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Improper Integral

  • Thread starter ttiger2k7
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This was a problem on one of my previous tests that I got wrong entirely. In preparing for my final, I'm attempting to redo it. I was wondering if someone could check my work.

Determine whether the following improper integral is convergent or divergent. If the integeral is convergent, find its value.

[tex]I = \int^{\infty}_{0}\frac{2x}{(x^{2}+1)^{1.2}}[/tex]

which, after carrying out the 1.2, becomes

[tex]\int^{\infty}_{0}\frac{2x}{(x^{2.4}+1)}[/tex]

That's where I get stuck. I know that the function diverges, but I don't know how to prove it.

I thought about trying to manipulate it somehow so that the function I'm comparing it to is in the form

[tex]\frac{1}{x^{p}}[/tex]

I started off by trying a direct comparison test and compared it to [tex]\frac{2x}{x^{2.4}}[/tex] which can simplify to

[tex]\frac{2}{x^{.4}}[/tex]

So, since the above function has [tex]p=.4<1[/tex], it diverges, by the Direct Comparison Test, [tex]I[/tex] diverges.
 
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Answers and Replies

  • #2
Mute
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Be careful! [itex](a + b)^p \neq (a^{p} + b^p)[/itex].

Furthermore, the function [itex]2x/x^{2.4}[/itex] is actually slightly larger than your original integrand, so the fact that the integral over this diverges doesn't actually tell you anything about the convergence/divergence of your original integrand.

The fact that there is a [itex]2x[/itex] in the integrand suggests you should make the substitution [itex]u = x^2 + 1[/itex]. Then your integral becomes

[tex]I = \int^{\infty}_{1}\frac{du}{u^{1.2}} = \int^{\infty}_{1}\frac{du}{u^{6/5}}[/tex]

This is an integral you can do, and you should be able to tell if it diverges or converges.
 
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  • #3
HallsofIvy
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This was a problem on one of my previous tests that I got wrong entirely. In preparing for my final, I'm attempting to redo it. I was wondering if someone could check my work.

Determine whether the following improper integral is convergent or divergent. If the integeral is convergent, find its value.

[tex]I = \int^{\infty}_{0}\frac{2x}{(x^{2}+1)^{1.2}}[/tex]

which, after carrying out the 1.2, becomes

[tex]\int^{\infty}_{0}\frac{2x}{(x^{2.4}+1)}[/tex]
No, it doesn't! In general, (x+ y)n is NOT xn+ yn!

That's where I get stuck. I know that the function diverges, but I don't know how to prove it.

I thought about trying to manipulate it somehow so that the function I'm comparing it to is in the form

[tex]\frac{1}{x^{p}}[/tex]

I started off by trying a direct comparison test and compared it to [tex]\frac{2x}{x^{2.4}}[/tex] which can simplify to

[tex]\frac{2}{x^{.4}}[/tex]

So, since the above function has [tex]p=.4<1[/tex], it diverges, by the Direct Comparison Test, [tex]I[/tex] diverges.
Let u= x2+ 1, so that du= 2x dx, in your original integral.
 

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