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Unformatted text preview: kim (tk5895) HW06 Henry (54974) 1 This printout should have 18 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 ln 2 5 e 2 x 4 e x dx . 1. I = 4 2. I = 5 3. I = 3 correct 4. I = 6 5. I = 7 Explanation: After division 5 e 2 x 4 e x = 5 e x 4 e x . Thus I = integraldisplay ln 2 braceleftBig 5 e x 4 e x bracerightBig dx = bracketleftBig 5 e x + 4 e x bracketrightBig ln 2 . On the other hand, e ln a = a, e ln a = 1 a . Consequently, I = parenleftBig 10 + 2 parenrightBig (5 + 4) = 3 . 002 10.0 points Determine the integral I = integraldisplay 3 x ( x + 4) dx . 1. I = 3 x ln  x + 4  + C 2. I = 3 ln  x + 4  + C 3. I = 6 x ln  x + 4  + C 4. I = 6 ln  x + 4  + C correct 5. I = 6 x ln  x + 4  + C 6. I = 3 x ln  x + 4  + C Explanation: Set u 2 = x . Then 2 u du = dx , so I = 6 integraldisplay 1 u + 4 du = 2 ln  u + 4  + C. Consequently, I = 6 ln  x + 4  + C with C an arbitrary constant. 003 10.0 points Evaluate the definite integral I = integraldisplay e 1 parenleftBig 2 x + 1 2 x parenrightBig 2 dx . 1. I = e 2 + 2 e 5 2 correct 2. I = 2 e 2 4 e + 3 3. I = 2 e 2 + 4 e 5 4. I = 1 2 e 2 + 2 e 3 2 5. I = 1 2 e 2 2 e + 5 2 6. I = e 2 2 e + 3 2 Explanation: kim (tk5895) HW06 Henry (54974) 2 After expansion, parenleftBig 2 x + 1 2 x parenrightBig 2 = 2 x + 2 + 1 2 x . Thus I = integraldisplay e 1 parenleftBig 2 x + 2 + 1 2 x parenrightBig dx = bracketleftBig x 2 + 2 x + 1 2 ln x bracketrightBig e 1 = e 2 + 2 e + braceleftBig 1 2 3 bracerightBig . Consequently, I = e 2 + 2 e 5 2 since ln e = 1 and ln1 = 0. keywords: definite integral, log integral, ex ponential number, properties of logs, 004 10.0 points Evaluate the integral I = integraldisplay 1 x 3 + 2 x 2 dx . 1. I = 1 4 ln 5 3 correct 2. I = ln 5 3 3. I = 1 2 ln 3 4. I = 1 2 ln 3 2 5. I = 1 4 ln 3 2 6. I = ln 3 Explanation: Set u = 3 + 2 x 2 ; then du = 4 x dx while x = 0 = u = 3 x = 1 = u = 5 . In this case, I = 1 4 integraldisplay 5 3 1 u du = 1 4 bracketleftBig ln  u  bracketrightBig 5 3 . Consequently, I = 1 4 (ln 5 ln 3) = 1 4 ln 5 3 . 005 10.0 points Evaluate the integral I = integraldisplay e 2 e 4 x ln x dx. 1. I = 4 2. I = 4(ln 2 1) 3. I = ln 3 4. I = 4 ln2 correct 5. I = 4( e 2 e ) Explanation: Since the integrand is of the form f ( x ) f ( x ) , f ( x ) = ln x up to a constant, the substitution u = ln x is suggested. For then du = 1 x dx, while x = e = u = 1 , x = e 2 = u = 2 . In this case I = 4 integraldisplay 2 1 1 u du = 4 bracketleftBig ln u bracketrightBig 2 1 . kim (tk5895) HW06 Henry (54974) 3 Consequently, I = 4 ln2 ....
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 Fall '08
 Cepparo
 Calculus

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