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# EVEN04 - CHAPTER Integration Section 4.1 Section 4.2...

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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...
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