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Homework 6

# Homework 6 - tgo72 Homework 6 Gompf(58370 This print-out...

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tgo72 – Homework 6 – Gompf – (58370) 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 5 t dt . 1. I = 4 5 (ln 4 + 1) 2. I = 1 5 (ln 2 + 1) 3. I = 1 5 (ln 4 - 1) 4. I = 4 5 (ln 4 - 1) correct 5. I = 8 5 (ln 2 + 1) 6. I = 8 5 (ln 2 - 1) Explanation: After integration by parts, I = 2 5 bracketleftBig t ln t bracketrightBig 4 1 - 2 5 integraldisplay 4 1 t parenleftBig 1 t parenrightBig dt = 4 5 ln 4 - 2 5 integraldisplay 4 1 1 t dt . But integraldisplay 4 1 1 t dt = 2 bracketleftBig t bracketrightBig 4 1 . Consequently, I = 4 5 ln 4 - 4 5 = 4 5 (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 = 4 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 = 2 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 .

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tgo72 – Homework 6 – Gompf – (58370) 2 keywords: integration by parts, log function 003 10.0 points Determine the integral I = integraldisplay ( x 2 - 4) cos 2 x dx . 1. I = 1 4 parenleftBig 2 x cos 2 x + (2 x 2 - 9) sin 2 x parenrightBig + C correct 2. I = 1 2 x 2 sin 2 x - x cos 2 x - 7 2 sin 2 x + C 3. I = 1 2 parenleftBig 2 x cos 2 x +(2 x 2 - 9) sin 2 x parenrightBig + C 4. I = 1 4 parenleftBig 2 x sin 2 x - (2 x 2 - 9) cos 2 x parenrightBig + C 5. I = - x 2 cos 2 x + x sin 2 x + 7 2 cos 2 x + C 6. I = 1 2 parenleftBig 2 x cos 2 x - (2 x 2 - 9) sin 2 x parenrightBig + C Explanation: After integration by parts, integraldisplay ( x 2 - 4) cos 2 x dx = 1 2 ( x 2 - 4) sin 2 x - 1 2 integraldisplay sin 2 x braceleftBig d dx ( x 2 - 4) bracerightBig dx = 1 2 ( x 2 - 4) sin 2 x - integraldisplay x sin 2 x dx . To evaluate this last integral we need to inte- grate by parts once again. For then integraldisplay x sin 2 x dx = - x cos 2 x 2 + integraldisplay cos 2 x 2 dx = - 1 2 x cos 2 x + 1 4 sin 2 x . Consequently, I = 1 4 parenleftBig 2 x cos 2 x + (2 x 2 - 9) sin 2 x parenrightBig + C with C an arbitrary constant. keywords: integration by parts, indefinite integral, trig function, integration by parts twice, 004 10.0 points Evaluate the definite integral I = integraldisplay 1 0 7 x 5 e - x 3 dx .
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Homework 6 - tgo72 Homework 6 Gompf(58370 This print-out...

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