# IU_CalculusI_Chap4(1).pdf - HA M Chapter 4 INTEGRATION T P...

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D. T. PHAMChapter 4: INTEGRATIONDuong T. PHAMCALCULUS IWinter 2017Duong T. PHAMSeptember 10, 20171 / 77
D. T. PHAMOutline1Area problems2The fundamental theorems of calculus3Indefinite integrals and the net change theorem4The substitution rule5Integration by parts6Additional techniques of integration7Approximate integration8Improper integralsDuong T. PHAMSeptember 10, 20172 / 77
D. T. PHAMArea problemsxyy=f(x)abS?x1x2x3· · ·xi-1xixi+1· · ·xn-1S1S2S3SiSnx1x2x3· · ·xi-1xixi+1· · ·xn-1Δxx*1x*2x*3x*ix*nRn=f(x*1) Δx+f(x*2) Δx+· · ·+f(x*i) Δx+· · ·+f(x*n) ΔxS= limn→∞RnDuong T. PHAMSeptember 10, 20173 / 77
D. T. PHAMDefinite integralsDef:Letf: [a,b]R.Divide [a,b] byx0,x1, . . . ,xnintonequal subintervals of width (b-a)/n(a=x0<x1< . . . <xn-1<xn=b).Letx*1,x*2, . . . ,x*nbe anysample points(x*i[xi-1,xi])Thedefinite integral offfromatobisZbaf(x)dx= limn→∞nXi=1f(x*i) Δxif this limit exists. In this case, we sayfis integrable on[a,b]The sumnXi=1f(x*i) Δxis calledRiemann sum.Duong T. PHAMSeptember 10, 20174 / 77
D. T. PHAMIntegrable functionsTheorem:Iffis continuous on [a,b], thenfis integrable on [a,b]Iffis continuous on [a,b], except at a finite number of points, thenfis integrable on [a,b]Theorem:Iffis integrable on [a,b], thenZbaf(x)dx= limn→∞nXi=1f(xi) Δx,where Δx=b-anandxi=a+iΔxDuong T. PHAMSeptember 10, 20175 / 77
D. T. PHAMDefinite IntegralsEx:Evaluate the Riemann sum forf(x) =x3-6xtaking the sample points tobe the right endpoints witha= 0,b= 3 andn= 6EvaluateR30(x3-6x)dxAns:(a)Forn= 6, the interval width isΔx=b-an=3-06=12=0.5, and theright endpoints arex1= 0.5,x2= 1,x3= 1.5,x4= 2,x5= 2.5,x6= 3.The Riemann sum isR6=6Xi=1f(xi) Δx= Δxf(0.5) +f(1) +f(1.5) +f(2) +f(2.5) +f(3)=12(-2.875-5-5.625-4 + 0.625 + 9 )=-3.9375Duong T. PHAMSeptember 10, 20176 / 77
D. T. PHAMDefinite integralsEx:EvaluateR30(x3-6x)dxAns:03x13nx22×3nxn-1(n-1)×3nWithnsubintervals, Δx=3n, andx0= 0,x1=3n,x2= 23n, . . . ,xi=i3n, . . . ,xn= 3.Z30(x3-6x)dx= limn→∞nXi=1f(xi) Δx= limn→∞nXi=1f3in3n= limn→∞3nnXi=1"3in3-63in#= limn→∞3nnXi=127i3n3-18in= limn→∞"81n4nXi=1i3-54n2nXi=1i#= limn→∞"81n4n(n+ 2)22-54n2n(n+ 1)2#= limn→∞"8141 +1n2-271 +1n#=814-27=-274Duong T. PHAMSeptember 10, 20177 / 77
D. T. PHAMThe Midpoint RuleMidpoint Rule:Zbaf(x)dxnXi=1fxi) Δx= Δx[fx1) +fx2) +· · ·+fxn)]where Δx=b-anand ¯xi=xi-1+xi2= midpoint of [xi-1,xi].Ex:Use Midpoint Rule withn= 5 to approximateZ211xdxAns:Δx= (2-1)/5 = 0.2. The endpoints of the subintervals arex0= 1,x1= 1.2,x2= 1.4,x3= 1.6,x4= 1.8andx5= 2.The midpoints are¯x1= 1.1,¯x2= 1.3,¯x3= 1.5,¯x4= 1.7,¯x5= 1.9.