# HW10-solutions - padilla(tp5647 HW10 cheng(57455 This...

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Unformatted text preview: padilla (tp5647) – HW10 – cheng – (57455) 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 1 integraldisplay 2 1 (4 x + 2 x 2 y ) dydx . 1. I = 1 2. I = 3 2 3. I = 3 correct 4. I = 5 2 5. I = 2 Explanation: The integral can be written in iterated form I = integraldisplay 1 parenleftBig integraldisplay 2 1 (4 x + 2 x 2 y ) dy parenrightBig dx . Now integraldisplay 2 1 (4 x + 2 x 2 y ) dy = bracketleftBig 4 xy + x 2 y 2 bracketrightBig 2 1 = 4 x + 3 x 2 . But then I = integraldisplay 1 (4 x + 3 x 2 ) dx = bracketleftBig 2 x 2 + x 3 bracketrightBig 1 . Consequently, I = 3 . keywords: definite integral, iterated integral, polynomial function, 002 10.0 points Evaluate the iterated integral I = integraldisplay 4 1 braceleftBig integraldisplay 4 1 parenleftBig x y + y x parenrightBig dy bracerightBig dx . 1. I = 4 ln 15 2 2. I = 15 ln4 correct 3. I = 15 2 ln15 4. I = 15 2 ln4 5. I = 4 ln15 6. I = 15 ln 15 2 Explanation: Integrating with respect to y keeping x fixed, we see that integraldisplay 4 1 parenleftbigg x y + y x parenrightbigg dy = bracketleftbigg x ln y + y 2 2 x bracketrightbigg 4 1 = (ln4) x + 15 2 parenleftbigg 1 x parenrightbigg . Thus I = integraldisplay 4 1 bracketleftbigg (ln4) x + 15 2 parenleftbigg 1 x parenrightbiggbracketrightbigg dx = bracketleftbiggparenleftbigg x 2 2 parenrightbigg ln4 + 15 2 ln x bracketrightbigg 4 1 . Consequently, I = 15ln4 . 003 10.0 points Determine the value of the double integral I = integraldisplay integraldisplay A 3 xy 2 4 + x 2 dA padilla (tp5647) – HW10 – cheng – (57455) 2 over the rectangle A = braceleftBig ( x, y ) : 0 ≤ x ≤ 3 ,- 3 ≤ y ≤ 3 bracerightBig , integrating first with respect to y . 1. I = 27 2 ln parenleftBig 13 4 parenrightBig 2. I = 27 2 ln parenleftBig 4 13 parenrightBig 3. I = 27 2 ln parenleftBig 13 8 parenrightBig 4. I = 27 ln parenleftBig 4 13 parenrightBig 5. I = 27 ln parenleftBig 13 4 parenrightBig correct 6. I = 27 ln parenleftBig 13 8 parenrightBig Explanation: The double integral over the rectangle A can be represented as the iterated integral I = integraldisplay 3 parenleftbiggintegraldisplay 3- 3 3 xy 2 4 + x 2 dy parenrightbigg dx , integrating first with respect to y . Now after integration with respect to y with x fixed, we see that integraldisplay 3- 3 3 xy 2 4 + x 2 dy = bracketleftBig xy 3 4 + x 2 bracketrightBig 3- 3 = 54 x 4 + x 2 . But integraldisplay 3 54 x 4 + x 2 dx = bracketleftBig 27 ln(4 + x 2 ) bracketrightBig 3 . Consequently, I = 27 ln parenleftBig 13 4 parenrightBig . 004 10.0 points Calculate the value of the double integral I = integraldisplay integraldisplay A x cos( x + y ) dxdy when A is the rectangle braceleftBig ( x, y ) : 0 ≤ x ≤ π 4 , ≤ y ≤ π 4 bracerightBig ....
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