EX_2[1] - Garza, Ernesto – Exam 2 – Due: Mar 27 2007,...

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Garza, Ernesto – Exam 2 – Due: Mar 27 2007, 11:00 pm – Inst: JEGilbert 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points The region R in the frst quadrant bounded by the x -axis and the graph oF y = ln(1 + 4 x - 2 x 2 ) is shown in R 1 2 Estimate the area oF R using Simpson’s Rule with 2 equal subintervals. 1. area ( R ) 2 ln 2 2. area ( R ) 4 3 ln 3 correct 3. area ( R ) 2 ln 3 4. area ( R ) 4 3 ln 5 2 5. area ( R ) 2 ln 5 2 6. area ( R ) 4 3 ln 2 Explanation: The area oF R is given by the defnite inte- gral I = Z 2 0 ln(1 + 4 x - 2 x 2 ) dx. Now by Simpson’s Rule with 2 equal subin- tervals, I h 3 n f (0) + 4 f (1) + f (2) o , h = 1 , where f (0) = 0 , f (1) = ln 3 , f (2) = 0 . Consequently, area R ≈ 4 3 ln 3 . keywords: Stewart5e, numerical integration, Simpson’s Rule, log Function, estimate area 002 (part 1 oF 1) 10 points Evaluate the integral I = Z π/ 4 0 sec 2 x { 1 - 4 sin x } dx. 1. I = 5 + 4 2 2. I = 5 - 4 2 correct 3. I = - 3 + 2 2 4. I = 5 + 2 2 5. I = - 3 - 4 2 6. I = - 3 - 2 2 Explanation: Since sec 2 x { 1 - 4 sin x } = sec 2 x - 4 sec x sin x cos x · , we see that I = Z π/ 4 0 { sec 2 x - 4 sec tan x } dx. But d dx tan x = sec 2 x,
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Garza, Ernesto – Exam 2 – Due: Mar 27 2007, 11:00 pm – Inst: JEGilbert 2 while d dx sec x = sec x tan x. Consequently, I = h tan x - 4 sec x i π/ 4 0 = 5 - 4 2 . keywords: defnite integral, tan integral, sec integral 003 (part 1 oF 1) 10 points Evaluate the integral I = Z 1 0 2 x 2 + 1 dx. 1. I = 2 ( 2 - 1 ) 2. I = 2 ln(1 + 2 ) 3. I = 2 ln(1 + 2 ) correct 4. I = 2 ln( 2 - 1 ) 5. I = 2 (1 + 2 ) 6. I = 2 ( 2 - 1 ) Explanation: Set x = tan u, then dx = sec 2 udu, x 2 + 1 = sec 2 u, while x = 0 = u = 0 , x = 1 = u = π 4 . In this case I = Z π/ 4 0 2 sec 2 u sec u du = Z π/ 4 0 2 sec udu. On the other hand, Z sec udu = ln | sec u + tan u | + C . Thus I = 2 ln | sec u + tan u | π/ 4 0 . Consequently, I = 2 ln(1 + 2 ) . keywords: 004 (part 1 oF 1) 10 points Evaluate the integral I = Z π 0 e x sin xdx. 1. I = - 1 2 ( e π + 1) 2. I = e π + 1 3. I = 1 2 ( e - π + 1) 4. I = e - π + 1 5. I = 1 2 ( e π + 1) correct 6. I = 1 2 (1 - e - π ) 7. I = 1 - e - π 8. I = e π - 1 Explanation: AFter integration by parts, I = h e x sin x i π 0 - Z π 0 e x cos xdx = 0 - Z π 0 e x cos xdx. To evaluate this last integral we integrate by parts once again. ±or then Z π 0 e x cos xdx = h e x cos x i π 0 + Z π 0 e x sin xdx = - e π - 1 + I ,
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Garza, Ernesto – Exam 2 – Due: Mar 27 2007, 11:00 pm – Inst: JEGilbert 3 in which case I = e π + 1 - I .
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This note was uploaded on 03/07/2009 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.

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EX_2[1] - Garza, Ernesto – Exam 2 – Due: Mar 27 2007,...

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