# ch07 - 7.1 SOLUTIONS 323 CHAPTER SEVEN Solutions for...

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Unformatted text preview: 7.1 SOLUTIONS 323 CHAPTER SEVEN Solutions for Section 7.1 1. 5 x 2. 5 2 x 2 3. 1 3 x 3 4. 1 3 t 3 + 1 2 t 2 5. x 5 5 . 6. t 8 8 + t 4 4 . 7. 5 q 3 3 . 8. 6( x 4 4 ) + 4 x = 3 x 4 2 + 4 x . 9. We break the antiderivative into two terms. Since y 3 is an antiderivative of 3 y 2 and- y 4 / 4 is an antiderivative of- y 3 , an antiderivative of 3 y 2- y 3 is y 3- y 4 4 . 10. 2 3 t 3 + 3 4 t 4 + 4 5 t 5 11. x 3 + 5 x . 12. 2 x 3- 4 x 2 + 3 x . 13. t 3 + 7 t 2 2 + t . 14. 2 3 z 3 2 15. 5 2 x 2- 2 3 x 3 2 16. Since ( √ z ) 3 = z 3 / 2 , an antiderivative of ( √ z ) 3 is z (3 / 2)+1 (3 / 2) + 1 = 2 5 z 5 / 2 . 17. t 4 4- t 3 6- t 2 2 18. ln | z | 19.- 1 t 20. y 5 5 + ln | y | 21. F ( x ) = x 7 7- 1 7 ( x- 5- 5 ) + C = x 7 7 + 1 35 x- 5 + C 22. P ( y ) = ln | y | + y 2 / 2 + y + C 23. e- 3 t- 3 =- e- 3 t 3 . 24.- cos t 25. G ( t ) = 5 t + sin t + C 26. G ( θ ) =- cos θ- 2 sin θ + C 27. f ( x ) = 3 , so F ( x ) = 3 x + C . F (0) = 0 implies that 3 · 0 + C = 0 , so C = 0 . Thus F ( x ) = 3 x is the only possibility. 324 Chapter Seven /SOLUTIONS 28. f ( x ) =- 7 x , so F ( x ) =- 7 x 2 2 + C . F (0) = 0 implies that- 7 2 · 2 + C = 0 , so C = 0 . Thus F ( x ) =- 7 x 2 / 2 is the only possibility. 29. f ( x ) = x 2 , so F ( x ) = x 3 3 + C . F (0) = 0 implies that 3 3 + C = 0 , so C = 0 . Thus F ( x ) = x 3 3 is the only possibility. 30. f ( x ) = x 1 / 2 , so F ( x ) = 2 3 x 3 / 2 + C . F (0) = 0 implies that 2 3 · 3 / 2 + C = 0 , so C = 0 . Thus F ( x ) = 2 3 x 3 / 2 is the only possibility. 31. f ( x ) = 2 + 4 x + 5 x 2 , so F ( x ) = 2 x + 2 x 2 + 5 3 x 3 + C . F (0) = 0 implies that C = 0 . Thus F ( x ) = 2 x + 2 x 2 + 5 3 x 3 is the only possibility. 32. Since d dx ( e x ) = e x , we take F ( x ) = e x + C. Now F (0) = e + C = 1 + C = 0 , so C =- 1 and F ( x ) = e x- 1 . 33. 3 x 2 2 + C 34. 2 t 2 + 7 t + C 35. 2 x 3 + C . 36. t 13 13 + C . 37. x 4 4- x 2 2 + C . 38. x 3 3 + x + C . 39. x 4 4 + 2 x 2 + 8 x + C . 40. 5 e z + C 41. q 3 3 + 5 q 2 2 + 2 q + C 42. x 6 6- 3 x 4 + C 43. 4 x 3 / 2 + C 44. x 3 3 + 2 x 2- 5 x + C 45.- 5 t- 3 t 2 + C 46. e 2 t 2 + C . 47. x 2 2 + 2 x 1 / 2 + C 48. t 3 3- 3 t 2 + 5 t + C . 49.- 1 . 05 e- . 05 t + C =- 20 e- . 05 t + C . 50. 2 x 4 + ln | x | + C . 51. Since d dx ( e- 3 t ) =- 3 e- 3 t , we have Z e- 3 t dt =- 1 3 e- 3 t + C. 52. sin θ + C 53.- cos t + C 54.- 150 e- . 2 t + C 55. 25 e 4 x + C 56. 2 x 2 + 2 e x + C 7.1 SOLUTIONS 325 57. 5 sin x + 3 cos x + C 58.- 1 3 cos(3 x ) + C 59. 1 4 sin(4 x ) + C 60. 2 sin(3 x ) + C 61. 10 x- 4 cos(2 x ) + C 62.- 6 cos(2 x ) + 3 sin(5 x ) + C 63. An antiderivative is F ( x ) = x 3 3 + x + C . Since F (0) = 5 , we have 5 = 0 + C , so C = 5 . The answer is F ( x ) = x 3 / 3 + x + 5 ....
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## This note was uploaded on 04/21/2008 for the course MATH 227 taught by Professor ? during the Winter '08 term at University of Cincinnati.

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ch07 - 7.1 SOLUTIONS 323 CHAPTER SEVEN Solutions for...

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