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Pre-Calc Homework Solutions 446

# Pre-Calc Homework Solutions 446 - 446 Cumulative Review 65...

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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.
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