Answer
$$\left( \frac{6}{15},\frac{1}{2}\right)$$
$$M_x=\frac{1}{15},\ \ \ \ \ M_y=\frac{1}{12} $$
Work Step by Step
given $$ y=x,\ \ \ \ \ \ y=x^2 ,\ \ \ \ \ 0\leq x\leq 1 $$
Since
$$ x\geq x^2 \ \ \ \text{for }\ 0\leq x\leq 1 $$
Then
\begin{align*}
M_x&=\frac{1}{2}\int_{0}^{1}[f(x)]^2-[g(x)]^2dx\\
&= \frac{1}{2}\int_{0}^{1}[ x^2-x^4] dx\\
&= \frac{1}{2}[\frac{1}{3}x^3-\frac{1}{5}x^5]\bigg|_{0}^{1}\\
&= \frac{1}{15}
\end{align*}
and
\begin{align*}
M_y&= \int_{0}^{1}x[[f(x)] -[g(x)] ]dx\\
&= \int_{0}^{1}[ x^2-x^3] dx\\
&= [\frac{1}{3}x^3-\frac{1}{4}x^4]\bigg|_{0}^{1}\\
&= \frac{1}{12}
\end{align*}
and
\begin{align*}
M &= \int_{0}^{1} [[f(x)] -[g(x)] ]dx\\
&= \int_{0}^{1}[ x -x^2] dx\\
&= [\frac{1}{2}x^2-\frac{1}{3}x^3]\bigg|_{0}^{1}\\
&= \frac{1}{6}
\end{align*}
Hence
\begin{align*}
\left( \bar{x},\bar{y}\right)&=\left(\frac{M_{x}}{M}, \frac{M_{y}}{M}\right)\\
&=\left( \frac{6}{15},\frac{1}{2}\right)
\end{align*}
