M408m-Luecke_HW9

# M408m-Luecke_HW9 - romasko(qrr58 – Assignment 9 –...

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Unformatted text preview: romasko (qrr58) – Assignment 9 – luecke – (55035) 1 This print-out should have 18 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 + 3 x 2 y ) dydx . 1. I = 9 2 2. I = 3 3. I = 4 4. I = 5 2 5. I = 7 2 correct Explanation: The integral can be written in iterated form I = integraldisplay 1 parenleftBig integraldisplay 2 1 (4 x + 3 x 2 y ) dy parenrightBig dx . Now integraldisplay 2 1 (4 x + 3 x 2 y ) dy = bracketleftBig 4 xy + 3 2 x 2 y 2 bracketrightBig 2 1 = 4 x + 9 2 x 2 . But then I = integraldisplay 1 (4 x + 9 2 x 2 ) dx = bracketleftBig 2 x 2 + 3 2 x 3 bracketrightBig 1 . Consequently, I = 7 2 . keywords: definite integral, iterated integral, polynomial function, 002 10.0 points Evaluate the iterated integral I = integraldisplay 2 1 braceleftBig integraldisplay 2 1 ( x + y ) 2 dx bracerightBig dy . 1. I = ln 3 2 correct 2. I = 2 ln 3 2 3. I = 1 2 ln 3 2 4. I = ln 2 3 5. I = 1 2 ln 2 3 6. I = 2 ln 2 3 Explanation: Integrating the inner integral with respect to x keeping y fixed, we see that integraldisplay 2 1 ( x + y ) 2 dx = bracketleftBig- 1 x + y bracketrightBig 2 = braceleftBig 1 y- 1 2 + y bracerightBig . In this case I = integraldisplay 2 1 braceleftBig 1 y- 1 2 + y bracerightBig dy = bracketleftBig ln y- ln(2 + y ) bracketrightBig 2 1 . Consequently, I = ln parenleftBig (2)(1 + 2) (2 + 2) parenrightBig = ln 3 2 . 003 10.0 points Evaluate the iterated integral I = integraldisplay 5 1 braceleftBig integraldisplay 5 1 parenleftBig x y + y x parenrightBig dy bracerightBig dx . romasko (qrr58) – Assignment 9 – luecke – (55035) 2 1. I = 12 ln5 2. I = 5 ln24 3. I = 24ln 12 4. I = 12 ln24 5. I = 24ln 5 correct 6. I = 5 ln12 Explanation: Integrating with respect to y keeping x fixed, we see that integraldisplay 5 1 parenleftbigg x y + y x parenrightbigg dy = bracketleftbigg x ln y + y 2 2 x bracketrightbigg 5 1 = (ln 5) x + 12 parenleftbigg 1 x parenrightbigg . Thus I = integraldisplay 5 1 bracketleftbigg (ln5) x + 12 parenleftbigg 1 x parenrightbiggbracketrightbigg dx = bracketleftbiggparenleftbigg x 2 2 parenrightbigg ln5 + 12 ln x bracketrightbigg 5 1 . Consequently, I = 24 ln5 . 004 10.0 points Evaluate the iterated integral I = integraldisplay ln 6 parenleftBigg integraldisplay ln 5 e 2 x- y dx parenrightBigg dy . 1. I = 10 correct 2. I = 9 3. I = 8 4. I = 7 5. I = 6 Explanation: Integrating with respect to x with y fixed, we see that integraldisplay ln 5 e 2 x- y dx = 1 2 bracketleftBig e 2 x- y bracketrightBig ln 5 = 1 2 parenleftBig e 2 ln 5- y- e- y parenrightBig = parenleftBig 5 2- 1 2 parenrightBig e- y ....
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