Double integral involving $e^{x^2}$ 
Given the function $f(x,y)$ such that
  $$\left\{\begin{array}
&e^{x^2}\quad x > y\\
e^{y^2}\quad y > x\end{array}\right.$$
What is $\iint_{[0,1]^2}f(x,y)\mathrm{d}A$

My approach:
$\int_0^1\int_0^xe^{x^2}\mathrm{d}x\mathrm{d}y=\int_0^1 e^{x^2}\mathrm{d}x\int_0^x\mathrm{d}y = \int_0^1xe^{x^2}\mathrm{d}x = \left[\dfrac{1}{2}e^{x^2}\right]_0^1 = \dfrac{e^{1/2}}{2}$
and then double this, because I'm thinking there is a symmetry involved, so $A = e^{1/2}$
But I'm not sure if this is correct...
 A: The integral is
$$\int_0^1 dx \, \int_0^x dy \, e^{x^2} + \int_0^1 dy \, \int_0^y dx \, e^{y^2} = 2 \int_0^1 dx \, x \, e^{x^2} = e-1$$
Note that the integrals are equal because $x$ and $y$ may be switched.
A: It is (almost) correct, but maybe less confusing if we write:
$$\int_0^1 \left( \int_0^xe^{x^2}\mathrm{d}y \right) \mathrm{d}x
=\int_0^1 e^{x^2} \left( \int_0^x\mathrm{d}y \right) \mathrm{d}x
=\int_0^1e^{x^2}x \; \mathrm{d}x
=\left[\dfrac12 e^{x^2}\right]_0^1
= \dfrac12 (e - 1)$$
Edit: I nearly overlooked the small error you made when you substituted $0$ and $1$ in, in the last bracket.
To make it clear, $\dfrac12 (e-1)$ is the integral over the "lower" triangle $0\le y \le x \le 1$ only. I absolutely agree with the method of the question of multiplying this partial result by two, by symmetry, to get the integral over the full unit square $[0,1]^2$. So I meant to say the final answer is $e-1$.
As I said in a comment, my reason for almost repeating the calculation of the original poster, was that his second expression, $\int_0^1 e^{x^2}\mathrm{d}x\int_0^x\mathrm{d}y$, contained an $x$ (in the upper limit of the second integral) which stands to the right of the $\mathrm{d}x$. I considered that confusing/wrong (depending on how free a notation one has for integrals) when this is considered as two "nested" or iterated integrals.
A: Your approach is correct, because
\begin{align}
\iint_R f(x,y)dA &=\int_0^1 \int_0^x e^{x^2}dydx + \int_0^1 \int_0^y e^{y^2}dxdy\\
&=\int_0^1 \int_0^x e^{x^2}dydx+\int_0^1 \int_0^x e^{x^2}dydx\\
&=2\int_0^1 \int_0^x e^{x^2}dydx.
\end{align}
