# 8.8 - 8.8 1 8.8 2 8.8 Improper Integrals Type I - Infinite...

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8.818.8Improper IntegralsType I - Infinite Limits of IntegrationConsider the following example.Example 1.Infinite Areas?Suppose thatb >1. Evaluate the followingdefinite integral.(1)integraldisplayb11x2dxy= 1/x21b1In other words, find the area under curve. Itfollows thatintegraldisplayb11x2dx=-1xb1= 1-1b(2)8.82Now, what happens in (2) asb→ ∞? In this casewe have the following.integraldisplay11x2dx= limb→∞integraldisplayb11x2dx= limb→∞1-1b= 1Motivated by the previous example, we adopt thefollowing definition.
8.83Definition.Improper Integrals - Type IDefinite integrals with infinite limits of integrationare calledimproper integrals of type I. Theyare defined as follows.1. Iff(x)is continuous on[a,), thenintegraldisplayaf(x)dx= limb→∞integraldisplaybaf(x)dx(3)2. Iff(x)is continuous on(-∞, b], thenintegraldisplayb-∞f(x)dx=lima→-∞integraldisplaybaf(x)dx(4)3. Iff(x)is continuous on(-∞,), thenintegraldisplay-∞f(x)dx=integraldisplayc-∞f(x)dx+integraldisplaycf(x)dx(5)for any real numberc. Notice that this lastcase is handled by combining the first two.8.84In all three cases, we say the improper integralconvergeswhenever the right-hand side is finite.Otherwise, the improper integraldiverges.Remark.So the improper integral in the firstexampleconvergedand its value was1. Also,sincef(x) = 1/x2>0we can say that the “areaunder the curve” is finite...in fact, the area is1.
8.85The following result is very useful.Proposition 1.Suppose thatfis continuouson[a,)andintegraltextaf(x)dxconverges. Thenintegraltextacf(x)dxalso converges for anycR.Proof.Letb > a. Then by the basic properties ofthe definite integralintegraldisplaybacf(x)dx=cintegraldisplaybaf(x)dxIt follows thatintegraldisplayacf(x)dx= limb→∞integraldisplaybacf(x)dx=climb→∞integraldisplaybaf(x)dx=cintegraldisplayaf(x)dx <8.86Example 2.Evaluate the improper integralintegraldisplay22x2-1dxNotice that the integrand is continuous on[2,)so by (3) we haveintegraldisplay22x2-1dx= limb→∞integraldisplayb22x2-1dxNow we can use the usual techniques (partialfractions in this case) to evaluate the integral.

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