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: momin (rrm497) Homework 08 cheng (58520) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine the indefinite integral I = integraldisplay 1 x 2 6 x 7 dx 1. I = sin 1 parenleftBig x 3 4 parenrightBig + C 2. I = ln vextendsingle vextendsingle vextendsingle x + 3 + radicalbig x 2 + 6 x 7 vextendsingle vextendsingle vextendsingle + C 3. I = sin 1 parenleftBig x 4 3 parenrightBig + C 4. I = ln vextendsingle vextendsingle vextendsingle x + 3 + radicalbig x 2 6 x 7 vextendsingle vextendsingle vextendsingle + C 5. I = ln vextendsingle vextendsingle vextendsingle x 3 + radicalbig x 2 6 x 7 vextendsingle vextendsingle vextendsingle + C correct 6. I = ln vextendsingle vextendsingle vextendsingle x 3 + radicalbig x 2 + 6 x 7 vextendsingle vextendsingle vextendsingle + C Explanation: By completing the square we see that x 2 6 x 7 = ( x 2 6 x + 9 ) 16 = ( x 3) 2 16 . This suggests the substitution x 3 = 4 sec , for then dx = 4 sec tan d , while ( x 3) 2 16 = 16tan 2 . In this case I = integraldisplay 4 sec tan 4 tan d = integraldisplay sec d = ln  sec + tan  + C . Now x 3 = 4 sec = tan = radicalbig ( x 3) 2 16 4 , so I = ln vextendsingle vextendsingle vextendsingle x 3 + radicalbig ( x 3) 2 16 4 vextendsingle vextendsingle vextendsingle + C . Consequently I = ln vextendsingle vextendsingle vextendsingle x 3 + radicalbig x 2 6 x 7 vextendsingle vextendsingle vextendsingle + C . 002 10.0 points Evaluate the definite integral I = integraldisplay 2 1 x 2 + 5 x + 1 dx . Correct answer: 2 . 93279. Explanation: After division x 2 + 5 x + 1 = ( x 2 1) + 6 x + 1 = x 2 1 x + 1 + 6 x + 1 = x 1 + 6 x + 1 . In this case I = integraldisplay 2 1 parenleftBig x 1 + 6 x + 1 parenrightBig dx = bracketleftBig 1 2 x 2 x + 6 ln  x + 1  bracketrightBig 2 1 = parenleftBig 1 1 2 parenrightBig + 6 parenleftBig ln 3 ln 2 parenrightBig . Consequently, I = 1 2 + 6 ln 3 2 = 2 . 93279 . momin (rrm497) Homework 08 cheng (58520) 2 003 10.0 points Evaluate the integral I = integraldisplay / 4 sec 2 x { 1 4 sin x } dx . 1. I = 3 4 2 2. I = 5 + 4 2 3. I = 3 2 2 4. I = 5 + 2 2 5. I = 3 + 2 2 6. I = 5 4 2 correct Explanation: Since sec 2 x { 1 4 sin x } = sec 2 x 4 sec x parenleftBig sin x cos x parenrightBig , we see that I = integraldisplay / 4 { sec 2 x 4 sec tan x } dx . But d dx tan x = sec 2 x , while d dx sec x = sec x tan x ....
View
Full
Document
This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin
 Calculus

Click to edit the document details