Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 0}^3 \left( {\mathop \smallint \limits_{x = \frac{2}{3}y}^{\frac{5}{3}y} \left( {1 - 2x} \right){\rm{d}}x} \right){\rm{d}}y = - \frac{{33}}{2}$
Work Step by Step
We have $f\left( {x,y} \right) = 1 - 2x$.
We can consider the domain as a horizontally simple region. The lower and the upper boundary are $y=0$ and $y=3$, respectively. Whereas, the left and right boundaries are the line $y = \frac{3}{2}x$ and $y = \frac{3}{5}x$, respectively. The line equations can be written as
$x = \frac{2}{3}y$, ${\ \ \ \ \ }$ $x = \frac{5}{3}y$
Thus, the domain description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 3,\frac{2}{3}y \le x \le \frac{5}{3}y} \right\}$
So, the double integral of $f\left( {x,y} \right)$ over ${\cal D}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 0}^3 \left( {\mathop \smallint \limits_{x = \frac{2}{3}y}^{\frac{5}{3}y} \left( {1 - 2x} \right){\rm{d}}x} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^3 \left( {\left( {x - {x^2}} \right)|_{\frac{2}{3}y}^{\frac{5}{3}y}} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^3 \left( {\frac{5}{3}y - \frac{{25}}{9}{y^2} - \frac{2}{3}y + \frac{4}{9}{y^2}} \right){\rm{d}}y$
$ = \left( {\frac{5}{6}{y^2} - \frac{{25}}{{27}}{y^3} - \frac{2}{6}{y^2} + \frac{4}{{27}}{y^3}} \right)|_0^3$
$ = \frac{{45}}{6} - 25 - 3 + 4$
$ = - \frac{{33}}{2}$
