1873_solutions

Integer and 0 then the series converges absolutely

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Unformatted text preview: imit is 2 then we should be saying that the number 2 is convergent. Of course such an interpretation is not intended. Instead, we mean that the limit exists and is finite. In b that sense, saying that Þ f x dx is convergent is not as precise as saying that f is improper a Riemann integrable on a, b . Some Exercises on the Comparison Test for Integrals 1. Determine the convergence or divergence of the following integrals: a. Þ1 Ý x2 x dx x1 Since lim xÝ and since Þ Ý 1 1 x 3/2 x2 x x1  1 1 x 3/2 dx is convergent, the integral Þ Ý 1 x2 x dx is convergent. x1 Þ0 1 b. 1 dx x  x2 Since 1 x x 2 1 x 0 x 1 lim and since Þ c. Þ1 Ý 1 0 1 x dx is divergent, the integral Þ sin 2 x dx x2 353 0 1 1 dx is divergent. x  x2 Since sin 2 x x2 0 for all x 1 and since the integral Þ 1 x2 dx is convergent, the integral Þ Ý1 1 x2 convergent. d. Þ0 Ý sin 2 x dx x2 x We break thisintegral into the two parts Þ Ý 1 sin 2 x dx is x2 sin 2 x dxand Þ Ý sin 2 x dx. 1 x2 x x2 x 1 0 Since sin 2 x x2 x lim 1 x 0 x and since the integral Þ 1 1 x 0 1 dx is convergent, the integral Þ 1 0 sin 2 x dx is convergent. x2 x Since sin 2 x x2 x 0 whenever x 1 and since the integral Þ Ý convergent. Therefore the given integral Þ e. Þ0 /2 1 x 5/2 1 Ý 0 1 x 5/2 dx is convergent, the integral Þ Ý 1 sin 2 x dx is x2 x sin 2 x dx is convergent. x2 x tan x dx From the fact that lim x /2 cos x  1  x 2 we see that tan x 1 x lim  2 x /2 and since the integral Þ /2 1  2 0 x dx is convergent, the integral Þ /2 0 tan x dx is convergent. Alternatively, we could have evaluated this integral directly. It is an elementary integral that can be evaluated by making the substitution u  tan x and then breaking the integral down with partial fractions. Þ1 2 f. 1 dx log x Since 1 log x lim 1 x 1 x1 and since theintegral Þ g. /2 Þ0 2 1 1 x1 1 dx is divergent, the integral Þ 2 1 1 dx is divergent. log x log sin x dx We observe that log sin x 0for each x 0,  2 and so the comparision test can’t be applied to this integral as it stands. On the other hand, we can apply the comparison test to the integral /2 log sin x dx. Since Þ0 354 lim log sin x x0 and since the integral Þ therefore the integral Þ h. Þ2 Ý 1 log x log x 1 0 /2 0 1 x 1 x 0 dx is convergent, the integral Þ /2 log sin x dx is convergent and 0 log sin xdx is convergent. dx Hint: This problem is the integral analogue of the problems that occurred earlier on the comparison test for series. For x sufficiently large we have 1 1  exp log x log log x log x log x Since the integral Þ i. Þ3 Ý 1 log x 2 1 dx log log x log x For x sufficiently large we have 1 1  exp log x log log log x log log x log x log x dx is convergent. Ý1 2 x2 dx is convergent, the integral Þ 1 exp 2 log x Ý 1 log log x 2 log x  1. x2 dx is convergent. Ý Þ 30 1 dx log log log x log x For x sufficiently large we have 1 1  exp log x log log log log x log log log x log x Since the integral Þ k. dx is convergent...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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