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Unformatted text preview: Version 093 EXAM 2 cheng (58520) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay 4 3 2 ( x 2)(8 x ) dx . 1. I = 1 6 ln parenleftbigg 7 4 parenrightbigg 2. I = 1 3 ln parenleftbigg 7 4 parenrightbigg 3. I = 1 3 ln parenleftbigg 5 2 parenrightbigg correct 4. I = 1 6 ln parenleftbigg 5 2 parenrightbigg 5. I = 2 ln parenleftbigg 5 2 parenrightbigg 6. I = 2 ln parenleftbigg 7 4 parenrightbigg Explanation: To find A, B so that 2 ( x 2)(8 x ) = A x 2 + B 8 x , we first bring the right hand side to a common denominator. In this case, 2 ( x 2)(8 x ) = A (8 x ) + B ( x 2) ( x 2)(8 x ) , and so A (8 x ) + B ( x 2) = 2 . To find the values of A, B , we can make par ticular choices of x since the equation holds for all x . When x = 2, therefore, A = 1 3 , and when x = 8, B = 1 3 . Thus I = integraldisplay 4 3 1 3 parenleftBig 1 x 2 + 1 8 x parenrightBig dx. Hence after integration, I = bracketleftbigg 1 3 { ln( x 2) ln(8 x ) } bracketrightbigg 4 3 = 1 3 bracketleftbigg ln parenleftbigg x 2 8 x parenrightbiggbracketrightbigg 4 3 . Consequently, I = 1 3 ln parenleftbigg 5 2 parenrightbigg . 002 10.0 points Evaluate the definite integral I = integraldisplay / 2 sin x 1 + cos 2 x dx . 1. I = 1 2. I = 1 2 3. I = 4. I = 1 2 5. I = 1 4 6. I = 1 4 correct Explanation: Let u = cos x . Then du = sin x dx , while x = 0 = u = 1 , x = 2 = u = 0 . In this case I = integraldisplay 1 1 1 + u 2 du = integraldisplay 1 1 1 + u 2 du = bracketleftBig tan 1 u bracketrightBig 1 . Version 093 EXAM 2 cheng (58520) 2 Consequently, I = 1 4 . 003 10.0 points Evaluate the definite integral I = integraldisplay 1 (9 x 2 4) e x dx . 1. I = 5 e 14 correct 2. I = 5 e + 14 3. I = 14 e 5 4. I = 5( e 1) 5. I = 14 e + 5 Explanation: After Integration by Parts once, integraldisplay 1 (9 x 2 4) e x dx = bracketleftBig (9 x 2 4) e x bracketrightBig 1 18 integraldisplay 1 xe x dx. To evaluate this last integral we Integrate by Parts once again. For then 18 integraldisplay 1 xe x dx = bracketleftBig 18 xe x bracketrightBig 1 18 integraldisplay 1 e x . Consequently, I = bracketleftBig (9 x 2 4 18 x + 18) e x bracketrightBig 1 , and so I = 5 e 14....
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This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin
 Calculus

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