高等数学英文版课件PPT 05 IntegralsPPT格式课件下载.pptx
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FirstapproximatetheregionSbypolygons,andthentakethelimitoftheareaofthesepolygons.(seethefollowingexample)Example1Findtheareaundertheparabolay=x2from0to1.ySolution:
Westartbydividingtheinterval0,1inton-subintervalswithequallength,andconsidertherectangleswhosebasesarethesesubintervalsandwhoseheightsarethevaluesofthefunctionattheright-handendpoints.o(1,1)1/n1xFigure2机动目录上页下页返回结束ThenthesumoftheareaoftheserectanglesisAsnincreases,Snbecomesabetterandbetterapproximationtotheareaoftheparabolicsegment.ThereforewedefinetheareaAtobethelimitofthesumsoftheareasoftheserectangles,thatis,ApplyingtheideaofExample1tothemoregeneralregionSofF.1,weintroducethedefinitionoftheareaasfollowing:
Step1:
Partition-Dividetheintervala,bintonsmallersubintervalsbychoosingpartitionpointsx0,x1,x2,.,xnsothata=x0x1x2xn=b机动目录上页下页返回结束Thissubdivisioniscalledapartitionofa,bandwedenoteitbyP.Letdenotethelengthofithsubintervalxi-1,xi,and|P|(thenormofP)denotesthelengthoflongestsubinterval.ThusStep2:
ApproximationBythepartitionabove,theareaofScanbeapproximatedbythesumofareasofnrectangles.UsingthepartitionPonecandividetheregionSintonstrips(seeF.3).Now,wechooseanumberineachsubinterval,theneachstripSicanbeapproximatedbyarectangleRi(seeF.4).Thesumofareasoftheserectanglesasanapproximationis机动目录上页下页返回结束y=f(x)SibxyoaFigure3XiXi-1S1S2SnR2Ribxyy=f(x)oaFigure4XiXi-1R1Rnapproximatedby机动目录上页下页返回结束Step3:
TakinglimitNoticethattheapproximationappearstobecomebetterandbetterasthestripsbecomethinnerandthinner.Sowedefinetheareaoftheregionasthelimitvalue(ifitexists)ofthesumofareasoftheapproximatingrectangles,thatis
(1)Remark1:
Itcanbeshownthatiffiscontinuous,thenthelimit
(1)doesexist.Remark2:
Instep1,wehavenoneedtodividedtheintervala,bintonsubintervalswithequallength.Butforpurposesofcalculation,itisoftenconvenienttotakeapartitionthatdividestheintervalintonsubintervalswithequallength.(Thisiscalledaregularpartition)机动目录上页下页返回结束tobetheright-handendpoint:
Letuschoosethepoint=xi=2i/nBydefinition,theareaisExample2:
Findtheareaundertheparabolay=x2+1from0to2.SolutionSincey=x2+1iscontinuous,thelimit
(1)mustexistforallpossiblepartitionPoftheintervala,baslongas|P|0.Tosimplifythingsletustakearegularpartition.Thenthepartitionpointsarex0=0,x1=2/n,x2=4/n,xi=2i/n,xn=2n/n=2SothenormofPis|P|=2/n机动目录上页下页返回结束Remark3:
Ifischosentobetheleft-handendpoint,onewillobtainthesameanswer.Example3Findtheareaunderthecosinecurvefrom0tob,whereSolutionWechoosearegularpartitionPsothat|P|=b/nandwechoosetobetheright-handendpointoftheithsub-interval:
=xi=ib/nSince|P|0asn,theareaunderthecosinecurvefrom0tobis机动目录上页下页返回结束/section5.2end机动目录上页下页返回结束5.3TheDefiniteIntegralInChapters6and8wewillseethatlimitofformoccursinawidevarietyofsituationsnotonlyinmathematicsbutalsoinphysics,Chemistry,BiologyandEconomics.Soitisnecessarytogivethistypeoflimitaspecialnameandnotation.1.DefinitionofaDefiniteIntegralIffisafunctiondefinedonaclosedintervala,b,letPbeapartitionofa,bwithpartitionpointsx0,x1,x2,.,xn,wherea=x0x1x20,=theareaunderthegraphofffromatob.Ingeneral,adefiniteintegralcanbeinterpretedasadifferenceofareas:
xyob+a-ofNote5:
Inthecaseofabanda=b,weextendthedefinitionasfollows:
Ifab,thenIfa=b,thenExample1Expressbyinterpretinginasanintegralontheinterval0,.Example2Evaluatetheintegraltermsofareas.机动目录上页下页返回结束SolutionWecomputetheintegralasthedifferenceoftheareasofthetwotriangles:
13o-1xyy=x-1A1A22.ExistenceTheoremTheoremIffiseithercontinuousormonotonicona,b,thenfisintegrableona,b;
thatis,thedefiniteintegralexists.Remark1:
Iffisdiscontinuousatsomepoints,thenmightexistoritmightnotexist.Butiffispiecewisecontinuous,thenfisintegrable.机动目录上页下页返回结束Remark2:
Itcanbeshownthatiffisintegrableona,b,thenfmustbeaboundedfunctionona,b.3.IntegralFormulasunderRegularPartitionIffisintegrableona,b,itisoftenconvenienttotakearegularpartition.Thenandtobetherightendpointineachsubinterval,Ifwechoosethen0asn,sothedefinitionSince|P|=(b-a)/n,wehave|P|gives机动目录上页下页返回结束TheoremIffisintegrableona,b,thenasanintegralontheExample3Expressinterval1,2.Answer:
Ifthepurposeistofindanapproximationtoanintegral,itisusuallybettertochoosewhic