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Now we set up the following inequalityESn≤24·15180n4≤11000Manipulating this inequality we haven4≥4003. Thus we write our final answerasLetnbe the smallesteveninteger greater than4q4003.Using a calculator, we see thatn= 4 subintervals is sufficient to guarantee thatour approximation is within11000of ln 2.Note.In our final answer, we had to write “Letnbe the smallest even integer...”Unlike the previous example, we had to specify thatnwas an eveninteger because we were using Simpson’s Rule, which requires an even numberof subintervals.8.2Improper IntegralsThere are two types of integrals that are described asimproper.I.f(x) is a bounded function, but the interval of integration is unbounded.For instance, in the graph below the interval is [a,∞).II.g(x) is an unbounded function, but the interval of integration is bounded.For instance, in the graph below the interval is [0,1], but the function hasa vertical asymptote atx= 0.46
Question: Can we define the area under the graph in either scenario?I. For anyc > a,Rcaf(x)dxis defined. IfRcaf(x)dxapproaches a value ascincreases without bound, then we defineZ∞af(x)dx= limc→∞Zcaf(x)dxII. Suppose we wish to evaluateRbag(x)dx,g(x) is continuous on (a, b], andg(x) has a vertical asymptote atx=a. TheRbcg(x)dxis defined for alla < c≤b. IfRacg(x)dxapproaches a value ascapproachesafrom above,then we defineZbag(x)dx=limc→a+Zbcg(x)dxIn either scenario, if the limit exists as a real number then we say the integralconverges. Otherwise itdiverges.8.2.1Type I Improper IntegralsWe will begin by investigating integrals in which our function is bounded, butwe are integrating over an unbounded interval.Example 8.3.Determine whetherZ∞11x2dxconverges or diverges. If it con-verges, to what value does it converge?We wish to find the area of the region shaded in the graph below. Is this aninfinite amount of area? Or is the amount of area under the curve bounded bya value as we let our upper bound increase?47
To determine whether or not this converges, we convert this integral to a limit,as below.Z∞11x2dx= limc→∞Zc11x2dx= limc→∞-1xc1= limc→∞-1c+ 1= 1Thus the integralconverges to 1.Note.No matter how large we letcbe, the amount of area under the curve on[1, c] is less than 1, but getting closer and closer to 1 as we letcbe larger. Thisis why we can conclude that there’s not an infinite amount of area under thecurve despite the interval being [1,∞).Example 8.4.Determine whetherZ∞11xdxconverges or diverges. If it con-verges, to what value does it converge?We approach this problem in the exact same way we approached the previousproblem.Z∞11xdx= limc→∞Zc11xdx= limc→∞lnxc1= limc→∞(lnc-0)=∞ThusR∞11xdxdiverges. There is an infinite amount of area under the graphoff(x) =1xon the interval [1,∞).