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Unformatted text preview: Version PREVIEW EXAM 2 Zheng (58355) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. CalC7g49b 001 10.0 points Determine if lim x (ln x ) 2 3 x + 6 ln x exists, and if it does, find its value. 1. limit = 2. none of the other answers 3. limit = 3 4. limit = 5. limit = 9 6. limit = 0 correct Explanation: Use of LHospitals Rule is suggested, Set f ( x ) = (ln x ) 2 , g ( x ) = 3 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 ) = 3 + 6 x , so lim x f ( x ) g ( x ) = lim x 2 ln x 3 x + 6 . We need to apply LHospital once again, for then lim x 2 ln x 3 x + 6 = lim x 2 x 3 = 0 . Consequently, the limit exists and lim x (ln x ) 2 3 x + 6 ln x = 0 . CalC8g01b 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 6 f ( x ) dx . 1. I 100 3 2. I 34 3. I 33 correct 4. I 98 3 5. I 101 3 Explanation: Simpsons Rule estimates the integral I = integraldisplay 6 f ( x ) dx Version PREVIEW EXAM 2 Zheng (58355) 2 by I 1 3 braceleftBig f (0) + 4 f (1) + 2 f (2) + 4 f (3) + 2 f (4) + 4 f (5) + f (6) bracerightBig , taking n = 6. Reading off the values of f from its graph we thus see that I 33 . keywords: definite integral, graph, Simpsons rule CalC8a11b 003 10.0 points Evaluate the definite integral I = integraldisplay 1 5 xe 3 x dx. 1. I = 5 3 e 3 2. I = 5 9 parenleftBig 3 e 3 + 1 parenrightBig 3. I = 5 3 parenleftBig 3 e 3 + 1 parenrightBig 4. I = 5 3 parenleftBig 2 e 3 + 1 parenrightBig 5. I = 5 9 parenleftBig 2 e 3 + 1 parenrightBig correct 6. I = 5 9 e 3 Explanation: After integration by parts, I = bracketleftBig 5 3 xe 3 x bracketrightBig 1 5 3 integraldisplay 1 e 3 x dx = bracketleftBig 5 3 xe 3 x 5 9 e 3 x bracketrightBig 1 . Consequently, I = 5 9 parenleftBig 2 e 3 + 1 parenrightBig . CalC7e64a 004 10.0 points Determine the integral I = integraldisplay / 2 5 cos 1 + sin 2 d . 1. I = 3 2 2. I = 2 3. I = 4. I = 7 4 5. I = 5 4 correct 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 = 5 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 ....
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 Spring '08
 Cepparo

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