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The FTC (e^-x^2)

  1. Nov 13, 2012 #1
    The function, p(x;y), of two variables is defi ned for x>y>0, and satisfi es

    We furthermore know that dp(x,y)/dx = (e^-x^2)

    and that p(y; y) = 0

    I now need to write p(x,y) as a definite integral of the form int (f(t)dt, with lower bound t=H and upper bound x.
    I suppose I need the info p(y; y) = 0 to get the bounds, but not quite sure how.
    Anyone can give me a hint :-)

    Dan
     
  2. jcsd
  3. Nov 13, 2012 #2

    mfb

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    For a fixed y, you know p(y, y) = 0. Based on that, how do you get p(x; y) = 0 with dp(x,y)/dx = (e^-x^2)?

    If that does not help, here is a slightly easier problem:
    For a fixed y, you can reduce it to a 1-dimensional problem: f(y)=0, df/dx=(e^-x^2). Can you write down f(x) with an integral?
     
  4. Nov 13, 2012 #3
    Dear MFB. Thanks so much for your answers. That's much appreciated.

    Well, with regards to your answer, I suppose (but do not know) I evaluate the definte integral of dp(x,y)/dx = (e^-x^2), with the bounds being an arbitrary constant and set it equal to zero?

    Is that correct, or am I still far off?

    best

    Dan
     
  5. Nov 13, 2012 #4

    mfb

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    I don't understand what you plan to do, or why.

    If you know the value of a one-dimensional function at one point and its derivative everywhere, how can you get the value of the function everywhere?
    You don't have to evaluate the integral.

    If g(0)=3 and g'(x)=x, how does g look like?
     
  6. Nov 13, 2012 #5
    Find the integral, and insert the "restriction", to find the constant!
     
  7. Nov 13, 2012 #6
    I do know that, my problem however, is it impossible to integrate the function as it stands there.

    best

    Dan
     
  8. Nov 13, 2012 #7

    mfb

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    That works, right.

    You do not have to integrate that expression. Just write down the integral, and you are done.
     
  9. Nov 13, 2012 #8

    Mark44

    Staff: Mentor

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