# Potential of a function

• Mar 28th 2007, 01:48 PM
pakman
Potential of a function
The function is given as a vector...

F(x,y) = (x*e^((-x)^2) - (-y)^2))i + (y*e^((-x)^2 - (-y)^2))j

Now according to the solutions provided, the potential of that function is the integral of df/dx and df/dy.

The answer came out to (-1/2)e^((-x)^2) - (-y)^2)) + C1(y) for df/dx. I think I'm getting confused about integrating the exp, where did the x go? Thanks.
• Mar 28th 2007, 04:19 PM
ThePerfectHacker
Here.
• Mar 28th 2007, 05:08 PM
pakman
I understand how the potential part works, I guess I'm just confused about the integral of an exp. I don't understand how the integral of x*e^(...) goes to -(1/2)*e(...)

Could you explain that? Thanks.
• Mar 28th 2007, 05:45 PM
topsquark
Quote:

Originally Posted by pakman
I understand how the potential part works, I guess I'm just confused about the integral of an exp. I don't understand how the integral of x*e^(...) goes to -(1/2)*e(...)

Could you explain that? Thanks.

Think about it in reverse. What's the derivative of e^{x^2}?

d/dx[e^{x^2}] = e^{x^2} * (2x) = 2x*e^{x^2}

by the chain rule. So
Int[x*e^{x^2} dx] = (1/2)*e^{x^2} + C

Alternately: Do a substitution.
Int[x*e^{x^2} dx]

Let y = x^2 ==> dy = 2x dx
Thus
Int[x*e^{x^2} dx] = Int[(1/2)e^y dy] = (1/2)*e^y + C = (1/2)*e^{x^2} + C

-Dan