Exam02vr200

# Exam02vr200 - Cheung Anthony Exam 2 Due Nov 7 2006 11:00 pm...

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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 7 2 f ( x ) dx. 1. I 28 2. I 53 2 correct 3. I 27 4. I 57 2 5. I 55 2 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 + 5 + 5 + 4 } + 3 i = 53 2 , reading o² the values oF f From the graph. keywords: trapezoidal rule, integral, graph 002 (part 1 oF 1) 10 points ±ind Z e 8 x 81 + e 16 x dx. 1. arcsin e 8 x + C 2. None oF these. 3. 1 72 arctan µ 1 9 e 8 x + C correct 4. 1 9 arcsin e 8 x + C 5. 1 9 + e 8 x + C 6. 1 9 arcsec e 8 x + e 8 x + C Explanation: Z e 8 x 81 + e 16 x dx = 1 8 Z 8 e 8 x dx (9) 2 + ( e 8 x ) 2 = 1 8 Z d ( e 8 x ) (9) 2 + ( e 8 x ) 2 = 1 8 · 1 9 arctan µ 1 9 e 8 x + C = 1 72 arctan µ 1 9 e 8 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 2 0 4 x - 8 x 2 - 2 x - 3 dx. 1. I = 2 ln 3 correct 2. I = - 2 ln 3 3. I = - 4 ln 3 4. I = 2 ln 4 5. I = 4 ln 4 6. I = - 2 ln 4 7. I = 4 ln 3 8. I = - 4 ln 4 Explanation: AFter Factorization x 2 - 2 x - 3 = ( x + 1)( x - 3) . But then by partial Fractions, 4 x - 8 x 2 - 2 x - 3 = 3 x + 1 + 1 x - 3 . Now Z 2 0 3 x + 1 dx = h 3 ln | ( x + 1) | i 2 0 = 3 ln 3 , while Z 2 0 1 x - 3 dx = h ln | ( x - 3) | i 2 0 = - ln 3 . Consequently, I = 2 ln 3 . 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 + 6 sin x cos 3 x dx. 1. I = 6 correct 2. I = 9 2 3. I = 9 4. I = 3 2 5. I = 0 Explanation: AFter division we see that 3 cos x + 6 sin x cos 3 x = 3 sec 2 x + 6 tan x sec 2 x = (3 + 6 tan x ) sec 2 x. Thus I = Z π/ 4 0 (3 + 6 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 + 6 u ) du = £ 3 u + 3 u 2 / 1 0 . Consequently,
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Exam02vr200 - Cheung Anthony Exam 2 Due Nov 7 2006 11:00 pm...

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