07_M_GOLD_9004_12_ch06-1 - Chapter 6 The Definite Integral...

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174 Chapter 6 The Definite Integral 6.1 Antidifferentiation 1. () 2 1 2 fx x Fx x C =⇒ = + 2. 29 9( ) f x x F xxC = + 3. 33 1 3 xx fx e Fx e C = + 4. 1 3 e C −− = + 5. f ( x ) = 3 F ( x ) = 3 x + C 6. 2 4( ) 2 fx x Fx =− + 7. 34 4 xd x 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. 23 4 1 4 xxd x x ⋅= 13. 2 21 1 2 2 2 1 2ln 4 x dx x dx dx xdx x C ⎛⎞ += + ⎜⎟ ⎝⎠ + 14. 1 1 ln 777 dx dx x C =⋅ = + 15. 32 52 2 5 d x x d x x C + 16. 12 2 2 4 4 3 4 4 3 x x x x x x d x C C += + + + 17. 2 1 2 3 121 ln 233 d x d x xdx x dx dx x x C −+ =−+ + + 18. 31 3 3 3 3 24 3 3 2 2 7 2 71 1 1 1 3 73 44 x x x d x x x xd x C C −= + + + 19. 3 3 2 ed x e C + 20. x e C + 21. edx ex C 22. 2 7 2 x dx e dx e C e + 23. ( ) 2 2 2 2 1 1 2 2 x x e dx e dx dx ex C C −+= + + 24. .5 .5 2. 5 5 2 1 2 3 5 3 x x e d x x x d x x eC + + + 25. 2 5 25 2 tt t d ke ke e k dt ⎡⎤ = ⇒ =− ⎣⎦ 26. /10 1 0 10 t d ke ke e k dt = 27. 41 1 42 2 x d ke ke e k dx =
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Section 6.1 Antidifferentiation 175 28. (3 1) (3 1) 31 (3 1) (3 1) 3 4 34 3 xx x dk d ke ke dx dx e ke e k −+ + ⎡⎤ == ⎢⎥ ⎣⎦ −=⇒ = 29. 12 22 (5 7) (5 7) (5) 5( 5 7 ) ( 5 7 ) 1 5 d kx dx x k −− −= =− = 30. 3/2 1/2 3 (1 ) ) ) 2 2 3 d x dx k += + = + = 31. 1 ln 4 ( 1) 44 4 1 k dx x x x k ⎡− = ==⇒ 32. 3 3 4 (8 ) (8 ) 3( 8 ) (1 ) 3 ( 8) 7 ( 8) 7 3 d dx dx x x k =−= −⇒ = 33. 54 (3 2) 5 (3 2) (3) 15 (3 2) (3 2) 1 15 d dx x k + =+ = + = 34. 43 3 (2 1) 8 (2 1 8 d x dx k = = 35. 3 [l n2 ] 3 k dx x x = ⇒= ++ 36. ln 2 3 ( 3) 23 35 5 23 23 3 dx x k k = = 37. 5/2 2 () 5 ft t f t t C =⇒= + 38. 4 () 4ln6 6 ft f t t C t + + + 39. () 0 f =⇒ = 40. 2 15 5 7 7 32 t f t t t tC =−−⇒ = − −+ 41. .2 .2 .2 0 () . 5 2 . 5 (0) 0 2.5 0 2.5 fx e e C fe C C −⋅ = + + =⇒ = Thus, .2 2 . 5 2 . 5 . x e + 42. 2 20 () 2 (0) 1 0 1 0 xe fx x e C C C =− ⇒ =+ + =⇒ + Thus, 2 . x 43. 2 1 2 fx x fx x C = + 2 1 (0) 3 0 3 3 2 fC C =⇒⋅ + =⇒ = Thus, 2 1 3 2 x . 44. 1/3 4/3 () 8 () 6 (1) 4 6 1 4 2 x x C C = + =⇒⋅ Thus, 2 x . 45. 2 1 3 2 (4) 0 4 4 0 3 8 84 0 33 fx x x xC CC = + + ++ =⇒ ⋅++ = ⇒ =− Thus, 8 x x . 46. 2 3 / 2 3 1 1 3 2 fx x x x x C C = + + =⇒⋅ +⋅ Thus, / 2 2 x x + . 47. 2 2 l n d x x ==+ 2 2 ln 1 2 2 C Thus, l n 2 . x 48. 11 d x + 1 (6) 3 (6) 3 1 3 C Thus, 1 1 . 3 x
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176 Chapter 6 The Definite Integral 49. () 2 11 ln ln (ln 1) 1 ln d Cx dx x x d xxxC x x dx ⎛⎞ += −≠ ⎜⎟ ⎝⎠ −+ = +−= 2 1l n (ln ) ln 2 dx xC x dx x The answer is (b). 50. 5/2 3/2 1/2 22 (1 ) ) 53 ) ) 1( 1 1) 1 d xx C dx x x +− ++ =+ −+ =++ = + 23 / 2 2 12 ) 21 ) ) 1 32 d C dx x x ⋅+ + + + + The answer is (a). 51. 52. 53. g ( x ) = f ( x ) + 3 1 () ( 5 ) ( 5 ) 4 gx f x g f =⇒== ′′ 54. h ( x ) = g ( x ) – f ( x ) = f ( x ) + 2 – f ( x ) = 2 () 0 hx = 55. a. 2 (96 32 ) 96 16 td t t t C −= + The initial height is 256 feet, so C = 256. Thus, 2 ( ) 16 96 256 st t t =− + + . b. Setting s ( t ) = 0, 2 16 96 256 0 tt 2 61 60 −−=⇒ ( t – 8)( t + 2) = 0. The only solution that is sensible is t = 8 seconds.
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This note was uploaded on 04/24/2010 for the course MATH 9374 taught by Professor Smith during the Spring '10 term at University of California, Berkeley.

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07_M_GOLD_9004_12_ch06-1 - Chapter 6 The Definite Integral...

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