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

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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 Simpson’s 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 0 ln(1 + 4 x - x 2 ) dx . Now by Simpson’s 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, 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 { 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 π/ 4 0 { 3 sec 2 x + 4 sec tan x } dx . But d dx tan x = sec 2 x ,
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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 π/ 4 0 = - 1 + 4 2 . keywords: definite integral, tan integral, sec integral 003 (part 1 of 1) 10 points Evaluate the integral I = Z 1 0 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 u du , x 2 + 1 = sec 2 u , while x = 0 = u = 0 , x = 1 = u = π 4 . In this case I = Z π/ 4 0 3 sec 2 u sec u du = Z π/ 4 0 3 sec u du . On the other hand, Z sec u du = ln | sec u + tan u | + C . Thus I = 3 ln | sec u + tan u | π/ 4 0 . Consequently, I = 3 ln(1 + 2 ) . keywords: 004 (part 1 of 1) 10 points Evaluate the integral I = Z π 0 e - x sin x dx . 1. I = e π - 1 2. I = 1 - e - π 3. I = 1 2 ( e - π + 1) correct 4. I = 1 2 ( e π + 1) 5. I = e - π + 1 6. I = 1 2 (1 - e - π ) 7. I = e π + 1 8. I = - 1 2 ( e π + 1) Explanation: After integration by parts, I = - h e - x sin x i π 0 + Z π 0 e - x cos x dx = 0 + Z π 0 e - x cos x dx . To evaluate this last integral we integrate by parts once again. For then Z π 0 e - x cos x dx = - h e - x cos x i π 0 - Z π 0 e x sin x dx = e - π + 1 - I ,
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Bryson, Spencer – Exam 2 – Due: Mar 27 2007, 11:00 pm – Inst: Mar Gonzalez 3 in which case I = e - π + 1 - I .
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