The area of region $R_x$ is

$A\={\int }_0^1\left(x-x^2\right) \mathit{\ⅆ}x$ = $1\/6$ 

Its centroid is given by

 $\stackrel{\&conjugate0;}{x}\=\frac{1}{1\/6}{\int }_0^{1}x \left(x-x^2\right) \mathit{\ⅆ}x$ = $1\/2$ 

and

$\stackrel{\&conjugate0;}{y}\=\frac{1}{1\/6}{\int }_0^1\left(x^2-{\left(x^2\right)}^2\right)\/2 \mathit{\ⅆ}x$ = $2\/5$ 

Calculate $A$, the area of region $R_x$ 

${\int }_0^1\left(x-x^2\right) \mathit{\ⅆ}x$ = $\frac{1}{6}$$\stackrel{\text{assign to a name}}{\to }$$A$

Calculate $\stackrel{\&conjugate0;}{x}$ 

$\frac{1}{A}{\int }_0^{1}x \left(x-x^2\right) \mathit{\ⅆ}x$ = $\frac{1}{2}$

Calculate $\stackrel{\&conjugate0;}{y}$ 

$\frac{1}{A}{\int }_0^1\left(x^2-{\left(x^2\right)}^2\right)\/2 \mathit{\ⅆ}x$ = $\frac{2}{5}$