Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + 1} \right){\rm{d}}A = 24$
Work Step by Step
We have $f\left( {x,y} \right) = x + 1$.
The domain is not a simple region. So, we divide it into two horizontally simple regions: ${{\cal D}_1}$ and ${{\cal D}_2}$ (please see the figure attached).
1. Consider region ${{\cal D}_1}$
The lower and upper boundaries are $y=1$ and $y=3$, respectively. Whereas, the left and right boundaries are the lines: $y - 1 = 2\left( {x - 1} \right)$ and $y - 1 = \frac{1}{2}\left( {x - 1} \right)$, respectively.
So,
Left boundary:
$y - 1 = 2\left( {x - 1} \right)$, ${\ \ \ \ }$ $y=2x-1$
$x = \frac{{y + 1}}{2}$
Right boundary:
$y - 1 = \frac{1}{2}\left( {x - 1} \right)$, ${\ \ \ \ }$ $y = \frac{1}{2}x + \frac{1}{2}$
$x = 2\left( {y - \frac{1}{2}} \right) = 2y - 1$
Thus, the domain description is ${{\cal D}_1} = \left\{ {\left( {x,y} \right)|1 \le y \le 3,\frac{{y + 1}}{2} \le x \le 2y - 1} \right\}$.
Evaluate the double integral of $f\left( {x,y} \right) = x + 1$ over ${{\cal D}_1}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 1}^3 \mathop \smallint \limits_{x = \left( {y + 1} \right)/2}^{2y - 1} \left( {x + 1} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 1}^3 \left( {\left( {\frac{1}{2}{x^2} + x} \right)|_{\left( {y + 1} \right)/2}^{2y - 1}} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 1}^3 \left( {\left( {\frac{1}{2}{{\left( {2y - 1} \right)}^2} + 2y - 1 - \frac{1}{2}{{\left( {\frac{{y + 1}}{2}} \right)}^2} - \frac{{y + 1}}{2}} \right)} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 1}^3 \left( {\frac{{15}}{8}{y^2} - \frac{3}{4}y - \frac{9}{8}} \right){\rm{d}}y$
$ = \left( {\frac{5}{8}{y^3} - \frac{3}{8}{y^2} - \frac{9}{8}y} \right)|_1^3$
$ = 11$
2. Consider region ${{\cal D}_2}$
The lower and upper boundaries are $y=3$ and $y=5$, respectively. Whereas, the left and right boundaries are the lines: $y - 1 = 2\left( {x - 1} \right)$ and $y - 3 = - 1\left( {x - 5} \right)$, respectively.
Notice that the left boundary is the same with that in ${{\cal D}_1}$. So,
Left boundary: $x = \frac{{y + 1}}{2}$.
Right boundary:
$y - 3 = - 1\left( {x - 5} \right)$, ${\ \ \ \ }$ $y=-x+8$
$x=8-y$
Thus, the domain description is ${{\cal D}_2} = \left\{ {\left( {x,y} \right)|3 \le y \le 5,\frac{{y + 1}}{2} \le x \le 8 - y} \right\}$.
Evaluate the double integral of $f\left( {x,y} \right) = x + 1$ over ${{\cal D}_2}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 3}^5 \mathop \smallint \limits_{x = \left( {y + 1} \right)/2}^{8 - y} \left( {x + 1} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 3}^5 \left( {\left( {\frac{1}{2}{x^2} + x} \right)|_{\left( {y + 1} \right)/2}^{8 - y}} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 3}^5 \left( {\left( {\frac{1}{2}{{\left( {8 - y} \right)}^2} + 8 - y - \frac{1}{2}{{\left( {\frac{{y + 1}}{2}} \right)}^2} - \frac{{y + 1}}{2}} \right)} \right){\rm{d}}y$$ = \mathop \smallint \limits_{y = 3}^5 \left( {\left( {\frac{1}{2}{{\left( {8 - y} \right)}^2} + 8 - y - \frac{1}{2}{{\left( {\frac{{y + 1}}{2}} \right)}^2} - \frac{{y + 1}}{2}} \right)} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 3}^5 \left( {\frac{3}{8}{y^2} - \frac{{39}}{4}y + \frac{{315}}{8}} \right){\rm{d}}y$
$ = \left( {\frac{1}{8}{y^3} - \frac{{39}}{8}{y^2} + \frac{{315}}{8}y} \right)|_3^5$
$ = 13$
Using the linearity properties of the double integral, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} f\left( {x,y} \right){\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} f\left( {x,y} \right){\rm{d}}A$
$ = 11 + 13$
$ = 24$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + 1} \right){\rm{d}}A = 24$.
