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Quick Integral question

  1. Aug 6, 2012 #1
    1. The problem statement, all variables and given/known data
    The question is as attached in the picture.


    3. The attempt at a solution

    I haven't got to the answer, but I wonder if this is OK:

    g(y) = ∫ f(x,y) dx

    = F(y) - F(0,y)

    (dg/dy) = f(y) - f(0,y)


    Since the 'x' is just a dummy variable where the final function g will be in terms of y. By integrating it with respect to x, then filling it in with 0 and y, by differentiating g does it reverse back the integration process and simply filling it with 0 and y?
     

    Attached Files:

  2. jcsd
  3. Aug 6, 2012 #2

    LCKurtz

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    I don't see any question in that picture.
     
  4. Aug 6, 2012 #3
    Sorry that was just part of the question, I am trying to find out if its OK to do the steps above.
     
  5. Aug 6, 2012 #4

    LCKurtz

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    I can't follow this. ##g(y) = \int f(x,y)\, dx## doesn't make any sense to me. It's an indefinite integral and surely there would be an ##x## on the left side. Then you have ##F(y)-F(0,y)##. I gather that is supposed to represent an antiderivative, but is ##F## a function of one or two variables? Then in your last line you have ##f(y) - f(0,y)##. But ##f## is a function of two variables, so ##f(y)## is meaningless.

    Anyway, whatever it is you are trying to do, try it with ##f(x,y) = y^2\cos(x)## or something like that. It has to make sense and work for something simple to have any chance of being true in general.
     
  6. Aug 6, 2012 #5
    Sorry the integral is meant to be a definite integral as in the question, I'm not sure how to use latex...

    How about if

    g(y) = ∫ f(x,y) dx (definite integral)

    = F(y,y) - F(0,y)

    then

    dg/dy = f(y,y) - f(0,y)

    Does this make more sense?
     
  7. Aug 6, 2012 #6
    You have the right idea, but the steps you're using are wrong. The function in the integral is not a function of x and y, it is solely a function of x, so your antiderivatives in the second part don't make any sense.
     
  8. Aug 6, 2012 #7
    Integrate the function in the integral with respect to x while keeping y constant.
     
  9. Aug 6, 2012 #8
    Yes that's right but are my steps right? I'm concerned about the conversion from 'F' to 'f' by differentiation. Since they are entirely in terms of y after integration.
     
  10. Aug 6, 2012 #9

    Ray Vickson

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    You don't need to use LaTeX (although some posters might complain if you don't), but you DO need to write things explicitly. For example, if you want to write
    [tex] \int_a^b f(x,y) \, dx [/tex] in plain text you can just write int( f(x,y) dx, x=a..b), for example.

    RGV
     
  11. Aug 6, 2012 #10
    Lleibniz integral rule:

    [tex]\frac {d} {dy} \int_{a(y)}^{b(y)} f(x, y)dx = \frac {db(y)} {dy} f(b(y), y) - \frac {da(y)} {dy} f(a(y), y) + \int_{a(y)}^{b(y)} \frac {\partial} {\partial y} f(x, y)dx[/tex]
     
  12. Aug 6, 2012 #11
    Would it apply, because the function isn't continuous over [0,y]?
     
  13. Aug 6, 2012 #12
    OP, try integrating by parts.

    EDIT: It worked for me. Integrate by parts and then fundamental theorem of calculus combined with the hint that f(0) = 0.
     
    Last edited: Aug 6, 2012
  14. Aug 6, 2012 #13
    And then the Leibniz rule :)
     
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