# HW8 - nguyen (kdn444) Assignment 8 guntel (55200) This...

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nguyen (kdn444) – Assignment 8 – guntel – (55200)1Thisprint-outshouldhave24questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001(part1of3)10.0points(i) Express the improper integralI=integraldisplay22xe-x/3dxas limt→ ∞Itwith eachIta proper integral.1.I=limt→ ∞integraldisplayt-t2xe-x/3dx2.I=limt→ ∞integraldisplayt22xe-x/3dxcorrect3.I=limt→ ∞integraldisplayt2xe-x/3dx4.I=limt→ ∞integraldisplay22xe-x/tdx5.I=limt→ ∞integraldisplay2+t2xe-x/3dxExplanation:The integralI=integraldisplay22xe-x/3dxis improper because of the infinite range ofintegration. To overcome this we restrict to afinite interval of integration and consider thelimitI=limt→ ∞It,It=integraldisplayt22xe-x/3dx.002(part2of3)10.0points(ii) Compute the value ofIt.4.It= 30e-236(t+ 3)e-t/3correct5.It= 30e-23+ 6(t+ 3)e-t/3Explanation:To evaluateIt=integraldisplayt22xe-x/3dxwe integrate by parts. ThenIt=bracketleftBig6xe-x/3bracketrightBigt2+ 6integraldisplayt2e-x/3dx=6bracketleftBig(x+ 3)e-x/3bracketrightBigt2.Consequently,It= 30e-236(t+ 3)e-t/3.003(part3of3)10.0points(iii) Determine iflimt→ ∞Itexists, and if it does,find its value.1.I= 6e232.I= 30e-23correct3.Idoes not exist4.I= 30e235.I= 6e-23Explanation:By L’Hospital’s Rule,limx→ ∞xme-x= 0for any powermofx. It follows thatlimt
nguyen (kdn444) – Assignment 8 – guntel – (55200)200410.0pointsDetermine if the improper integralI=integraldisplay-5-∞14vdvis convergent or divergent, and if convergent,find its value.1.I= 32.I= 63.I=924.I=325.Iis divergentcorrectExplanation:The integral is improper because of the in-finite interval of integration. It will convergeiflimt→ ∞integraldisplay-5-t14vdvexists. Nowintegraldisplay-5-t14vdv=bracketleftBig24vbracketrightBig-5-t=6 + 24 +t .But we know thatlimt→ ∞4 +t=.Consequently,Iis divergent.00510.0pointsDetermine if the improper integralI=integraldisplaye1x(ln 4x)2dxconverges, and if it does, compute its value.1.Idoes not converge2.I=4ln 4e3.I= ln 4e4.I= 45.I=1ln 4ecorrect6.I=14eExplanation:The integral is improper because of the in-finite interval of integration, so we writeI=limt→ ∞It,It=integraldisplayte1x(ln 4x)2dx,whenever the limit exists. To evaluateIt, firstsetu= ln 4x. Thenintegraldisplay1x(ln 4x)2dx=integraldisplay1u2du=1u+C ,and soIt=bracketleftBig1ln 4xbracketrightBigte=parenleftBig1ln 4e1ln 4tparenrightBig.On the other hand,limt→ ∞1ln 4t= 0.Consequently, limt→ ∞Itexists, andI=limt→ ∞parenleftBig1ln 4e1ln 4tparenrightBig=1ln 4e.00610.0pointsDetermine if the improper integralI=integraldisplay12 tan-1x1 +x2dxconverges, and if it does, find its value.
nguyen (kdn444) – Assignment 8 – guntel – (55200)31.I=316π2correct2.I=3163.I=144.I

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