#### Answer

$$ \left(\bar{x},\bar{y}\right)= \left(\frac{4}{5},\frac{2}{7}\right) $$

#### Work Step by Step

Since
\begin{aligned}
M_{y} &=\delta \int_{a}^{b} x f(x) d x \\
&=\int_{0}^{1} x\left(x^{3}\right) d x \\
&=\int_{0}^{1} x^{4} d x \\
&=\left.\left(\frac{x^{5}}{5}\right)\right|_{0} ^{1} \\
&=\frac{1}{5}
\end{aligned}
and
\begin{aligned}
M_{x} &=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x \\
&=\frac{1}{2} \int_{0}^{1}\left(x^{3}\right)^{2} d x \\
&=\frac{1}{2} \int_{0}^{1} x^{6} d x \\
&=\left.\frac{1}{2}\left(\frac{x^{7}}{7}\right)\right|_{0} ^{1} \\
&=\frac{1}{14}
\end{aligned}
and
$$M=\delta A=\int_{0}^{1} x^{3} d x=\left.\left(\frac{x^{4}}{4}\right)\right|_{0} ^{1}=\frac{1}{4}$$
Then
$$\bar{x}=\frac{M_{y}}{M}=\frac{4}{5} \quad \text { and } \quad \bar{y}=\frac{M_{x}}{M}=\frac{2}{7}$$