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Determine the centroid $(\overline{x}, \overline{y})$ of the shaded area.

$A=\int_{0}^{1} {d A}=\int_{0}^{1} y d x=\int_{0}^{1} x^{3} d x$

$=\frac{1}{4} m^{2}$

$\overline{x}=\frac{\int_{0}^{1} \tilde{x} d A}{d A}=\frac{\int_{0}^{1}(y d x)}{\frac{1}{4}}=\frac{\int_{0}^{1} x\left(x^{3}\right) d x}{\frac{1}{4}}=\frac{\int_{0}^{1} x^{4} d x}{\frac{1}{4}}=4 / 5 m$

$\overline{y}=\frac{\int_{0}^{1} \tilde{x} d A}{d A}= \frac{\int_{0}^{1}\left(\frac{y}{2}\right)(y d A)}{\frac{1}{4}}= \frac{\int_{0}^{1}\left(\frac{x^{6}}{2}\right) d x}{\frac{1}{4}}=\frac{2}{7} m$

Locate the centroid $\overline{x}$ of the area.

$\int d A=\int_{0}^{1}\left(y_{2}-y_{1}\right) d x$

$=\int_{0}^{1}\left(x^\frac{1}{2}-x^{2}\right) d x= \left[\frac{x^\frac{3}{2}}{\frac{3}{2}}-\frac{x^{3}}{3}\right]_{0}^{1}$

$∴ \quad=\quad \frac{1}{3} \quad m^{2}$

$\int \tilde{x} d A=\int_{0}^{1} x\ \left(x^\frac{1}{2}-x^{2}\right) d x=\int_{0}^{1}\left(x^\frac{3}{2}-x^{3}\right) d x$

$=\left[\frac{x^\frac{5}{2}} {\frac{5}{2}}-\frac{x^{4}}{4}\right]_{0}^{1}=\frac{2}{5}-\frac{1}{4}=\frac{3}{20} m^{3}$

$\overline{x}=\frac{\int \tilde{x} d A}{\int d A} =\frac{\frac{3}{20}}{\frac{1}{3}}=\frac{9}{20}=0.45 \mathrm{m}$