chapter 6 - Chapter 6 Exercises 6.1 1. F( x ) = 12 x +C 2 9...

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203 Chapter 6 Exercises 6.1 1. F ( x ) = 1 2 x 2 + C 2. F ( x ) = x 9 + C 3. F ( x ) = 1 3 e 3 x + C 4. F ( x ) =− 1 3 e 3 x + C 5. F ( x ) = 3 x + C 6. F ( x ) 2 x 2 + C 7. 34 4 x dx x C =+ 8. 2 11 336 x dx xdx x C == + ∫∫ 9. 77 dx x C 10. 22 kd x kx C 11. 2 2 x dx xdx x C cc c + 12. 1 4 x dx x C 13. 21 1 2 x dx x dx xx ⎛⎞ ⎛ += + ⎜⎟ ⎜ ⎝⎠ ⎝ 2 1 2ln 4 x xC + 14. 1 1 ln 777 dx dx x C =⋅ = + 15. 35 2 5 x xdx x dx x C + 16. 2 2 x dx x x dx x ⎛⎞ +=+ ⎜⎟ ⎝⎠ 13 3 2 44 33 x x xC =+ + = + + 17. 1 x xd x x x −+ = −+ d x 23 121 ln 233 x C = + 18. 1 3 3 3 3 2 2 x dx x x dx x −= − 4 2 3 73 x = 19. 3 3 2 ed x e C −− = 20. x e C = 21. edx ex C = + 22. 1 x dx e dx e C e = + 23. () 1 2 2 x e x C + + 2 2 x ex C = −−+ 24. 2 7 24 2 x dx e dx e C e = + 25. d dt [ ke 2 t ] 2 ke 2 t = 5 e 2 t ; k 5 2 26. d dt [ ke t /10 ] = 1 10 ke t /10 = 3 e t /10 k = 30 27. d dx [ ke 4 x 1 ] = 4 ke 4 x 1 = 2 e 4 x 1 k = 1 2 28. d dx k e 3 x + 1 = d dx [ ke (3 x + 1) ] 3 ke x + = 4 e x + 1) k 4 3 29. d dx [ k (5 x 7) 1 ] k x 7) 2 (5) 5 k x 7) 2 = x 7) 2 k 1 5
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Chapter 6: The Definite Integral ISM: Calculus & Its Applications, 11e 204 30. d dx [ k ( x + 1) 3/2 ] = 3 2 k ( x + 1/2 = ( x + k = 2 3 31. d dx k ln4 x [] = k 4 x ( 1) = k 4 x = 1 4 x k = –1 32. d dx k (8 x ) 3 = d dx [ k x ) 3 ] =− 3 k x ) 4 ( = 3 k x ) 4 = 7(8 x ) 4 k = 7 3 33. d dx [ k (3 x + 2) 5 ] = 5 k x + 2) 4 (3) = 15 k x + 4 = x + 2) 4 k = 1 15 34. d dx [ k (2 x 4 ] = 8 k x 1) 3 = x 3 k = 1 8 35. 3 [l n2 ] 22 dk kx dx x x += = ++ 3 k = 36. 35 n2 3] ) 23 k dx x x x −= = = −− 5 3 k 37. f ( t ) = t ; f ( t ) = 2 5 t 5/2 + C 38. f ( t ) = 4 6 + t f ( t ) = 4ln6 + t + C 39. f ( t ) = 0; f ( t ) = C 40. f ( t ) = t 2 5 t 7; f ( t ) = 1 3 t 3 5 2 t 2 7 t + C 41. .2 .2 () . 5 ; () 2 . 5 xx f xef x e == C + 42. 2 () 2 ; f e f x x e =+ + C 43. f ( x ) = x f ( x ) = 1 2 x 2 + C f (0) = 3 = 1 2 0 2 + C ; C = 3 Thus, f ( x ) = 1 2 x 2 + 3. 44. f ( x ) = 8 x 1/3 f ( x ) = 6 x 4/3 + C f (1) = 4 = 6 + C ; C = –2 Thus, f ( x ) = 6 x 2 . 45. f ( x ) = x + 1; f ( x ) = 2 3 x + x + C f (4) = 0 = 2 3 4 + 4 + C = 2 3 8 + 4 + C = 28 3 + C ; C 28 3 ; Thus, f ( x ) = 2 3 x + x 28 3 . 46. f ( x ) = x 2 + x f ( x ) = 1 3 x 3 + 2 3 x 3/ 2 + C f (1) = 3 = 1 3 + 2 3 + C = 1 + C ; C = 2 Thus, f ( x ) = 1 3 x 3 + 2 3 x + 2. 47. f ( x ) = 2 x dx = 2ln x + C f = 2 = 2ln1 + C , C = 2, and f ( x ) = x + 2 48. f ( x ) = 1 3 dx = 1 3 x + C f (6) = 3 = 1 3 (6) + C C = 1, and f ( x ) = 1 3 x + 1 49. d dx 1 x + C 1 x 2 ln x , d dx x ln x x + C = (ln x + 1) 1 = ln x , d dx 1 2 (ln x ) 2 + C = ln x x ln x so the answer is b.
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ISM: Calculus & Its Applications, 11e Chapter 6: The Definite Integral 205 50. d dx 2 5 ( x + 1) 5/2 2 3 ( x + 1) 3/2 + C 1/2 (1 ) (1 ) 1( 1 1 xx x x =+ −+ =++ −= + 23 / 2 2 12 ) 21 ) ) 32 1 d C dx x x ⎛⎞ ⋅++ ⎜⎟ ⎝⎠ =++ + ≠+ The answer is a. 51. x y y = F ( x ) y = G ( x ) 2 2 –2 –3 4 6 4 52. y x 53. g ( x ) = f ( x ) + 3, g ( x ) =′ f ( x ), g (5) = 1 4 54. h ( x ) = g ( x ) – f ( x ) = f ( x ) + 2 – f ( x ) = 2; h ( x ) = 0 55. a. The initial height is 256 feet, so C = 256. Thus, s . (96 32 t ) dt = 96 t 16 t 2 + C ( t ) =− 16 t 2 + 96 t + 256 b. Setting s ( t ) = 0, 2 16 96 256 0 tt ++ = ; ( t – 8)( t + 2) = 0. The only solution that is sensible is t = 8 seconds.
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This note was uploaded on 04/04/2010 for the course MATH MATH 16B taught by Professor Unknown during the Spring '10 term at UC Davis.

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chapter 6 - Chapter 6 Exercises 6.1 1. F( x ) = 12 x +C 2 9...

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