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Integral test correction

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Took a test today, one question I am not sure I got right.

    [itex]\int_{0}^{\infty} te^{-at}dt[/itex], when a>0

    2. Relevant equations

    3. The attempt at a solution

    I set let infinity be b, then took the limit as b went to infinity of the integral with new bounds from 0 to b. My solution claimed that it converges to 1/a^2. Is this correct or should I expect to miss that question?
     
  2. jcsd
  3. Oct 18, 2011 #2

    gb7nash

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    List your steps and we can tell you if you made a mistake.
     
  4. Oct 18, 2011 #3
    Ok, I'll take that as an "it's incorrect".

    I'll redo the problem now and post it here in a little while.
     
  5. Oct 18, 2011 #4

    gb7nash

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    I'm not saying that. The onus is on you to show your work so we can see what you did and if your method/answer is correct.
     
    Last edited: Oct 18, 2011
  6. Oct 18, 2011 #5
    It's no problem. I just got home so I will redo it, I made sure to memorize the question after I turned it in because it was giving me a hard time with it's recursive limit.

    2dtty77.jpg

    I keep on L'hopitaling, and it keeps on repeating! Finally I somehow worked it out to tend towards zero, so the solution was that it converges to [itex]\frac{1}{a^{2}}[/itex]. Unless I made another mistake that is, and provided, that limit actually does tend to zero. I think my logic was that numerator went to zero, and the denominator went to zero, so really it didn't matter if it was a type o/o indeterminate, I believe that it what I was thinking anyway.

    Here is the limit that I don't know how to solve.
    [tex]lim_{b \to \infty} \frac{be^{-ab}}{-a}[/tex]
     
    Last edited: Oct 18, 2011
  7. Oct 18, 2011 #6

    Dick

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    The only part you need to care about is limit b->inf b*exp(-ab). Take the log. So you've got log(b)-ab. Factor it into b*(log(b)/b-a). Use l'Hopital to show limit log(b)/b goes to zero. So you are left with b*(-a) since a>0. Now goes to -infinity, yes? If log goes to -infinity then b*exp(-ab) goes to zero. Still ok?
     
  8. Oct 18, 2011 #7
    But it's [itex]lim_{b \to \infty}be^{-ab}[/itex] ?
     
  9. Oct 18, 2011 #8

    Dick

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    I mean exp(-ab) to be e^(-ab). Not sure what the question is? Sorry about the failure to correctly TeX things up.
     
  10. Oct 18, 2011 #9
    I read that as [itex]b^{-ab}[/itex], I didn't know that pxp(m) means p^m, sorry.

    I don't understand what you mean when you say to take the log? It looks like you just take the log of the numerator?
     
  11. Oct 19, 2011 #10

    Dick

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    I was just writing exp(x) instead of e^x. Don't worry if you haven't seen it before. Yes, you can just work with the numerator. The denominator is just a constant.
     
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