Answer
The center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( {2,1} \right)$.
Work Step by Step
We have ${\cal D}$, the region bounded by $y=4-x$, $x=0$, $y=0$; and the mass density $\delta \left( {x,y} \right) = x$.
We describe ${\cal D}$ as a vertically simple region:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 4,0 \le y \le 4 - x} \right\}$
Step 1. Evaluate the mass of ${\cal D}$:
$M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = 0}^{4 - x} x{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^4 x\left( {y|_0^{4 - x}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^4 \left( {4x - {x^2}} \right){\rm{d}}x$
$ = \left( {2{x^2} - \frac{1}{3}{x^3}} \right)|_0^4 = \left( {32 - \frac{{64}}{3}} \right) = \frac{{32}}{3}$
Step 2. Evaluate the $x$-coordinate of the center of mass:
${x_{CM}} = \frac{{{M_y}}}{M} = \frac{3}{{32}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\delta \left( {x,y} \right){\rm{d}}A = \frac{3}{{32}}\mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = 0}^{4 - x} {x^2}{\rm{d}}y{\rm{d}}x$
$ = \frac{3}{{32}}\mathop \smallint \limits_{x = 0}^4 {x^2}\left( {y|_0^{4 - x}} \right){\rm{d}}x$
$ = \frac{3}{{32}}\mathop \smallint \limits_{x = 0}^4 \left( {4{x^2} - {x^3}} \right){\rm{d}}x$
$ = \frac{3}{{32}}\left( {\frac{4}{3}{x^3} - \frac{1}{4}{x^4}} \right)|_0^4 = \frac{3}{{32}}\left( {\frac{{256}}{3} - \frac{{256}}{4}} \right) = 2$
Step 3. Evaluate the $y$-coordinate of the center of mass:
${y_{CM}} = \frac{{{M_x}}}{M} = \frac{3}{{32}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y\delta \left( {x,y} \right){\rm{d}}A = \frac{3}{{32}}\mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = 0}^{4 - x} xy{\rm{d}}y{\rm{d}}x$
$ = \frac{3}{{64}}\mathop \smallint \limits_{x = 0}^4 x\left( {{y^2}|_0^{4 - x}} \right){\rm{d}}x$
$ = \frac{3}{{64}}\mathop \smallint \limits_{x = 0}^4 \left( {16x - 8{x^2} + {x^3}} \right){\rm{d}}x$
$ = \frac{3}{{64}}\left( {8{x^2} - \frac{8}{3}{x^3} + \frac{1}{4}{x^4}} \right)|_0^4 = \frac{3}{{64}}\left( {128 - \frac{{512}}{3} + 64} \right) = 1$
Thus, the center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( {2,1} \right)$.
