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Homework 6

# Homework 6 - ejm753 Homework 6 Cepparo(58400 This print-out...

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ejm753 – 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 2 t dt . 1. I = 4(ln 2 + 1) 2. I = 1 2 (ln 2 + 1) 3. I = 1 2 (ln 4 - 1) 4. I = 2(ln 4 - 1) correct 5. I = 2(ln 4 + 1) 6. I = 4(ln 2 - 1) Explanation: After integration by parts, I = bracketleftBig t ln t bracketrightBig 4 1 - integraldisplay 4 1 t parenleftBig 1 t parenrightBig dt = 2 ln 4 - integraldisplay 4 1 1 t dt . But integraldisplay 4 1 1 t dt = 2 bracketleftBig t bracketrightBig 4 1 . Consequently, I = 2 ln 4 - 2 = 2(ln 4 - 1) . keywords: integration by parts, logarithmic functions 002 10.0 points Determine the integral I = integraldisplay 5 x (ln x ) 2 dx . 1. I = 5 x 2 parenleftBig (ln x ) 2 - ln x + 1 2 parenrightBig + C 2. I = 5 2 x 2 parenleftBig (ln x ) 2 - ln x - 1 2 parenrightBig + C 3. I = 5 x 2 parenleftBig (ln x ) 2 + ln x + 1 2 parenrightBig + C 4. I = 5 2 x 2 parenleftBig (ln x ) 2 + ln x + 1 2 parenrightBig + C 5. I = 5 x 2 parenleftBig (ln x ) 2 + ln x - 1 2 parenrightBig + C 6. I = 5 2 x 2 parenleftBig (ln x ) 2 - ln x + 1 2 parenrightBig + C correct 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 = 5 2 x 2 parenleftBig (ln x ) 2 - ln x + 1 2 parenrightBig + C .

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ejm753 – Homework 6 – Cepparo – (58400) 2 keywords: integration by parts, log function 003 10.0 points Determine the integral I = integraldisplay ( x 2 - 2) sin 2 x dx . 1. I = 1 4 parenleftBig 2 x cos 2 x +(2 x 2 - 5) sin 2 x parenrightBig + C 2. I = - x 2 cos 2 x + x sin 2 x - 3 2 cos 2 x + C 3. I = 1 2 x 2 sin 2 x - x cos 2 x + 3 2 sin 2 x + C 4. I = 1 2 parenleftBig 2 x sin 2 x - (2 x 2 - 5) cos 2 x parenrightBig + C 5. I = 1 4 parenleftBig 2 x sin 2 x +(2 x 2 - 5) cos 2 x parenrightBig + C 6. I = 1 4 parenleftBig 2 x sin 2 x - (2 x 2 - 5) cos 2 x parenrightBig + C correct Explanation: After integration by parts, integraldisplay ( x 2 - 2) sin 2 x dx = - 1 2 ( x 2 - 2) cos 2 x + 1 2 integraldisplay cos 2 x braceleftBig d dx ( x 2 - 2) bracerightBig dx = - 1 2 ( x 2 - 2) 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 sin 2 x 2 - integraldisplay sin 2 x 2 dx = 1 2 x sin 2 x + 1 4 cos 2 x . Consequently, I = 1 4 parenleftBig 2 x sin 2 x - (2 x 2 - 5) cos 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 6 x 3 e - x 2 dx .
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Homework 6 - ejm753 Homework 6 Cepparo(58400 This print-out...

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