Question

Evaluate
${\int }_0^1{x}^2\sqrt{1-x}dx$

Answer 1

$I={\int }_0^1{x}^2\sqrt{1-x}dx$

]$={\int }_0^1(1-x{)}^2\sqrt{1-(1-x)}dx$ $[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(b+a-x)dx]$

$={\int }_0^1(1+x^2-2x)\sqrt{x}dx$

$={\int }_0^1(x^{1/2}+x^{5/2}-2x^{3/2})dx$

$={[{\frac{2x}{3}}^{3/2}+{\frac{2x}{7}}^{7/2}-{\frac{4x}{5}}^{5/2}]}_0^1$

$=\frac{2}{3}+\frac{2}{7}-\frac{4}{5}=\frac{16}{105}$