HW07-solutions - kim(tk5895 – HW07 – Henry –(54974 1...

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Unformatted text preview: kim (tk5895) – HW07 – Henry – (54974) 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 2 x ln x dx . 1. I = x 2 ln x- 2 x 2 + C 2. I = x 2 ln x- 1 2 x 2 + C correct 3. I = 2 x 2 ln x- x 2 + C 4. I = 2 x 2 ln x- 1 2 x 2 + C 5. I = 2 x 2 ln x + 1 2 x 2 + C 6. I = x 2 ln x + 1 2 x 2 + C Explanation: After integration by parts, I = x 2 ln x- integraldisplay x 2 parenleftBig 1 x parenrightBig dx = x 2 ln x- integraldisplay x dx . Consequently, I = x 2 ln x- 1 2 x 2 + C with C an arbitrary constant. 002 10.0 points Evaluate the definite integral I = integraldisplay e 1 20 ln x x 5 dx. 1. I = 5 6 parenleftBig 1 + 5 e 6 parenrightBig 2. I = 5 6 parenleftBig 1- 5 e 6 parenrightBig 3. I = 5 4 parenleftBig 1- 5 e 4 parenrightBig correct 4. I =- 5 e 4 5. I = 5 4 parenleftBig 1 + 5 e 4 parenrightBig Explanation: After integration by parts integraldisplay e 1 4 ln x x 5 dx = bracketleftBig- ln x x 4 bracketrightBig e 1 + integraldisplay e 1 1 x 5 dx. But bracketleftBig- ln x x 4 bracketrightBig e 1 =- 1 e 4 , since ln e = 1 and ln1 = 0, while integraldisplay e 1 1 x 5 dx = 1 4 parenleftbigg 1- 1 e 4 parenrightbigg . Thus I = 5 4 parenleftbigg 1- 5 e 4 parenrightbigg . 003 10.0 points Evaluate the integral I = integraldisplay 4 1 ln t 4 √ t dt . 1. I = ln 4 + 1 2. I = 2(ln 2- 1) 3. I = 1 4 (ln2 + 1) 4. I = 2(ln 2 + 1) 5. I = 1 4 (ln4- 1) kim (tk5895) – HW07 – Henry – (54974) 2 6. I = ln 4- 1 correct Explanation: After integration by parts, I = 1 2 bracketleftBig √ t ln t bracketrightBig 4 1- 1 2 integraldisplay 4 1 √ t parenleftBig 1 t parenrightBig dt = ln4- 1 2 integraldisplay 4 1 1 √ t dt . But integraldisplay 4 1 1 √ t dt = 2 bracketleftBig √ t bracketrightBig 4 1 . Consequently, I = ln4- 1 . keywords: integration by parts, logarithmic functions 004 10.0 points Determine the integral I = integraldisplay ( x 2 + 1) cos2 x dx . 1. I =- x 2 cos 2 x + x sin 2 x- 3 2 cos 2 x + C 2. I = 1 4 parenleftBig 2 x sin2 x- (2 x 2 +1) cos2 x parenrightBig + C 3. I = 1 4 parenleftBig 2 x cos 2 x +(2 x 2 +1) sin 2 x parenrightBig + C correct 4. I = 1 2 parenleftBig 2 x cos2 x- (2 x 2 +1) sin2 x parenrightBig + C 5. I = 1 2 x 2 sin 2 x- x cos2 x + 3 2 sin 2 x + C 6. I = 1 2 parenleftBig 2 x cos2 x +(2 x 2 +1) sin2 x parenrightBig + C Explanation: After integration by parts, integraldisplay ( x 2 + 1) cos2 x dx = 1 2 ( x 2 + 1) sin 2 x- 1 2 integraldisplay sin2 x braceleftBig d dx ( x 2 + 1) bracerightBig dx = 1 2 ( x 2 + 1) sin2 x- integraldisplay x sin 2 x dx ....
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HW07-solutions - kim(tk5895 – HW07 – Henry –(54974 1...

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