Answer
We prove that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} 1{\rm{d}}A = \mathop \smallint \limits_a^b \left( {{g_1}\left( x \right) - {g_2}\left( x \right)} \right){\rm{d}}x$
Work Step by Step
Suppose that ${\cal D}$ is the region between two curves $y = {g_1}\left( x \right)$ and $y = {g_2}\left( x \right)$ with ${g_2}\left( x \right) \le {g_1}\left( x \right)$ for $a \le x \le b$. Then, ${\cal D}$ is a vertically simple region whose description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|a \le x \le b,{g_2}\left( x \right) \le y \le {g_1}\left( x \right)} \right\}$
Using Eq. (3), we evaluate the area of the domain ${\cal D}$ as an iterated integral:
$Area\left( {\cal D} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} 1{\rm{d}}A = \mathop \smallint \limits_{x = a}^b \left( {\mathop \smallint \limits_{y = {g_2}\left( x \right)}^{{g_1}\left( x \right)} {\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = a}^b \left( {y|_{{g_2}\left( x \right)}^{{g_1}\left( x \right)}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = a}^b \left( {{g_1}\left( x \right) - {g_2}\left( x \right)} \right){\rm{d}}x$
Hence, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} 1{\rm{d}}A = \mathop \smallint \limits_a^b \left( {{g_1}\left( x \right) - {g_2}\left( x \right)} \right){\rm{d}}x$.
