#### Answer

$$
(\bar{x},\bar{y})=(1,2)
$$

#### Work Step by Step

Since
\begin{aligned}
M_{y} &=\delta \int_{a}^{b} x f(x) d x \\
&=\int_{0}^{3} x(6-2 x) d x \\
&=\int_{0}^{3}\left(6 x-2 x^{2}\right) d x \\
&=\left.\left(3 x^{2}-\frac{2}{3} x^{3}\right)\right|_{0} ^{3} \\
&=9
\end{aligned}
and
\begin{aligned}
M_{x} &=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x \\
&=\frac{1}{2} \int_{0}^{3}(6-2 x)^{2} d x \\
&=\frac{1}{2} \int_{0}^{3}\left(36-24 x+4 x^{2}\right) d x \\
&=\left.\frac{1}{2}\left(36 x-12 x^{2}+\frac{4}{3} x^{3}\right)\right|_{0} ^{3} \\
&=18
\end{aligned}
and
$$M=\delta A=\int_{0}^{3}(6-2 x) d x=\left.\left(6 x-x^{2}\right)\right|_{0} ^{3}=9$$
Then
$$
\bar{x}=\frac{M_{y}}{M}=\frac{9}{9}=1 \quad \text { and } \quad \bar{y}=\frac{M_{x}}{M}=\frac{18}{9}=2
$$