# The FTC (e^-x^2)

1. Nov 13, 2012

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. Nov 13, 2012

### Staff: Mentor

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?

3. Nov 13, 2012

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

4. Nov 13, 2012

### Staff: Mentor

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?

5. Nov 13, 2012

Find the integral, and insert the "restriction", to find the constant!

6. Nov 13, 2012

I do know that, my problem however, is it impossible to integrate the function as it stands there.

best

Dan

7. Nov 13, 2012

### Staff: Mentor

That works, right.

You do not have to integrate that expression. Just write down the integral, and you are done.

8. Nov 13, 2012