Answer
$$\left(\frac{\pi}{2} ,\frac{\pi}{8}\right)$$
Work Step by Step
Given $$ f(x)=\sin x, \quad[0, \pi]$$
Since
\begin{align*}
M_{y}&=\rho \int_{a}^{b} x f(x) d x\\
&=\int_{0}^{\pi} x \sin x d x
\end{align*}
Let
\begin{aligned}
&u=x
&d v=\sin x\\
&d u=1 & v=-\cos x
\end{aligned}
Then
\begin{aligned}
M_{y}=\int_{0}^{\pi} x \sin x d x&=-x \cos x\bigg|_{0}^{\pi} +\int_{0}^{\pi} \cos x \, d x \\
&=-x \cos x+\sin x\bigg|_{0}^{\pi} \\
&= \pi
\end{aligned}
Since
\begin{aligned}
M_{x}=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x &=\frac{1}{2} \int_{0}^{\pi} \sin ^{2} x d x \\
&=\frac{1}{2} \int_{0}^{\pi} \frac{1-\cos 2 x}{2} \\
&=\frac{1}{4} \int_{0}^{\pi} 1 d x-\frac{1}{4} \int_{0}^{\pi} \cos 2 x d x \\
&=\left.\frac{1}{4}(x)\right|_{0} ^{\pi}-\frac{1}{4}\frac{\sin 2x}{2} \bigg|_{0}^{\pi} \\
&=\frac{\pi}{4}
\end{aligned}
and
\begin{align*}
M=\rho A&=\int_{0}^{\pi} \sin x d x\\
&=\left.(-\cos x)\right|_{0} ^{\pi}\\
&=2
\end{align*}
Then
\begin{align*}
\left(\bar{x}, \bar{y}\right) &=\left(\frac{M_{x}}{M}, \frac{M_{y}}{M}\right)=\left(\frac{\pi}{2} ,\frac{\pi}{8}\right)
\end{align*}
