1. ## 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

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

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.

3. ## 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?

4. ## 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