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Integral Calculus Exam 2

# Integral Calculus Exam 2 - Panjwani Sameer Exam 2 Due 1:00...

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Panjwani, Sameer – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: James Rath 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 3 0 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 Z 3 0 2 x + 1 dx = h 2 ln | ( x + 1) | i 3 0 = 2 ln 4 , while Z 3 0 1 x - 4 dx = h ln | ( x - 4) | i 3 0 = - 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 = Z 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 = h x 3 ln x i e 1 - Z e 1 x 2 dx = e 3 - Z e 1 x 2 dx , since ln e = 1 and ln 1 = 0. But Z 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

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

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