227_Q1_Solutions_Spring_2009[1]

227_Q1_Solutions_Spring_2009[1] - Z e 2 x 3 p 1 + e 2 x dx:...

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Math 227.04 Quiz 1 Key February 3, 2009 Name Instructions. You must show your work for full credit. Please write complete mathematical sentences, display your answers clearly, and erase or cross out any work you wish to be ignored. 1. Transform the following integrals using the substitution u = 1 + x 2 : Do not evaluate the integrals. (a) (3 pts) Z x 3 ln 1 + x 2 ± dx Solution. du = 2 xdx , so dx = du 2 x ; and x 2 = u 1 : Thus Z x 3 ln 1 + x 2 ± dx = Z x 3 [ln u ] du 2 x = Z x 2 2 ln u du = 1 2 Z ( u 1) ln u du: (b) (2 pts) Z 2 0 x 3 ln 1 + x 2 ± dx = 1 2 Z 5 1 ( u 1) ln u du:
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Unformatted text preview: Z e 2 x 3 p 1 + e 2 x dx: Solution. Let u = 1 + e 2 x , so du = 2 e 2 x dx . Thus Z e 2 x 3 p 1 + e 2 x dx = 1 2 Z 3 p 1 + e 2 x (2 e 2 x ) dx = 1 2 Z 3 p udu = 1 2 3 4 u 4 = 3 + C = 3 8 & 1 + e 2 x 4 = 3 + C: Remark. You have the choice of using the "crutch" of solving for " dx " in terms of " du " (as indictated in the solution of 1( a ) ) or determining the conversion factor mentally (as in the solution for 2). 1...
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This note was uploaded on 09/08/2011 for the course MATH 227 taught by Professor Cheung during the Spring '09 term at S.F. State.

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