Here.
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.
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