#### Answer

$$\left(\frac{8}{5}, \frac{16}{7}\right)$$

#### Work Step by Step

Since
\begin{align*}
M_{x}&=\frac{1}{2} \rho \int_{0}^{2}\left(f(x)\right)^{2} d x\\
&=\frac{1}{2} \rho \int_{0}^{2}\left(x^{3}\right)^{2} d x\\
&= \frac{1}{2} \frac{x^7}{7}\bigg|_{0}^{2}\\
&=\frac{64 \rho}{7}
\end{align*}
and
\begin{aligned}
M_{y}&=\rho \int_{a}^{b} x f(x) d x \\
&=\rho \int_{0}^{2} x^{4} d x \\
&=\left.\rho\left(\frac{x^{5}}{5}\right)\right|_{0} ^{2} \\
&=\frac{32\rho}{5}
\end{aligned}
and
\begin{align*}
M&=\rho A\\
&=\rho \int_{0}^{2} x^{3} d x\\
&=\left.\rho\left(\frac{x^{4}}{4}\right)\right|_{0} ^{2}\\
&=4 \rho
\end{align*}
Then
\begin{align*}
(\bar{x}, \bar{y})&=\left(\frac{M_{y}}{M}, \frac{M_{x}}{M}\right)\\
&=\left(\frac{8}{5}, \frac{16}{7}\right)
\end{align*}