Exam02vr32 - Cheung, Anthony Exam 2 Due: Nov 7 2006, 11:00...

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Cheung, Anthony – Exam 2 – Due: Nov 7 2006, 11:00 pm – Inst: David Benzvi 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 IF f is the Function whose graph on [0 , 10] is given by 2 4 6 8 2 4 6 8 use the Trapezoidal Rule with n = 5 to esti- mate the defnite integral I = Z 8 3 f ( x ) dx. 1. I 51 2 2. I 47 2 3. I 25 correct 4. I 49 2 5. I 24 Explanation: The Trapezoidal Rule estimates the defnite integral I = Z 8 3 f ( x ) dx by I 1 2 h f (3) + 2 { f (4)+ ··· + f (7) } + f (8) i when n = 5. ±or the given f , thereFore, I 1 2 h 7 + 2 { 6 + 6 + 5 + 3 } + 3 i = 25 , reading o² the values oF f From the graph. keywords: trapezoidal rule, integral, graph 002 (part 1 oF 1) 10 points ±ind Z e 10 x 49 + e 20 x dx. 1. arcsin e 10 x + C 2. 1 7 arcsin e 10 x + C 3. None oF these. 4. 1 70 arctan µ 1 7 e 10 x + C correct 5. 1 7 arcsec e 10 x + e 10 x + C 6. 1 7 + e 10 x + C Explanation: Z e 10 x 49 + e 20 x dx = 1 10 Z 10 e 10 x dx (7) 2 + ( e 10 x ) 2 = 1 10 Z d ( e 10 x ) (7) 2 + ( e 10 x ) 2 = 1 10 · 1 7 arctan µ 1 7 e 10 x + C = 1 70 arctan µ 1 7 e 10 x + C keywords: exponential Function, inverse trig Function 003 (part 1 oF 1) 10 points
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2 Evaluate the defnite integral I = Z 1 0 5 x - 7 x 2 - x - 2 dx. 1. I = - 3 ln 2 2. I = 3 ln 3 3. I = - 5 ln 2 4. I = - 3 ln 3 5. I = - 5 ln 3 6. I = 5 ln 3 7. I = 5 ln 2 8. I = 3 ln 2 correct Explanation: AFter Factorization x 2 - x - 2 = ( x + 1)( x - 2) . But then by partial Fractions, 5 x - 7 x 2 - x - 2 = 4 x + 1 + 1 x - 2 . Now Z 1 0 4 x + 1 dx = h 4 ln | ( x + 1) | i 1 0 = 4 ln 2 , while Z 1 0 1 x - 2 dx = h ln | ( x - 2) | i 1 0 = - ln 2 . Consequently, I = 3 ln 2 . keywords: defnite integral, rational Function, partial Fractions, natural log 004 (part 1 oF 1) 10 points Evaluate the defnite integral I = Z π/ 4 0 cos x + 5 sin x cos 3 x dx. 1. I = 9 4 2. I = - 1 4 3. I = - 3 2 4. I = 6 5. I = 7 2 correct Explanation: AFter division we see that cos x + 5 sin x cos 3 x = sec 2 x + 5 tan x sec 2 x = (1 + 5 tan x ) sec 2 x. Thus I = Z π/ 4 0 (1 + 5 tan x ) sec 2 xdx. Let u = tan x ; then du = sec 2 xdx, while x = 0 = u = 0 , x = π 4 = u = 1 . In this case I = Z 1 0 (1 + 5 u ) du = u + 5 2 u 2 1 0 . Consequently,
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This note was uploaded on 01/27/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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Exam02vr32 - Cheung, Anthony Exam 2 Due: Nov 7 2006, 11:00...

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