m 408l hw 6

# m 408l hw 6 - wwm364 Homework 6 Cepparo (58400) 1 This...

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Unformatted text preview: wwm364 Homework 6 Cepparo (58400) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay 4 1 ln t 7 t dt . 1. I = 8 7 (ln2 + 1) 2. I = 4 7 (ln4- 1) correct 3. I = 8 7 (ln2- 1) 4. I = 4 7 (ln4 + 1) 5. I = 1 7 (ln4- 1) 6. I = 1 7 (ln2 + 1) Explanation: After integration by parts, I = 2 7 bracketleftBig t ln t bracketrightBig 4 1- 2 7 integraldisplay 4 1 t parenleftBig 1 t parenrightBig dt = 4 7 ln 4- 2 7 integraldisplay 4 1 1 t dt . But integraldisplay 4 1 1 t dt = 2 bracketleftBig t bracketrightBig 4 1 . Consequently, I = 4 7 ln4- 4 7 = 4 7 (ln 4- 1) . keywords: integration by parts, logarithmic functions 002 10.0 points Determine the integral I = integraldisplay 4 x (ln x ) 2 dx . 1. I = 4 x 2 parenleftBig (ln x ) 2 + ln x- 1 2 parenrightBig + C 2. I = 2 x 2 parenleftBig (ln x ) 2- ln x- 1 2 parenrightBig + C 3. I = 4 x 2 parenleftBig (ln x ) 2 + ln x + 1 2 parenrightBig + C 4. I = 2 x 2 parenleftBig (ln x ) 2- ln x + 1 2 parenrightBig + C correct 5. I = 2 x 2 parenleftBig (ln x ) 2 + ln x + 1 2 parenrightBig + C 6. I = 4 x 2 parenleftBig (ln x ) 2- ln x + 1 2 parenrightBig + C Explanation: After integration by parts, integraldisplay x (ln x ) 2 dx = 1 2 x 2 (ln x ) 2- integraldisplay x 2 1 x ln x dx = 1 2 x 2 (ln x ) 2- integraldisplay x ln x dx. But after integration by parts once again, integraldisplay x ln x dx = 1 2 x 2 ln x- 1 2 integraldisplay x 2 1 x dx = 1 2 x 2 ln x- 1 2 integraldisplay x dx = 1 2 x 2 ln x- 1 4 x 2 + C. Thus integraldisplay x (ln x ) 2 dx = 1 2 x 2 (ln x ) 2- 1 2 x 2 ln x + 1 4 x 2 + C. Consequently, I = 2 x 2 parenleftBig (ln x ) 2- ln x + 1 2 parenrightBig + C . wwm364 Homework 6 Cepparo (58400) 2 keywords: integration by parts, log function 003 10.0 points Determine the integral I = integraldisplay ( x 2- 3) sin2 x dx . 1. I = 1 4 parenleftBig 2 x sin2 x +(2 x 2- 7) cos2 x parenrightBig + C 2. I = 1 2 x 2 sin 2 x- x cos2 x + 5 2 sin 2 x + C 3. I = 1 4 parenleftBig 2 x sin 2 x- (2 x 2- 7) cos 2 x parenrightBig + C correct 4. I = 1 4 parenleftBig 2 x cos2 x +(2 x 2- 7) sin2 x parenrightBig + C 5. I =- x 2 cos 2 x + x sin 2 x- 5 2 cos 2 x + C 6. I = 1 2 parenleftBig 2 x sin2 x- (2 x 2- 7) cos2 x parenrightBig + C Explanation: After integration by parts, integraldisplay ( x 2- 3) sin2 x dx =- 1 2 ( x 2- 3) cos2 x + 1 2 integraldisplay cos 2 x braceleftBig d dx ( x 2- 3) bracerightBig dx =- 1 2 ( x 2- 3) cos 2 x + integraldisplay x cos 2 x dx . To evaluate this last integral we need to inte- grate by parts once again. For then integraldisplay x cos 2 x dx = x sin2 x 2- integraldisplay sin 2 x 2 dx = 1 2 x sin 2 x + 1 4 cos 2 x ....
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## This note was uploaded on 03/24/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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m 408l hw 6 - wwm364 Homework 6 Cepparo (58400) 1 This...

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