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140Bquiz14asols

# 140Bquiz14asols - 1 MATH 140B QUIZ#14 Solutions 1 Let u = 3...

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1 MATH 140B QUIZ#14 Solutions Dec 14, 2007 1. Let u = 3 - x so du = - dx . When x = - 1 , u = 4 and when x = 1 , u = 2 . Therefore, integraldisplay 1 - 1 dx 3 - x = - integraldisplay 2 4 1 u du = - ln u bracketrightbig 2 4 = - (ln2 - ln4) = ln4 - ln2 = 2ln2 - ln2 = ln2 . Therefore, C) is the answer. 2. Let u = x 3 - 1 so du = 3 x 2 dx . When x = 0 , u = - 1 and when x = 1 , u = 0 . Therefore, integraldisplay 1 0 9 x 2 ( x 3 - 1) 5 dx = 3 integraldisplay 0 - 1 u 5 du = 1 2 u 6 bracketrightbigg 0 - 1 = 1 2 bracketleftbig 0 - ( - 1) 6 bracketrightbig = - 1 2 . Therefore, E) is the answer. 3. Let u = x 2 - 2 so du = 2 x dx and 2 du = 4 x dx. When x = 0 , u = - 2 and when x = 1 , u = - 1. Hence integraldisplay 1 0 4 x ( x 2 - 2) 3 dx
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