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10.7IMPROPER INTEGRALSIn the definition of the definite integralbaf(x)dxit is assumed that [a,b] is a bounded interval andfis a bounded function. In thissection we use the limit process to investigate integrals in which either the intervalis unbounded or the integrand is an unbounded function. Such integrals are calledimproper integrals.Integrals over Unbounded IntervalsWe begin with a functionfcontinuous on an unbounded interval [a,∞). For eachnumberb>awe can form the definite integralbaf(x)dx.(Figure 10.7.1)If, asbtends to∞, this integral tends to afinite limitL,yfxabFigure 10.7.1limb→∞baf(x)dx=L,then we write∞af(x)dx=L
10.7IMPROPER INTEGRALS623and say thatthe improper integral∞af(x)dxconverges to L.Otherwise, we say thatthe improper integral∞af(x)dxdiverges.In a similar manner,improper integralsb−∞f(x)dxarise as limits of the formlima→−∞baf(x)dx.Example 1(a)∞0e−2xdx=12.(b)∞1dxxdiverges.(c)∞1dxx2=1.(d)1−∞cosπx dxdiverges.VERIFICATION(a)∞0e−2xdx=limb→∞b0e−2xdx=limb→∞−e−2x2b0=limb→∞12−12e2b=12.(b)∞1dxx=limb→∞b1dxx=limb→∞lnb= ∞.(c)∞1dxx2=limb→∞b1dxx2=limb→∞−1xb1=limb→∞1−1b=1.(d)Notefirst that1acosπx dx=1πsinπx1a= −1πsinπa.Asatends to−∞, sinπaoscillates between−1 and 1. Therefore the integraloscillates between 1/πand−1/πand does not converge.The usual formulas for area and volume are extended to the unbounded case bymeans of improper integrals.Example 2Letpbe a positive number. Ifis the region between thex-axis and thegraph off(x)=1xp,x≥1,(Figure 10.7.2)yxΩ1f(x) =xpFigure 10.7.2thenarea of=1p−1,ifp>1∞,ifp≤1.
624CHAPTER 10SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALSThis comes about from settingarea of=limb→∞b1dxxp=∞1dxxp.Forp=1,∞1dxxp=limb→∞b1dxxp=limb→∞11−p(b1−p−1)=1p−1,ifp>1∞,ifp<1.Forp=1,∞1dxxp=∞1dxx= ∞,as you have seen already.For future reference we record the following: Letp>0,(10.7.1)∞1dxxpconverges forp>1 and diverges forp≤1.Example 3We know that the region below the graph off(x)=1/x,x≥1, hasinfinite area. Suppose that this region with infinite area is revolved about thex-axis(see Figure 10.7.3). What is the volumeVof the resulting solid? It may surprise yousomewhat, but the volume is not infinite. In fact, it isπ. Using the disc method tocalculate the volume (Section 6.2), we haveV=∞1π[f(x)]2dx=π∞1dxx2=πlimb→∞b1dxx2=πlimb→∞−1xb1=π·1=π.