Find the centroid of the region bounded by the given curves. y=x^{3},x+y=2,y=

${\displaystyle f\left(x\right)=2-x}$, ${\displaystyle g\left(x\right)=x^3}$
or ${\displaystyle x=2-y}$, ${\displaystyle x=y^{\frac{1}{3}}}$
They intersect at (1,1)
Since integrating with respect to x would mean we need to do two separate integrals for everything (from ${\displaystyle x=0}$ to l and from ${\displaystyle x=1}$ to 2), we could alternatively integrate with respect to y, where ${\displaystyle x=2}$ — y is the "top" function and ${\displaystyle x=y^{\frac{1}{3}}}$ is the "bottom",

Find the area of the bounded region.
${\displaystyle A={\int }_0^1((2-y)-y^{\frac{1}{3}})dy}$
${\displaystyle ={[2y-\frac{1}{2}{y}^2-\frac{3}{4}{y}^{\frac{4}{3}}]}_0^1}$
${\displaystyle =2-\frac{1}{2}-\frac{3}{4}-(0-0-0)}$
${\displaystyle =\frac{8-2-3}{4}}$
${\displaystyle =\frac{3}{4}}$
If we change all the x to y in the formula to find ${\displaystyle \bar{x}}$, it will give the y coordinate of the centroid.
${\displaystyle \bar{y}=\frac{1}{\frac{3}{4}}{\int }_0^{1}y((2-y)-y^{\frac{1}{3}})dy=\frac{4}{3}{\int }_0^1(2y-y^2-y^{\frac{4}{3}})dy}$
${\displaystyle =\frac{4}{3}{[y^2-\frac{1}{3}{y}^3-\frac{3}{7}{y}^{\frac{7}{3}}]}_0^1}$
${\displaystyle =\frac{4}{3}[1-\frac{1}{3}-\frac{3}{7}-(0-0-0)]}$
${\displaystyle =\frac{4}{3}\left[\frac{21-7-9}{21}\right]}$
${\displaystyle =\frac{4}{3}\left[\frac{5}{21}\right]}$
${\displaystyle =\frac{20}{63}}$
Likewise, if we change all the x to y in the formula to find ${\displaystyle \bar{y}}$, it will give the x coordinate of the centroid.

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