# Test 1 Review Sheet - Math 113 Calculus II Test 1 Review 1...

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Math 113, Calculus IITest 1: Review1Review of the materialChapter 5 is an introduction tointegral calculus.In this chapter we study techniques forevaluating both definite and indefinite integrals of continuous functions.1.1SIGMA NOTATIONA convenient shorthand for writing long (and even infinite) sums is the so-called “sigma nota-tion”. The expressiona1+a2+a3+· · ·+anis abbreviatedni=1ai, i.e.,nXi=1ai=a1+a2+a3+· · ·+an,but this sigma-representation of the sum is not unique. There are importantformulas:ni=11 =n,ni=1i=n(n+ 1)2,ni=1i2=n(n+ 1)(2n+ 1)6,ni=1i3=n(n+ 1)22,andrules:ni=1kf(i) =kni=1f(i),ni=1[f(i)±g(i)] = [f(i)]±[g(i)],ni=1f(i) =nj=1f(j) =nk=1f(k),mi=1f(i) +ni=m+1f(i) =ni=1f(i).1.2RIEMANN SUMS and DEFINITE INTEGRALSOur introduction to definite integrals begins with Riemann sums. First, we divide the interval[a, b] intonsubintervals, each of widthΔx=b-an.Next, we determined the endpoints of these subintervals to bexj=a+jΔxforj= 0,1,2, . . . , n. In particular,x0=aandxn=b. Then aRiemann sumfor a functionfonthe interval [a, b] is defined to benXi=1f(x*ix=f(x*1x+f(x*2x+· · ·+f(x*nx,
Math 113, Calculus IITest 1: Reviewwherex*iis some point in thei-th subinterval [xi-1, xi], fori= 1, . . . , n. We defined thedefiniteintegral offon[a, b] to beZbaf(x)dx= limn→∞nXi=1f(x*ix.When we compute definite integrals using the limit definition above, we often takex*ito be theright-hand endpoint,xi, of thei-th subinterval.(Signed) area: The definite integralRbaf(t)dtis thesignedarea of the region in the planebounded by the graph ofy=f(x) and the linesy= 0,x=a, andx=b. Remember that areaabovethex-axis is positive, but areabelowthex-axis is negative.The definite integralRbaf(x)dxmay also be interpretted as thetotal changein the function overthe interval [a, b].Properties of Definite Integrals:Rba[f(t) +g(t)]dt=Rbaf(t)dt+Rbag(t)dtandRbakf(t)dt=kRbaf(t)dtRaaf(t)dt= 0;Rabf(t)dt=-Rbaf(t)dt;andRbaf(t)dt=Rcaf(t)dt+Rbcf(t)dtIfmf(x)Mon [a, b], thenm(b-a)Rbaf(t)dtM(b-a)Evaluation Theorem:Supposef