HW04-solutions.pdf

# HW04-solutions.pdf - young(toy68 HW04 villafuerte...

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young (toy68) – HW04 – villafuerte altu – (53615) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Evaluate the integral I = integraldisplay e 1 2 x ln( x ) dx . 1. I = e + 1 2. I = 1 2 ( e - 1) 3. I = e 2 + 1 4. I = 1 2 ( e 2 + 1) correct 5. I = 1 2 ( e 2 - 1) 6. I = e - 1 Explanation: After integration by parts, I = bracketleftBig x 2 ln( x ) bracketrightBig e 1 - integraldisplay e 1 x 2 parenleftBig 1 x parenrightBig dx = e 2 ln( x ) - integraldisplay e 1 x dx . Consequently, I = e 2 - bracketleftBig 1 2 x 2 bracketrightBig e 1 = 1 2 ( e 2 + 1) . 002 10.0points Evaluate the integral I = integraldisplay 4 1 ln( t ) 6 t dt . 1. I = 2 3 (ln(4) - 1) correct 2. I = 4 3 (ln(2) - 1) 3. I = 2 3 (ln(4) + 1) 4. I = 1 6 (ln(4) - 1) 5. I = 4 3 (ln(2) + 1) 6. I = 1 6 (ln(2) + 1) Explanation: After integration by parts, I = 1 3 bracketleftBig t ln( t ) bracketrightBig 4 1 - 1 3 integraldisplay 4 1 t parenleftBig 1 t parenrightBig dt = 2 3 ln(4) - 1 3 integraldisplay 4 1 1 t dt . But integraldisplay 4 1 1 t dt = 2 bracketleftBig t bracketrightBig 4 1 . Consequently, I = 2 3 ln(4) - 2 3 = 2 3 (ln(4) - 1) . keywords: integration by parts, logarithmic functions 003 10.0points Determine the integral I = integraldisplay ( x 2 + 2) cos(2 x ) dx . 1. I = 1 2 parenleftBig 2 x cos(2 x ) - (2 x 2 +3) sin(2 x ) parenrightBig + C 2. I = 1 4 parenleftBig 2 x cos(2 x )+(2 x 2 +3) sin(2 x ) parenrightBig + C correct 3. I = 1 4 parenleftBig 2 x sin(2 x ) - (2 x 2 +3) cos(2 x ) parenrightBig + C

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young (toy68) – HW04 – villafuerte altu – (53615) 2 4. I = 1 2 parenleftBig 2 x cos(2 x )+(2 x 2 +3) sin(2 x ) parenrightBig + C 5. I = - x 2 cos(2 x )+ x sin(2 x ) - 5 2 cos(2 x )+ C 6. I = 1 2 x 2 sin(2 x ) - x cos(2 x )+ 5 2 sin(2 x )+ C Explanation: After integration by parts, integraldisplay ( x 2 + 2) cos(2 x ) dx = 1 2 ( x 2 + 2) sin(2 x ) - 1 2 integraldisplay sin(2 x ) braceleftBig d dx ( x 2 + 2) bracerightBig dx = 1 2 ( x 2 + 2) 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 + 3) sin(2 x ) parenrightBig + C with C an arbitrary constant. keywords: integration by parts, indefinite integral, trig function, integration by parts twice, 004 10.0points Determine the integral I = integraldisplay 2 xe 3 x dx . 1. I = 2 3 parenleftBig x - 1 3 parenrightBig e 3 x + C correct 2. I = 2 9 parenleftBig x - 1 3 parenrightBig e 3 x + C 3. I = 2 9 parenleftBig x + 1 3 parenrightBig e 3 x + C 4. I = 2 3 xe 3 x + C 5. I = 2 9 xe 3 x + C 6. I = 2 3 parenleftBig x + 1 3 parenrightBig e 3 x + C Explanation: After integration by parts, I = 2 3 xe 3 x - 2 3 integraldisplay e 3 x dx . Consequently, I = 2 3 parenleftBig x - 1 3 parenrightBig e 3 x + C . 005 10.0points Evaluate the definite integral I = integraldisplay 1 0 4 x 5 e x 3 dx . 1. I = 4 3 parenleftBig e + 2 e parenrightBig 2. I = 4 5 ( e + 2) 3. I = 4 5 parenleftBig e - 2 e parenrightBig 4. I = 4 3 parenleftBig e - 2 e parenrightBig correct 5. I = 4 3 ( e - 2) 6. I = 4 5 parenleftBig e + 2 e parenrightBig Explanation: Set u = x 3 . Then du = 3 x 2 dx ,
young (toy68) – HW04 – villafuerte altu – (53615) 3 while x = 0 = u = 0 , x = 1 = u = 1 .

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• Fall '09
• GOGOLEV
• Cos, dx

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