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Unformatted text preview: So (dds785) HW06 Radin (54915) 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 4 3 e 2 x 4 e x dx . 1. I = 7 2. I = 13 2 3. I = 11 2 4. I = 6 correct 5. I = 15 2 Explanation: After division 3 e 2 x 4 e x = 3 e x 4 e x . Thus I = integraldisplay ln 4 braceleftBig 3 e x 4 e x bracerightBig dx = bracketleftBig 3 e x + 4 e x bracketrightBig ln 4 . On the other hand, e ln a = a, e ln a = 1 a . Consequently, I = parenleftBig 12 + 1 parenrightBig (3 + 4) = 6 . 002 10.0 points Determine the integral I = integraldisplay 2 x ( x + 6) dx . 1. I = 2 x ln  x + 6  + C 2. I = 4 x ln  x + 6  + C 3. I = 4 ln  x + 6  + C correct 4. I = 2 ln  x + 6  + C 5. I = 4 x ln  x + 6  + C 6. I = 2 x ln  x + 6  + C Explanation: Set u 2 = x . Then 2 u du = dx , so I = 4 integraldisplay 1 u + 6 du = 2 ln  u + 6  + C. Consequently, I = 4 ln  x + 6  + C with C an arbitrary constant. 003 10.0 points Evaluate the definite integral I = integraldisplay e 1 parenleftBig x 1 x parenrightBig 2 dx . 1. I = e 2 2 e + 3 2 2. I = e 2 + 2 e 5 2 3. I = 1 2 e 2 + 2 e 3 2 4. I = 2 e 2 + 4 e 5 5. I = 2 e 2 4 e + 3 6. I = 1 2 e 2 2 e + 5 2 correct Explanation: So (dds785) HW06 Radin (54915) 2 After expansion, parenleftBig x 1 x parenrightBig 2 = x 2 + 1 x . Thus I = integraldisplay e 1 parenleftBig x 2 + 1 x parenrightBig dx = bracketleftBig 1 2 x 2 2 x + ln x bracketrightBig e 1 = 1 2 e 2 2 e + braceleftBig 1 + 3 2 bracerightBig . Consequently, I = 1 2 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 5 + 3 x 2 dx . 1. I = 1 3 ln 5 3 2. I = 1 6 ln 8 5 correct 3. I = ln 5 2 4. I = 1 6 ln 5 3 5. I = 1 3 ln 5 2 6. I = ln 8 5 Explanation: Set u = 5 + 3 x 2 ; then du = 6 x dx while x = 0 = u = 5 x = 1 = u = 8 . In this case, I = 1 6 integraldisplay 8 5 1 u du = 1 6 bracketleftBig ln  u  bracketrightBig 8 5 . Consequently, I = 1 6 (ln 8 ln 5) = 1 6 ln 8 5 . 005 10.0 points Evaluate the integral I = integraldisplay e 2 e 2 x ln x dx. 1. I = 2 ln2 correct 2. I = 2( e 2 e ) 3. I = 2 4. I = 2(ln 2 1) 5. I = ln 3 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 = 2 integraldisplay 2 1 1 u du = 2 bracketleftBig ln u bracketrightBig 2 1 . So (dds785) HW06 Radin (54915) 3 Consequently, I = 2 ln2 ....
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 Fall '08
 Cepparo
 Calculus

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