A1W08-138Sol

A1W08-138Sol - Math 138 Assignment 1 Solutions Fall 2008 1....

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Unformatted text preview: Math 138 Assignment 1 Solutions Fall 2008 1. Evaluate each of the following indefinite integrals: a) Z x 2 ( x 3 + 5) 9 dx Solution Let u = x 3 + 5, then du = 3 x 2 dx , and by substitution, we obtain: Z ( u ) 9 du 3 = Z 1 3 u 9 du = 1 30 u 10 + C = 1 30 ( x 3 + 5) 10 + C . b) Z e x sin( e x ) dx Solution Let u = e x , then du = e x dx , and by substitution, we obtain: Z e x sin( e x ) dx = Z sin u du =- cos u + C =- cos( e x ) + C . c) Z tan- 1 x 1 + x 2 dx Solution Let u = tan- 1 x , then du = 1 1+ x 2 dx , and, by substitution, we obtain: Z tan- 1 x 1 + x 2 dx = Z u du = 1 2 u 2 + C = 1 2 (tan- 1 x ) 2 + C . d) Z x 1 + x 4 dx Solution Let u = x 2 , then du = 2 xdx , and, by substitution, we obtain: Z x 1 + x 4 dx = Z 1 1 + u 2 du 2 = 1 2 tan- 1 u + C = 1 2 tan- 1 ( x 2 ) + C . Math 138 Assignment 1 Solutions Fall 2008 2. Evaluate Z 1 2 sin- 1 x 1- x 2 dx . Solution 1 Let u = sin- 1 x , then du = 1 1- x 2 , dx . Also u = 0 when x = 0, and u = 6 when x = 1 2 . Sub- stitution leads to:...
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A1W08-138Sol - Math 138 Assignment 1 Solutions Fall 2008 1....

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