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©Page|1(A)WHATISINTEGRATION?Integrationistheprocessoffindingtheantiderivative(orintegral)ofagivenfunction.Asstatedinthefundamentaltheoremofcalculus,itisthereverseofdifferentiation.Whiledifferentiationisconcernedwithratesofchange,integrationisbasedontheconceptofsummation.Itoriginatedfromtheproblemoffindingtheareaofregionwithacurvedboundary.Today,integralcalculusisusedtofindvolumes,areasandsolutionstodifferentialequations.Definition:Afunction()Fxisanantiderivativeofafunctionf(x)if()()Fxfxforallxinthedomainoff.Thesetofalltheantiderivativesoffistheindefiniteintegraloffwithrespecttox,denotedby()fxdx(readas‘theintegraloffwithrespecttox‘)Thesymbolisanintegralsign.ItisintroducedbyLeibnizwhichanelongatedletter“S”representingsumma,Latinforsum.()()fxdxFxCForExample:(a)2()2dxxdx2xdx2xC(b)32()3dxxdx(c)43()4dxxdxFromthefigure,thegradientofthefunctiony=x2+Catanypoint(x,y)isalwaysconstant.Sothesolutionof22xdxxCisthefamilyofcurve.(fig1)*Note:Foreachintegration,anarbitraryconstantmustinclude.Figure1Example1:Integrandarbitraryconstant(1,2)©Page|2Giventhaty=(x–4)2x,finddydxandhencefind2xdxx.[solution]12(4)(2)yxxdydx(B)StandardIntegralsBydifferentiation,since11nndxxdxn,1,11nnxxdxcnn(1)Example2:(i)352xdxx=3552()xdxxx(ii)21xdxx(iii)2223()xdxx(iv)2()xxdx=25(2)xxdx=412()4xx+C=4112CxxExample3:Giventhatthegradientofacurve,atthepoint(x,y)onthecurveisgivenby2-3x+4x2.Giventhatthecurvepassesthroughthepoint(1,1),findtheequationofcurve.[solution:]Given2234dyxxdxy=2(234)xxdx=2234dxxdxxdxy=2x-233423xxcsincethecurvepassesthrough(1,1),whenx=1,y=1=2(1)-2334(1)(1)23c=1c=56theequationofthecurveisy=233452236xxx.Example4:Hint:SomepropertiesofIndefiniteIntegral:1.2.,kisconstant,k03.©Page|3Giventhat32(1)(32)dyxxxdx,andy=2whenx=1,findyintermsofx.[y=x4–x3+x2–2x+3]Example5:Example6:Thegradientofacurve,atthepoint(x,y)Foracurvey=f(x),22dydx=6x-2,giventhaty=11onthecurve,isgivenby224xx.Giventhatanddydx=10whenx=2.Findthecoordinatesofthethecurvepassesthroughthepoint(2,7),findpointatwhichthecurvepassingthroughtheyaxis.theequationofthecurve.[y=x+4x+3][Ans:(0,3)]C.IntegrationofafunctioninvolvingaLinearFactorSince1()(1)().(1)()nnndaxbnaxbaanaxbdx,a,bareconstant1(1)()()nnanaxbdxaxba(n+1)()naxbdx=(ax+b)n+111()()(1)nnaxbdxaxbcan,cisarbitraryconstant.(2),n-1,a0©Page|4Example7:Integratethefollowingwithrespecttox:(a)6(27)xdx(b)54xdx(c)3423dxx(d)353xdx=7(27)7(2)xc=12(54)xdx=364(23)dxx=71(27)14xc=32(54)3()(5)2x+c=364(23)xdx=322(54)15xc=2(23)64[](2)(2)xc=216(23)cxExample8:Integratethefollowingwithrespecttox:(a)3(23)xdx(b)335(12)dxx(c)()mpxqdx(d)455(32)dxxExample9:Ifdydx=2249(1)(2)xxandy=0whenx=0,findywhenx=7.[0]Example10:Givendydx=ax–8,andthegradientoftangenttothecurvey=f(x)atpoint(4,7)is8,findthevalueofaandhencefindtheequationofthecurve.©Page|5Revision:1.Integratethefollowingwithrespecttox:(a)23213xx(b)332xx(c)3xxx(d)21()xx(e)(3)xx(f)2(2)xx(g)32435xx(h)3214xxx2.Findthefollowingindefiniteintegrals:(a)352xdxx(b)2223()xdxx(c)4(7)(7)xxdxx(d)223(1)xxdxx(e)1334(37)xxdx(f)5(23)(4)xxdxx(g)(1)(2)xxdxx(h)24(25)xdxx©Page|63.Findthefollowingindefiniteintegral:(form()naxbdx)4.Given221yx,showthat2221dyxdxx.Henceorotherwise,find221xdxx.5.Acurveissuchthat2(4),0dyxxdxx.Giventhatthecurvepassesthroughthepoint(9,30),findtheequationofthecurve.6.Giventhatthegradientfunctionofthecurveat(x,y)is(1)(35)dyxxdx,findtheequationofthecurvewhichpassingthroughthepoint(1,0).(a)5(32)xdx(b)23(41)dxx(c)pxqdx(d)311dxx(e)723dxx(f)31()91dxx(g)3(31)xdx2337dxx©Page|77.Acurvehasagradientof10atthepoint(2,5)and2264dyxdx.Findtheequationofthecurve.8.Thegradientofthenormaltoacurveis2392xx.Giventhatthecurvepassesthroughthepoint(3,5).Findtheequationofthecurve.9.Acurveforwhich22dykxdxwherekisaconstant,passesthrough(-1,3/5).Thenormalat(-1,3/5)isparallelto4y=5x-3.Findthevalueofkandhencefindtheequationofthecurve.10.Giventhatdydxisdirectlyproportionaltox2-1,andthaty=3anddydx=9whenx=2,findyintermsofx.11.TherateofchangeofV,thevolumeofasolidobject,isgivenby26(21)1dVtdt.GiventhatV=4whent=1,findVasafunctionoft.©Page|812.AparticlemovingalongastraightlinepassesthroughafixedpointOwithavelocityof24m/s.Itsacceleration,am/s2,isgivenbya=18t–36,wheretisthetimeinsecondsafterpassingthroughO.FindthedistanceoftheparticlefromOafter3seconds.13.Evaluatethefollowingdefiniteintegrals.(a)23211(82)2xdxx(b)92112xdxx(c)41(63)xxdx(d)01(2)(2)xxxdx(e)4021xdx(f)120(23)xxdx14.Giventhat2(32)6mxdx,wherem-2,findthevalueofm.(ans:m=4)15.Given4611()8,()fxdxfxdx=20and82()10gxdx,find(i)6842()3()2gxfxdxdx;(ii)if413[()]114fxkxdx,findthevalueofk.[CHAPTER9:INTEGRATION]©Page|9
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