Answer
We prove:
If $\frac{{{\partial ^2}F}}{{\partial x\partial y}} = f\left( {x,y} \right)$, then
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = F\left( {b,d} \right) - F\left( {a,d} \right) - F\left( {b,c} \right) + F\left( {a,c} \right)$
where ${\cal R} = \left[ {a,b} \right] \times \left[ {c,d} \right]$.
Work Step by Step
We have ${\cal R} = \left[ {a,b} \right] \times \left[ {c,d} \right]$.
Suppose that $\frac{{{\partial ^2}F}}{{\partial x\partial y}} = f\left( {x,y} \right)$.
Using Fubini's Theorem we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = a}^b \left( {\mathop \smallint \limits_{y = c}^d \frac{{{\partial ^2}F}}{{\partial x\partial y}}{\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = a}^b \frac{\partial }{{\partial x}}\left( {\mathop \smallint \limits_{y = c}^d \frac{{\partial F}}{{\partial y}}{\rm{d}}y} \right){\rm{d}}x$
Let us consider the inner integral $\mathop \smallint \limits_{y = c}^d \frac{{\partial F}}{{\partial y}}{\rm{d}}y$.
By the Fundamental Theorem of Calculus, Part I (Section 5.4), we get
$\mathop \smallint \limits_{y = c}^d \frac{{\partial F}}{{\partial y}}{\rm{d}}y = F\left( {x,y} \right)|_{y = c}^d = F\left( {x,d} \right) - F\left( {x,c} \right)$
Thus,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = a}^b \frac{\partial }{{\partial x}}\left( {F\left( {x,d} \right) - F\left( {x,c} \right)} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = a}^b \left( {\frac{{\partial F\left( {x,d} \right)}}{{\partial x}} - \frac{{\partial F\left( {x,c} \right)}}{{\partial x}}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = a}^b \frac{{\partial F\left( {x,d} \right)}}{{\partial x}}{\rm{d}}x - \mathop \smallint \limits_{x = a}^b \frac{{\partial F\left( {x,c} \right)}}{{\partial x}}{\rm{d}}x$
Again, using the Fundamental Theorem of Calculus, Part I, we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = F\left( {x,d} \right)|_{x = a}^b - F\left( {x,c} \right)|_{x = a}^b$
Hence,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = F\left( {b,d} \right) - F\left( {a,d} \right) - F\left( {b,c} \right) + F\left( {a,c} \right)$
where ${\cal R} = \left[ {a,b} \right] \times \left[ {c,d} \right]$.
