EXAM 2 Review-solutions

# EXAM 2 Review-solutions - Van Ligten(hlv63 EXAM 2 Review...

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Van Ligten (hlv63) – EXAM 2 Review – Gilbert – (56650) 2 2. I = 5 ln 12 3. I = 24 ln 12 4. I = 12 ln 5 5. I = 12 ln 24 6. I = 5 ln 24 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 (ln 5) x + 12 parenleftbigg 1 x parenrightbiggbracketrightbigg dx = bracketleftbigg parenleftbigg x 2 2 parenrightbigg ln 5 + 12 ln x bracketrightbigg 5 1 . Consequently, I = 24 ln 5 . 004 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A xy 2 x 2 + 3 dxdy when A = braceleftBig ( x, y ) : 0 x 1 , - 3 y 3 bracerightBig . 1. I = 9 2 ln parenleftbigg 4 3 parenrightbigg 2. I = 9 ln (2) 3. I = 9 2 ln (2) 4. I = 27 4 ln parenleftbigg 4 3 parenrightbigg 5. I = 27 4 ln (2) 6. I = 9 ln parenleftbigg 4 3 parenrightbigg correct Explanation: Since A = braceleftBig ( x, y ) : 0 x 1 , - 3 y 3 bracerightBig is a rectangle with sides parallel to the coor- dinate axes, the double integal can be inter- preted as the iterated integral integraldisplay 3 3 parenleftbiggintegraldisplay 1 0 xy 2 x 2 + 3 dx parenrightbigg dy . But to integrate integraldisplay 1 0 xy 2 x 2 + 3 dx with respect to x with y fixed we use the substitution u = x 2 + 3. For then du = 2 x dx while x = 0 = u = 3 , x = 1 = u = 4 . In this case integraldisplay 1 0 xy 2 x 2 + 3 dx = 1 2 integraldisplay 4 3 y 2 u du = y 2 2 bracketleftBig ln u bracketrightBig 4 3 .
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