EXAM 2 - Portillo, Tom Exam 2 Due: Nov 7 2006, 11:00 pm...

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Portillo, Tom – 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 7 2 f ( x ) dx. 1. I 59 2 2. I 31 3. I 61 2 4. I 63 2 5. I 30 correct Explanation: The Trapezoidal Rule estimates the defnite integral I = Z 7 2 f ( x ) dx by I 1 2 h f (2) + 2 { f (3)+ ··· + f (6) } + f (7) i when n = 5. ±or the given f , thereFore, I 1 2 h 8 + 2 { 7 + 6 + 6 + 5 } + 4 i = 30 , reading o² the values oF f From the graph. keywords: trapezoidal rule, integral, graph 002 (part 1 oF 1) 10 points ±ind Z e 5 x 4 + e 10 x dx. 1. 1 2 + e 5 x + C 2. 1 2 arcsec e 5 x + e 5 x + C 3. 1 2 arcsin e 5 x + C 4. None oF these. 5. arcsin e 5 x + C 6. 1 10 arctan µ 1 2 e 5 x + C correct Explanation: Z e 5 x 4 + e 10 x dx = 1 5 Z 5 e 5 x dx (2) 2 + ( e 5 x ) 2 = 1 5 Z d ( e 5 x ) (2) 2 + ( e 5 x ) 2 = 1 5 · 1 2 arctan µ 1 2 e 5 x + C = 1 10 arctan µ 1 2 e 5 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 3 0 3 x - 7 x 2 - 3 x - 4 dx. 1. I = - 3 ln 5 2. I = ln 5 3. I = 3 ln 5 4. I = - ln 4 5. I = ln 4 correct 6. I = - 3 ln 4 7. I = - ln 5 8. I = 3 ln 4 Explanation: AFter Factorization x 2 - 3 x - 4 = ( x + 1)( x - 4) . But then by partial Fractions, 3 x - 7 x 2 - 3 x - 4 = 2 x + 1 + 1 x - 4 . Now Z 3 0 2 x + 1 dx = h 2 ln | ( x + 1) | i 3 0 = 2 ln 4 , while Z 3 0 1 x - 4 dx = h ln | ( x - 4) | i 3 0 = - ln 4 . Consequently, I = ln 4 . 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 3 cos x + 2 sin x cos 3 x dx. 1. I = 2 2. I = 7 2 3. I = 5 4. I = 4 correct 5. I = 5 2 Explanation: AFter division we see that 3 cos x + 2 sin x cos 3 x = 3 sec 2 x + 2 tan x sec 2 x = (3 + 2 tan x ) sec 2 x. Thus I = Z π/ 4 0 (3 + 2 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 (3 + 2 u ) du = £ 3 u + u 2 / 1 0 . Consequently,
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This note was uploaded on 02/18/2010 for the course MATH 408L taught by Professor Benzvi during the Spring '06 term at North Texas.

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EXAM 2 - Portillo, Tom Exam 2 Due: Nov 7 2006, 11:00 pm...

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