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[DEFINITEINTEGRALANDITSAPPLICATIONS]--------------------------1---------------------------------DefiniteIntegralInthischapter,youshould:Understandingtheideathattheareaunderacurveisthelimitofasumoftheareasofrectangles.Definethedefiniteintegralasalimitofsumsandinterpretitastheareaunderacurve.Thepropertiesofdefiniteintegrals.Findplaneareaofaregionboundedcurve(s)andlines.Findthevolumeofrevolutionaboutthex-ory-axisofaregionboundedbycurve(s)andlines.heDefiniteIntegralhaswiderangingapplicationsinmathematics,thephysicalsciencesandengineering.Manythingswewanttoknowcanbecalculatedwithintegrals:thevolumesofsolids,thelengthofcurves,theplaneareaundercurve,theamountofworkittakestopumpliquidsfrombelowgroundandthecoordinatesofthepointswheresolidobjectsbalance.1.1Areaboundedbycurveandthex–axisConsideraregionAboundedbythex-axis,linesx=a,x=bandacurve()yfxwhichiscontinuousintheintervalaxb.HowdowefindtheareaoftheregionA(infigure1)?Theareaunderacurvecanbeapproximatedbythesumofrectangles.Thefigure1.1ontheleftshowsinscribedrectangleswhilethefigure1.2ontherightshowscircumscribedrectangles.Figure1Figure1.1LowerSumofRectangles(Left-EndPoints)Figure1.2:UpperSumofRectangles(Right-EndPoints)TA[DEFINITEINTEGRALANDITSAPPLICATIONS]--------------------------2---------------------------------Example1:Findtheareaenclosedby2yx,thexaxisandx=1bydividingtheregioninto6equalrectangles.[Solution:]Dividetheregioninto6subdivisionsofwidthx=Thediagramalongsideshowslowerrectangleswithtopedgesattheminimumvalueofcurveonthatinterval.Theareaoflowerrectangles,AL=Thenextdiagramshowsupperrectangles,whicharerectangleswithtopedgesatthemaximumvalueofthecurveonthatinterval.Theareaofupperrectangles,AU=Fromtheabove,wediscoveredthatALAAUIfwedividetheinterval[0,1]intoaverylargenumberofparts,thesumoftheareameasuresoftheshadedrectanglesbecomesverysmall.Asn,ALandAUapproachthesamelimit,A.Lookthefollowingexample:Fig1.3whenn=20andx=120Fig1.4whenn=40,andx=140ix016263646561iy[DEFINITEINTEGRALANDITSAPPLICATIONS]--------------------------3---------------------------------1.2DefiniteIntegrals:Letfbeafunctionwhichiscontinuousontheclosedinterval[a,b].TheDefiniteIntegralofffromatobisdefinedtobethelimit1()lim()nbianifxdxfxxwhere1()niifxxisaRiemannSumoffon[a,b].LetPmovetoapositionT,wherethex-coordinateisx+x,andletthecorrespondingycoordinatebeyy.Also,lettheincreaseinarea,PQRT,beA.Fromfigure1.5,WeseethatareaofPQRSAareaofUQRTi.e.()yxAyyxyAxyyAsx0,y0andy+yy.Thus,0limxdAAdxx=yIntegrate,withrespecttox,A=ydx=()fxdx=F(x)+Cwhere()()Fxfxandcisanyrealnumber.figure1.5WhenPisatpointM,x=a,andA=0.0=F(a)+c=c=-F(a).Hence,whenx=b,A=F(b)–F(a)Example2:Evaluatethefollowingdefiniteintegral:(i)120(1)xxdx(ii)21221()2xdxx(iii)92132xdxx(iv)1512(21)xdx(ans:1/30)(ans:528)(ans:0)(ans:1/12)PropertiesofDefiniteIntegral:(wherecisconstant)1.()0aafxdx4.()()()ccbabafxdxfxdxfxdx(cba)2.()()abbafxdxfxdx5.[()()]()()bbbaaafxgxdxfxdxgxdx3.()()bbaacfxdxcfxdx6.()()bbaafxdxftdtxx+PQRST(x+)U(x,y)y=f(x)yxMNabThisiscalledthefundamentaltheoremofcalculus.[DEFINITEINTEGRALANDITSAPPLICATIONS]--------------------------4---------------------------------Example3:Example4:Giventhaty=32(12)x,showthat123(12).dyxdxGiventhaty=329x,showthat23329dyxdxx.Hence,evaluate1420(12)x.Henceevaluate223029xdxx.(ans:2/3)[solution]y=32(12)x12123(12)22=3(1+2x)dyxdx134422001344220033223(12)[(12)]1(12)[(12)]31=[(18)(10)]3126=(271)33xdxxxdxxExample5:Findthevalueofkif122(32)4axdxxdx.(ans:a=-1,4)Example6:Giventhat1301()()8fxdxfxdx,find(a)302()fxdx;(b)03112()5()fxdxfxdx;(c)thevalueofkforwhich30[()]7fxkxdx.(ans:(a)32;(b)24;(c)k=2)[DEFINITEINTEGRALANDITSAPPLICATIONS]--------------------------5---------------------------------1.3FindingtheareabyusingdefiniteintegralFromtheabove,wefoundthattheareaboundedbythecurvey=f(x),thex-axisandtheordinatesx=a,x=bis()bafxdxorbaydx(i)whenthecurveisabovethe“x–axis“(ii)whenthecurveisbelowthe“x-axis”(iii)whenpartofthearearequiredisabovethex-axisandpartofitisbelowthexaxis:Important:Alwayssketchthegraphfirstcd=[DEFINITEINTEGRALANDITSAPPLICATIONS]--------------------------6---------------------------------Example7:Example8:Findtheareaenclosedbythecurvey=x3–4x2+3xFindtheareaoftheregionboundedbythecurvey=3x,theandthex-axis.(ans:3712sq.units)y-axisandtheliney=0andy=3.(ans:43sq.units)Example9:Sketchthecurvey=4x-x2.Findtheareaoftheregionboundedbythecurveandthex-axis.(ans:1023units2)Example10:FindtheareaoftheregionRboundedbythey=x(3-x)andthexaxis.Ifthestraightliney=kxdividestheregionRintotwopartswithequalarea,findthevalueofk.(ans:k=343(1)2[DEFINITEINTEGRALANDITSAPPLICATIONS]--------------------------7---------------------------------1.4Areaboundedbycurves1.Areaofonestrip:Areabetweenthetwocurves,2linesx=a,x=b=2.Areaofonestrip=(y1–y2)x(Notethaty2isnegative)Areabetween2curves&2linesx=a,x=b=Example11:Findtheareaoftheregionenclosedbythecurvey=2-x2andtheliney+x=0.(ans:4.5units2)[solution]3.Areaofonestrip=(
本文标题:Chapter 10 Definite Integral
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