Question

$\underset{0}{\overset{2}{\int }}x\sqrt{2-x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{2}x\sqrt{2-x}dx \\ ={\int }_0^2\left(2-x\right)\sqrt{2-2+x}dx \\ ={\int }_0^2\left(2-x\right)\sqrt{x}dx \\ ={\int }_0^2\left(2\sqrt{x}-x\sqrt{x}\right)dx \\ ={\int }_0^2\left(2x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)dx \\ ={\left[2\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{x^{{\displaystyle \frac{5}{2}}}}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}_0^2 \\ ={\left[\frac{4}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}\right]}_0^2 \\ =\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}=\frac{16\sqrt{2}}{15} \end{gathered}$$