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Integral equation

  1. Jan 21, 2009 #1
    Hello,

    Could you please help me to solve this equation:

    G(x)= INTEGRAL[F(t)dt]

    lower integral limit: x-m,
    upper integral limit: x,
    m is a constant,
    G(x) is a known function,
    F(x) is unknown and should be found.

    Thank you very much!
     
  2. jcsd
  3. Jan 21, 2009 #2
    Differentiate both sides?

    You should get G'(x) = F(x) - F(x-m).
     
  4. Jan 21, 2009 #3
    so easy? are you sure?
     
  5. Jan 21, 2009 #4
    Nope, I'm not in your class, I don't know what you are learning nor what the teacher is trying to show you with that exercise but I do like giving false leads so people would do bad on their homework.
     
  6. Jan 21, 2009 #5
    I was thinking that there should be a more complex expression while differentiating th integral by a parameter x.
     
  7. Jan 21, 2009 #6
    Maybe it is, consult FTC
     
  8. Jan 21, 2009 #7
    what's FTS?
     
  9. Jan 21, 2009 #8
    Not sure what FTS is but FTC is Fundamental Theorem of Calculus
     
  10. Jan 21, 2009 #9

    In the Fundamental Theorem of Calculus, the f(x) under integral and the final F(x) are different functions, namely f(x)=F'(x). Therefore, in my case it does not help, since my function under the integral is unknown.

    This equation is easy, but not that easy.

    I am now trying to solve it by reformulating through series.
     
  11. Jan 21, 2009 #10
    http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part

    Your G(x) is their F(x) therefore your G'(x) is their F'(x). Your F(t) is their f(t). The only difference is their integral goes from a to x and yours goes from x-m to x therefore you can generalize their version to

    [tex] F(x) = \int_{\alpha(x)}^{\beta(x)} f(t) dt \Rightarrow F'(x) = f(\beta(x)) \cdot \beta'(x) - f(\alpha(x)) \cdot \alpha'(x) [/tex]
     
  12. Jan 21, 2009 #11
    nope, ...but may be you are right.
    [​IMG]
     
    Last edited: Jan 21, 2009
  13. Jan 21, 2009 #12
    nope what?
     
  14. Jan 21, 2009 #13
    I wasn't claiming it's THAT easy to solve for F(x) it is not, in what I told you, I don't explicitly solve for F(x) I just give an exprsession that involves it and if your G(x) is "nice" enough you can "guess" what F(x) would have to be.
     
  15. Jan 21, 2009 #14
    I have just derived another solution for my problem throught series, and it reads like this:

    F(x)=G(x)- SUM(F(t)) {summation for t from x-m to x-1}
     
  16. Jan 21, 2009 #15
    Your solution is probably right, and then these two solutions should be equivalent. If my solution through series is right.
     
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