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Help me solve this question I have also provided a snippet of the explanation. 1. Determine Whether the following integTals converge. Explain Why each converges

Help me solve this question I have also provided a snippet of the explanation.

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1. Determine Whether the following integTals converge. Explain Why each converges or diverges using the eonwrgenee test for I? f(:.c)d5: on page 404 of the textbook. (a) m Do do 1 u l 113.2 [W1+3in26d9 92 D {b} (C) 304 mm mm Notice that we rst looked at the behavior of the integrand as a: > 00. This is useful because the convergence or divergence of the integral is determined by what happens as :1: > 00. The Comparison Tbs! for f\" f (:23) da: Assume f (1') is positive. Making a comparison involves two stages: 1. Guess, by looking at the behavior of the integrand for large 2:, whether the integral con- verges or not. {This is the \"behaves like" principle.) 2. Conrm the guess by comparison with a positive function 9(3): 0 If [(1') 5 9(3) and 1:\" 9(1'] deconverges, then If\" r) tin: converges. o If 9(3) 5 f(.1:) and 1:\" g(:r] dry: diverges, then I? f (3:) d2: diverges. \"0 dt Example 2 Decide whether / W converges or diverges. 4 _ Solution Since lnt grows without bound as t > 00. the 1 is eventually going to be insignicant in compar- ison to In t. Thus, as far as convergence is concerned, 0 1 0 1 [4 W (it behaves like A E d1. Does LWU/ Int) dt converge or diverge? Since lnt grows very slowly. 1/ int goes to zero very slowly. and so the integral probably does not converge. We know that (int) 1 e, we take reciprocals: l 1 1 f>.>. o 1'\" J4 'IIET I. \"(Inn1 Int Does f 400(1 / in t) (it converge or diverge? Since lnt grows very slowly. 1/ Int goes to zero very slowly. and so the integral probably does not converge. We know that [in t) 1 e, we take reciprocals: 1 > 1 > 1 (Int) 1 In: t' Since f4(1/t)dt diverges. we conclude that m 1 d ' - A (111:) _ 1 tdlverges How Do We Know What to Compare With? In Examples 1 and 2, we investigated the convergence of an integral by comparing it with an easier integral. How did we pick the easier integral? This is a matter of trial and error, guided by any infor- mation we get by looking at the original integrand as a: -+ co. We want the comparison integrand to be easy and. in particular. to have a simple antiderivative. Useful Integrals for Comparison on a / 33de converges for p > 1 and diverges for p g l. 1 1 a ] Iipdx converges for p

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