Answer
The mass of the lamina is m=6.
The center of mass is $\left(\frac{3}{4}, \frac{3}{2}\right).
$
Work Step by Step
The triangle is shown in the Figure. (Note that the equation of the upper boundary is $y=3-x .)$ The mass of the lamina is
$$
\begin{aligned}
&\begin{aligned}
m &=\iint_{D} \rho(x, y) d A
\\
&=\int_{0}^{2} \int_{x / 2}^{3-x}(x+y) d y d x \\
&=\int_{0}^{2}\left[x y+\frac{1}{2} y^{2}\right]_{y=x / 2}^{y=3-x} d x \\
&=\int_{0}^{2}\left[x\left(3-\frac{3}{2} x\right)+\frac{1}{2}(3-x)^{2}-\frac{1}{8} x^{2}\right] d x \\
&=\int_{0}^{2}\left(-\frac{9}{8} x^{2}+\frac{9}{2}\right) d x \\
&=\left[-\frac{9}{8}\left(\frac{1}{3} x^{3}\right)+\frac{9}{2} x\right]_{0}^{2} \\
&=6,
\end{aligned}
\\
\\
M_{y} &=
\iint_{D} x \rho(x, y) d A\\
&=\int_{0}^{2} \int_{x / 2}^{3-x}\left(x^{2}+x y\right) d y d x \\
&=\int_{0}^{2}\left[x^{2} y+\frac{1}{2} x y^{2}\right]_{y=x / 2}^{y=3-x} d x \\
&=\int_{0}^{2}\left(\frac{9}{2} x-\frac{9}{8} x^{3}\right) d x \\
&=\frac{9}{2},\\
\\
M_{x} &=\iint_{D} y \rho(x, y) d A\\
&=\int_{0}^{2} \int_{x / 2}^{3-y}\left(x y+y^{2}\right) d y d x \\
&=\int_{0}^{2}\left[\frac{1}{2} x y^{2}+\frac{1}{3} y^{3}\right]_{y=x / 2}^{y=3-x} d x \\
&=\int_{0}^{2}\left(9-\frac{9}{2} x\right) d x\\
&=9.\\
&\text { Hence, the mass of the lamina is } m=6, \text { and the center of mass is }\\
(\bar{x}, \bar{y}) & =\left(\frac{M_{y}}{m}, \frac{M_{x}}{m}\right) \\
&=\left(\frac{3}{4}, \frac{3}{2}\right)
\end{aligned}
$$
