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Math Help - Integral of (e^-x^2)

  1. #1
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    Integral of (e^-x^2)

    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. I suppose I need the info p(y; y) = 0 to get thebound, but not quite sure how.
    Anyone can give me a hint :-)

    Dan
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  2. #2
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    Re: Integral of (e^-x^2)

    Hey IsomaDan

    Consider that p(x,y) will be the integral of dp(x,y)/dx with respect to x (or dx) plus some constant (if you are looking at an indefinite integral) and if p(y,y) = 0 then this constant needs to take into account the nature of this condition.
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    Re: Integral of (e^-x^2)

    Hi Chiro.

    Thanks so much for your response. However, I am still wondering, how I should calculate that constant, because I am not actually evaluating the integral; It is only gonna go into the bounds - but how?
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  4. #4
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    Re: Integral of (e^-x^2)

    Your integral goes from (y,y) to (x,y). Since it is zero at (y,y), if you set the lower limit at y, the constant is zero. So you have:
    p(x,y)=\int_y^xe^{-t^2}dt
    I used t for the integration variable since I want to use x for the upper limit.

    You can see that p(y,y)=0. And \frac{dp}{dx}=e^{-x^2} by the Fundamental Theorem of Calculus.

    - Hollywood
    Thanks from IsomaDan
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