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# EVNREV04 - Review Exercises for Chapter 4 483 Review...

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Unformatted text preview: Review Exercises for Chapter 4 483 Review Exercises for Chapter 4 2. y 4. f′ u du 3x 3 dx 3 x 2 dx 3x 2 3 3x 13 3 dx 3x 23 C f 6. x3 2x2 x2 1 dx x 12 x 2 2 2x x 1 x 2 dx C 8. 5 cos x 2 sec2 x dx 5 sin x 2 tan x C 10. f x fx 6x 1 6x 1 dx 3x 1 2 12. 45 mph C1 30 mph at 66 ft sec 44 ft sec a 66 ft sec. 0. Since the slope of the tangent line at 2, 1 is 3, it follows that f 2 3 C1 3 when C1 0. fx fx f2 fx 1 x 3x 3x C2 1 3 1 2 at 66 since v 0 a2 st t 66t since s 0 2 Solving the system vt s(t we obtain t 55 12 t s 72 5 at a2 t 2 66 66t 44 264 vt 1 2 dx x 1 3 C2 1 when C2 0. 24 5 and a 55 12. We now solve 66 0 and get t 72 5. Thus, 55 12 72 2 5 2 66 72 5 475.2 ft. Stopping distance from 30 mph to rest is 475.2 14. a t vt st (a) v t (b) s 4.08 (c) v t (d) s 2.04 9.8 m sec2 9.8t 4.9t2 9.8t v0 40t 40 9.8t s0 40 0 40 9.8 4.08 sec. 264 211.2 ft. 0 when t 81.63 m 9.8t 40 20 when t 20 9.8 2.04 sec. 61.2 m 484 16. x1 (a) (b) Chapter 4 2, x2 15 x 5i 1 i 5 i 5 1 Integration 1, x3 5, x4 1 1 1 5 22 1 5 1 3 2 2 3, x5 3 1 7 2 7 16 5 37 210 2 1 1 1 3 2 1 2 5 1 2 xi2 xi 7 1 xi 2xi (c) i 5 1 25 5 7 5 3 2 23 5 3 2 27 7 2 56 (d) i 2 xi 1 5 18. y 9 12 x, x 4 1 9 1 4 4 1, n 4 1 9 4 1 16 4 1 25 4 S4 9 9 9 22.5 s4 1 9 1 9 4 9 1 16 4 9 1 25 4 9 9 14.5 2 n f ci i n n→ 1 y 12 10 20. y x2 3, x n right endpoints x 2 8 6 Area n→ lim lim i 1 2i n 3 3 2 n 4 2 x 1 2 2 n 4i2 lim n→ n i 1 n2 n→ lim lim 2 4 nn n n2 4n 3 2 n n 1 2n 6 1 1 3n 6 8 3 6 26 3 n→ 1 2n n2 22. y 13 x, x 4 n→ y 20 Area lim i n 1 f ci 1 2 14 x 2i n 3 15 10 n→ lim i 2 n 24i2 n2 3i2 n2 i3 n3 3 nn n2 1 2n 6 1 1 n2 n 1 n3 4 2 5 x 1n lim 8 n→ 2n i 1 n→ 24i n 3i n 8i3 n3 1 2 3 4 lim 4n 1 ni 1 4 n n 6 4 n→ lim 3nn 1 n 2 1 15 4 R eview Exercises for Chapter 4 b 4 b 4 b 4 n 485 24. (a) S s (b) S n m m m bi n b n b n 2b 4 b 4 b 4 b 4 m n i n i 1 1 m 3b 4 b 4 b 4 m m b n m 4b 4 b 4 i b 4 mb2 1 16 mb2 1 16 2 2 3 3 4 3mb2 8 1 5mb2 8 y y = mx m0 2b 4 b n b n 3b 4 2n i 1 f i n 1 1 0 mbi n mb2 n n 1 n2 2 1 mb2 n 2n mb2 n 2n sn i bi f n bi m n 0 n→ b m n 1 2n i i 0 mb2 n 1 n n2 2 1 b mb 2 1 x=b x (c) Area b n→ lim mb2 n 2n 12 mx 2 b 0 1 lim mb2 n 2n 12 mb 2 1 base height 2 (d) 0 mx dx 12 mb 2 3 n 26. lim → i 1 3ci 9 ci 2 xi 1 3x 9 x2 dx 28. 6 5 3 2 1 y x − 4 −3 −2 −1 −2 1 2 3 4 4 16 4 x2 dx 1 2 4 2 8 (semicircle) 6 3 6 30. (a) 0 f x dx 0 3 f x dx 3 6 f x dx 1 4 1 1 3 (b) 6 4 f x dx 3 f x dx 0 6 (c) 4 f x dx 6 (d) 3 10 f x dx 10 3 f x dx 10 1 10 3 32. 1 12 dx x3 12x 2 2 3 1 6 x2 x5 5 32 5 52 15 3 1 6 9 2x3 3 16 3 6 16 , (d) 3 1 34. 1 t2 2 dt t3 3 1 2t 1 14 3 2 2 36. 2 x4 2x2 5 dx 5x 2 10 32 5 16 3 10 2 38. 1 1 x2 1 dx x3 2 x 1 2 x 3 dx 1 x 1 2x2 2 1 1 2 1 8 1 1 2 1 8 4 4 40. 4 sec2 t dt tan t 4 1 1 2 486 2 Chapter 4 Integration 2 2 42. 0 x y 7 6 5 4 4 dx x2 2 4x 0 10 44. 1 x2 x 2 dx x3 3 8 3 10 3 7 6 x2 2 2 9 2 4 2 2x 1 1 3 1 2 2 2 1 −2 −1 x 1 2 3 4 5 2 3 y −2 −1 −2 x 1 3 1 3 46. 0 x1 x dx x1 2 x3 2 2 dx 1 2 0 48. Area 0 3 sec2 x dx 3 0 23 x 3 2 3 y 25 x 5 4 15 tan x y 5 4 3 2 5 1 2 x 1 π 6 π 3 x 50. 1 2 0 x3 x 2 x3 dx 0 x4 8 2 y 2 0 8 6 2 3 2 4 2 ( 3 2 , 2) x 1 2 52. F x 1 x2 1 x 2 54. F x csc2 x 56. x dx x3 3 x2 2x 2 1 x x 2 dx C R eview Exercises for Chapter 4 58. u x3 x2 x3 3, du 3 dx 3x2 dx 1 3 5, du 5 2 487 x3 3 12 3x2 dx 23 x 9 3 32 C 60. u x2 x2 x 6x 3 6x 2x x2 6 dx 2x 6 dx 6x 5 2 12 x 2 6x 5 1 1 2 C 2 x2 1 6x 5 C 62. x sin 3x2 dx 1 6 sin 3x2 6x dx 1 cos 3x2 6 C 64. cos x dx sin x sin x 12 cos x dx 2 sin x 12 C 2 sin x C 66. sec 2x tan 2x dx 1 2 sec 2x tan 2x 2 dx 1 sec 2x 2 15 cot 5 1 x3 12 C 68. cot4 csc2 d cot 4 csc2 d C 1 70. 0 x2 x3 1 3 dx 1 3 1 1 x3 0 1 3 3x2 dx 1 4 0 1 16 12 6 1 5 4 6 72. 3 x 3 x2 x 8 1, x dx 1 6 u 6 x2 3 8 12 2x dx 12 x 3 8 12 3 1 27 3 1 74. u 1, dx du 0, u 1 When x 0 1, u x2 x 1 0. When x 1 dx 2 0 1 1. 1 2 2 u u5 0 2 u du 2 2 2u3 u1 2 du 2 27 u 7 2 45 u 5 2 23 u 3 1 2 0 32 105 4 76. 4 sin 2x dx 0 since sin 2x is an odd function. 78. u 1 x, x a, u b 1 1 u, dx du b, u 1155 32 3u5 2 When x Pa, b a. When x x 2 32 1 1 1 b b. 1 a a 1155 3 x1 32 1 1 b dx 2 u 3 u3 2 du 1155 2 11 2 u 32 11 1 b a 1155 32 u9 a 3u7 u3 2 du 29 u 3 2 67 u 7 2 25 u 5 1 2 1 b a 1 b a 1155 2u5 2 105u3 32 1155 (a) P0, 0.25 (b) P0.5, 1 u5 2 16 105u3 385u2 495u 231 231 1 0.75 u5 2 105u3 16 2.5% 73.6% 385u2 495u 231 1 385u2 385u2 495u 495u 0.025 1 0 u5 2 105u3 16 231 0.5 0.736 488 2 Chapter 4 t 2 Integration 2 t 2 2 80. 0 1.75 sin Increase is 7 dt 1.75 cos 2 1.75 1 1 7 2.2282 liters 5.1 1.9 0.6048 liters. 1 82. Trapezoidal Rule n Simpson’s Rule n 4: 0 1 x3 3 x3 2 2 x x 2 2 dx 1 0 8 1 0 12 21 432 3 1 42 41 432 3 1 42 21 232 3 1 22 21 232 3 1 22 23 432 3 3 42 43 432 3 3 42 1 2 1 2 0.172 0.166 4: 0 3 dx Graphing utility: 0.166 84. Trapezoidal Rule n Simpson’s Rule n 4: 0 1 sin2 x dx 3.820 4 : 3.820 Graphing utility: 3.820 Problem Solving for Chapter 4 x 2. (a) F x 2 sin t2 dt 0 0.8048 1 x 2 1.9 0.6106 0.75 x→2 x x Fx 1.0 0.4945 1.5 0.0265 1.9 0.0611 2.0 0 2.1 0.0867 2.5 0.3743 3.0 0.0312 4.0 0.0576 5.0 0.2769 (b) G x x Gx x→2 sin t2 dt 2 1.95 0.6873 1.99 0.7436 2.01 0.7697 2.05 0.8174 2.1 0.8671 lim G x lim (c) F 2 Fx x 1 x 2 F2 2 x x→2 lim sin t2 dt 2 x→2 lim G x sin x2, F 2 0.7568 sin 4 x→2 Since F x Note: sin 4 lim G x . Problem Solving for Chapter 4 4. Let d be the distance traversed and a be the uniform acceleration. We can assume that v 0 0 and s 0 0. Then at vt st st a at 12 at . 2 2d . a a 0. 2ad 0 ad . 2 2d a 2ad. 489 d when t The highest speed is v The lowest speed is v The mean speed is 1 2 The time necessary to traverse the distance d at the mean speed is t d ad 2 2d a which is the same as the time calculated above. 6. (a) 100 80 60 40 20 x 0.2 0.4 0.6 0.8 1.0 y (b) v is increasing (positive acceleration) on 0, 0.4 and 0.7, 1.0 . (c) Average acceleration v 0.4 0.4 v0 0 60 0 0.4 150 mi hr2 (d) This integral is the total distance traveled in miles. 1 v t dt 0 1 0 10 2 20 2 60 2 40 2 40 65 385 10 38.5 miles (e) One approximation is a 0.8 v 0.9 0.9 v 0.8 0.8 50 0.1 40 100 mi hr2 (other answers possible) 490 x Chapter 4 Integration x x x x 8. 0 ft x Thus, d dx x t dt 0 x f t dt 0 t f t dt x x 0 f t dt 0 x t f t dt f t dt 0 ft x 0 t dt xf x 0 f t dt xf x Differentiating the other integral, d dx x 0 x x f v dv dt 0 0 f v dv. Thus, the two original integrals have equal derivatives, x x t ft x 0 t dt 0 0 f v dv dt 0. 1 2 0 C Letting x 1 0, we see that C 23 x 3 10. Consider 0 x dx 2 . The corresponding 3 Riemann Sum using right-hand endpoints is Sn 1 n 1 n3 2 Thus, lim n→ 1 n 1 2 n3 3 2 n 2 ... 2 ... ... n n n n 2 . 3 1 3 12. (a) Area 3 9 2 9x 2 27 x2 dx x3 3 9 3 0 2 0 9 x2 dx 10 8 7 6 5 4 3 2 1 y 36 9. Area 2 bh 3 b2 2 69 3 36. −4 −2 −1 (b) Base 6, height x 12 45 (c) Let the parabola be given by y ba a2x2, a, b > 0. Area 2 0 b2 b2x b a a2 a2x2 dx y 2 x3 3 ba b2 0 3 2 b2 2 Base b3 a a2 b 3a 1 b3 3a b2 4 b3 3a − b a b a x 2b , height a Archimedes’ Formula: Area 2 2b 2 b 3a 4 b3 3a Problem Solving for Chapter 4 14. (a) 1 (b) n 491 i 3 1 3i 3i 3i 1 1 3i2 i n i3 ⇒ 1 1 i 3 i 3 i3 3i2 3i 1 3i2 3i2 i 1 i3 1 3 i3 33 1 23 ... n 1 3 i 1 23 n n 13 1 3 n3 Hence, n (c) n ⇒ i 1 1 3 i n 1 3i2 3i2 i 1 3i 3i 1 1 n 1 3i2 i 1 1 n 3 3n n 2 n 2n 1 n 3i2 1 n3 2n3 2n3 nn 3n2 6n2 3n2 2 1 2n 2 1 2n 6 3n 6n 2 n 1 1 3n n 1 2 3n2 3n ⇒ i n i2 1 nn 20 16. (a) C (b) C 0.1 8 18 12 sin 12 sin 10 t8 12 t8 12 dt 6 dt 14.4 cos 14.4 t8 12 cos 3 2 20 8 14.4 18 1 1 \$9.17 0.1 t8 12 10.8 0.6t 10 14.4 Savings 9.17 b 14.4 3 2 6 \$3.14 3.14 fx fb \$6.03. 18. (a) Let A 0 fx x dx. dx. Let u 0 b fb fb fb x, du fb u fb u fb x u fu u fu x fx A b b 0 b 0 du du dx Then, b 2A 0 b fx 1 dx fx fb b. b x dx 0 fb fb x x fx dx 0 Thus, A (b) b b . 2 1 1⇒ 0 sin 1 sin x dx x sin x 1 2 ...
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