HW15-solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 9

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online