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Problem integrating WKBJ

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data
    I am having a problem integrating in a WKBJ semi-classic integral. Well it's this : I have to integrate

    [itex]\int_{0}^{\sqrt{m}E}\sqrt{E-\frac{x}{\sqrt{m}}}dx[/itex]

    2. Relevant equations
    Actually I don't have that much experience at integrating, so could you somehow show me how to integrate when you have a square root? Step by step this particular one, for example.

    3. The attempt at a solution
    I have tried setting the square root equal to a variable, t, and saying that the integral goes like
    [itex]\int_{0}^{\sqrt{m}E}t^2dt[/itex] but it didn't seem to work out later on, plus I am almost sure this isn't correct.
     
  2. jcsd
  3. Feb 6, 2009 #2

    turin

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    I would just look it up in a table, as I usually do. I'm sure there's some trick that I was taught in calc 2, but you know, in my experience, most of those tricks are almost never useful anywhere besides a calc 2 test. And for such a simple integral, you can definitely find it in a table. Any integral of a squareroot of a 2nd order polynomial will be in even a modest table of integrals.
     
  4. Feb 6, 2009 #3

    turin

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    OK, now I feel dumb. Yes, that is such an easy substitution. You just screwed up your limits. I'm guessing that you defined t as the squareroot. So, what is t when x=0 and what is t when x=\sqrt{m}E? Also, I think you get some additional factor.
     
  5. Feb 7, 2009 #4
    That integral can be solved with a simple substitution. Hint: look at the quantity under the radical sign.
     
  6. Feb 7, 2009 #5
    Is this what I should be getting from a table?

    When i need to integrate [itex]\int (ax+b) dx[/itex] I set the square root equal to S and proceed to [itex]\int_{0}^{\sqrt{m}E}S dx=\frac{2S^3}{3a}[/itex] if [itex]S=\sqrt{ax + b}[/itex]?
     
  7. Feb 7, 2009 #6

    Redbelly98

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    Yes, though you probably meant to say

    [itex]
    \int \sqrt{ax+b} \ dx
    [/itex]

    for the integral.
     
  8. Feb 7, 2009 #7
    Yes indeed, my mistake! Well thanks for the help, I will work on it now :)
     
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