227_sample_exam_solutions_spring_2010

227_sample_exam_solutions_spring_2010 - Math 227 Sections 4...

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Math 227 Sections 4 and 5 Fall 2010 Solutions to Sample Exam 1 double-sided sheet of notes. R 1 0 e 2 x p 1 + e 4 x dx using the substitution u = e 2 x : Make sure Solution: u = e 2 x ) du = 2 e 2 x dx , so e 2 x dx = 1 2 du: Also, e 4 x = ( e 2 x ) 2 = u 2 : Thus Z 1 0 e 2 x p 1 + e 4 x dx = Z x =1 x =0 p 1 + u 2 du 2 = 1 2 Z e 2 1 p 1 + u 2 du: 2. (10 points each) Find each of the following. (a) R 6 x x 2 9 dx Solution: 6 x x 2 9 = 6 x ( x + 3)( x 3) = A x + 3 + B x 3 = 18 3 3 x + 3 + 18 3+3 x 3 = 3 x + 3 + 3 x 3 : Thus Z 6 x x 2 9 dx = Z [ 3 x + 3 + 3 x 3 ] dx = 3 ln j x + 3 j + 3 ln j x 3 j + C = 3 ln x 2 9 + C: (b) R 3 1 x 2 ln x dx Solution: apply parts: u = ln x; dv = x 2 dx du = 1 x dx; v = 1 3 x 3 to get Z 3 1
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This note was uploaded on 10/16/2011 for the course MATH 227 taught by Professor Cheung during the Spring '09 term at S.F. State.

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227_sample_exam_solutions_spring_2010 - Math 227 Sections 4...

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