Answer
$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 x{{\rm{e}}^{xy}}{\rm{d}}x{\rm{d}}y \simeq 0.718282$
Work Step by Step
We use Fubini's Theorem to change the order of integration and re-write the integral:
$I = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{x = 0}^1 x{{\rm{e}}^{xy}}{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = 0}^1 x{{\rm{e}}^{xy}}{\rm{d}}y} \right){\rm{d}}x$
We evaluate first the inner integral with respect to $y$.
$S\left( x \right) = \mathop \smallint \limits_{y = 0}^1 x{{\rm{e}}^{xy}}{\rm{d}}y$
The antiderivative of $x{{\rm{e}}^{xy}}$ with respect to $y$ is ${{\rm{e}}^{xy}}$ because
$\frac{\partial }{{\partial y}}\left( {{{\rm{e}}^{xy}}} \right) = x{{\rm{e}}^{xy}}$
So,
$S\left( x \right) = \mathop \smallint \limits_{y = 0}^1 x{{\rm{e}}^{xy}}{\rm{d}}y = {{\rm{e}}^{xy}}|_{y = 0}^1 = {{\rm{e}}^x} - 1$
Thus,
$I = \mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = 0}^1 x{{\rm{e}}^{xy}}{\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {{{\rm{e}}^x} - 1} \right){\rm{d}}x$
$ = {{\rm{e}}^x}|_0^1 - \mathop \smallint \limits_{x = 0}^1 {\rm{d}}x$
$ = {\rm{e}} - 1 - 1$
$ = e - 2$
$ \simeq 0.718282$
So, $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 x{{\rm{e}}^{xy}}{\rm{d}}x{\rm{d}}y \simeq 0.718282$.
