227-04T1solnsSp11

# 227-04T1solnsSp11 - Math 227.04 Exam 1 Solutions Z =4...

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Math 227.04 Exam 1 Solutions Z 4 0 sin(2 x ) 1 + cos 2 (2 x ) dx using the substitution u = cos(2 x ) : Let u = cos(2 x ) , then du = 2 sin(2 x ) : Thus 4 Z 0 sin(2 x ) 1 + cos 2 (2 x ) dx = 1 2 Z cos( 2) cos(0) du 1 + u 2 = 1 2 Z 0 1 du 1 + u 2 1 = 1 2 Z 1 0 du 1 + u 2 2. (a) Z x p 4 x 2 dx can be solved using the substitution u = 4 x 2 , du = 2 xdx: Thus Z x p 4 x 2 dx = Z 1 2 u 1 = 2 du = 1 2 ± 2 3 u 3 = 2 + C = 1 3 4 x 2 ± 3 = 2 + C: (b) Z x p x 2 dx can be solved using the substitution u = x 2 ; so x = u + 2 and dx = du: Thus Z x p x 2 dx = Z ( u + 2) u 1 = 2 du = Z ² u 3 = 2 + 2 u 1 = 2 ³ du = 2 5 u 5 = 2 + 4 3 u 3 = 2 + C = 2 5 ( x 2) 5 = 2 + 4 3 ( x 2) 3 = 2 + C: (c) Z 2 1 x 3 ln x dx can be found using the method of Parts with u = ln x; du = 1 x dx and dv = x 3 dx; v = 1 2 x 2 : Thus 2 Z

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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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227-04T1solnsSp11 - Math 227.04 Exam 1 Solutions Z =4...

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