sol16-7 - = 1 Quite a surprise, no? but . .. you know that...

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SOLUTIONS FOR HOMEWORK PROBLEMS 16.7 MATH 132 - WI00 4 R 5 0 ( - 3 x ) dx = ( - 3) x 2 2 | 5 0 = ( - 3) 5 2 2 - ( - 3) 0 2 2 = ( - 3) * 12 . 5 = - 37 . 5 12 R 2 1 x - 2 2 dx = 1 2 R 2 1 x - 2 dx = 1 2 x - 1 - 1 | 2 1 = 1 2 ( 2 - 1 - 1 - 1 - 1 - 1 ) = 1 2 ( - 1 2 + 1) = 1 2 * 1 2 = 1 4 24 R e - 1 0 1 x +1 dx = ln ( x + 1) | e - 1 0 = ln e - ln 1 = 1 - 0 = 1 In case you’re not sure how to get ln there, the trick is to use u=x+1 as your substitution, and work on from here. 30 R 1 - 1 q p q 2 + 3 dq Use substitution u = q 2 + 3, du = 2 q dq etc . .. you’ll get: 1 2 ( q 2 +3) 3 2 3 2 | 1 - 1 = ... = 0 40 R 4 3 e ln x x dx = R 4 3 x x dx = R 4 3 1 dx = x | 4 3
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Unformatted text preview: = 1 Quite a surprise, no? but . .. you know that e ln x = x and ln e x = x (ln and e annihilate each other when composed together . .. and they give you x !). You can of course use substitution here, namely say u = ln x , du = 1 x dx ... but this takes longer a bit. Date : 02/29/2000. 1...
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