Answer
$$
\left(\bar{x},\bar{y} \right)=\left(\frac{93}{35},\frac{45}{56} \right)
$$
Work Step by Step
Since
\begin{aligned}
M_{y} &=\delta \int_{a}^{b} x f(x) d x \\
&=\int_{1}^{4} x \sqrt{x} d x \\
&=\int_{1}^{4} x^{\frac{3}{2}} d x \\
&=\left.\left(\frac{2}{5} x^{\frac{5}{2}}\right)\right|_{1} ^{4} \\
&=\left(\frac{64}{6}-\frac{2}{5}\right) \\
&=\frac{62}{5}
\end{aligned}
and
\begin{aligned}
M_{x} &=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x \\
&=\frac{1}{2} \int_{1}^{4}(\sqrt{x})^{2} d x \\
&=\frac{1}{2} \int_{1}^{4} x d x \\
&=\left.\frac{1}{2}\left(\frac{x^{2}}{2}\right)\right|_{1} ^{4} \\
&=\frac{1}{2}\left(8-\frac{1}{2}\right) \\
&=\frac{15}{4}
\end{aligned}
and
$$
M=\delta A=\int_{1}^{4} \sqrt{x} d x=\int_{1}^{4} x^{\frac{1}{2}} d x=\left.\left(\frac{2}{3} x^{\frac{3}{2}}\right)\right|_{1} ^{4}=\left(\frac{16}{3}-\frac{2}{3}\right)=\frac{14}{3}
$$
Then
$$
\bar{x}=\frac{M_{y}}{M}=\frac{93}{35} \quad \text { and } \quad \bar{y}=\frac{M_{x}}{M}=\frac{45}{56}
$$
