Does the integral converge
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Does the integral converge

[From: ] [author: ] [Date: 11-09-26] [Hit: ]
Note that for x > 1,(x + 1)/(x^5 + x^2)^(1/4) > (x + 0)/(x^5 + 15x^5)^(1/4) = 1/(2x^(1/4)).[I put the 15 there so the constant comes out nicely.Since ∫(x = 1 to ∞) dx/(2x^(1/4)) = (1/2) * (4/3)x^(3/4) {for x = 1 to ∞} = ∞ (divergent),we conclude by the Comparison Test that the original integral also diverges.I hope this helps!......
Does the integral ∫ (x+1) / ( (x^5) + (x^2) )^1/4 , between 1 and infinity, converge or diverge? I'm supposed to solve using the comparison test but I can't seem to find the right integral to compare it too. I get a different answer each time.

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Ignoring the 'smaller terms', the integrand behaves like x/(x^5)^(1/4) = 1/x^(1/4), which produces a divergent integral when integrated from 1 to ∞.

Let's tidy this up.

Note that for x > 1, we have
(x + 1)/(x^5 + x^2)^(1/4) > (x + 0)/(x^5 + 15x^5)^(1/4) = 1/(2x^(1/4)).
[I put the 15 there so the constant comes out nicely.]

Since ∫(x = 1 to ∞) dx/(2x^(1/4)) = (1/2) * (4/3)x^(3/4) {for x = 1 to ∞} = ∞ (divergent),
we conclude by the Comparison Test that the original integral also diverges.

I hope this helps!
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