CalcExam2Solutions - Version 010 EXAM 2 Odell (58340) 1...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 010 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 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 ....
View Full Document

Page1 / 9

CalcExam2Solutions - Version 010 EXAM 2 Odell (58340) 1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online