Draw the area enclosed by the curves, and estimate the position of the center of...

Consider the following curves:
${\displaystyle y=f\left(x\right)=e^x,y=0,x=0,x=5}$
Find the exact coordinates of the centroid
Calculate the total are as follows:
${\displaystyle A={\int }_a^{b}f\left(x\right)dx}$
${\displaystyle ={\int }_0^5{e}^{x}dx}$
${\displaystyle ={\left(e^x\right)}_0^5}$
${\displaystyle =e^5-1}$

Calculate the y-coordinates of the centroid as follows:
${\displaystyle \bar{y}=\frac{M_x}{A}}$
Here ${\displaystyle M_x=\frac{1}{2}{\int }_a^b{f}^2\left(x\right)dx}$
${\displaystyle =\frac{1}{2}{\int }_0^5{\left(e^x\right)}^{2}dx}$
${\displaystyle =\frac{1}{2}{\int }_0^5{e}^{2x}dx}$
${\displaystyle =\frac{1}{2}{\left(\frac{e^{2x}}{2}\right)}_0^5}$
${\displaystyle =\frac{1}{4}(e^{10}-1)}$
Then, ${\displaystyle \bar{y}=\frac{M_x}{A}}$
$=\frac{\frac{1}{4}(e^{10}-1)}{e^5-1}$
$=\frac{1}{4}\times \frac{(e^5{)}^2-1^2}{e^5-1}$
$=\frac{1}{4}\times \frac{(e^5-1)(e^5+1)}{e^5-1}$
$=\frac{1}{4}(e^5+1)$
$\approx 37.35$

Calculate the x-coordinates of the centroid as follows:
$\bar{x}=\frac{M_y}{A}$
Here $M_y={\int }_a^{b}xf(x)dx$
$={\int }_0^{5}x{e}^{x}dx$
$=5e^5-e^5+1$
$4e^5+1$
Then, $\bar{x}=\frac{M_y}{A}$
$\frac{4e^5+1}{e^5-1}$
$\approx 4.0$