Integral Calculus Exam 2*

# Integral Calculus Exam 2* - Panjwani, Sameer – Exam 2 –...

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Unformatted text preview: 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 = 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.

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Integral Calculus Exam 2* - Panjwani, Sameer – Exam 2 –...

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