Answer
We show that $\frac{{{\partial ^2}F}}{{\partial x\partial y}} = y{{\rm{e}}^{xy}}$
Using the result of Exercise 50, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} y{{\rm{e}}^{xy}}{\rm{d}}A \simeq 4.64356$
Work Step by Step
We have $F\left( {x,y} \right) = {x^{ - 1}}{{\rm{e}}^{xy}}$. So, the partial derivatives are
$\frac{{\partial F}}{{\partial y}} = {{\rm{e}}^{xy}}$, ${\ \ \ \ }$ $\frac{{{\partial ^2}F}}{{\partial x\partial y}} = y{{\rm{e}}^{xy}}$
Recall from the result of Exercise 50:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = F\left( {b,d} \right) - F\left( {a,d} \right) - F\left( {b,c} \right) + F\left( {a,c} \right)$
where $f\left( {x,y} \right) = \frac{{{\partial ^2}F}}{{\partial x\partial y}}$ and ${\cal R} = \left[ {a,b} \right] \times \left[ {c,d} \right]$.
We have $f\left( {x,y} \right) = \frac{{{\partial ^2}F}}{{\partial x\partial y}} = y{{\rm{e}}^{xy}}$.
Since ${\cal R} = \left[ {1,3} \right] \times \left[ {0,1} \right]$, using the result of Exercise 50 we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} y{{\rm{e}}^{xy}}{\rm{d}}A$
$ = F\left( {3,1} \right) - F\left( {1,1} \right) - F\left( {3,0} \right) + F\left( {1,0} \right)$
$ = \frac{{{{\rm{e}}^3}}}{3} - {\rm{e}} - \frac{1}{3} + 1$
$ = \frac{{{{\rm{e}}^3}}}{3} - {\rm{e}} + \frac{2}{3}$
$ \simeq 4.64356$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} y{{\rm{e}}^{xy}}{\rm{d}}A \simeq 4.64356$.
