§Appendix C Sigma Notation - College of Arts and Sciences



§Appendix C Sigma NotationLet a1, a2, …, an,… be a sequence of numbers.a2 is the second number in the sequenceai is the ith number in the sequencedefinei=1nai=a1+a2+…+anwhere i is the index of summation1 is the starting value of i in this examplen is the final value of i in this exampleExample. Let ai=ii=14i=1+2+3+4=10■Example. Let ai=i2i=14i2=12+22+32+42=1 +4 +9 +16=30■Four Basic Properties of SumsLet a1, a2,… and b1, b2,…be sequences of numbers and let c be a constant.i=1nc=nci=1nc ai=ci=1naii=1nai+bi=i=1nai+i=1nbii=1nai-bi=i=1nai-i=1nbiWhy?i=13c=c+c+c=3ci=13c ai=ca1+ca2+ca3=c(a1+a2+a3)=ci=1nai i=1nai+bi=a1+b1+a2+b2+(a3+b3)=a1+a2+a3+(b1+b2+b3)=i=1nai+i=1nbi similar to 3. ■Two Sums of Powersi=1ni=n(n+1)2i=1ni2=nn+1(2n+1)6Proof of (A)1 + 2 + … + n-1 + nn + n-1 + … + 2 + 1n+1+n+1+…+n+1+(n+1)=n(n+1)This is equal to 2i=1ni, so we must divide by 2 to get the final result.Proof of (B)recall x+y3=x3+3x2y+3xy2+y3Consider the following telescoping sumi=1n[1+i3-i3]=23-13+33-23+… +n3-(n-1)3+(n+13-n3) =n+13-1 =n3+3n2+3n[1]on the other handi=1n[1+i3-i3]=i=1n1+3i+3i2 =i=1n1+i=1n3i+i=1n3i2 =i=1n1+3i=1ni+3i=1ni2 =n+3n(n+1)2+3S[2]where S is the sum we seek. Equating [1] with [2]n+32n2+n+3S=n3+3n2+3n 3S=n3+32n2+12n S=2n3+3n2+n6 =n(2n2+3n+1)6=nn+1(2n+1)6■§5.1 Areas and DistancesThe Area ProblemFind the area of the region S lying under the graph of y=f(x) on the interval from x=a to x=b.…a…b…x- , …, curve, f, SApprox. the area under f by a sum of rectangles.a…b…,…,f, four rectangles, x1*, x2*, x3*, x4*It’s easy to compute the area of a rectanglerectangle, width Δx of subinterval, height f(xi*), xi* sample point for iTH rectangleTake the limit as the number of rectangles increases to ∞.Example. Estimate the area A under the graph of fx=x from x=0 to x=1.0…1, 0…1, fx=xDivide the interval [0,1] into four subintervals. addEstimate the area under the graph over each subinterval by a rectangle of height f(xi*), with sample point xi* the right hand endpoint of each subinterval.0…1, 0…1, fx=x, rectanglesThe width of each rectangle is Δx=1/4. Area of rectanglesR4=Δx?14+Δx?12+Δx?34+Δx?1 =Δx?14+12+34+ 1 =Δx?1+2+3+44 =14?104=1016=58 Clearly A<R4Alternatively, let xi* be the left hand endpoint of each subinterval.0…1, 0…1, fx=x, rectanglesArea of rectangles L4= Δx?04+Δx?14+Δx?12+Δx?34 =Δx?0+1+2+34 =14?64=616=38 ClearlyL4<A<R438<A<58Improve estimate by dividing area into 8 stripsL8<A<R8716<A<916Example. Obtain the exact area under fx=x from x=0 to x=1 by taking the limit as n increases.Let the sample points be right hand endpoints of each subinterval.0…1, …, fx=x, rectangles of heights 1n,2n,3n, …1Rn=Δx?1n+Δx?2n+…+Δx?nn =Δx?1n(1+2+…+n) =1n2 (1+2+…+n) =1n2n(n+1)2=n+12n Exact areaA=limn→∞Rn =limn→∞n+12n =12 Alternatively, let the sample points be left hand endpoints of each subinterval.heights of rectangles =0n,1n,2n…,n-1nLn=Δx?0n+Δx?1n+…+Δx?n-1n =Δx?1n0+1+…+n-1 =1n2 0+1+…+n-1 =1n2(n-1)n2=n-12n Exact AreaA=limn→∞Ln =limn→∞n-12n=12 ■Fact: We can take the height of the iTh rectangle as f(xi*) for any number xi* in the iTh subinterval.A=limn→∞fx1*Δx+fx2*Δx+…+fxn*ΔxA=limn→∞i=1nfxi*ΔxThis formula works for any continuous function f…a…b…, 0…, iTh rect., width Δx, xi*, height f(xi*)The Distance ProblemFind the distance traveled by an object if its velocity is knownIf the velocity is constantdistance=velocity×timeExample120 miles=60 mileshour×2 hours■What if the velocity varies?…a…b…time t-, … velocity v-, curvedefine time subintervals of length Δt add 4pick sample time ti* in each subinterval addapproximate the velocity in each subinterval as v(ti*)d=distance traveled≈vt1*Δt+vt2*Δt+vt3*Δt+vt4*Δt=i=14vti*ΔtThe exact distance is obtained in the limit as we use more and more subintervals.d=limn→∞i=1nvti*Δt§5.2 The Definite Integral…a…b…x-, …, curve fdefine the definite integralStepsDivide [a,b] into n subintervals of equal widthΔx=(b-a)/n add subinterval ixi=a+iΔx add x0, xi-1, xi, xnChoose sample points in these subintervalsx1*, x2*,…xn* add xi*Construct the Riemann Sumi=1nfxi*ΔxTake the limit n→∞ to obtain the definite integralabfxdx=limn→∞i=1nfxi*ΔxExample. The distance problemd=abvtdt ■Notation integral sign, lower limit of integration a, upper limit b, integrand, ghost of ΔxNote that integral does not depend on “x”abfxdx=abftdtx and t are dummy variables.Example. ExpressI=limn→∞i=1n11+in2?1nas an integral on [0,1].Let Δx=1n andfxi=11+xi*2 where xi*=in.Then I=0111+x2dx ■?? Example. Determine a region whose area is equal to the give limit. limn→∞i=1ntanπ i4 nπ4n■Interpretation of the Definite Integralabfxdx=limn→∞n=1∞fxi*ΔxFor f positive and b>a, the Riemann sum approximates the area under the curve.…a…b…x-, …, graph of f, area A, 4 rectanglesabfxdx=A=area under f from a to bSuppose f can be positive or negative…a…b…x-, …, f, A1 under f, A2 above f, rectangleshere fxi*>0…∑fxi*Δx≈A1, here fxi*<0…∑fxi*Δx is a negative quantity whose magnitude ≈A2abfxdx=A1-A2=net area under or above f from a to bEvaluating Integrals Recalli=1nc=nci=1nc ai=c i=1naii=1nai+bi=i=1nai+i=1nbii=1nai-bi=i=1nai-i=1nbiwhere a1, a2, … and b1, b2, … are sequences of numbers and c is a constant (an expression that does not depend on the index of summation i).Recall two sums of powersi=1ni=12n(n+1)i=1ni2=16nn+1(2n+1)Example. Prove that abx dx=12(b2-a2)…a…b…x-, partition points, Δx, Δx=b-anchoose sample points in each subintervalan easy choice is the right hand endpointsx1*=a+Δxx2*=a+2Δxxi*=a+iΔxBegin with the definitionabx dx=limn→∞i=1nfxi*Δx=limn→∞i=1nxi*Δx=limn→∞i=1na+iΔxΔx=limn→∞i=1naΔx+iΔx2=limn→∞(i=1naΔx+i=1niΔx2)abx dx=limn→∞(i=1naΔx+i=1niΔx2)=limn→∞(aΔxi=1n1+Δx2i=1ni)=limn→∞ab-ann+b-an212nn+1=limn→∞ab-a+b-a212n+1n=ab-a+12b-a2limn→∞n+1n=ab-a+12b-a2=b-aa+12b-a=12b-ab+a=12(b2-a2) we are done! ■Example. Prove that 01x2dx=130…1…x-, 0…1…y-, y=x2width of subintervals Δx=1nsample points xi*=in add01x2dx=limn→∞i=1nxi*2Δx =limn→∞i=1nin21n =limn→∞1n3i=1ni2 =limn→∞1n316 nn+12n+1 =limn→∞2n3+…6n3 =13 ■Properties of the definite integralReversing limits of integrationabfxdx=-bafxdxExample. Suppose f>0 and b>a…a…b…,…,fabfxdx>0 and bafxdx<0 ■Corollary. Let b=aaafxdx=0Picture …a…, fThe “area” under a point is zero!Four basic properties of integralsabc dx=c(b-a)abc fxdx=cabfxdxabfx+gxdx=abfxdx+abgxdxabfx-gxdx=abfxdx-abgxdxNotes2. constant multiple rule3. integral of sum is sum of integrals4. integral of difference is difference of integrals.Property 1 has a simple interpretation…a…b…,…c…,function c, area =c(b-a)Example. Evaluate I=035-2xdx use the difference ruleI=035 dx-032xdx property #1 and constant multiple rule =53-0-203x dx rule that abx dx=12(b2-a2) =15-2 129-0 =6 ■Addition Property wrt Interval of IntegrationFor any real nos. a, b and cacfxdx+ cbfxdx=abfxdxPicture for a<c<b and f>0…a…c…b…x-, …, f, A, A1, A2Example: Evaluate I=34fxdx+13fxdx+41fxdx by addition property reverse limits so I=0 ■Comparison Property of IntegralsIf fx≥0 and b>a thenabfxdx≥0 If fx≥g(x) and b>a thenabfxdx≥ abgxdx If m≤fx≤M thenmb-a≤ abfxdx≤M(b-a) Picture for M>m>0:…a…b… , …m…M… function m, function M, fExample. Show that 25x2-1 dx≤10.5 without evaluating the integral.Solution. For x>1:x2-1<x2=x Then by property 7:251-x2 dx≤ 25x dx =12(52-22) =1225-4=212 =10.5■Example. Show that π6<0π3cosxdx≤π3without evaluating the integral.0…π3…π2…x-, 0…1…y-, cos?(x)on the interval [0,π3]12≤cosx≤1by property 812π3-0<0π3cosxdx<1π3-0 which gives the result we are trying to show ■§5.3 The Evaluation TheoremIf f is continuous on the closed interval [a,b], thenabfxdx=Fb-F(a)where F is any antiderivative of f, that is F'=f.Example. abx dx=Fb-F(a)where Fx=12x2+C. Fb=12b2+C, Fa=12a2+C abx dx=12(b2-a2) Notice that the constant of integration C cancels. We may as well set C=0. ■Example. 01x2dx=F1-F(0)where Fx=13x3F1=13 , F0=001x2dx=13 ■Example. The distance problem.v velocitys positiond distance traveleda starting timeb ending timewe have seen previouslyd=abvtdtrecall s't=v(t)By the Evaluation Theoremabvtdt=sb-sa=distance traveled! ■Recall the Mean Value Theorem. Let f be continuous on [a,b] and differentiable on (a,b). Then there is a no. c in (a,b) such that f'c=fb-f(a)b-aProof of the Evaluation Theorem.Partition [a,b] into n subintervals of equal width.…a…b…x-, x0, x1, x2, …, xn-1, xnwidth of each subinterval Δx=(b-a)/nLet F be any antiderivative of f. Write Fb-F(a) as a telescoping sumFb-Fa=Fxn-F(x0)=Fxn-Fxn-1+Fxn-1-Fxn-2+…+Fx2-Fx1+Fx1-Fx0Apply the Mean Value Theorem to F on an arbitrary subinterval i: [xi-1,xi]Fxi-Fxi-1=F'xi*xi-xi-1=fxi*Δx,where xi* is some no. on the interval (xi-1,xi).ThenFb-Fa=fxn*Δx+fxn-1*Δx+…fx1*Δx=i=1nfxi*Δxthe last expression is a Riemann sum!Let n→∞limn→∞Fb-Fa=limn→∞i=1nfxi*ΔxFb-F(a) does not depend on n and the limit on the right is the integral. Therefore we getFb-Fa=abfxdx■Notation Fb-Fa=FxabThen the Evaluation Theorem may be writtenabfxdx=Fxabwhere F is any antiderivative of F.Example. Evaluate 01x37 dxFor fx=xn, an antiderivative of f isFx=1n+1xn+1For n=3/7Fx=x107107=710x107Thus01x37 dx=710x10701 =7101107-7100107 =710-0 =710 ■Example. Evaluate I=0π/4sec2θdθRecall ddθtanθ=sec2(θ)Then I=tanθ0π4=tanπ4-tan0=1■The Evaluation Theorem in ApplicationsInterpret derivatives as rates of changes positionv velocityQ electric chargeI=dQdt electric currenty=h(x) height of trail dydx=h'(x)slope of trail x miles from startrewrite the evaluation theorem in terms of rate of changeNet Change TheoremThe integral of a rate of change is a net changeabF'xdx=Fb-F(a)where F'(x) is the rate of change of F wrt x and Fb-F(a) is the net change in F.Note that the word “net” connotes that there can be positive and negative contributions to the rate of changeExamples.The integral of velocity gives the net change in positionabvtdt=sb-s(a) The integral of current gives the net charge passing through a wireabItdt=Qb-Q(a) The integral of the slope of trail gives the net change in height of trailabh'xdx=hb-h(a) ■Particle Motion (back and forth) along a straight lines(t)position at time tv(t) velocity at time t|vt|speed at time tConsider the following particle path0…4…s-, t1 at 0, t2 at 4, t3 at 3the particle’s displacement (distance traveled)st3-st1=3=t1t3vtdt the total distance the particle travels5=4+1=t1t3vtdt =t1t2vtdt+t2t3-vt dtExample. The velocity function for a particle moving along a line is vt=4-2t meters/secFind (a) the displacement and (b) the total distance traveled over the time interval 0≤t≤3 seconds(a) 03vtdt=034-2tdt =4t-t203 =12-9-(0-0) =3 meters(b) 03vtdt= 02vtdt+23-vtdt =024-2tdt+232t-4dt =(4t-t2)02+t2-4t23 =8-4-0-0+9-12-4-8 =4-0+-3--4 =4+1 =5 meters ■Indefinite IntegralsThe symbol∫fxdxis called the indefinite integral and means the general antiderivative of f(x).If F is any antiderivative of f then∫fxdx=Fx+CC is the constant of integration.Antiderivative Formulas using indefinite integral notationLet k be a constant and let f and g be functions∫k dx=kx+C∫k fxdx=kfxdx+C∫fx+gxdx=∫fxdx+∫gxdx∫fx-gxdx=∫fxdx-∫gxdxwe also have formulas for specific functions∫xn dx=1n+1xn+1+C valid for n≠-11x dx=ln|x|+C∫exdx=ex+C∫ax dx=axlna+C∫sinxdx=-cosx+C∫cosxdx=sin?(x)+C∫sec2xdx=tanx+C∫secxtanxdx=secx+C11+x2 dx=tan-1x+C11-x2 dx=sin-1x+CExample. Find the indefinite integralI=∫(cosx-2sin?(x)) dxsolutionI=cosxdx-2sinxdx =cosxdx-2 sinxdx =sinx+2cosx+C ■Example. Find the indefinite integralI=∫(15x+3secxtan(x)) dxsolutionI=∫15xdx +∫3secxtan(x)) dx =151x dx+3 ∫secxtanxdx =15ln|x|+3sec(x)+C ■ §5.4 The Fundamental Theorem of CalculusThe “Area so far” function…a…x…b…t-, 0…, f(t), area g(x) from a to xgx=axftdt If f>0, g is the “area so far” under fExample. Let a=1 and ft=1t.gx=1x1t dt =lnt1x =lnx-ln?(1) where x>0 =ln(x)Notice that g'x=1x=f(x) This is not a coincidence! ■The Fundamental Theorem of Calculus, part 1 (FTC 1)If f is continuous on [a,b], then the function g defined by gx=axftdt a≤x≤bis an antiderivative of f, that isg'x=f(x) for a≤x≤b.Plausibility Argument. Why is g'x=f(x)?…a…x…x+h…b…, …, area g(x) from a to xg'x=limh→0gx+h-g(x)h from the graphgx+h-gx≈fxh this becomes more accurate as h→0g'x=limh→0fxh+small correctionh =f(x) ■Our text gives a rigorous proof.Using Leibniz notation, the FTC 1 becomesddx axftdt=f(x) Example. Find the derivative of gx=-1xt3+1 dt Solution ft=t3+1 is continuous on [-1,∞).g'x=x3+1 ■Example. Find the derivative of y=x2πsin?(t)t dt Solution. ft=sin?(t)t is continuous for t>0.Reverse the limits of integration.y=-πx2sin?(t)t dt =-πusin?(t)t dtwhere u=x2. By the FTC 1dydu=-sin(u)u By the chain ruledydx=dydududx =-sin?(u)ududx =-sin?(x2)x2?2x = -2xsin(x2) ■The Fundamental Theorem of Calculus (FTC)Suppose f is continuous on [a, b]ddx axftdt=f(x)abfxdx=Fb-F(a)where F is any antiderivative of f. This is the Evaluation Theorem!Part 2 can be rewrittenabF'xdx=Fb-F(a) Roughly speaking, the FTC says that differentiation and integration are inverse processes!Recall the Mean Value Theorem:If f is continuous on [a,b] and differentiable on (a,b), then there is a no. c in (a,b) such thatf'c=fb-f(a)b-aMean Value Theorem for IntegralsIf f is continuous on [a,b], then there is a no. c in (a,b) such that fc=abfxdxb-aProof. Let F be the “Area so far” functionFx=axftdtF'x=f(x) by FTC 1. F is continuous on [a,b] and differentiable on (a,b).By the Mean Value Theorem, there is a no. c in [a,b] such thatF'c=Fb-F(a)b-aBy FTC 2 (the Evaluation Theorem), this becomesfc=abfxdxb-awhich is what we want to prove! ■Geometrical Interpretation…a…b…x-, …, f add c, f(c), rectangle abfxdx=fc(b-a) area under curve = area of rectangleNote: fc=fave=1b-aabfxdx is the average value of f on [a,b]ExampleFind the average value of fx=x3 on [0,1]fave=11-001x3 dx=14x401=14Find c such that fave=f(c)fc=c3=14c=1413≈0.63■STOP §5.5 The Substitution RuleDifferential Notation revisitedConsider y=f(x). By the definition of derivative:dydx=limh→0fx+h-f(x)h this is not exactly a ratio…Let dx be any real no. and definedy=f'x dx another real no. This allows us to interpret the left hand side ofdydx=f'(x) as a true ratio.Practice. Let’s write u=g(x). Then du=g'xdxLet u=x2. Then du=2x dx?? Let u=x. Then du=?? Let u=ln|x|. Then du=?? Let u=tan?(θ). Then du=Integration by SubstitutionRecall the indefinite integral∫F'xdx=Fx+CRecall the chain rule let Fx=fgxF'x=f'gx g'(x)Thus∫f'gxg'xdx=fgx+CExample. Let Fx=sin(x2) F'x=cosx2 2x so ∫cosx22x dx=sinx2+C ■This is taking the chain rule backwards!How to recognize f'gx g'(x) in an integrand?Trick – “Integration by Substitution”Suppose you haveI=∫f'gxg'xdxbut you might not recognize the form.Guess the inner function of the compositionu=g(x)form the differentialdu=g'xdxsubstitute into I I=∫f'udu =fu+C =fgx+CExample. EvaluateI=2xcosx2dxguess the inner functionu=x2form the differentialdu=2x dxsubstitute into II=∫cosudu =sinu+ C =sinx2+C ■Example. EvaluateI=∫3x+4 dxguess the inner functionu=3x+4form the differentialdu=3 dxsolve for dxdx=13 dusubstitute into II=u12 13 du =13?23 u32+ C =29 3x+432+ C can check by differentiationddx29 3x+432+ C=29?32 3x+4123+0=3x+412■Example. EvaluateI=∫tan2xsec2xdxguess the inner functionu=tan(x)form the differentialdu=sec2xdxsubstitute into II=∫u2 du =13 u3+ C =13tan3x+ C■Example. EvaluateI=∫x1+x4 dxguess the inner functionu=x2form the differentialdu=2x dxsolve for x dxx dx=12 dusubstitute into II=∫11+u212 du =12tan-1u+C =12tan-1x2+C■Evaluating Definite Integrals By SubstitutionRecall the Fundamental Theorem of Calculus, part 2abF'xdx=Fb-F(a)Method I Find the indefinite integral by substitution and then apply the FTC, part 2.Example. Evaluate-103x+4 dxletu=3x+4, du=3dx, 13 du=dxthen∫3x+3 dx=∫u12?13 du=13?23u32+C =29 3x+432+ CNow let Fx=29 (3x+4). Then by FTC 1I=F0-F-1 =29?432-29?132 =29 8-1 =149 ■Method II. Transform limits of integration while substituting and apply FTC 2.This method is better once you get used to it because it involves less writingExample. EvaluateI=-103x+4 dxLetu=3x+4, du=3dx, 13 du=dxAlso note that if x=-1 then u=1 if x=0 then u=4thenI=14u1213 du =13?23 u3214 =29 432-132 =29 8-1=149 ■Example. EvaluateI=-π4π4tan2xsec2xdxLetu=tan(x)du=sec2(x)Also note that if x=-π4 then u=-1 if x=π4 then u=1thenI=-11u2 du =13 u3-11 =13 13--13 =23■Example. EvaluateI=12cosπxx2 dxLet u=πx=πx-1. Then du=π-x-2 dx and -1π du=x-2dx.If x=1 then u=π x=2 u=π2ThenI=ππ2cosu -1π du =1ππ2πcosudu =1πsinuπ2π =1πsinπ-sinπ2 =1π0-1 =-1/π ■Example. EvaluateI=241xln(x) dxLet u=lnx.Then du=1x dxIf x=2 then u=ln2 x=4 u=ln4=ln22=2ln2Then I=ln2ln41u du =lnuln2ln4 =ln(ln4))-ln(ln(2)) =ln(2ln2))-ln(ln(2)) =ln2+ln(ln2)-ln(ln(2)) =ln2■ ................
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