2i n 2i n 4i n 2 n 4i2 n2 4 nn n2 n 1 2n n2 4n 1 2n 6 1 1 2n n2 1 1 6n 1 ni 1 6 n n 6 3 4nn 1 n 2 12 6 n n 1 4 3x 2 dx 1 n→ lim 12 12 n 1 n 8 26 6 8. y 3x 2 n 2 on f ci xi 1, 2 . Note: x n 2 n 3 n 1 3 ; n → 0 as n → f i 1 1 1 i 1 3i n 3n 3 ni 1 3n 31 ni 1 3 3n n 15 2 3i n 6i n 2 2 9i2 n2 2 27 n n n2 1 2n n2 27 n 2 1 2n 6 1 1 2n n2 1 1 2n 18 n n 1 n 2 27 n 1 n 15 27 4 27 n 2 n n 15 1 3x 2 1 2 dx n→ lim 27 27 15 n n 10. lim →0 i 6ci 4 1 ci 2 xi 0 6x 4 x 2 dx 12. lim →0 i 1 3 ci2 3 xi 1 3 dx x2 on the interval 0, 4 . 2 2 on the interval 1, 3 . 1 14. 0 4 2x dx 16. 0 x2 dx 18. 1 1 x 2 4 1 dx 20. 0 tan x dx 464 2 Chapter 4 Integration 22. 0 y 2 2 dy 24. Rectangle A A a y 26. Triangle 24 a A A 0 y bh a 1 bh 2 4 1 42 2 4 4 dx 8a x dx 2 5 3 2 3 2 1 Triangle Rectangle 1 x 1 2 3 4 a x −1 −a 28. Triangle A A 0 30. Triangle 1 88 2 x dx 32 32 A A a y 32. Semicircle 1 2a a 2 a x dx a2 A A r y 1 bh 2 8 1 bh 2 a 12 r 2 r 8 r2 x 2 dx 12 r 2 y a r Semicircle Triangle 10 8 6 4 2 x 2 4 6 8 10 −r −a a x −r r x 4 4 4 In Exercises 34– 40, 2 2 x3 dx 60, 2 x dx 6, 2 dx 2. 4 4 34. 2 x3 dx 0 36. 2 15 dx 15 2 dx 15 2 30 4 4 4 4 4 4 4 38. 2 x3 4 dx 2 x3 dx 4 2 dx 60 42 68 40. 2 6 2x x3 dx 6 2 dx 2 2 x dx 2 x3 dx 36 62 6 3 6 0 1 26 1 60 42. (a) 0 f x dx 0 3 f x dx 3 6 f x dx 1 4 1 1 3 44. (a) 1 1 f x dx 1 0 f x dx 0 f x dx 5 30 35 5 0 15 0 10 5 5 (b) 6 3 f x dx 3 f x dx 0 6 (b) 0 1 f x dx 1 f x dx 1 (c) 3 f x dx 6 (c) 1 1 3f x dx 3f x dx 0 3 1 1 f x dx f x dx 0 (d) 3 5f x dx 5 3 f x dx 5 1 5 (d) 3 S ection 4.3 5 5 5 3 Riemann Sums and Definite Integrals 5 465 46. (a) 0 5 fx 2 dx 0 5 f x dx 0 2 dx 4 10 14 (b) 2 5 fx 2 dx 0 f x dx 4 Let u x 2. (c) 5 f x dx 2 0 f x dx 24 8 f even (d) 5 f x dx 0 f odd 48. The right endpoint approximation will be less than the actual area: < 50. The average of Exercise 39 and Exercise 40 consists of a trapezoidal approximation, and is greater than the exact area: > 54. 4 3 2 1 x 1 4 1 2 3 4 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 1 1 y x dx x is integrable, sketch a graph of the region bounded by fx x x and the x-axis for 1 ≤ x ≤ 1. You see that the integral equals 0. y 1 b. A 4 3 square units 2 1 x 1 2 −2 −2 3 56. 9 8 7 6 5 4 3 2 1 y 58. 0 5 x2 1 4 dx n Ln 123456789 x 8 7.0855 6.2470 5.3980 12 6.8062 7.2460 5.6812 16 6.6662 6.2457 5.8225 20 6.5822 6.2455 5.9072 7.9224 6.2485 4.5474 Mn Rn c. Area 3 27. 60. 0 x sin x dx n Ln Mn Rn 4 2.8186 3.1784 3.1361 8 2.9985 3.1277 3.1573 12 3.0434 3.1185 3.1493 16 3.0631 3.1152 3.1425 20 3.0740 3.1138 3.1375 62. False 1 1 1 64. True x dx 0 0 66. False 4 x x dx 0 x dx 2 x dx 6 466 68. f x x0 Chapter 4 sin x, 0, 2 0, x1 4 , x2 Integration 3 , x3 , x4 2 , x4 3 3 2 f 3 2 x1 c1 4 6 4 , x2 12 3 xi , c3 , x3 , c2 f ci 2 ,c 34 f 1 2 6 4 x1 x2 f i 1 2 3 3 2 x3 2 3 f 3 2 1 x4 0.708 3 2 12 70. To find y 2 0 x dx, use a geometric approach. 3 2 1 x 1 −1 2 3 −1 Thus, 2 x dx 0 12 1 1. Section 4.4 2. f x cos x 0 The Fundamental Theorem of Calculus 2 4. f x 2 x2 x2 x −2 5 cos x dx 0 0 x dx is negative. 2 2 −2 −5 7 7 6. 2 3 dv 3v 2 37 32 15 5 8. 2 3v 4 dv 32 v 2 5 4v 2 75 2 3 20 6 8 39 2 5 2 3 10. 1 3x 2 5x 4 dx x3 5x 2 2 4x 1 27 38 45 2 12 1 4 1 12. 1 1 t3 9t dt 14 t 4 u2 2 92 t 2 1 u 1 1 1 4 1 2 9 2 1 4 9 2 1 2 0 14. 2 u 1 du u2 1 2 1 2 2 S ection 4.4 3 The Fundamental Theorem of Calculus 467 16. 3 8 v1 3 dv 2 dx x 2 0 1 34 v 4 2 3 3 3 8 3 4 12 3 3 4 3 3 8 4 0 8 18. 1 2 x 1 2 dx 2 2 x1 2 1 2 2x 1 8 22 2 20. t t dt 0 2t1 1 2 1 2 t3 2 dt 43 t 3 2 25 t 5 2 2 0 tt 20 15 6t 0 22 20 15 12 16 2 15 22. 8 x x2 dx 23 x x2 8 3 3 x5 38 x 8 3 3 dx 1 8 135 x 25 4 x5 3 24 80 4 1 15x 8 1 39 80 3 dx 32 144 80 4569 80 3 24. 1 3 1x 31 dx 1 3 3 x dx 1 x 4 3 dx 3 3 x 6 3 x dx x2 2 4 3 x2 2 9 2 4 4 1 3 6x 1 1 2 16 18 24 9 2 3 dx 8 13 2 3 18 9 2 4 26. 0 x2 4x 3 dx 0 x2 x3 3 1 3 4 3 0 4 4x 2x2 2 3 4 3 x2 1 4x 3x 3 dx 3 3 x2 x3 3 2x2 64 3 4x 4 3 dx split up the integral at the zeros x 1, 3 1 3x 0 x3 3 18 0 4 2x2 9 3x 3 1 9 4 3 4 1 3 2 3 32 12 9 18 9 4 28. 0 2 1 sin2 cos2 d 0 d 0 2 4 0 4 30. 4 2 2 csc2 x dx 2x cot x 2 2 2 2 2 1 2 1 2 1 2 2 32. 2 2t 2 cos t dt t2 sin t 2 4 2 1 4 2 34. P 2 0 1 sin d 2 cos 0 0 1 63.7% 2 36. A 1 1 x 4 dx x x2 2 15 x 5 1 1 8 5 2 2 38. A 1 1 dx x2 1 x 2 1 1 2 1 1 2 40. A 0 x sin x dx cos x 0 2 2 4 2 468 Chapter 4 Integration 42. Since y ≥ 0 on 0. 8 , 8 Area 0 1 x1 3 dx x 34 x 4 8 3 0 8 3 16 4 20 3 44. Since y ≥ 0 on 0, 3 , 3 46. 1 9 dx x3 fc 3 1 9 2x2 3 1 1 2 9 2 4 A 0 3x x2 dx 32 x 2 x3 3 3 0 9 . 2 4 2 9 2 3 9 c3 c3 c 9 2 1.6510 3 3 48. 3 cos x dx fc sin x 3 3 3 33 2 ± 0.5971 2 50. 1 3 1 3 1 4 x2 1 dx x2 3 2 1 1 1 3 x 2 dx 16 3 2x 1 x 3 1 3 3 cos c c 23 52. 1 2 2 0 cos x dx 0 2 sin x 0 2 1.5 2 (0.881, π ( Average value cos x x 7 2 0 − 0.5 2.71 2 0.881 7 f x dx Sum of the areas A1 1 3 2 8 A2 1 A3 1 1 2 A4 2 1 2 2 1 31 (b) Average value (c) A 8 62 1 54. (a) 1 f x dx 7 20 20 6 1 8 6 4 3 Average value y 7 6 10 3 y 4 5 A1 3 4 A2 2 1 A3 A4 3 2 1 x 1 x 2 3 4 5 6 7 1 2 3 4 5 6 7 S ection 4.4 6 6 2 2 The Fundamental Theorem of Calculus 2 469 56. 2 f x dx area or region B 0 f x dx 0 f x dx 58. 0 2f x dx 2 0 f x dx 2 1.5 3.0 3.5 60. Average value 64. P (a) 5 t P t 1 155 30 2 1 6 6 1.5 f x dx 0 5.0 1 3.5 6 0.5833 62. 1 R 0 R kR 0 2 r dr 2 k2 Rr R r 3 3R 0 2kR 3 2 3 158.660 157.071 2 4 160 5 161.180 160 954.061 6 6 162.247 161.180 159.010 162.247 954.158 6 159.026 157.071 1 155 6 30 dt Average profit (b) 1 6 6.5 158.660 6.5 5 0.5 t 2 1 5 t3 6 3 30t 5.0 (c) The definite integral yields a better approximation. 4 66. (a) R (b) 2.33t 4 100 14.67t3 3.67t2 70.67t (c) 0 R t dt 2.33t5 5 181.957 14.67t4 4 3.67t3 3 70.67t2 2 4 0 −1 − 20 5 68. (a) histogram 18 16 14 12 10 8 6 4 2 N t 123456789 (b) 6 7 9 12 15 14 11 7 2 60 83 60 4980 customers (c) Using a graphing utility, you obtain Nt (d) 16 0.084175t3 0.63492t2 0.79052 4.10317. −2 −2 10 9 (e) 0 N t dt 85.162 5110. 7 The estimated number of customers is 85.162 60 (f) Between 3 P.M. and 7 P.M., the number of customers is approximately 3 N t dt 60 50.28 60 3017. Hence, 3017 240 12.6 per minute. 470 Chapter 4 x Integration t4 4 x4 4 x4 4 x 70. F x 2 t3 2t 2 dt t2 x2 x2 2t 2 2x 2x 4 4 4 4 2 F2 F5 F8 72. F x 4 625 4 84 4 x 2 4 4 25 4 10 16 x 0 4 4 2t 2 Note: F 2 2 t3 2t 2 dt 0 167.25 1068 64 2 t3 1 4 1 4 1 4 dt 0 3 dt 1 t2 x 2 1 x2 1 4 F2 F5 F8 1 4 1 25 1 64 x 21 100 15 64 0.21 x 74. F x 0 sin d 1 1 1 x cos 0 cos x cos 0 1 cos x F2 F5 F8 cos 2 cos 5 cos 8 1.4161 0.7163 1.1455 x 76. (a) 0 t t2 d 14 x dx 4 x 1 dt 0 t3 x3 x t dt x x2 14 t 4 1 12 t 2 x 0 14 x 4 12 x 2 x2 2 x 4 2 (b) 12 x 2 23 t 3 16 3 t2 t2 1 1 x 2 4 78. (a) 4 t dt d 23 x dx 3 x 2 23 x 3 2 2 16 3 x 23 x 3 x 2 x 8 80. (a) 3 sec t tan t dt d sec x dx 2 sec t 3 sec x 2 (b) x1 (b) sec x tan x 82. F x 1 dt 84. F x Fx 4 x 4 1 x t dt 86. F x 0 sec3 t dt sec3 x Fx x2 x 2 x Fx S ection 4.4 x The Fundamental Theorem of Calculus 471 88. F x Fx 0 t 3 dt x t4 4 x 0 x Alternate solution x Fx x 0 t3 dt x t3 dt x x 0 t3 dt x t3 dt 0 0 t3 dt 1 x3 0 Fx x 2 x 3 90. F x 2 t 3 dt t 2 x2 2 2 1 2t2 x2 3 x2 2 1 2x4 2x 5 1 ⇒F x 8 2x 5 Alternate solution: F x x 2 2x 92. F x 0 sin sin x 2 2d 2 Fx 94. (a) 2x 2x sin x 4 (b) 6 4 2 y x gx 1 1 2 2 3 0 4 2 6. 5 4 6 6 7 3 8 0 9 3 10 6 (c) Minimum of g at 6, x −2 2 4 6 10 12 (d) Minimum at 10, 6 . Relative maximum at 2, 2 . (e) On 6, 10 g increases at a rate of (f) Zeros of g: x 4 t2 4 4 3, x 8. 12 4 3. −4 −6 x 96. (a) g t t→ 4 (b) A x 1 4 4t 4x2 4 t 4 dt t2 x lim g t 4x 1 4 x 4x x 4 x 8 8 1 2 Horizontal asymptote: y 8x x x→ 4 x→ lim A x lim 4x 0 8 The graph of A x does not have a horizontal asymptote. 98. True 100. Let F t be an antiderivative of f t . Then, vx vx f t dt ux Ft ux Fvx Fux Fux d dx vx f t dt ux d Fvx dx F vx v x f vx v x F ux u x f ux u x. 472 Chapter 4 x s Integration f t dt ds 0 0 s s 102. G x 0 s (a) G 0 0 s 0 f t dt ds x 0 (b) Let F s Gx s 0 f t dt. x (c) G x (d) G 0 x 0 fx 0 f t dt 0 F s ds 0 x f0 0 f t dt 0 Gx G0 Fx 0 x 0 f t dt 0 0 0 f t dt 104. x t xt t 3t2 1t 14t 3 2 t3 7t 2 15t 9 15 Using a graphing utility, 5 Total distance 0 x t dt 27.37 units Section 4.5 f g x g x dx Integration by Substitution u gx du g x dx 2. 4. 6. x2 x3 1 dx x3 2x sin x 1 3x2 dx 2 dx cos x dx sec 2x tan 2x dx cos x dx sin2 x x2 4 4 8. x2 Check: 9 3 2x dx 9 4 C 4 x2 4 9 3 d x2 9 dx 4 C 2x x2 9 3 2x 10. 1 2x2 13 4x dx 2x2 43 3 1 4 C 2x2 3 4 43 C 2x2 13 Check: d3 1 dx 4 4 1 3 4x 1 2x2 13 4x 12. x2 x3 Check: 5 4 dx 1 3 5 x3 C 5 4 3x2 dx 1 x3 5 3 5 x3 5 C 5 4x2 x3 15 5 5 C d x3 5 dx 15 5 x3 5 4 3x2 15 14. x 4x2 Check: 3 3 dx 1 8 4 4x2 C 3 3 8x dx 3 32 1 4x2 3 8 4 3 4 C 3 3 4x2 3 32 4 C d 4x2 3 dx 32 4 4x2 8x x 4x2 Section 4.5 1 4 5 1 t4 5 4 32 5 12 32 Integration by Substitution 473 16. t3 t4 Check: 5 dt d14 t dt 6 t4 32 5 12 4t3 dt 1 6 C t4 14 t 6 5 12 5 t3 32 C C 34 t 2 4t3 18. u2 u3 Check: 2 du 1 3 u3 32 2 C 12 3u2 du 2 9 33 u 2 1 u3 2 3 32 2 12 32 C u3 2 u3 9 2 12 2 32 C d 2 u3 2 du 9 1 4 3u2 u2 20. x3 dx 1 x4 2 Check: 1 x4 C 2 4x3 dx 1 1 4 x4 2 1 1 4 4x3 x4 1 C 1 41 x4 C d 1 4 dx 4 1 x 1 3 x3 1 x4 1 16 3 2 22. x2 16 Check: x3 2 dx 16 C x3 1 3 2 3x2 dx 1 16 x3 2 x3 1 16 1 C x2 x3 2 1 3 16 x3 C 1 d dx 3 16 x3 1 4 1 2 x4 3x2 24. x3 dx 1 x4 Check: d dx 1 3x 1 x4 C 12 4x3 dx 1 1 2 x4 1 1 x4 4 12 12 12 C x3 1 x4 x3 3 1 2 x4 C 1 2 1 x 9 C 4x3 26. x2 Check: 2 dx 1 x 9 1 x2 2 dx 1 x 9 x3 3 2 1x1 91 x2 1 3x 2 C 1 9x C 3x4 1 9x C d 13 x dx 3 1 x 2 x x2 28. 1 2x dx d dx 12 dx 1 1 x1 2 212 2x 23 t 3 t1 2 C x C Check: C 30. t 2t2 t dt 2 t1 2 2t3 2 2 dt 2 45 t 5 2t3 2 2 C t 2t2 t 23 t 15 2 5 6t C Check: d 23 t dt 3 1 dt 4t2 45 t 5 C 32. t3 3 Check: 13 t 3 1 4t C 1 t 4 2 dt 1 t4 34 1 4t2 1t1 41 C 14 t 12 1 4t C d 14 t dt 12 13 t 3 474 Chapter 4 Integration 27 y 7 27 y 7 2 34. 2 y8 Check: y3 2 dy 2 y3 8y 2 y5 C 2 dy d 2 dy 2 4y2 4y2 2 C C 4 y2 14 7 16 y y3 2 C 2y8 y3 2 d 4 y2 14 dy 7 10x2 dx 1 x3 2 y5 2 36. y 10 3 38. y 12 x x2 1 2 x2 4 8x 8x 1 1 dx 12 1 x3 3x2 dx C 2x C 8 dx 10 1 x3 3 12 20 3 40. (a) 3 12 1 x2 2 x2 8x 1 12 8x 1 12 1 x3 C dy dx y C y (b) x cos x2, 0, 1 x cos x2 dx 1 sin x2 2 C C⇒C 1 1 1 cos x2 2x dx 2 −5 −2 −3 x 5 0, 1 : 1 y 1 sin 0 2 1 sin x2 2 42. 4x3 sin x4 dx sin x4 4x3 dx cos x4 C 44. cos 6x dx 1 6 cos 6x 6 dx 1 sin 6x 6 C 46. x sin x2 dx 1 2 sin x2 2x dx 1 cos x2 2 C 48. sec 1 x tan 1 x dx sec 1 x tan 1 x 1 dx sec 1 x C 50. tan x sec2 x dx tan x 3 32 2 C 2 tan x 3 32 C 52. sin x dx cos3 x x dx 2 cos x 3 sin x dx cos x 2 x 2 2 C 1 2 cos2 x C 1 sec2 x 2 C 54. csc2 2 csc2 x 2 1 dx 2 2 cot C 56. f x sec x tan x dx 1 sec 1. 3 sec x C, C C 1. Thus Since f 1 3 fx sec x Section 4.5 1 u 2 1 dx 1 4 1 du 2 1 32 Integration by Substitution 475 58. u 2x 1, x 1 , dx 1 u 2 u 60. u 2 x x, x 1 2 2 u, dx x dx du 3 3u 2u3 2u3 5 2 x 2x u u 1 du 2 du 2 u 12 u du u 32 12 du C C 2 3 x C C 125 u 45 2 23 u 3 5 1 1 1 32 2 C 25 u 5 u 32 2 u3 2 3u 30 1 2x 30 1 2x 30 1 2x 15 62. Let u 2x x x 1 dx 4 4, x u 2u 4, du 4 u 2 C 3 2x 6x 3x 1 2 1 5 C C C 5 x x 2 2 5 2 2 5 5 x 32 32 32 dx. 1 du du C C 4 13 21 C C 64. u t 4, t t3 t u 4 dt 4, dt u u4 37 u 7 3u4 7 3 t 7 3 t 7 3 3 du 4 u1 3 du 4u1 3u4 u 4 4 7 43 3 3 2u1 43 u 3 21 u 3 2 3 2 3 66. Let u 4 2 7u 14u1 2 12 du C C 2 3 2u 21 x x 42x 4 2x t t 4 3 7 C C 43 x3 8, du 8 2 dx 1 3 3x2 dx. 4 x2 x3 2 x3 2 8 8 3 2 3x2 dx 8 8 3 1 x3 8 3 3 34 2 1 64 9 x2, du x2 dx 41,472 68. Let u 1 1 x1 0 2x dx. 1 2 1 1 0 x2 12 2x dx 1 1 3 1 x2 32 0 0 1 3 1 3 70. Let u 2 0 1 x 1 2x2, du 2x2 dx 4x dx. 1 4 2 1 0 2x2 12 4x dx 1 2 2 1 2x2 0 3 2 1 2 1 476 Chapter 4 4 2 Integration 2x dx. 1 2 2 72. Let u x2, du x2 dx x3 4 0 4 0 x2 13 2x dx 3 4 8 2 x2 43 0 34 8 8 3 44 3 6 33 4 2 3.619 74. Let u When x 5 1 2x 1, du 1, u 2 dx, x 1 u 2 5, u 1. 9. 1 4 9 1. When x 9 x 2x 1 dx 1 1 2u 11 du u 2 9 u1 1 2 u 12 du 123 u 43 1 4 16 3 2 2 2u1 2 1 2 27 3 23 2 3 2 76. 3 x cos x dx x2 2 2 2 2 sin x 3 8 1 18 3 2 52 72 2 2 3 78. u x 2, x u 2, u 6 2, dx du 6, u 8 When x Area 0. When x 2 dx 8. 8 x2 3 x 2 u 0 2 23 u du 0 u7 3 4u4 3 4u1 3 du 3 10 u 10 3 12 7 u 7 8 3 3u4 3 0 4752 35 80. A 0 sin x cos 2x dx cos x 1 sin 2x 2 2 0 82. Let u 2x, du 4 2 dx. csc 2x cot 2x dx 12 Area 1 2 4 csc 2x cot 2x 2 dx 12 1 csc 2x 2 4 12 1 2 2 5 2 84. 0 x3 x 20 2 dx 7.581 86. 1 50 x2 x 1 dx 67.505 88. 0 2 sin 2x dx 1.0 0 0 0 2 1 0 5 2 −1 90. sin x cos x dx sin x cos x dx cos2 x 2 C2 sin x 1 cos x dx cos x 1 1 sin2 x 2 C1 cos2 x 2 C2 1 2 C2 sin x dx C2 1 . 2 sin2 x 2 C1 sin2 x 2 They differ by a constant: C2 Section 4.5 92. f x 2 Integration by Substitution 477 sin2 x cos x is even. 2 94. f x 2 sin x cos x is odd. sin x cos x dx 2 sin2 x cos x dx 2 0 sin2 x cos x dx 2 2 3 sin3 x 3 2 0 0 4 96. (a) 4 sin x dx 4 0 since sin x is symmetric to the origin. 4 4 (b) 2 cos x dx 4 2 0 2 cos x dx cos x dx 0 2 sin x 0 2 2 since cos x is symmetric to the y-axis. 2 0 (c) 2 cos x dx 2 2 2 sin x x cos x (d) sin x cos x dx 2 0 since sin sin x cos x and hence, is symmetric to the origin. 98. sin 3x cos 3x dx sin 3x dx cos 3x dx 0 2 0 cos 3x dx 2 sin 3x 3 0 0 100. If u 5 x2, then du 2x dx and x5 x2 3 dx 1 2 5 x2 3 2x dx 13 u du. 2 102. dQ dt Qt Q 100 Qt Q0 Thus, Q t $250,000. k 100 k 100 C 0 k 100 3 k 100 3 t 2 104. R k 100 3 (a) t 3 3.121 8 2.399 sin 0.524t 1.377 t 2 dt C 0 12 0 t 3 3 2,000,000 ⇒ k t 3. Relative minimum: 6.4, 0.7 or June 6 (b) 0 Relative maximum: 0.4, 5.5 or January 12 2 100 When t 50, Q 50 R t dt 1 3 12 37.47 inches 1 13 3 4.33 inches (c) R t dt 9 24 106. (a) 70 (b) Volume 0 R t dt 1272 (5 thousand of gallons) 0 0 24 Maximum flow: R 61.713 at t 9.36. 18.861, 61.178 is a relative maximum. 478 Chapter 4 4 Integration 108. (a) 0 g 9.4 (b) g is nonnegative because the graph of f is positive at the beginning, and generally has more positive sections than negative ones. f −4 (c) The points on g that correspond to the extrema of f are points of inflection of g. (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. (e) 0 4 9.4 −4 The graph of h is that of g shifted 2 units downward. t gt 0 f x dx 2 t f x dx 0 2 f x dx 2 ht. 110. False x x2 1 2 dx 1 2 x2 1 2x dx 12 x 4 1 2 C 112. True b b b 2 sin x dx a cos x a cos b cos a cos b 2 cos a a sin x dx 114. False sin2 2x cos 2x dx 1 2 x f x dx a a 0 a sin 2x 2 2 cos 2x dx 1 sin 2x 2 3 3 C 13 sin 2x 6 C 116. Because f is odd, f a 0 f x . Then a f x dx a f x dx f x dx. 0 f x dx 0 Let x When x 1 u, dx 0, u du in the first integral. 0. When x a a, u a a. f x dx 0 f x dx a 0 f a u du a f u du 0 0 f x dx 0 Section 4.6 Numerical Integration 479 Section 4.6 1 Numerical Integration x2 2 x2 2 x2 2 1 dx 1 dx 1 dx x3 6 1 1 8 1 1 12 2 1 2. Exact: 0 1 x 0 7 6 14 2 2 1.1667 1 2 Trapezoidal: 0 1 2 4 2 1 2 12 2 2 1 2 2 1 4 34 2 2 1 2 12 2 1 12 2 1 1 75 64 7 6 1.1719 1.1667 Simpson’s: 0 14 2 12 2 34 2 2 4. Exact: 1 2 1 dx x2 1 dx x2 1 dx x2 1 x 1 1 8 1 1 12 0.5000 1 Trapezoidal: 1 2 2 4 4 5 4 5 2 2 2 4 6 2 4 6 2 2 2 4 7 4 4 7 2 1 4 2 0.5090 0.5004 Simpson’s: 1 1 4 8 6. Exact: 0 8 3 x dx x dx x dx 34 x 4 1 0 2 1 0 3 8 3 0 12.0000 2 4 23 2 23 2 23 3 43 3 23 4 23 4 23 5 43 5 23 6 23 6 23 7 43 7 2 2 11.7296 11.8632 Trapezoidal: 0 8 3 Simpson’s: 0 3 3 8. Exact: 1 3 4 4 1 3 x2 dx x2 dx x2 dx 4x 1 3 4 1 3 6 12 x 3 1 0 4 1 0 6 x3 3 3 3 1 11 3 3 2 2 2 3 20 0.6667 24 25 4 5 2 5 2 Trapezoidal: Simpson’s: 1 24 44 9 4 5 0.6667 0.7500 4 0 44 2 2 10. Exact: 0 2 x x2 x x2 0 2 1 dx 1 dx 1 dx 1 2 4 32 0 13 5 3 12 12 2 2 1 21 21 3.393 12 12 1 1 2 4 3 2 3 2 32 32 2 Trapezoidal: Simpson’s: 0 1 2 1 2 1 1 1 1 2 22 2 22 1 1 3.457 3.392 x x2 2 2 2 12. Trapezoidal: 0 2 1 1 1 1 x3 x 3 dx dx 1 1 4 1 1 6 2 4 1 1 1 1 12 3 12 3 2 2 1 1 1 1 13 1 3 2 4 1 1 1 1 32 3 32 3 1 3 1 3 1.397 1.405 Simpson’s: 0 Graphing utility: 1.402 480 Chapter 4 Integration 5 5 sin 8 8 5 5 sin 8 8 2 3 3 sin 4 4 3 3 sin 4 4 4 7 7 sin 8 8 7 7 sin 8 8 0 14. Trapezoidal: 2 x sin x dx x sin x dx 2 16 24 2 2 1 4 2 2 2 0 1.430 1.458 Simpson’s: Graphing utility: 1.458 4 16. Trapezoidal: 0 tan x2 dx 4 8 0.271 tan 0 2 tan 4 4 2 2 tan 4 2 2 2 tan 3 4 4 2 2 tan 4 4 Simpson’s: 0 tan x2 dx 4 12 0.257 tan 0 4 tan 4 4 2 2 tan 4 2 2 4 tan 3 4 4 2 2 tan 4 Graphing utility: 0.256 2 18. Trapezoidal: 0 2 1 1 0 cos2 x dx cos2 x dx 16 24 2 2 21 41 cos2 cos2 8 8 21 21 cos2 cos2 4 4 21 41 cos2 3 cos2 3 8 8 1 1 1.910 1.910 Simpson’s: Graphing utility: 1.910 sin x dx x sin x dx x 2 sin 4 4 sin 4 4 2 sin 2 4 2 sin 2 2 2 2 sin 3 4 34 4 sin 3 4 34 20. Trapezoidal: 0 8 12 1 1 0 0 1.836 1.852 Simpson’s: 0 Graphing utility: 1.852 22. Trapezoidal: Linear polynomials Simpson’s: Quadratic polynomials 1 x x x x x 1 1 1 2 1 6 1 24 1 5 3 2 24. fx fx fx fx f 4 4 x (a) Trapezoidal: Error ≤ 10 12 42 2 2 1 96 0. 0.01 since f x is maximum in 0, 1 when x (b) Simpson’s: Error ≤ since f 4 1 05 24 180 44 1 1920 0.0005 0. x is maximum in 0, 1 when x S ection 4.6 2 1 x 3 Numerical Integration 481 26. f x in 0, 1 . 0 and f 0 2. 130. (a) f x is maximum when x Trapezoidal: Error ≤ f (b) f 4 4 1 2 < 0.00001, n2 > 16,666.67, n > 129.10; let n 12n2 in 0, 1 0 and f 4 x 24 1 x 5 x is maximum when x 0 24. 12. (In Simpson’s Rule n must be even.) Simpson’s: Error ≤ 1 24 < 0.00001, n4 > 13,333.33, n > 10.75; let n 180n4 28. f x (a) f x x 1 23 2 9x 1 43 in 0, 2 . 0 and f 0 2 . 9 122. f x is maximum when x Trapezoidal: Error ≤ (b) f 4 82 12n4 9 10 3 < 0.00001, n2 > 14,814.81, n > 121.72; let n x 4 81 x 56 1 in 0, 2 0 and f 4 f x is maximum when x 0 56 . 81 12. (In Simpson’s Rule n must 32 56 Simpson’s: Error ≤ 180n4 81 be even.) 30. f x (a) f x sin x2 2 2x2 sin x2 < 0.00001, n4 > 12,290.81, n > 10.53; let n cos x2 in 0, 1 . 1 and f 1 3 f x is maximum when x Trapezoidal: Error ≤ (b) f 4 2.2853. 139. x 4 16x4 10 2.2853 < 0.00001, n2 > 19,044.17, n > 138.00; let n 12n2 12 sin x2 48x2 cos x2 in 0, 1 0.852 and f 4 f x is maximum when x 0.852 28.4285. 12. Simpson’s: Error ≤ 1 05 28.4285 < 0.00001, n4 > 15,793.61, n > 11.21; let n 180n4 32. The program will vary depending upon the computer or programmable calculator that you use. 34. f x n 4 8 10 12 16 20 1 Ln 0.8739 0.8350 0.8261 0.8200 0.8121 0.8071 x2 on 0, 1 . Mn 0.7960 0.7892 0.7881 0.7875 0.7867 0.7864 Rn 0.6239 0.7100 0.7261 0.7367 0.7496 0.7571 Tn 0.7489 0.7725 0.7761 0.7783 0.7808 0.7821 Sn 0.7709 0.7803 0.7818 0.7826 0.7836 0.7841 482 Chapter 4 Integration 36. f x n 4 8 10 12 16 20 sin x on 1, 2 . x Ln 0.7070 0.6833 0.6786 0.6754 0.6714 0.6690 Mn 0.6597 0.6594 0.6594 0.6594 0.6594 0.6593 Rn 0.6103 0.6350 0.6399 0.6431 0.6472 0.6496 Tn 0.6586 0.6592 0.6592 0.6593 0.6593 0.6593 Sn 0.6593 0.6593 0.6593 0.6593 0.6593 0.6593 38. Simpson’s Rule: n 2 8 3 6 17.476 1 22 sin 0 3 4 1 22 sin 3 16 2 1 22 sin 3 8 ... 1 22 sin 3 2 83 0 1 22 sin d 3 40. (a) Trapezoidal: 2 f x dx 0 2 4.32 28 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 Simpson’s: 2 f x dx 0 2 4.32 38 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 (b) Using a graphing utility, y 1.3727x3 2 4.0092x2 12.53 0.6202x 4.2844 Integrating, 0 y dx 42. Simpson’s Rule: n 1 6 dx 4 1 36 3.14159 1 4 16 2 4 0 1 1 x2 1 2 26 2 1 4 36 2 1 2 46 2 1 4 56 2 1 2 44. Area 120 75 2 12 7435 sq m 2 81 2 84 2 76 2 67 2 68 2 69 2 72 2 68 2 56 2 42 2 23 0 46. The quadratic polynomial px x x1 x2 x x2 x1 x3 y x3 1 x x2 x1 x x1 x2 x3 y x3 2 x x3 x1 x x1 x3 x2 y x2 3 passes through the three points. ...
View Full Document

Ask a homework question - tutors are online