Answer
$$\left(\frac{7}{6}, 3\right)$$
Work Step by Step
From the following figure, it is clear that $y=3$ in the center.
We calculate the moments as follows:
\begin{align*}
M_{x}&=\frac{1}{2} \int_{0}^{2}\left((f(x))^{2}-(g(x))^{2}\right) d x\\
&=\frac{1}{2} \int_{0}^{2}\left((x+4)^{2}-(2-x)^{2}\right) d x\\
&=\frac{1}{2}\left(\frac{1}{3}(x+4)^{3}+\frac{1}{3}(2-x)^{3}\right) \bigg|_{0}^{2}\\
&=24
\end{align*}
and
\begin{align*}
M_{y}&=\int_{0}^{2} x(f(x)-g(x)) d x\\
&=\int_{0}^{2} x((x+4)-(2-x)) d x\\
&=\int_{0}^{2} (2x^2+2x) d x\\
&=\frac{2}{3}x^3+x^2\bigg|_{0}^{2}\\
&=\frac{28}{3}
\end{align*}
and
\begin{align*}
A&=\int_{0}^{2}((x+4)-(2-x)) d x\\
&=\frac{1}{2}(x+4)^2+\frac{1}{2}(2-x)^2\bigg|_{0}^{2}\\
&=8
\end{align*}
Hence
$$(\bar{x},\bar{y})= \left(\frac{7}{6}, 3\right)$$
