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Unformatted text preview: Panjwani, Sameer – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: James Rath 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 = 3 3 x 7 x 2 3 x 4 dx . 1. I = 3 ln 5 2. I = 3 ln 5 3. I = ln 5 4. I = ln 5 5. I = ln 4 correct 6. I = ln 4 7. I = 3 ln 4 8. I = 3 ln 4 Explanation: After factorization x 2 3 x 4 = ( x + 1)( x 4) . But then by partial fractions, 3 x 7 x 2 3 x 4 = 2 x + 1 + 1 x 4 . Now 3 2 x + 1 dx = 2 ln  ( x + 1)  3 = 2 ln 4 , while 3 1 x 4 dx = ln  ( x 4)  3 = ln 4 . Consequently, I = ln 4 . keywords: definite integral, rational function, partial fractions, natural log 002 (part 1 of 1) 10 points Evaluate the definite integral I = e 1 3 x 2 ln x dx. 1. I = (2 e 3 1) 2. I = (2 e 3 + 1) 3. I = 1 3 (2 e 3 + 1) correct 4. I = 1 3 (2 e 3 1) 5. I = 2 3 e 3 Explanation: After integration by parts, I = x 3 ln x e 1 e 1 x 2 dx = e 3 e 1 x 2 dx , since ln e = 1 and ln 1 = 0. But e 1 x 2 dx = 1 3 ( e 3 1) . Consequently, I = e 3 1 3 ( e 3 1) = 1 3 (2 e 3 + 1) . keywords: integration by parts, log function 003 (part 1 of 1) 10 points Panjwani, Sameer – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: James Rath 2 Evaluate the integral I = π/ 4 (1 4 sin 2 θ ) dθ . 1. I = 1 4 π 1 2. I = 1 2 π 1 2 3. I = 1 2 π 4. I = π 5. I = 1 1 4 π correct 6. I = π Explanation: Since sin 2 θ = 1 2 1 cos 2 θ , the integral can be rewritten as I = π/ 4 2 cos 2 θ 1 dθ = sin 2 θ θ π/ 4 . Consequently I = 1 1 4 π . keywords: definite integral, trig function, double angle formula 004 (part 1 of 1) 10 points Evaluate the definite integral I = 1 t (2 t ) 2 dt . 1. I = 2(2 ln 3) 2. I = 1 + ln 2 3. I = 2 ln 3 4. I = 2(1 + ln 2) 5. I = 1 ln 2 correct 6. I = 2(1 ln 2) Explanation: Set u = 2 t . Then du = dt , while t = 0 = ⇒ u = 2 , t = 1 = ⇒ u = 1 . Then I = 1 2 (2 u ) u 2 du = 2 1 (2 u ) u 2 du = 2 1 2 u 2 1 u du = 2 u + ln  u  2 1 ....
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This note was uploaded on 04/30/2009 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.
 Fall '08
 Cepparo
 Calculus

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