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# solution8_pdf - So (dds785) HW07 radin (54915) 1 This...

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Unformatted text preview: So (dds785) HW07 radin (54915) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine the indefinite integral I = integraldisplay 8 x ln x dx . 1. I = 4 x 2 ln x- 2 x 2 + C correct 2. I = 8 x 2 ln x- 2 x 2 + C 3. I = 4 x 2 ln x- 8 x 2 + C 4. I = 8 x 2 ln x- 4 x 2 + C 5. I = 4 x 2 ln x + 2 x 2 + C 6. I = 8 x 2 ln x + 2 x 2 + C Explanation: After integration by parts, I = 4 x 2 ln x- 4 integraldisplay x 2 parenleftBig 1 x parenrightBig dx = 4 x 2 ln x- 4 integraldisplay x dx . Consequently, I = 4 x 2 ln x- 2 x 2 + C with C an arbitrary constant. 002 10.0 points Evaluate the definite integral I = integraldisplay e 1 12 ln x x 3 dx. 1. I =- 3 e 2 2. I = 3 2 parenleftBig 1 + 3 e 4 parenrightBig 3. I = 3 parenleftBig 1 + 3 e 2 parenrightBig 4. I = 3 2 parenleftBig 1- 3 e 4 parenrightBig 5. I = 3 parenleftBig 1- 3 e 2 parenrightBig correct Explanation: After integration by parts integraldisplay e 1 2 ln x x 3 dx = bracketleftBig- ln x x 2 bracketrightBig e 1 + integraldisplay e 1 1 x 3 dx. But bracketleftBig- ln x x 2 bracketrightBig e 1 =- 1 e 2 , since ln e = 1 and ln1 = 0, while integraldisplay e 1 1 x 3 dx = 1 2 parenleftbigg 1- 1 e 2 parenrightbigg . Thus I = 3 parenleftbigg 1- 3 e 2 parenrightbigg . 003 10.0 points Evaluate the integral I = integraldisplay 4 1 ln t 6 t dt . 1. I = 1 6 (ln2 + 1) 2. I = 2 3 (ln4 + 1) 3. I = 2 3 (ln4- 1) correct 4. I = 4 3 (ln2- 1) 5. I = 1 6 (ln4- 1) So (dds785) HW07 radin (54915) 2 6. I = 4 3 (ln2 + 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 ln4- 2 3 = 2 3 (ln 4- 1) . keywords: integration by parts, logarithmic functions 004 10.0 points Determine the integral I = integraldisplay ( x 2 + 4) cos2 x dx . 1. I = 1 2 parenleftBig 2 x cos2 x- (2 x 2 +7) sin2 x parenrightBig + C 2. I = 1 4 parenleftBig 2 x cos 2 x +(2 x 2 +7) sin 2 x parenrightBig + C correct 3. I = 1 4 parenleftBig 2 x sin2 x- (2 x 2 +7) cos2 x parenrightBig + C 4. I =- x 2 cos 2 x + x sin 2 x- 9 2 cos 2 x + C 5. I = 1 2 x 2 sin 2 x- x cos2 x + 9 2 sin 2 x + C 6. I = 1 2 parenleftBig 2 x cos2 x +(2 x 2 +7) sin2 x parenrightBig + C Explanation: After integration by parts, integraldisplay ( x 2 + 4) cos2 x dx = 1 2 ( x 2 + 4) sin 2 x- 1 2 integraldisplay sin2 x braceleftBig d dx ( x 2 + 4) bracerightBig dx = 1 2 ( x 2 + 4) sin2 x- integraldisplay x sin 2 x dx ....
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## This note was uploaded on 10/25/2010 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas at Austin.

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solution8_pdf - So (dds785) HW07 radin (54915) 1 This...

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