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Unformatted text preview: Version 010 EXAM 2 Odell (58340) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine if lim x (ln x ) 2 4 x + 6 ln x exists, and if it does, find its value. 1. none of the other answers 2. limit = 3. limit = 0 correct 4. limit = 10 5. limit = 6. limit = 4 Explanation: Use of LHospitals Rule is suggested, Set f ( x ) = (ln x ) 2 , g ( x ) = 4 x + 6 ln x . Then f, g have derivatives of all orders and lim x f ( x ) = , lim x g ( x ) = . Thus LHospitals Rule applies: lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) . But f ( x ) = 2 ln x x , g ( x ) = 4 + 6 x , so lim x f ( x ) g ( x ) = lim x 2 ln x 4 x + 6 . We need to apply LHospital once again, for then lim x 2 ln x 4 x + 6 = lim x 2 x 4 = 0 . Consequently, the limit exists and lim x (ln x ) 2 4 x + 6 ln x = 0 . 002 10.0 points The graph of a function f is shown in 2 4 6 8 10 2 4 6 8 Use Simpsons Rule with n = 6 to estimate the integral I = integraldisplay 7 1 f ( x ) dx . 1. I 88 3 2. I 89 3 correct 3. I 30 4. I 29 5. I 86 3 Explanation: Simpsons Rule estimates the integral I = integraldisplay 7 1 f ( x ) dx Version 010 EXAM 2 Odell (58340) 2 by I 1 3 braceleftBig f (1) + 4 f (2) + 2 f (3) + 4 f (4) + 2 f (5) + 4 f (6) + f (7) bracerightBig , taking n = 6. Reading off the values of f from its graph we thus see that I 89 3 . keywords: definite integral, graph, Simpsons rule 003 10.0 points Evaluate the definite integral I = integraldisplay 1 6 xe 3 x dx. 1. I = 2 3 e 3 2. I = 2 3 parenleftBig 3 e 3 + 1 parenrightBig 3. I = 2 3 parenleftBig 2 e 3 + 1 parenrightBig correct 4. I = 2 parenleftBig 2 e 3 + 1 parenrightBig 5. I = 2 e 3 6. I = 2 parenleftBig 3 e 3 + 1 parenrightBig Explanation: After integration by parts, I = bracketleftBig 2 xe 3 x bracketrightBig 1 2 integraldisplay 1 e 3 x dx = bracketleftBig 2 xe 3 x 2 3 e 3 x bracketrightBig 1 . Consequently, I = 2 3 parenleftBig 2 e 3 + 1 parenrightBig . 004 10.0 points Determine the integral I = integraldisplay / 2 4 cos 1 + sin 2 d . 1. I = 7 4 2. I = 5 4 3. I = 3 4 4. I = correct 5. I = 3 2 Explanation: Since d d sin = cos , the substitution u = sin is suggested. For then du = cos d , while = 0 = u = 0 , = 2 = u = 1 , so that I = 4 integraldisplay 1 1 1 + u 2 du , which can now be integrated using the fact that d du tan 1 u = 1 1 + u 2 . Consequently, I = 4 bracketleftBig tan 1 u bracketrightBig 1 = since tan 1 1 = 4 . Version 010 EXAM 2 Odell (58340) 3 keywords: definite integral, trig substitution, inverse tan integral 005 10.0 points Determine the integral I = integraldisplay / 2 (2 cos + cos 3 ) d ....
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 Fall '08
 Cepparo

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