# Thomas' Calculus: Early Transcendentals

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Mat104 Fall 2002, Improper Integrals From Old ExamsFor the following integrals, state whether they are convergent or divergent, and give your reasons.(1)0dxx3+ 2converges. Break it up as10dxx3+ 2+1dxx3+ 2. The first of these is properand finite. The second behaves like the integral of 1/x3on [1,) and thus converges.(2)10dxx+xconverges. Asx0,xgoes to 0 much more slowly thanxdoes. (Thinkabout the graphs.) Therefore whenxis very close to 0, the denominatorx+xx.So this integral will behave like the integral of 1/xon [0,1], and this integral converges.(3)11 +xx3converges. Asxgoes to, the integrand behaves likexx3=1x5/2.(4)0x2x3+ 1dxdiverges. Break it up into two integrals10x2x3+ 1dx+1x2x3+ 1dx. Thefirst integral is proper and finite. The second can be compared to the integral of 1/xon[1,) which diverges.(5)10lnx dxconverges to-1. Here we can compute directly since integration by parts tellsus thatlnx dx=xlnx-x+C. Evaluating at thex= 1 endpoint gives ln 1-1 =-1.For the other endpoint we have to take the limit asxgoes to 0. For this we need L’Hˆopital’srule.limx0xlnx= limx0lnx1/x= limx01/x-1/x2= limx0-x= 0.So evaluating at thex= 0 endpoint gives 0.(6)10dxex-1diverges. The only difficulty is that the denominator is 0 whenx= 0. There area couple of approaches we could take. The easiest is to use the Taylor series forex. Thenwe know thatex-1 =x+ higher powers ofxand asxgoes to zero, the higher powers ofxwill vanish much more rapidly. So this function behaves essentially like 1/xwhenxisclose to 0. Sincedx/xdiverges, this integral will also.Alternatively, we could compute the integral, making the substitutionu=exand thenuse partial fractions.(7)0dxx2+ 2x+ 2converges. The only difficulty is that we have an infinite endpoint. Theintegrand is asymptotic to 1/x2asxgoes to infinity.Since1dx/x2converges, thisintegral will as well. (To compare these we should break up the integral. First integratefrom 0 to 1, which gives a finite value. Then integrate further from 1 out to. This givesa finite value as well by comparison to 1/x2.)(8)1x3lnx+x4dxdiverges. Again the only problem is that we have an infinite endpoint. Asxgoes to infinity,x4grows much faster than lnx. Thus the integrand will be asymptotictox3/x4= 1/xasxgoes to infinity.

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