Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 1/2}^{\pi /2} \mathop \smallint \limits_{y = 1}^{2x} \cos \left( {2x + y} \right){\rm{d}}y{\rm{d}}x \simeq - 0.41611$
Work Step by Step
We have $f\left( {x,y} \right) = \cos \left( {2x + y} \right)$ and the domain ${\cal D} = \left\{ {\left( {x,y} \right)|\frac{1}{2} \le x \le \frac{\pi }{2},1 \le y \le 2x} \right\}$.
The domain ${\cal D}$ is a vertically simple region.
We evaluate the integral of $f\left( {x,y} \right) = \cos \left( {2x + y} \right)$ over ${\cal D}$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 1/2}^{\pi /2} \mathop \smallint \limits_{y = 1}^{2x} \cos \left( {2x + y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 1/2}^{\pi /2} \left( {\sin \left( {2x + y} \right)|_1^{2x}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 1/2}^{\pi /2} \left( {\sin \left( {4x} \right) - \sin \left( {2x + 1} \right)} \right){\rm{d}}x$
$ = \left( { - \frac{1}{4}\cos 4x + \frac{1}{2}\cos \left( {2x + 1} \right)} \right)|_{1/2}^{\pi /2}$
$ = - \frac{1}{4} + \frac{1}{2}\cos \left( {\pi + 1} \right) + \frac{1}{4}\cos 2 - \frac{1}{2}\cos 2$
$ = - \frac{1}{4} + \frac{1}{2}\cos \left( {\pi + 1} \right) - \frac{1}{4}\cos 2$
$ \simeq - 0.41611$
