Riemann integral.docx
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Riemann integral.docx
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Riemannintegral
Riemannintegral
FromWikipedia,thefreeencyclopedia
Theintegralastheareaofaregionunderacurve.
AsequenceofRiemannsumsoveraregularpartitionofaninterval.Thenumberontopisthetotalareaoftherectangles,whichconvergestotheintegralofthefunction.
Thepartitiondoesnotneedtoberegular,asshownhere.Theapproximationworksaslongasthewidthofeachsubdivisiontendstozero.
Inthebranchofmathematicsknownasrealanalysis,theRiemannintegral,createdbyBernhardRiemann,wasthefirstrigorousdefinitionoftheintegralofafunctiononaninterval.[1]Formanyfunctionsandpracticalapplications,theRiemannintegralcanbeevaluatedbythefundamentaltheoremofcalculusorapproximatedbynumericalintegration.
TheRiemannintegralisunsuitableformanytheoreticalpurposes.SomeofthetechnicaldeficienciesinRiemannintegrationcanberemediedwiththeRiemann–Stieltjesintegral,andmostdisappearwiththeLebesgueintegral.
Contents
∙1Overview
∙2Definition
o2.1Partitionsofaninterval
o2.2Riemannsums
o2.3Riemannintegral
∙3Examples
∙4Similarconcepts
∙5Properties
o5.1Linearity
∙6Integrability
∙7Generalizations
∙8Seealso
∙9Notes
∙10References
∙11Externallinks
Overview
Letfbeanon-negativereal-valuedfunctionontheinterval[a,b],andlet
betheregionoftheplaneunderthegraphofthefunctionfandabovetheinterval[a,b](seethefigureonthetopright).WeareinterestedinmeasuringtheareaofS.Oncewehavemeasuredit,wewilldenotetheareaby:
ThebasicideaoftheRiemannintegralistouseverysimpleapproximationsfortheareaofS.Bytakingbetterandbetterapproximations,wecansaythat"inthelimit"wegetexactlytheareaofSunderthecurve.
Notethatwherefcanbebothpositiveandnegative,thedefinitionofSismodifiedsothattheintegralcorrespondstothesignedareaunderthegraphoff:
thatis,theareaabovethex-axisminustheareabelowthex-axis.
Definition
Partitionsofaninterval
Apartitionofaninterval[a,b]isafinitesequenceofnumbersoftheform
Each[xi,xi+1]iscalledasubintervalofthepartition.Themeshornormofapartitionisdefinedtobethelengthofthelongestsubinterval,thatis,
Ataggedpartition
ofaninterval[a,b]isapartitiontogetherwithafinitesequenceofnumbers
subjecttotheconditionsthatforeachi,
.Inotherwords,itisapartitiontogetherwithadistinguishedpointofeverysubinterval.Themeshofataggedpartitionisthesameasthatofanordinarypartition.
Supposethattwopartitions
and
arebothpartitionsoftheinterval[a,b].Wesaythat
isarefinementof
ifforeachintegeri,with
thereexistsaninteger
suchthat
andsuchthat
forsomejwith
.Saidmoresimply,arefinementofataggedpartitionbreaksupsomeofthesubintervalsandaddstagstothepartitionwherenecessary,thusit"refines"theaccuracyofthepartition.
Wecandefineapartialorderonthesetofalltaggedpartitionsbysayingthatonetaggedpartitionisgreaterorequaltoanotheriftheformerisarefinementofthelatter.
Riemannsums
Chooseareal-valuedfunctionfwhichisdefinedontheinterval[a,b].TheRiemannsumoffwithrespecttothetaggedpartitionx0,…,xntogetherwitht0,…,tn−1is:
[2]
Eachterminthesumistheproductofthevalueofthefunctionatagivenpoint,andthelengthofaninterval.Consequently,eachtermrepresentsthe(signed)areaofarectanglewithheightf (ti)andwidthxi+1−xi.TheRiemannsumisthe(signed)areaofalltherectangles.
Riemannintegral
Looselyspeaking,theRiemannintegralisthelimitoftheRiemannsumsofafunctionasthepartitionsgetfiner.Ifthelimitexiststhenthefunctionissaidtobeintegrable(ormorespecificallyRiemann-integrable).TheRiemannsumcanbemadeascloseasdesiredtotheRiemannintegralbymakingthepartitionfineenough.[3]
Oneimportantrequirementisthatthemeshofthepartitionsmustbecomesmallerandsmaller,sothatinthelimit,itiszero.Ifthiswerenotso,thenwewouldnotbegettingagoodapproximationtothefunctiononcertainsubintervals.Infact,thisisenoughtodefineanintegral.Tobespecific,wesaythattheRiemannintegraloffequalssifthefollowingconditionholds:
Forallε>0,thereexistsδsuchthatforanytaggedpartition
and
whosemeshislessthanδ,wehave
Unfortunately,thisdefinitionisverydifficulttouse.ItwouldhelptodevelopanequivalentdefinitionoftheRiemannintegralwhichiseasiertoworkwith.Wedevelopthisdefinitionnow,withaproofofequivalencefollowing.OurnewdefinitionsaysthattheRiemannintegraloffequalssifthefollowingconditionholds:
Forallε>0,thereexistsataggedpartition
and
suchthatforanytaggedpartition
and
whichisarefinementof
and
wehave
Bothofthesemeanthateventually,theRiemannsumoffwithrespecttoanypartitiongetstrappedclosetos.Sincethisistruenomatterhowclosewedemandthesumsbetrapped,wesaythattheRiemannsumsconvergetos.Thesedefinitionsareactuallyaspecialcaseofamoregeneralconcept,anet.
Aswestatedearlier,thesetwodefinitionsareequivalent.Inotherwords,sworksinthefirstdefinitionifandonlyifsworksintheseconddefinition.Toshowthatthefirstdefinitionimpliesthesecond,startwithanε,andchooseaδthatsatisfiesthecondition.Chooseanytaggedpartitionwhosemeshislessthanδ.ItsRiemannsumiswithinεofs,andanyrefinementofthispartitionwillalsohavemeshlessthanδ,sotheRiemannsumoftherefinementwillalsobewithinεofs.
Toshowthattheseconddefinitionimpliesthefirst,itiseasiesttousetheDarbouxintegral.FirstoneshowsthattheseconddefinitionisequivalenttothedefinitionoftheDarbouxintegral;forthisseethearticleonDarbouxintegration.NowwewillshowthataDarbouxintegrablefunctionsatisfiesthefirstdefinition.Fixε,andchooseapartition
suchthatthelowerandupperDarbouxsumswithrespecttothispartitionarewithin
ofthevaluesoftheDarbouxintegral.Let
Ifr=0,thenfisthezerofunction,whichisclearlybothDarbouxandRiemannintegrablewithintegralzero.Thereforewewillassumethatr>0.Ifm>1,thenwechooseδsuchthat
Ifm=1,thenwechooseδtobelessthanone.Chooseataggedpartition
and
withmeshsmallerthanδ.WemustshowthattheRiemannsumiswithinεofs.
Toseethis,chooseaninterval
.Ifthisintervaliscontainedwithinsome
then
wheremjandMjarerespectively,theinfimumandthesupremumoffon
.Ifallintervalshadthisproperty,thenthiswouldconcludetheproof,becauseeachtermintheRiemannsumwouldbeboundedacorrespondingtermintheDarbouxsums,andwechosetheDarbouxsumstobenears.Thisisthecasewhenm=1,sotheproofisfinishedinthatcase.
Therefore,wemayassumethatm>1.Inthiscase,itispossiblethatoneofthe
isnotcontainedinany
.Instead,itmaystretchacrosstwooftheintervalsdeterminedby
.(Itcannotmeetthreeintervalsbecauseδisassumedtobesmallerthanthelengthofanyoneinterval.)Insymbols,itmayhappenthat
(Wemayassumethatalltheinequalitiesarestrictbecauseotherwiseweareinthepreviouscasebyourassumptiononthelengthofδ.)Thiscanhappenatmostm−1times.
Tohandlethiscase,wewillestimatethedifferencebetweentheRiemannsumandtheDarbouxsumbysubdividingthepartition
at
.Theterm
intheRiemannsumsplitsintotwoterms:
Suppose,withoutlossofgenerality,that
.Then
sothistermisboundedbythecorrespondingtermintheDarbouxsumforyj.Toboundtheotherterm,noticethat
Itfollows:
Sincethishappensatmostm−1times,thetotalofallthetermswhicharenotboundedbytheDarbouxsumisatmost
.ThereforethedistancebetweentheRiemannsumandsisatmost ε.
Examples
Let
bethefunctionwhichtakesthevalue1ateverypoint.AnyRiemannsumoffon[0,1]willhavethevalue1,thereforetheRiemannintegraloffon[0,1]is1.
Let
betheindicatorfunctionoftherationalnumbersin[0,1];thatis,IQtakesthevalue1onrationalnumbersand0onirrationalnumbers.ThisfunctiondoesnothaveaRiemannintegral.Toprovethis,wewillshowhowtoconstructtaggedpartitionswhoseRiemannsumsgetarbitrarilyclosetobothzeroandone.
Tostart,let
and
beataggedpartition(eachtiisbetweenxiand
).Chooseε>0.Thetihavealreadybeenchosen,andwecan'tchangethevalueoffatthosepoints.Butifwecutthepartitionintotinypiecesaroundeachti,wecanminimizetheeffectoftheti.Then,bycarefullychoosingthenewtags,wecanmakethevalueoftheRiemannsumturnouttobewithinεofeitherzeroorone—ourchoice!
Ourfirststepistocutupthepartition.Therearenoftheti,andwewanttheirtotaleffecttobelessthanε.Ifweconfineeachofthemtoanintervaloflengthlessthan
thenthecontributionofeachtitotheRiemannsumwillbeatleast
andatmost
.Thismakesthetotalsumatleastzeroandatmostε.Soletδbeapositivenumberlessthan
.Ifithappensthattwoofthetiarewithinδofeachother,chooseδsmaller.Ifithappensthatsometiiswithinδofsomexj,andtiisnotequaltoxj,chooseδsmaller.Sincethereareonlyfinitelymanytiandxj,wecanalwayschooseδsufficientlysmall.
Nowweaddtwocutstothepartitionforeachti.Oneofthecutswillbeat
andtheotherwillbeat
.Ifoneoftheseleavestheinterval[0,1],thenweleaveitout.tiwillbethetagcorrespondingtothesubinte
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