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# Exam 02 - Su Yung Exam 2 Due 1:00 am Inst Shinko Harper...

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Su, Yung – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: Shinko Harper 1 This print-out should have 16 questions. Multiple-choice 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 1 0 x - 8 x 2 - x - 2 dx . 1. I = ln 3 2. I = - ln 2 3. I = - ln 3 4. I = 5 ln 3 5. I = - 5 ln 2 6. I = 5 ln 2 correct 7. I = ln 2 8. I = - 5 ln 3 Explanation: After factorization x 2 - x - 2 = ( x + 1)( x - 2) . But then by partial fractions, x - 8 x 2 - x - 2 = 3 x + 1 - 2 x - 2 . Now Z 1 0 3 x + 1 dx = h 3 ln | ( x + 1) | i 1 0 = 3 ln 2 , while Z 1 0 2 x - 2 dx = h 2 ln | ( x - 2) | i 1 0 = - 2 ln 2 . Consequently, I = 5 ln 2 . keywords: definite integral, rational function, partial fractions, natural log 002 (part 1 of 1) 10 points Evaluate the definite integral I = Z e 1 2 x 3 ln x dx. 1. I = 1 8 (3 e 4 - 1) 2. I = 1 2 (3 e 4 - 1) 3. I = 3 8 e 4 4. I = 1 8 (3 e 4 + 1) correct 5. I = 1 2 (3 e 4 + 1) Explanation: After integration by parts, I = h 1 2 x 4 ln x i e 1 - 1 2 Z e 1 x 3 dx = 1 2 e 4 - 1 2 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 = 1 2 e 4 - 1 8 ( e 4 - 1) = 1 8 (3 e 4 + 1) . keywords: integration by parts, log function 003 (part 1 of 1) 10 points

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Su, Yung – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: Shinko Harper 2 Evaluate the integral I = Z π/ 4 0 (1 - 4 sin 2 θ ) dθ . 1. I = - π 2. I = 1 4 π - 1 3. I = 1 - 1 4 π correct 4. I = - 1 2 π 5. I = 1 2 π - 1 2 6. I = π Explanation: Since sin 2 θ = 1 2 1 - cos 2 θ · , the integral can be rewritten as I = Z π/ 4 0 n 2 cos 2 θ - 1 o = h sin 2 θ - θ i π/ 4 0 . 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 = Z 2 0 t (3 - t ) 2 dt . 1. I = 3 - ln 4 2. I = 2(2 - ln 3) 3. I = 2 + ln 3 4. I = 2 - ln 3 correct 5. I = 2(3 - ln 4) 6. I = 2(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 - u ) u 2 du = Z 3 1 (3 - u ) u 2 du = Z 3 1 n 3 u 2 - 1 u o du = - h 3 u + ln | u | i 3 1 . Consequently, I = - (1 - 3) - ln 3 = 2 - ln 3 . keywords: substitution, integral 005 (part 1 of 1) 10 points To which one of the following does the inte- gral I = Z x x 2 - 1 dx reduce after an appropriate trig substitution. 1. I = Z sec 2 θ dθ correct 2. I = Z tan θ sec 2 θ dθ 3. I = Z sin 2 θ dθ 4. I = Z sin 2 θ sec θ dθ
Su, Yung – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: Shinko Harper 3 5. I = Z tan 2 θ dθ 6. I = Z sin θ sec 2 θ dθ Explanation: Set x = sec θ . Then

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Exam 02 - Su Yung Exam 2 Due 1:00 am Inst Shinko Harper...

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