0678 18004 01479 34 x 4 x1 3 3 b 2 77 a b t dt 3 3 16

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Unformatted text preview: 3 t2 6t 2 12t 4t 3t 5 Integration by Substitution 197 9t 9 3 1 2 3t Total distance x t dt 0 5 3t 0 1 3t 4t 20 1 dt 3 5 3 0 t2 4 3 dt 3 1 t2 4t 3 dt 3 3 t2 4t 3 dt 4 28 units 4 105. Total distance 1 4 x t dt v t dt 1 4 1 1 t 4 2 1 dt 2t1 22 1 2 units Section 4.5 f g x g x dx Integration by Substitution u gx du g x dx 1. 3. 5. 5x2 x x 2 1 2 10x dx 5x2 x2 tan x 5 1 1 10x dx 2x dx sec2 x dx 1 dx tan2 x sec2 x dx 7. 1 Check: 2x 4 2 dx d 1 2x dx 5 x2 12 1 5 5 2x C 21 x2 32 C 32 C 9 2x 4 9. 9 Check: 2x dx x2 32 C 2 3 3 9 2 2 9 3 x2 12 x2 32 C 9 x2 2x d2 9 dx 3 2x 198 Chapter 4 Integration 11. x3 x4 Check: 3 2 dx 1 4 3 x4 C 3 2 4x3 dx 2 1 x4 3 4 3 4x3 3 C 3 2 x4 12 x3 3 3 C d x4 3 dx 12 1 3 3 x4 3 12 x4 13. x2 x3 Check: 1 4 dx x3 5 1 4 3x2 dx 1 x3 1 3 5 5 C 1 4 x3 15 1 5 C d x3 1 dx 15 C 5 x3 1 4 3x2 15 x2 x3 15. t t2 Check: 2 dt d dt t2 1 2 2 3 t2 32 2 C 12 2t dt 3 2 t2 1 t2 2 2 32 2 3 12 32 C t2 2 t2 3 1 2t 2 32 C 2t 17. 5x 1 Check: x2 d dx 13 dx 5 2 x2 1 43 x2 C 13 2x dx 15 8 4 1 3 5 2 x2 1 x2 43 2x 43 C 5x 1 x2 15 1 8 13 x2 43 C x2 15 1 8 1 2 13 5x 3 1 19. x 1 Check: x2 3 dx 1 C x2 3 2x dx 21 x2 11 2 3 x2 2 2x 2 C x 1 x2 3 1 41 x2 2 C 1 d dx 4 1 x2 1 3 1 31 2 1 4 21. 1 x2 dx x3 2 d dx 1 x3 C 2 3x2 dx 1 3 11 3 11 x3 x3 1 2 1 C 3x2 1 31 2 x3 C Check: x3 x2 1 x3 12 23. x 1 Check: x2 d dx 1 t 3 dx 1 1 2 x2 1 12 x2 C 12 2x dx 1 1 2 x2 1 1 x2 2 12 12 C x 1 4 1 x2 C 2x x2 25. 1 Check: 1 dt t2 1 1t 4 4 1 C 1 t 3 1 dt t2 1 41 4 1 t 1 3 1t 4 1 t2 C 1 1 t2 1 t 3 d dt 27. 1 dx 2x Check: d dx 1 2 2x 2x C 12 2 dx 1 2x 2 1 2x 1 2 2 12 12 C 1 2x 2x C 2 S ection 4.5 Integration by Substitution 199 29. x2 3x x 7 dx 2 x3 2x3 2 2 3x1 14x1 2 7x C 12 dx 25 x 5 3x x 2 2x3 2 14x1 2 C 2 5 x x2 5x 35 C Check: d 25 x dx 5 2 dt t d 14 t dt 4 2 x2 7 31. t2 t Check: t3 t2 C 2t dt t3 14 t 4 2t t2 t2 t C 2 t 33. 9 Check: y y dy d 23 y dy 5 2 9y1 15 y 2 y3 C 2 dy 9 23 y 3 2 2 25 y 5 2 2 C 9y1 23 y 5 2 2 15 2 y 9 C y y d 6y3 dy 25 y 5 C y3 35. y 4x 4 x dx 4 x2 2 2 4x dx 16 x2 2 16 16 x2 12 12 37. y 2x dx C 1 2 x x2 x2 2x 1 3 2 dx 2 2x 2x 1 1 2x 3 3 2x C C 2 dx x2 12 x2 1 x2 2 2 x2 dy dx y 1 2x2 4 16 C 3 39. (a) 3 y (b) x4 x4 1 2 2 4 3 x2, 2, 2 x2 dx x2 1 4 3 1 4 3 2 1 2 32 4 x2 12 2x dx x2 32 x −2 −1 2 C 32 1 4 3 C⇒C 2 C 2, 2 : 2 y 22 x2 2 32 −2 2 −1 41. sin x dx cos x C 43. sin 2x dx 1 2 sin 2x 2x dx 1 cos 2x 2 C 45. 1 2 cos 1 d cos 1 1 2 d sin 1 C 200 Chapter 4 Integration 47. sin 2x cos 2x dx sin 2x cos 2x dx sin 2x cos 2x dx 1 2 1 2 1 2 sin 2x 2 cos 2x dx cos 2x 1 sin 2x 2 2 2 C 2 12 sin 2x 4 C1 C2 C OR C1 OR 2 sin 2x dx 1 2 1 cos 2x 2 2 sin 4x dx 1 cos2 2x 4 2 sin 2x cos 2x dx 1 cos 4x 8 49. tan4 x sec2 x dx tan5 x 5 C 15 tan x 5 C 51. csc2 x dx cot3 x cot x cot x 2 2 3 csc2 x dx C 1 2 cot2 x C 12 tan x 2 C 1 sec2 x 2 1 C 1 sec2 x 2 C1 53. cot2 x dx csc2 x 1 dx cot x x C 55. f x Since f 0 fx cos x dx 2 3 2 sin 2 sin 0 x 2 C C, C 3. Thus, 2 sin x 2 3. 57. u x 2, x xx u 2 dx 2, dx u u3 25 u 5 2 2 du 2 u du 2u1 43 u 3 2 2 du C C 2 4 10 C C 2u3 2 3u 15 2 x 15 2 x 15 59. u 1 x, x 1 x dx u, dx 1 u1 23 u 3 2 10 32 2 2 3x 3x 32 du u 2 x2 1 u du 2 2u3 45 u 5 u5 2 du 2 2 2 27 u 7 C C x 8 15 1 C x 2 2u3 2 35 105 2 1 105 2 1 105 x x 42u 32 15u2 42 1 12x 35 15x2 C 32 S ection 4.5 Integration by Substitution 201 61. u 2x 1, x 1 u 2 1 , dx 12 u 1 u 8 1 8 1 du 2 1 u u2 2u1 43 u 3 10u 2 x2 1 dx 2x 1 11 du 2 1 3u 6u1 45 1 2 12 12 2u 2 4 du du C C 10 2x 52 13 1 C C 45 C u3 2 125 u 85 2 2 2 u1 2 2 3u 60 2x 1 3 2x 60 1 60 1 15 63. u x x 1, x 1 u x x 1 1, dx dx 2x 2x 1 12x2 1 3x2 8x 2x du u u u u 1 u u x x x where C1 u 2u1 2u 1 2x 2x 1 1 du u 1 u 12 u 1 du C C 1 du 2 2x 1 1 C. 1 1 C C C1 65. Let u 1 x2 x...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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