ODDREV06 - 40 Chapter 6 Applications of Integration Review...

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Unformatted text preview: 40 Chapter 6 Applications of Integration Review Exercises for Chapter 6 5 1. A 1 y 1 dx x2 1 x 5 1 4 5 1 3. A 1 1 x2 1 1 dx arctan x 1 1 (1, 1) 4 5, , 2 3 4 4 y 2 1 25 x (1, 0) (5, 0) 2 1, 1 2 1 1, 1 (−1, 0) 1 2 x 1 (1, 0) 1 2 5. A 2 0 x 12 x 2 x3 dx 14 x 4 1 7. A 0 e2 xe 2 ex dx 2 2 1 2 ex 0 0 e2 y y 1 (0, e 2) 6 (2, e 2) 1 (1, 1) x 4 1 (0, 0) 1 1 (0, 1) x 1 1 2 3 ( 1, 1) 5 4 8 9. A 4 sin x cos x 1 2 4 2 cos x d x 5 4 4 11. A 0 8 3 16x 0 8x x2 x2 8x 3 dx sin x 1 2 1 2 1 2 2x 2 d x 23 x 3 8 0 8x 2 20 512 3 170.667 22 −4 (8, 3) (0, 3) 10 − 16 Review Exercises for Chapter 6 2 2 41 13. y A 1 1 x 1 x dx 2x 1 4 32 x 3 2 2 (0, 1) −1 0 1 2 (1, 0) 1 0 x dx 12 x 2 1 0 −1 x 1 6 0.1667 15. x A y2 0 2y ⇒ x 1 1 1 1 2y d y y 1 1 2 ⇒y 1± 0 x 1 3 y x y2 x 2 1 dx 1 2x y2 13 y 3 1 dx (0, 2) 2 0 2 A 0 0 2y 0 y2 dy 4 3 1 (0, 0) −2 −1 −1 x 2 2 17. A 0 2 0 1 x dx 2 1 3 x 2 3 2 3 y dx 2 1 x 2 dx 3 2 x dx 2y (0, 1) (3, 1) y y A 1 x 1 x ⇒x 2 2⇒x y 2 2 y 2 32 y 2 1 0 (2, 0) x 2, y 1 1 3 2y d y 3 2 0 1 3y d y 0 19. Job 1 is better. The salary for Job 1 is greater than the salary for Job 2 for all the years except the first and 10th years. 21. (a) Disk 4 (b) Shell x2 dx 0 V y 4 3 x3 3 4 0 64 3 4 V y 4 3 2 0 x2 dx 23 x 3 4 0 128 3 2 1 x 1 2 3 4 2 1 x 1 2 3 4 —CONTINUED— 42 Chapter 6 Applications of Integration 21. —CONTINUED— (c) Shell 4 (d) Shell 4 V 2 0 4 4 4x 0 x x dx x2 dx x3 3 4 0 V 2 0 4 6 6x 0 x x dx x2 dx 13 x 3 4 0 2 2 y 4 3 2 64 3 y 5 4 3 2x 2 2 3x 2 160 3 2 1 x 1 2 3 2 1 x 1 −1 2 3 4 5 23. (a) Shell 4 (b) Disk x 0 V 4 3 3 4 4 16 2 16 3 x2 dx 4 V 64 0 2 0 3 4 2 16 x3 3 4 x2 48 0 dx 1 2 y 4 x2 32 9 8 16x y 4 2 1 −3 −2 −1 −2 −4 x 1 2 3 −3 −2 −1 2 1 x 1 2 3 −2 −4 25. Shell 1 y V 2 0 1 0 x x4 2x x2 2 1 dx dx 1 1 1 2 0 x 1 arctan x 0 2 4 4 Review Exercises for Chapter 6 27. Shell u x dx V u2 2u du 6 43 y 3 2 1 x 1 −1 2 3 4 5 6 x 2 2 2 2 2 x 1 u3 1 x 2u du u 12 u 2 0 2 2 dx 4 4 0 2 u2 1 u 2 2u du u 3 3 1 4 20 3 u du 9 ln 3 42.359 −2 −3 4 0 u2 0 4 13 u 3 3u 3 ln 1 u 0 29. Since y ≤ 0, A 1 xx 1 d x. 31. From Exercise 23(a) we have: V 1 V 4 y0 64 ft 3 16 16 1 9 u x dx A x u du 1 1 Disk: 3 1 1 16 9 9 y0 y2 dy y2 dy 13 y 3 27 y0 3 u 0 1 u du 0 u3 2 4 15 u1 2 du 1 9 9 3 2 52 u 5 y 2 32 u 3 1 0 9y 9y0 13 y 30 y03 9 27 9 0 −1 x 27y0 By Newton’s Method, y0 −1 1.042 and the depth of the 1.958 ft. gasoline is 3 1.042 33. fx fx 1 fx 2 4 54 x 5 x1 4 1 1 u 2u 4 35. y x x 1 2 y 300 cosh x 2000 280, 2000 ≤ x ≤ 2000 3 x sinh 20 2000 2000 u x dx s s 2000 1 1 20 2000 3 x sinh 20 2000 400 9 sinh2 2 dx 1 du 3 2000 x dx 2000 4018.2 ft (by Simpson’s Rule or graphing utility) uu 1 du 1 0 3 x dx 2 1 2 1 u3 2 2 52 u 5 u1 2 du 2 32 u 3 3 1 2 4 32 u 3u 15 3 5 1 8 1 15 63 6.076 44 Chapter 6 3 x 4 y 1 y 2 Applications of Integration 37. y 39. F 3 4 25 16 4 kx k1 4x 5 5 4 F W 4x d x 0 2x 2 0 S y 4 2 0 3 x 4 25 dx 16 15 8 x2 4 2 0 15 50 in lb 4.167 ft lb (4, 3) 3 2 1 x 1 2 3 4 41. Volume of disk: Weight of disk: 62.4 Distance: 175 W 62.4 9 150 1 3 2 y 1 3 2 43. Weight of section of chain: 5 x Distance moved: 10 y W 62.4 9 ton y2 2 150 0 10 x x dx 5 10 2 250 ft lb 10 5 0 10 x 2 0 y 175 y dy 163.4 ft 175y 0 104,000 ft b lb 45. W a 4 F x dx a x 2dx 0 80 a ax3 3 4 0 64 a 3 3 80 64 a 15 4 3.75 a 47. A 0 a 1 A x y 6 a2 6 a2 a x 2 dx 0 a 2 a x1 2 x dx ax 4 3 a x3 2 12 x 2 a 0 a2 6 x 0 a a a 0 x 2 dx 4 6 a2 dx 6ax a ax 0 2 a x3 2 x2 dx a y 61 a2 2 3 a2 a x ( x, y ) a2 0 4a 3 2x 1 2 8 32 32 ax 3 4a 1 2x 3 2 8 12 52 ax 5 x2 dx 13 x 3 a 0 a x 32 ax a2 x, y aa , 55 3ax 2 a 5 R eview Exercises for Chapter 6 49. By symmetry, x 1 45 0. x2 dx 2 a 2x x3 3 a 0 y A 1 A y 2 0 a2 4a 3 3 a2 3 4a 3 31 4a 3 2 6 8a 3 a a ( x, y ) a2 a x2 2 dx x4 dx 15 x 5 15 a 5 a 0 −a a x a4 0 2a 2x 2 2a 2 3 x 3 25 a 3 6 a 4x 8a 3 6 a5 8a 3 x, y 0, 2a 2 5 2a 2 5 51. y 0 by symmetry 4 3 2 y 1 y= 6x+1 For the trapezoid: m My 0 6 0 (6, 2) 46 6 16 1 x 6 12 x 3 1 2x d x 18 1 x 6 x3 9 1 x2 0 1 x 1 2 3 4 5 7 x dx 6 −1 −2 −3 −4 (6, −2) 60 For the semicircle: m My 6 1 2 8 2 x 2 2 8 4 x u 6 2 4 3 4 6 2 4 x 6 2 dx 6, u u2 du 16 3 2 6 x4 x 8, u 6 2 dx 2. Let u My x 2 6, then x 2 6 and dx u2 du 2 du. When x 2 0. When x 2 u 0 2 0 u4 2 4 2 12 0 4 44 3 u2 du 9 2 Thus, we have: x 18 2 1 2 u2 32 0 12 12 60 x 180 44 3 44 3 29 3 9 9 29 49 ,0 . 9 1 29 3 49 9 The centroid of the blade is 46 Chapter 6 Applications of Integration weight per cubic volume. g y dy d y 53. Let D F c surface of liquid; d D d y fy D Dfy c g y dy c yfy d g y dy d g c f ( x, y ) d yfy fy g y dy D c d g y dy x c c fy y g y dy Area D Area depth of centroid Problem Solving for Chapter 6 1. T R 0 12 cc 2 c 13 c 2 x2 dx cx2 2 x3 3 c 0 cx c3 2 c3 3 c3 6 T lim c→0 R 13 c 2 lim c→0 1 c3 6 1 3 3. (a) 1 V 2 2 0 1 1 41 y2 2 2 1 y2 1 4 y2 2 dy y2 1 y2 dy 4 0 1 y2 41 8 0 1 2 y2 dy 2 Integral represents 1 4 area of circle 4 R± 2 2 8 (b) x 1 V 2 R 2 r 4 y2 R ⇒V r2 ⇒ x r2 y2 r2 R y2 r2 y2 2 dy 0 r 4R r2 0 y2 dy 22 4R V 2 r 22 12 r 4 rR rR 5. V 22 r2 h2 4 x r2 22 r 3 h3 8 r x2 dx r r2 h2 4 2 r2 − h 4 r x2 + y2 = r2 2 4 3 x2 32 h 2 h3 which does not depend on r! 6 Problem Solving for Chapter 6 7. (a) Tangent at A: y y x3, y 1 y To find point B: x3 x Tangent at B: y y 1 2 47 3x2 3x 3x 1 2 x3 3x 0 0⇒B 2, 8 2 (b) Tangent at A a, a3 : y a3 3a2 x y 3a2x 2a3 2a 2a, 12a2 x 12a2x 16a3 x 4a a 2a 3 To find point B: x3 x 3a2x a 2 0 0⇒ 8a3 2a 16a3 0 0⇒ x B 3x x 2 2 3x2 Tangent at B: y 8a3 y x3, y 8 y 12 x 12x x3 2 16 12x 0 0⇒C 27 4 108 16 4, 64 16 To find point C: x3 x 12a2x 2a 2 To find point C: x3 x 1 C a 4a, 64a3 2a3 dx 3 12x 2 2 16 4 2 dx x3 dx area S area R Area of R 2a 4a x3 12a x 2a 2 3a2x 16a x 27 4 a 4 108a 4 Area of R 2 4 x3 12x 2 3x 16 Area of S Area of S x dx 3 Area of S Area of S 16 area of R 16 area of R x 9. s x (a) s x (b) ds ds (c) s x 2 1 ds dx 1 1 x f t 2 dt 1 fx 2 f x 2 dx fx 1 2 dx 31 t 2 2 1 2 dy dx x 2 dx 1 2 dx 2 dy 2 2 dt 1 1 2 9 t dt 4 22 27 1 to x 22 2. b (d) s 2 1 1 9 t dt 4 8 1 27 9 t 4 32 2 1 13 27 13 2.0858 This is the length of the curve y x3 2 from x 11. (a) y My 0 by symmetry 6 (b) m 1 x3 1 x2 6 2 1 6 1 dx x3 b2 b2 2b b 1 1 2b b 1 2 2b b 2, 0 1 x 1 6 1 x3 1 dx x3 12 7 dx 1 6 1 2 dx x2 2 1 x 6 1 5 3 My x 2 1 1 dx x2 1b 1 b2 m x y 3 2 1 2 1 35 36 12 ,0 7 2b b2 b→ x, y x, y ,0 53 35 36 x, y (c) lim x b→ lim 2b b 1 x −1 −2 −3 2 3 4 5 6 48 Chapter 6 area area Applications of Integration 2 3 4 1 6 1 12 2 1 2 7 1 2 15. Point of equilibrium: 50 x P0, x0 10, 80 80 13. (a) W (b) W 0.5x 80, p 0.125x 10 Consumer surplus 0 80 50 10 0 0.5x 0.125x dx 10 dx 400 1600 Producer surplus 17. (a) Wall at shallow end From Exercise 22: F (b) Wall at deep end From Exercise 22: F (c) Side wall From Exercise 22: F1 4 62.4 2 4 20 9984 lb 62.4 4 8 20 39,936 lb 20 15 y 62.4 2 4 40 y 10y d y y2 dy 624 4y 2 19,968 lb y=8 10 x = 40 F2 62.4 0 4 8 8y 0 5 624 y3 3 4 5 10 15 20 25 40 45 1 y = 10 x x 0 26,624 lb Total force: F1 F2 46,592 lb ...
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