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Comprison test for Integrals

  1. Apr 12, 2008 #1
    Hey guys. I'm a little new on the comparison test, if you could just check that i'm on the right track, it would be great.

    1. The problem statement, all variables and given/known data
    Using comparison test to decide whether or not [tex]\frac{\sqrt{x}}{\ln{x} + x^2}[/tex] converges.

    2. Relevant equations
    (Check 3.)


    3. The attempt at a solution
    Okay, so first off, I guess that this converges. So find a function that is larger than the one that is given and test if that converges, yes?

    Would: [tex]\frac{\sqrt{x}}{\ln{x} + x^2} < \frac{\sqrt{x}}{x^2}[/tex] be a good choice?

    So would I be right in saying that as [tex]x^{-3/2}[/tex] converges, that also the one given converges?


    I'm not sure if i've gotten this backwards or not. Could someone please clear it up for me?

    Thanks.
     
  2. jcsd
  3. Apr 12, 2008 #2

    HallsofIvy

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    First of all, it doesn't make any sense to ask "whether [tex]\frac{\sqrt{x}}{\ln{x} + x^2}[/tex] converges". Do you mean to determine whether or not an integral of that converges? If so then you still need to specify whether you are asking about the improper integral [tex]\int_a^\infty \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex] with a> 0 or [tex]\int_0^a \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex], again with a> 0.
     
  4. Apr 12, 2008 #3
    ... Whoops. So sorry, forgot to include the integral.

    But yes, check if the Integral converges. Sorry about that.

    [tex]\int_1^\infty \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex]

    Not sure why, but my upper bound won't stay as infinity. (Today's the first time i've used LaTeX.)

    Bounds are 1 to Infinity.

    Oh, and whenever I said in the first post whether or not things converge, it's when taking the limit as x approaches infinity. Sorry about not making that clear in the first post.
     
    Last edited: Apr 12, 2008
  5. Apr 12, 2008 #4

    Dick

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    Your choice of x^(-3/2) to compare with is a good one. And yes, it works. You might want to pound that point in by saying why the inequality is true.
     
  6. Apr 12, 2008 #5
    Alright, thanks. :)

    I'll add in an extra step saying that x^2 < ln(x) + x^2 and explaining it. Thanks.
     
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