HW05-solutions - wei (jw35975) HW05 kalahurka (55230) 1...

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Unformatted text preview: wei (jw35975) HW05 kalahurka (55230) 1 This print-out should have 20 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 6 1 ln t t 2 dt . 1. I = 1 + 1 6 ln 6 2. I = 5 6- 1 6 ln 6 correct 3. I = 5 6 + 1 6 ln 6 4. I = 1- 1 6 ln 6 5. I = 6- 1 6 ln 6 6. I = 6 + 1 6 ln 6 Explanation: After integration by parts, integraldisplay 6 1 ln t t 2 dt = bracketleftbigg- ln t t bracketrightbigg 6 1 + integraldisplay 6 1 1 t 2 dt = bracketleftbigg- ln t t- 1 t bracketrightbigg 6 1 = 5 6 + parenleftBig ln1- ln 6 6 parenrightBig . But ln1 = 0. Consequently, I = 5 6- 1 6 ln 6 . 002 10.0 points Determine the integral I = integraldisplay ( x 2- 3) sin2 x dx . 1. I = 1 4 parenleftBig 2 x cos2 x +(2 x 2- 7) sin2 x parenrightBig + C 2. I =- x 2 cos 2 x + x sin 2 x- 5 2 cos 2 x + C 3. I = 1 2 parenleftBig 2 x sin 2 x- (2 x 2- 7) cos 2 x parenrightBig + C 4. I = 1 4 parenleftBig 2 x sin2 x- (2 x 2- 7) cos2 x parenrightBig + C correct 5. I = 1 2 x 2 sin2 x- x cos 2 x + 5 2 sin 2 x + C 6. I = 1 4 parenleftBig 2 x sin 2 x +(2 x 2- 7) cos 2 x parenrightBig + C Explanation: After integration by parts, integraldisplay ( x 2- 3) sin 2 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) cos2 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 sin2 x- (2 x 2- 7) cos2 x parenrightBig + C with C an arbitrary constant. keywords: integration by parts, indefinite integral, trig function, integration by parts twice, 003 10.0 points Evaluate the integral I = integraldisplay 1 5 xe 2 x dx. wei (jw35975) HW05 kalahurka (55230) 2 1. I = 5 4 parenleftBig 2 e 2 + 1 parenrightBig 2. I = 5 2 parenleftBig e 2 + 1 parenrightBig 3. I = 5 2 parenleftBig 2 e 2 + 1 parenrightBig 4. I = 5 4 e 2 5. I = 5 4 parenleftBig e 2 + 1 parenrightBig correct 6. I = 5 2 e 2 Explanation: After integration by parts, I = bracketleftBig 5 2 xe 2 x bracketrightBig 1- 5 2 integraldisplay 1 e 2 x dx = bracketleftBig 5 2 xe 2 x- 5 4 e 2 x bracketrightBig 1 . Consequently, I = 5 4 parenleftBig e 2 + 1 parenrightBig . 004 10.0 points Evaluate the integral I = integraldisplay 1 2 e 6 x dx . 1. I = 2 3 parenleftBig 1- 7 e 6 parenrightBig 2. I = 2 3 parenleftBig 1 + 7 e 6 parenrightBig 3. I = 1 9 parenleftBig 1- 7 e 6 parenrightBig correct 4. I = 1 9 parenleftBig 1- 6 e 6 parenrightBig 5. I = 1 9 parenleftBig 1 + 7 e 6 parenrightBig Explanation: First we use the substitution u = x to eliminate the square root in the integrand....
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This note was uploaded on 03/26/2012 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas at Austin.

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HW05-solutions - wei (jw35975) HW05 kalahurka (55230) 1...

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