exam2'06 - Bryson Spencer Exam 2 Due 11:00 pm Inst Mar...

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Unformatted text preview: Bryson, Spencer Exam 2 Due: Mar 27 2007, 11:00 pm Inst: Mar Gonzalez 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 The region R in the first quadrant bounded by the x-axis and the graph of y = ln(1 + 4 x- x 2 ) is shown in R 2 4 Estimate the area of R using Simpsons Rule with 2 equal subintervals. 1. area ( R ) 4 ln 4 2. area ( R ) 8 3 ln 5 correct 3. area ( R ) 8 3 ln 9 2 4. area ( R ) 4 ln 9 2 5. area ( R ) 8 3 ln 4 6. area ( R ) 4 ln 5 Explanation: The area of R is given by the definite inte- gral I = Z 4 ln(1 + 4 x- x 2 ) dx. Now by Simpsons Rule with 2 equal subin- tervals, I h 3 n f (0) + 4 f (2) + f (4) o , h = 2 , where f (0) = 0 , f (2) = ln 5 , f (4) = 0 . Consequently, area R 8 3 ln 5 . keywords: Stewart5e, numerical integration, Simpsons Rule, log function, estimate area 002 (part 1 of 1) 10 points Evaluate the integral I = Z / 4 sec 2 x { 3 + 4 sin x } dx. 1. I = 7 + 2 2 2. I = 7 + 4 2 3. I =- 1 + 4 2 correct 4. I = 7- 2 2 5. I =- 1- 2 2 6. I =- 1- 4 2 Explanation: Since sec 2 x { 3 + 4 sin x } = 3 sec 2 x + 4 sec x sin x cos x , we see that I = Z / 4 { 3 sec 2 x + 4 sec tan x } dx. But d dx tan x = sec 2 x, Bryson, Spencer Exam 2 Due: Mar 27 2007, 11:00 pm Inst: Mar Gonzalez 2 while d dx sec x = sec x tan x. Consequently, I = h 3 tan x + 4 sec x i / 4 =- 1 + 4 2 . keywords: definite integral, tan integral, sec integral 003 (part 1 of 1) 10 points Evaluate the integral I = Z 1 3 x 2 + 1 dx. 1. I = 2 ( 2- 1 ) 2. I = 2 ln(1 + 2 ) 3. I = 2 (1 + 2 ) 4. I = 3 ( 2- 1 ) 5. I = 3 ln(1 + 2 ) correct 6. I = 3 ln( 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 = 4 . In this case I = Z / 4 3 sec 2 u sec u du = Z / 4 3 sec udu. On the other hand, Z sec udu = ln | sec u + tan u | + C . Thus I = 3 ln | sec u + tan u | / 4 . Consequently, I = 3 ln(1 + 2 ) . keywords: 004 (part 1 of 1) 10 points Evaluate the integral I = Z e- x sin xdx. 1. I = e - 1 2. I = 1- e- 3. I = 1 2 ( e- + 1) correct 4. I = 1 2 ( e + 1) 5. I = e- + 1 6. I = 1 2 (1- e- ) 7. I = e + 1 8. I =- 1 2 ( e + 1) Explanation: After integration by parts, I =- h e- x sin x i + Z e- x cos xdx = 0 + Z e- x cos xdx....
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This homework help was uploaded on 04/16/2008 for the course CALC 303L taught by Professor Cheng during the Fall '07 term at University of Texas.

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exam2'06 - Bryson Spencer Exam 2 Due 11:00 pm Inst Mar...

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