# HW10-solutions - chao(nc23585 HW10 pool(53705 This...

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chao (nc23585) – HW10 – pool – (53705)1Thisprint-outshouldhave25questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsDetermine the value of the iterated integralI=integraldisplay20integraldisplay21(1 + 2xy)dx dy .1.I= 102.I= 143.I= 124.I= 165.I= 8correctExplanation:Integrating with respect toxand holdingyfixed, we see thatintegraldisplay21(1 + 2xy)dx=bracketleftBigx+x2ybracketrightBigx=2x=1.ThusI=integraldisplay20(1 + 3y)dy=bracketleftBigy+32y2bracketrightBig20.Consequently,I= 2 + 6 = 8.keywords:00210.0pointsEvaluate the double integralI=integraldisplay integraldisplayA(3x+ 2y)dxdywhenA=braceleftBig(x, y) : 0x2,0y1bracerightBig.1.I= 122.I= 113.I= 104.I= 95.I= 8correctExplanation:SinceA=braceleftBig(x, y) : 0x2,0y1bracerightBigis a rectangle with sides parallel to the coor-dinate axes, the value ofIcan be found byinterpreting the double integral as the iter-ated integralI=integraldisplay10integraldisplay20(3x+ 2y)dxdy .But now after integration with respect toxkeepingyfixed, we see thatintegraldisplay20(3x+ 2y)dx=bracketleftBig32x2+ 2xybracketrightBig20.ThusI=integraldisplay20(6 + 4y)dy=bracketleftBig6y+ 2y2bracketrightBig10.Consequently,I= 8.00310.0pointsEvaluate the iterated integralI=integraldisplay41integraldisplay401(x+y)2dx dy .1.I=12lnparenleftbigg5parenrightbigg
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chao (nc23585) – HW10 – pool – (53705)22.I=12lnparenleftbigg127parenrightbigg3.I= lnparenleftbigg127parenrightbigg4.I= 2 lnparenleftbigg127parenrightbigg5.I= 2 lnparenleftbigg52parenrightbigg6.I= lnparenleftbigg52parenrightbiggcorrectExplanation:Integrating the inner integral with respecttoxkeepingyfixed, we see thatintegraldisplay401(x+y)2dx=bracketleftBig-1x+ybracketrightBig40=1y-14 +y.In this caseI=integraldisplay41parenleftbigg1y-14 +yparenrightbiggdy=bracketleftBigln(y)-ln(4 +y)bracketrightBig4.1.I= 3 ln (2)correct2.I= 2 ln (3)3.I= 2 lnparenleftbigg32parenrightbigg4.I=32ln (3)5.I= 3 lnparenleftbigg32parenrightbigg6.I=32ln (2)Explanation:Integrating with respect toykeepingxfixed, we see thatintegraldisplay21parenleftbiggxy+yxparenrightbiggdy=bracketleftbiggxln(y) +y22xbracketrightbigg21= (ln(2))x+32parenleftbigg1xparenrightbigg.ThusI=integraldisplay21bracketleftbigg(ln(2))x+32parenleftbigg1xparenrightbiggbracketrightbiggdx=bracketleftbiggparenleftbiggx22parenrightbiggln(2) +32ln(x)bracketrightbigg21.Consequently,I= 3 ln(2).005After integration with respect tox,I=integraldisplay32bracketleftbigex-ybracketrightbig20dy=integraldisplay3(e2-y-e-y)dy .
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Consequently,I= lnparenleftBig(4)(1 + 4)(4 + 4)parenrightBig= lnparenleftbigg52parenrightbigg.00410.0pointsEvaluate the iterated integralI=integraldisplay21integraldisplay21parenleftBigxy+yxparenrightBigdydx .
10.0pointsEvaluate the double integralI=integraldisplay32integraldisplay20ex-ydxdy .1.I=e-3-e-2-e-1-12.I=e-3-e-2-e-1+ 1correct3.I=e-3+e-2-e-1+ 14.I=e-3-e-2+e-1+ 1Explanation:
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chao (nc23585) – HW10 – pool – (53705)3But then, after integrating next with respecttoywe see thatI=bracketleftbig-e2-y+e-ybracketrightbig32=-e-1+e-3-(-1 +e-2).Consequently,I=e-3-e-2-e-1+ 1.00610.0pointsEvaluate the double integralI=integraldisplay integraldisplayAxy2x2+ 4dx dywhenA=braceleftBig(x, y) : 0x2,0y3bracerightBig.1.I= 9 ln (2)2.I=274lnparenleftbigg52parenrightbigg3.I= 9 lnparenleftbigg52parenrightbigg4.I=92lnparenleftbigg52parenrightbigg5.I=274ln (2)6.I=92ln (2)correctExplanation:SinceA=braceleftBig(x, y) : 0x2,0y3bracerightBigis a rectangle with sides parallel to the coor-dinate axes, the double integal can be inter-preted as the iterated integralintegraldisplay30integraldisplay20xy2x2+ 4dx dy .But to integrateintegraldisplay20xy2x2+ 4dxwith respect toxwithyfixed we use thesubstitutionu=x2+ 4. For thendu= 2x dxwhilex= 0=u= 4,x= 2=u= 8.