HW15-solutions

# HW15-solutions - husain(aih243 – HW15 – Gilbert...

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Unformatted text preview: husain (aih243) – HW15 – Gilbert – (56215) 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 e 1 x ln x dx . 1. I = 1 4 ( e- 1) 2. I = 1 2 ( e- 1) 3. I = 1 4 ( e 2 + 1) correct 4. I = 1 2 ( e + 1) 5. I = 1 2 ( e 2 + 1) 6. I = 1 4 ( e 2- 1) Explanation: After integration by parts, I = bracketleftBig 1 2 x 2 ln x bracketrightBig e 1- 1 2 integraldisplay e 1 x 2 parenleftBig 1 x parenrightBig dx = 1 2 e 2 ln x- 1 2 integraldisplay e 1 x dx . Consequently, I = 1 2 e 2- bracketleftBig 1 4 x 2 bracketrightBig e 1 = 1 4 ( e 2 + 1) . 002 10.0 points Evaluate the integral I = integraldisplay 4 e ln x x 2 dx . 1. I = 4 e + 1 2 parenleftBig ln4 + 1 parenrightBig 2. I = 2 e + 1 4 parenleftBig ln4 + 1 parenrightBig 3. I = 4 e- 1 2 parenleftBig ln 4 + 1 parenrightBig 4. I = 4 e- 1 2 parenleftBig ln 4- 1 parenrightBig 5. I = 2 e- 1 4 parenleftBig ln 4- 1 parenrightBig 6. I = 2 e- 1 4 parenleftBig ln 4 + 1 parenrightBig correct Explanation: After integration by parts, I =- bracketleftBig 1 x ln x bracketrightBig 4 e + integraldisplay 4 e 1 x 2 dx . Thus I =- bracketleftBig 1 x ln x + 1 x bracketrightBig 4 e . Consequently, I = 2 e- 1 4 parenleftBig ln4 + 1 parenrightBig . 003 10.0 points Evaluate the definite integral I = integraldisplay 5 te − t dt . 1. I = 1- 6 e 5 correct 2. I = 1- 5 e 5 3. I = 1 + 5 e 6 4. I = 1 + 6 e 5 5. I = 1 + 6 e 6 6. I = 1- 5 e 6 Explanation: husain (aih243) – HW15 – Gilbert – (56215) 2 After Integration by Parts, I = bracketleftBig- te − t bracketrightBig 5 + integraldisplay 5 e − t dt = bracketleftBig- te − t- e − t bracketrightBig 5 . Consequently, I =- 5 e − 5- e − 5 + 1 = 1- 6 e 5 . 004 10.0 points Evaluate the definite integral I = integraldisplay 1 (4 x 2 + 1) e x dx . 1. I = 5 e- 9 correct 2. I = 5 e + 9 3. I = 4 e + 9 4. I = 4 e + 7 5. I = 5 e- 7 6. I = 4 e- 9 Explanation: After integration by parts, I = bracketleftBig (4 x 2 + 1) e x bracketrightBig 1- integraldisplay 1 e x braceleftBig d dx (4 x 2 + 1) bracerightBig dx = bracketleftBig (4 x 2 + 1) e x bracketrightBig 1- 8 integraldisplay 1 xe x dx . To evaluate this last integral, we integrate by parts again. For then integraldisplay 1 xe x dx = bracketleftBig xe x bracketrightBig 1- integraldisplay 1 e x dx = bracketleftBig ( x- 1) e x bracketrightBig 1 . Thus I = bracketleftBig (4 x 2 + 1) e x- 8( x- 1) e x bracketrightBig 1 = bracketleftBig (4 x 2- 8 x + 9) e x bracketrightBig 1 ....
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## This note was uploaded on 11/30/2010 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.

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HW15-solutions - husain(aih243 – HW15 – Gilbert...

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