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# Exam_2_Answers - Miller Kierste Exam 2 Due 11:00 pm Inst...

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Miller, Kierste – Exam 2 – Due: Mar 27 2007, 11:00 pm – Inst: Gary Berg 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 - 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 5 2 2. area ( R ) 4 3 ln 3 correct 3. area ( R ) 4 3 ln 2 4. area ( R ) 2 ln 3 5. area ( R ) 4 3 ln 5 2 6. area ( R ) 2 ln 2 Explanation: The area of R is given by the definite inte- gral I = 2 0 ln(1 + 4 x - 2 x 2 ) dx . Now by Simpson’s Rule with 2 equal subin- tervals, I h 3 f (0) + 4 f (1) + f (2) , 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 = π / 4 0 sec 2 x { 3 - 2 sin x } dx . 1. I = 5 + 2 2. I = 5 + 2 2 3. I = 1 - 2 4. I = 5 - 2 2 correct 5. I = 1 - 2 2 6. I = 1 + 2 Explanation: Since sec 2 x { 3 - 2 sin x } = 3 sec 2 x - 2 sec x sin x cos x , we see that I = π / 4 0 { 3 sec 2 x - 2 sec tan x } dx . But d dx tan x = sec 2 x ,

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

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