Kathy Williams

2021-12-21

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

Determine the precise coordinates of the centroid
$\left(x,y\right)=$

Thomas Nickerson

Expert

Consider the following curves:

Find the exact coordinates of the centroid
Calculate the total are as follows:
$A={\int }_{a}^{b}f\left(x\right)dx$
$={\int }_{0}^{5}{e}^{x}dx$
$={\left({e}^{x}\right)}_{0}^{5}$
$={e}^{5}-1$

Raymond Foley

Expert

Calculate the y-coordinates of the centroid as follows:
$\stackrel{―}{y}=\frac{{M}_{x}}{A}$
Here ${M}_{x}=\frac{1}{2}{\int }_{a}^{b}{f}^{2}\left(x\right)dx$
$=\frac{1}{2}{\int }_{0}^{5}{\left({e}^{x}\right)}^{2}dx$
$=\frac{1}{2}{\int }_{0}^{5}{e}^{2x}dx$
$=\frac{1}{2}{\left(\frac{{e}^{2x}}{2}\right)}_{0}^{5}$
$=\frac{1}{4}\left({e}^{10}-1\right)$
Then, $\stackrel{―}{y}=\frac{{M}_{x}}{A}$
$=\frac{\frac{1}{4}\left({e}^{10}-1\right)}{{e}^{5}-1}$
$=\frac{1}{4}×\frac{\left({e}^{5}{\right)}^{2}-{1}^{2}}{{e}^{5}-1}$
$=\frac{1}{4}×\frac{\left({e}^{5}-1\right)\left({e}^{5}+1\right)}{{e}^{5}-1}$
$=\frac{1}{4}\left({e}^{5}+1\right)$
$\approx 37.35$

RizerMix

Expert

Calculate the x-coordinates of the centroid as follows:
$\stackrel{―}{x}=\frac{{M}_{y}}{A}$
Here ${M}_{y}={\int }_{a}^{b}xf\left(x\right)dx$
$={\int }_{0}^{5}x{e}^{x}dx$
$=5{e}^{5}-{e}^{5}+1$
$4{e}^{5}+1$
Then, $\stackrel{―}{x}=\frac{{M}_{y}}{A}$
$\frac{4{e}^{5}+1}{{e}^{5}-1}$
$\approx 4.0$

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