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Convergent or Divergent Integral: Comparison Theorem

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine if the following is improper and convergent, improper and divergent, or proper
    [tex]\int \frac{dx}{\sqrt[3]{x^2 - 7}}[/tex]
    from 8 to infinity


    3. The attempt at a solution
    Since I don't know how to integrate it, I believe I would use the comparison theorem. This is where I have trouble. I understand how it works in regards to seeing if it is convergent or not, but I have trouble determining whether I should get a function larger or smaller. Similarly, I have trouble creating that function. I don't have many examples on it.

    So, I used: [tex]\frac{1}{\sqrt[3]{x^{3}}}[/tex] [tex]\leq[/tex] [tex]\frac{dx}{\sqrt[3]{x^2 - 7}}[/tex], in which case, since it is ln(abs(x)), it diverges, and therfore the other must also diverge.

    I'm not quite sure if I have that right, and if it was ok to just pick a function that was smaller than the other, without somehow deriving it from the first.
     
    Last edited: Feb 20, 2010
  2. jcsd
  3. Feb 20, 2010 #2

    Mark44

    Staff: Mentor

    I can't tell what your integrand is.
    [tex]\int \frac{dx}{\sqrt[3]{x^2} - 7}[/tex]

    [tex]\int \frac{dx}{\sqrt[3]{x^2 - 7}}[/tex]

    Is it one of these? If so, which one?
     
  4. Feb 20, 2010 #3
    There's no reason for the comparison function to be related in any way to the function being studied, except that it satisfy the inequality in the region in question.
     
  5. Feb 20, 2010 #4
    Sorry, I couldn't get the script right. Its
    [tex]\int \frac{dx}{\sqrt[3]{x^2 - 7}}[/tex] from 8 to infinity
     
  6. Feb 20, 2010 #5
    Ok, thanks. I'm still unsure if I did the question properly. I know the answer is right, but I'm just not sure whether the work is right or not.
     
  7. Feb 20, 2010 #6

    Mark44

    Staff: Mentor

    A more obvious choice would be 1/x2/3

    [tex]\int \frac{dx}{\sqrt[3]{x^2 - 7}} \geq \int \frac{dx}{\sqrt[3]{x^2}}[/tex]

    Since the denominator of the expression on the left is clearly smaller than that in the right, the overall fraction is larger that that on the right, hence the integral on the left is larger than the one on the right.

    When you evaluate the definite integrals, be sure to use a limit.

    Whether to pick a function that is smaller or larger depends on whether you think the improper integral converges or converges. If you believe that the integral converges, you want to choose a function that is larger. If your integrand is smaller than that of a convergent integral, your improper integral converges. If your integrand is larger than that of a divergent integral, your improper integral diverges.

    Note that there are two cases you don't want: your integral is larger than a convergent integral; and your integral is smaller than a divergent integral. You can't tell anything from these cases.
     
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