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Unformatted text preview: Kang, Danby Exam 2 Due: Oct 31 2007, 1:00 am Inst: Diane Radin 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the definite integral I = Z 2 x 11 x 2 2 x 3 dx. 1. I = 5 ln 3 correct 2. I = 5 ln 4 3. I = 5 ln 3 4. I = ln 3 5. I = 5 ln 4 6. I = ln 3 7. I = ln 4 8. I = ln 4 Explanation: After factorization x 2 2 x 3 = ( x + 1)( x 3) . But then by partial fractions, x 11 x 2 2 x 3 = 3 x + 1 2 x 3 . Now Z 2 3 x + 1 dx = h 3 ln  ( x + 1)  i 2 = 3 ln 3 , while Z 2 2 x 3 dx = h 2 ln  ( x 3)  i 2 = 2 ln 3 . Consequently, I = 5 ln 3 . keywords: definite integral, rational function, partial fractions, natural log 002 (part 1 of 1) 10 points Evaluate the definite integral I = Z e 1 3 x 3 ln xdx. 1. I = 3 4 (3 e 4 1) 2. I = 9 16 e 4 3. I = 3 16 (3 e 4 1) 4. I = 3 16 (3 e 4 + 1) correct 5. I = 3 4 (3 e 4 + 1) Explanation: After integration by parts, I = h 3 4 x 4 ln x i e 1 3 4 Z e 1 x 3 dx = 3 4 e 4 3 4 Z e 1 x 3 dx, since ln e = 1 and ln 1 = 0. But Z e 1 x 3 dx = 1 4 ( e 4 1) . Consequently, I = 3 4 e 4 3 16 ( e 4 1) = 3 16 (3 e 4 + 1) . keywords: integration by parts, log function 003 (part 1 of 1) 10 points Kang, Danby Exam 2 Due: Oct 31 2007, 1:00 am Inst: Diane Radin 2 Evaluate the integral I = Z / 4 (1 4 sin 2 ) d . 1. I = 1 1 4 correct 2. I = 3 2 3. I = 1 2 4. I = 5. I = 1 4  1 6. I = 3 4  1 Explanation: Since sin 2 = 1 2 1 cos 2 , the integral can be rewritten as I = Z / 4 n 2 cos 2  1 o d = h sin 2  i / 4 . Consequently I = 1 1 4 . keywords: definite integral, trig function, double angle formula 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 2 3 t (3 t ) 2 dt. 1. I = 3 ln 4 2. I = 3(3 ln 4) 3. I = 3(2 ln 3) correct 4. I = 2 ln 3 5. I = 2 + ln 3 6. I = 3(2 + ln 3) Explanation: Set u = 3 t . Then du = dt , while t = 0 = u = 3 , t = 2 = u = 1 . Then I = Z 1 3 3(3 u ) u 2 du = Z 3 1 3(3 u ) u 2 du = Z 3 1 n 9 u 2 3 u o du = h 9 u + 3 ln  u  i 3 1 ....
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 Fall '08
 Cepparo
 Calculus

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