M408L_EXAM02 - Version 104 EXAM 2 Berg (58460) 1 This...

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Unformatted text preview: Version 104 EXAM 2 Berg (58460) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay 5 4 7 ( x- 3)(9- x ) dx . 1. I = 7 6 ln parenleftbigg 5 2 parenrightbigg correct 2. I = 7 6 ln parenleftbigg 7 4 parenrightbigg 3. I = 7 ln parenleftbigg 7 4 parenrightbigg 4. I = 1 6 ln parenleftbigg 5 2 parenrightbigg 5. I = 1 6 ln parenleftbigg 7 4 parenrightbigg 6. I = 7 ln parenleftbigg 5 2 parenrightbigg Explanation: To find A, B so that 7 ( x- 3)(9- x ) = A x- 3 + B 9- x , we first bring the right hand side to a common denominator. In this case, 7 ( x- 3)(9- x ) = A (9- x ) + B ( x- 3) ( x- 3)(9- x ) , and so A (9- x ) + B ( x- 3) = 7 . To find the values of A, B , we can make par- ticular choices of x since the equation holds for all x . When x = 3, therefore, A = 7 6 , and when x = 9, B = 7 6 . Thus I = integraldisplay 5 4 7 6 parenleftBig 1 x- 3 + 1 9- x parenrightBig dx. Hence after integration, I = bracketleftbigg 7 6 { ln( x- 3)- ln(9- x ) } bracketrightbigg 5 4 = 7 6 bracketleftbigg ln parenleftbigg x- 3 9- x parenrightbiggbracketrightbigg 5 4 . Consequently, I = 7 6 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 4 2. I = 1 3. I = 1 2 4. I = 5. I = 1 2 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 104 EXAM 2 Berg (58460) 2 Consequently, I = 1 4 . 003 10.0 points Evaluate the definite integral I = integraldisplay 1 (8 x 2- 3) e x dx . 1. I = 5( e- 1) 2. I = 13 e- 5 3. I = 13 e + 5 4. I = 5 e + 13 5. I = 5 e- 13 correct Explanation: After Integration by Parts once, integraldisplay 1 (8 x 2- 3) e x dx = bracketleftBig (8 x 2- 3) e x bracketrightBig 1- 16 integraldisplay 1 xe x dx. To evaluate this last integral we Integrate by Parts once again. For then 16 integraldisplay 1 xe x dx = bracketleftBig 16 xe x bracketrightBig 1- 16 integraldisplay 1 e x . Consequently, I = bracketleftBig (8 x 2- 3- 16 x + 16) e x bracketrightBig 1 , and so I = 5 e- 13....
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M408L_EXAM02 - Version 104 EXAM 2 Berg (58460) 1 This...

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