Answer
$$M=\frac{8}{15},\ \ I_x=\frac{64}{105}$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{2} \int_{-y}^{y-y^{2}}(x+y) d x d y\\
&=\int_{0}^{2}\left[\frac{x^{2}}{2}+x y\right]_{-y}^{y-y^{2}} d y\\
&=\int_{0}^{2}\left(\frac{y^{4}}{2}-2 y^{3}+2 y^{2}\right) d y\\
&=\left[\frac{y^{5}}{10}-\frac{y^{4}}{2}+\frac{2 y^{3}}{3}\right]_{0}^{2}\\
&=\frac{8}{15}
\end{align*}
and
\begin{align*}
I_{x}&=\int_{0}^{2} \int_{-y}^{y-y^{2}} y^{2}(x+y) d x d y\\
&=\int_{0}^{2}\left[\frac{x^{2} y^{2}}{2}+x y^{3}\right]_{-y}^{y-y^{2}} d y\\
&=\int_{0}^{2}\left(\frac{y^{6}}{2}-2 y^{5}+2 y^{4}\right) d y\\
&=\frac{64}{105}
\end{align*}
