# The FTC (e^-x^2)

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 P: 10 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
 Mentor P: 11,631 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?
 P: 10 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
 Mentor P: 11,631 The FTC (e^-x^2) 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?
 P: 10 Find the integral, and insert the "restriction", to find the constant!
 P: 10 I do know that, my problem however, is it impossible to integrate the function as it stands there. best Dan
Mentor
P: 11,631
 Quote by IsomaDan Find the integral, and insert the "restriction", to find the constant!
That works, right.

 I do know that, my problem however, is it impossible to integrate the function as it stands there.
You do not have to integrate that expression. Just write down the integral, and you are done.
 Mentor P: 21,214 I'm locking this thread as it has been cross-posted. The other thread is here: http://www.physicsforums.com/showthread.php?t=651819.