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# Business Calc Homework w answers_Part_90 - 446 Cumulative...

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60. Use integration by parts. u 5 e 2 x dv 5 cos 2 x dx du 5 2 e 2 x dx v 5 } 1 2 } sin 2 x E e 2 x cos 2 x dx 5 } 1 2 } e 2 x sin 2 x 1 E } 1 2 } e 2 x sin 2 x dx Now let u 5 e 2 x dv 5 } 1 2 } sin 2 x dx du 5 2 e 2 x dx v 5 2 } 1 4 } cos 2 x Then E e 2 x cos 2 x dx 5 } 1 2 } e 2 x sin 2 x 2 } 1 4 } e 2 x cos 2 x 2 } 1 4 } E e 2 x cos 2 x dx so E e 2 x cos 2 x dx 5 } e 2 5 x } (2 sin 2 x 2 cos 2 x ) 1 C 61. } x 2 2 x 1 5 x 2 2 6 } 5 } ( x 1 x 1 1 )( x 2 2 6) } 5 } x 1 A 1 } 1 } x 2 B 6 } x 1 2 5 A ( x 2 6) 1 B ( x 1 1) 5 ( A 1 B ) x 1 ( B 2 6 A ) Solving A 1 B 5 1, B 2 6 A 5 2 yields A 5 2 } 1 7 } , B 5 } 8 7 } so } x 2 2 x 1 5 x 2 2 6 } 5 } 7( x 8 2 6) } 2 } 7( x 1 1 1) } . Then E } x 2 2 x 1 5 x 2 2 6 } dx 5 E 1 } 7( x 8 2 6) } 2 } 7( x 1 1 1) } 2 dx 5 } 8 7 } ln ) x 2 6 ) 2 } 1 7 } ln ) x 1 1 ) 1 C 5 } 1 7 } ln } ( x ) x 2 1 6 1 ) ) 8 } 1 C 62. Area < } 5 2 } [3 1 2(8.3) 1 2(9.9) 1 1 2(8.3) 1 3] 5 359; Volume < 25 3 359 5 8975 ft 3 63. y 5 2 ( t 1 1) 2 1 2 } 1 2 } e 2 2 t 1 C ; y (0) 5 2 1 2 } 1 2 } 1 C 5 2, so C 5 } 7 2 } and y 5 2 } t 1 1 1 } 2 } 1 2 } e 2 2 t 1 } 7 2 } . 64. y 9 5 2 } 1 2 } cos 2 u 2 sin u 1 C 1 , and y 9 1 } p 2 } 2 5 0 y 9 5 2 } 1 2 } cos 2 u 2 sin u 1 } 1 2 } . y 5 2 } 1 4 } sin 2 u 1 cos u 1 } 1 2 } u 1 C 2 , and y 1 } p 2 } 2 5 0 y 5 2 } 1 4 } sin 2 u 1 cos u 1 } 1 2 } u 2 } p 4 } 65. Use integration by parts. u 5 x 2 dv 5 sin x dx du 5 2 x dx v 5 2 cos x E x 2 sin x dx 5 2 x 2 cos x 1 E 2 x cos x dx Now let u 5 x dv 5 2 cos x dx du 5 dx v 5 2 sin x E x 2 sin x dx 5 2 x 2 cos x 1 2 x sin x 2 E 2 sin x dx 52 x 2 cos x 1 2 x sin x 1 2 cos x 1 C 5 (2 2 x 2 ) cos x 1 2 x sin x 1 C The graph of the slope field of the differential equation } d d y x } 5 x 2 sin x and the antiderivative y 5 (2 2 x 2 ) cos x 1 2 x sin x is shown below. [ 2 5, 5] by [ 2 10, 10] 66. Use integration by parts. u 5 x dv 5 e x dx du 5 dx v 5 e x E x e x dx 5 xe x 2 E e x dx 5 xe x 2 e x 1 C 5 e x ( x 2 1) 1 C Confirm by differentiation: } d d x } [ e x ( x 2 1) 1 C ] 5 e x 1 ( x 2 1) e x 5 xe x 67. (a) y 5 Ce kt , with 6,000 5 Ce k (2) and 10,000 5 Ce k (5) . Then } 1 6 0 , , 0 0 0 0 0 0 } 5 e k (5 2 2) , so } 5 3 } 5 e 3 k and therefore k 5 < 0.170. Furthermore, C 5 } 6, e 0 2 0 k 0 } < 4268. The approximate number of bacteria is given by y 5 4268 e 0.170 t . (b) About 4268 ln 1 } 5 3 } 2 } 3 446 Cumulative Review

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68. Let t be the time in minutes where t 5 0 represents right now, and let T ( t ) be the number of degrees above room temperature. Then we may write T ( t ) 5 T 0 e 2 kt where T (0) 5 50 and T ( 2 15) 5 65, giving T 0 5 50 and k 5 } 1 1 5 } ln } 1 1 3 0 } < 0.0175. (a) 50 e 2 k (120) < 6.13°C above room temperature. (b) Solving 5 5 50 e 2 kt gives t 5 } ln 2 0 k .1 } < 131.6 minutes, or about 2 hours and 12 minutes from now. 69. } d d y x } 5 0.08 y 1 1 2 } 50 y 0 } 2 } y (5 5 0 0 0 0 2 dy y ) } 5 0.08 dx } (5 y 0 ( 0 50 2 0 y 2 ) 1 y ) y } dy 5 0.08 dx 1 } 1 y } 1 } 500 1 2 y } 2 dy 5 0.08 dx Integrate both sides. ln ) y ) 2 ln ) 500 2 y ) 5 0.08 x 1 C 1 } 500 y 2 y } 5 C 2 e 0.08 x y ? (1 1 C 2 e 0.08 x ) 5 500 C 2 e 0.08 x y 5 } 1 1 C 50 e 0 2 0.08 x } 70. } d d y x } 5 ( y 2 4)( x 1 3) } y d 2 y 4 } 5 ( x 1 3) dx E } y d 2 y 4 } 5 E ( x 1 3) dx ln ) y 2 4 ) 5 } x 2 2 } 1 3 x 1 C 1 y 2 4 5 e C 1 e ( x 2 /2) 1 3 x 1 4 y 5 Ce ( x 2 /2) 1 3 x 1 4 71. Use EULERT.
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