Pre-Calc Homework Solutions 363

Pre-Calc Homework Solutions 363 - Chapter 8 Review 68 For x...

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66. x 5 sin u , dx 5 cos u d u , 2 } p 2 } # u # } p 2 } 1 2 x 2 5 cos 2 u E } (1 4 2 x 2 x d 2 x ) 3/2 } 5 E } 4 si ) n co 2 s u 3 c u o ) s u } d u 5 E } 4(1 c 2 os c 2 o u s 2 u ) } d u 5 E (4 sec 2 u 2 4) d u 5 4 tan u 2 4 u 1 C 5 } ˇ 1 w 4 2 w x x w 2 w } 2 4 sin 2 1 x 1 C Use Figure 8.18(b) with a 5 1. 67. For x $ 0, y $ 0 on (0, 1]. Volume 5 E 1 0 p ( 2 ln x ) 2 dx 5 p E 1 0 (ln x ) 2 dx 5 p lim b →0 1 E 1 b (ln x ) 2 dx Evaluate E (ln x ) 2 dx by using integration by parts. u 5 (ln x ) 2 dv 5 dx du 5 } 2 l x n x } dx v 5 x E (ln x ) 2 dx 5 x (ln x ) 2 2 E 2 ln x dx Evaluate E 2 ln x dx by using integration by parts. u 5 2 ln x dv 5 dx du 5 } 2 x } dx v 5 x E 2 ln x dx 5 2 x ln x 2 E 2 dx 5 2 x ln x 2 2 x 1 C E (ln x ) 2 dx 5 x (ln x ) 2 2 2 x ln x 1 2 x 1 C Area 5 p lim b →0 1 3 x (ln x ) 2 2 2 x ln x 1 2 x 4 5 p lim b →0 1 [2 2 b (ln b ) 2 1 2 b ln b 2 2 b ] 5 2 p 2 lim b →0 1 } p ( 1 ln / b b ) 2 } 1 2 lim b →0 1 } p 1 l / n b b } 5 2 p 2 lim b →0 1 } 2 p ( 2 ln 1 b / b )( 2 1/ b ) } 1 2 lim b →0 1 } 2 p 1 / / b b 2 } 5 2 p 2 lim b →0 1 } 2 p 2 ( 1 ln / b b ) } 1 2 lim b →0 1 ( 2 p b ) 5 2 p 2 lim b →0 1 } 2 1 p / b / 2 b } 1 2 p 2 lim b →0 1 2 p b 5 2 p 68. For x $ 0, y $ 0 on [0, ). Area 5 E 0 xe 2 x dx 5 lim b E b 0 xe 2 x dx Evaluate E xe 2 x dx by using integration by parts. u 5 x dv 5 e 2 x dx du 5 dx v 5 2 e 2 x E xe 2 x dx 5 2 xe 2 x 1 E e 2 x dx 5 2 xe 2 x 2 e 2 x 1 C
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