Answer
$$(\bar{x},\bar{y})=\left(\frac{1-5 e^{-4}}{1-e^{-4}},\frac{1-e^{-8}}{4-4 e^{-4}}\right)$$
Work Step by Step
Since
\begin{align*}
M_{y}&=\delta \int_{a}^{b} x f(x) d x\\
&=\int_{0}^{4} x e^{-x} d x\\
&=-xe^{-x}\bigg|_{0}^{4}+ \int_{0}^{4} e^{-x} d x\\
&=-e^{-x}(x+1)\bigg|_{0}^{4}\\
&= 1-5e^{-4}
\end{align*}
and
\begin{aligned}
M_{x}=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x &=\frac{1}{2} \int_{0}^{4} e^{-2 x} d x \\
&=\left.\frac{1}{2}\left(-\frac{1}{2} e^{-2 x}\right)\right|_{0} ^{4} \\
&=\frac{1}{2}\left(-\frac{1}{2} e^{-8}+\frac{1}{2}\right) \\
&=\frac{1}{4}\left(1-e^{-8}\right)
\end{aligned}
and
\begin{align*}
M&=\delta A\\
&=\int_{0}^{4} e^{-x} d x\\
&=\left.\left(-e^{-x}\right)\right|_{0} ^{4}=\left(1-e^{-4}\right)
\end{align*}
Then
$$\bar{x}=\frac{M_{y}}{M}=\frac{1-5 e^{-4}}{1-e^{-4}} \quad \text { and } \quad \bar{y}=\frac{M_{x}}{M}=\frac{1-e^{-8}}{4-4 e^{-4}}$$
