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EXAM2 - Version 093 EXAM 2 cheng(58520 This print-out...

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Version 093 – EXAM 2 – cheng – (58520) 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 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 parenrightbigg bracketrightbigg 4 3 . Consequently, I = 1 3 ln parenleftbigg 5 2 parenrightbigg . 002 10.0 points Evaluate the definite integral I = integraldisplay π/ 2 0 sin x 1 + cos 2 x dx . 1. I = 1 2. I = 1 2 3. I = π 4. I = 1 2 π 5. I = 1 4 6. I = 1 4 π correct Explanation: Let u = cos x . Then du = - sin x dx , while x = 0 = u = 1 , x = π 2 = u = 0 . In this case I = - integraldisplay 0 1 1 1 + u 2 du = integraldisplay 1 0 1 1 + u 2 du = bracketleftBig tan 1 u bracketrightBig 1 0 .

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Version 093 – EXAM 2 – cheng – (58520) 2 Consequently, I = 1 4 π . 003 10.0 points Evaluate the definite integral I = integraldisplay 1 0 (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 0 (9 x 2 - 4) e x dx = bracketleftBig (9 x 2 - 4) e x bracketrightBig 1 0 - 18 integraldisplay 1 0 xe x dx. To evaluate this last integral we Integrate by Parts once again. For then 18 integraldisplay 1 0 xe x dx = bracketleftBig 18 xe x bracketrightBig 1 0 - 18 integraldisplay 1 0 e x . Consequently, I = bracketleftBig (9 x 2 - 4 - 18 x + 18) e x bracketrightBig 1 0 , and so I = 5 e - 14. 004 10.0 points Evaluate the definite integral I = integraldisplay π/ 3 0 sec x tan x 3 + 2 sec x dx . 1. I = 2 ln 7 6 2. I = 1 2 ln 5 3 3. I = - 2 ln 7 6 4. I = - 2 ln 5 3 5. I = 1 2 ln 7 5 correct 6. I = - 1 2 ln 7 5 Explanation: Since d dx (sec x ) = sec x tan x , use of the substitution u = 3 + 2 sec x is suggested. For then du = 2 sec x tan xdx , while x = 0 = u = 5 , x = π 3 = u = 7 .
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