int cal exam 2 - Version 084 – EXAM 2 – Radin –...

Info iconThis preview shows pages 1–3. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 084 – EXAM 2 – Radin – (57410) 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 2 x + 6 ln x exists, and if it does, find its value. 1. limit = 2 2. none of the other answers 3. limit = 0 correct 4. limit = 8 5. limit =-∞ 6. limit = ∞ Explanation: Use of L’Hospital’s Rule is suggested, Set f ( x ) = (ln x ) 2 , g ( x ) = 2 x + 6 ln x . Then f, g have derivatives of all orders and lim x →∞ f ( x ) = ∞ , lim x →∞ g ( x ) = ∞ . Thus L’Hospital’s Rule applies: lim x →∞ f ( x ) g ( x ) = lim x →∞ f ′ ( x ) g ′ ( x ) . But f ′ ( x ) = 2 ln x x , g ′ ( x ) = 2 + 6 x , so lim x →∞ f ′ ( x ) g ′ ( x ) = lim x →∞ 2 ln x 2 x + 6 . We need to apply L’Hospital once again, for then lim x →∞ 2 ln x 2 x + 6 = lim x →∞ 2 x slashBig 2 = 0 . Consequently, the limit exists and lim x →∞ (ln x ) 2 2 x + 6 ln x = 0 . 002 10.0 points Evaluate the definite integral I = integraldisplay π/ 3 sec x tan x 5 + sec x dx . 1. I =- ln 7 6 2. I = ln 6 5 3. I = ln 7 10 4. I = ln 7 6 correct 5. I =- ln 6 5 6. I =- ln 7 10 Explanation: Since d dx (sec x ) = sec x tan x , use of the substitution u = 5 + sec x is suggested. For then du = sec x tan xdx , while x = 0 = ⇒ u = 6 , x = π 3 = ⇒ u = 7 . Thus I = integraldisplay 7 6 1 u du = bracketleftBig ln u bracketrightBig 7 6 . Version 084 – EXAM 2 – Radin – (57410) 2 Consequently, I = ln 7 6 . 003 10.0 points Evaluate the integral I = integraldisplay 1 1 √ 4- x 2 dx . 1. I = 1 3 2. I = 1 6 3. I = 1 4 π 4. I = 1 4 5. I = 1 6 π correct 6. I = 1 3 π Explanation: Set x = 2 sin u ; then dx = 2 cos u du and 4- x 2 = 4(1- sin 2 u ) = 2 cos 2 u , while x = 0 = ⇒ u = 0 , x = 1 = ⇒ u = π 6 . In this case I = integraldisplay π/ 6 cos u cos u du = integraldisplay π/ 6 du . Consequently I = 1 6 π . 004 10.0 points Determine the integral I = integraldisplay 2 x (3 + 2 ln x ) 3 dx. 1. I =- 1 4 (3 + 2 ln x ) 4 + C 2. I =- 1 4 ln x (3 + 2 ln x ) 2 + C 3. I = 1 2 ln x (3 + 2 ln x ) 2 + C 4. I = 1 4 ln x (3 + 2 ln x ) 2 + C 5. I = 1 4 (3 + 2 ln x ) 4 + C correct 6. I =- 1 2 (3 + 2 ln x ) 4 + C 7. I = 1 2 (3 + 2 ln x ) 4 + C 8. I =- 1 2 ln x (3 + 2 ln x ) 2 + C Explanation: Set u = 3 + 2 ln x . Then du = 2 x dx, so I = integraldisplay u 3 du = 1 4 u 4 + C ....
View Full Document

This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

Page1 / 9

int cal exam 2 - Version 084 – EXAM 2 – Radin –...

This preview shows document pages 1 - 3. Sign up to view the full document.

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