solution Exam 2_pdf

# solution Exam 2_pdf - Version 040 EXAM 2 Odell (58340) 1...

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Unformatted text preview: Version 040 EXAM 2 Odell (58340) 1 This print-out should have 16 questions. Multiple-choice 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 3 x + 7 ln x exists, and if it does, find its value. 1. none of the other answers 2. limit = 10 3. limit =- 4. limit = 5. limit = 3 6. limit = 0 correct Explanation: Use of LHospitals Rule is suggested, Set f ( x ) = (ln x ) 2 , g ( x ) = 3 x + 7 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 + 7 x , so lim x f ( x ) g ( x ) = lim x 2 ln x 3 x + 7 . We need to apply LHospital once again, for then lim x 2 ln x 3 x + 7 = lim x 2 x 3 = 0 . Consequently, the limit exists and lim x (ln x ) 2 3 x + 7 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 9 3 f ( x ) dx . 1. I 83 3 2. I 85 3 correct 3. I 86 3 4. I 28 5. I 29 Explanation: Simpsons Rule estimates the integral I = integraldisplay 9 3 f ( x ) dx Version 040 EXAM 2 Odell (58340) 2 by I 1 3 braceleftBig f (3) + 4 f (4) + 2 f (5) + 4 f (6) + 2 f (7) + 4 f (8) + f (9) bracerightBig , taking n = 6. Reading off the values of f from its graph we thus see that I 85 3 . keywords: definite integral, graph, Simpsons rule 003 10.0 points Evaluate the definite integral I = integraldisplay 1 4 xe 3 x dx. 1. I = 4 9 parenleftBig 2 e 3 + 1 parenrightBig correct 2. I = 4 3 e 3 3. I = 4 9 e 3 4. I = 4 3 parenleftBig 3 e 3 + 1 parenrightBig 5. I = 4 9 parenleftBig 3 e 3 + 1 parenrightBig 6. I = 4 3 parenleftBig 2 e 3 + 1 parenrightBig Explanation: After integration by parts, I = bracketleftBig 4 3 xe 3 x bracketrightBig 1- 4 3 integraldisplay 1 e 3 x dx = bracketleftBig 4 3 xe 3 x- 4 9 e 3 x bracketrightBig 1 . Consequently, I = 4 9 parenleftBig 2 e 3 + 1 parenrightBig . 004 10.0 points Determine the integral I = integraldisplay / 2 4 cos 1 + sin 2 d . 1. I = 3 4 2. I = 3 2 3. I = 1 2 4. I = correct 5. I = 5 4 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 ....
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## This note was uploaded on 09/23/2009 for the course M 57420 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.

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solution Exam 2_pdf - Version 040 EXAM 2 Odell (58340) 1...

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