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Unformatted text preview: Math 1B Quiz 2 Solutions Section 107 September 12, 2009 1. (4 points) R 3 2 4 x x 2 dx Method 1: Complete the square to obtain R 3 2 p 4 ( x 2) 2 dx . Then substitute u = x 2, du = dx to obtain R 1 4 u 2 du . (For the limits of integration, x = 2 when u = 0; and x = 3 when u = 1). Now use the trig substitution u = 2sin , du = 2cos d for / 2 / 2 to obtain Z 1 p 4 u 2 du = Z / 6 p 4 4sin 2 (2cos ) d. For the limits of integration, u = 0 when = 0; and u = 1 = 2sin when = / 6. We have p 4 4sin 2 = 2  cos  = 2cos since cos is positive for / 2 / 2. Therefore, Z / 6 p 4 4sin 2 (2cos ) d = Z / 6 4cos 2 d = 2 Z / 6 1 + cos2 d by the trig identity cos 2 = 1 2 (1 + cos2 ). Now we can evaluate the integral: Z 3 2 p 4 x x 2 dx = 2 Z / 6 1 + cos2 d = 2 + sin2 2  / 6 = 2 6 + sin( 3 ) 2 = 3 + 3 2 ....
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This note was uploaded on 03/25/2011 for the course MATH 1B taught by Professor Reshetiken during the Fall '08 term at University of California, Berkeley.
 Fall '08
 Reshetiken
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