∮▒e^y dx+2xe^y dy= ∬_0^x▒e^y dydx
I have a square described by the following eq: x=0, x=1, y=0 and y=1.
I'm computing the Green's Theorem for the function (e^{y})dx+(2xe^{y})dy.
So...
Line Integral of (e^{y})dx+(2xe^{y})dy = Double Integral of (dQ/dX - dP/dY)dA = Double Integral of e^{y}dA
Since I'm on a square, shouldn't the bounces of my double integral be 0 and 1 for X and 0 and X for Y?
The answer should be e-2 right?