This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 040 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 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 ....
View
Full
Document
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.
 Fall '09
 RAdin
 Calculus

Click to edit the document details