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Let $I=\int_{0}^{2} x \sqrt{2-x} d x$

$I=\int_{0}^{2}(2-x) \sqrt{x} d x$       $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$=\int_{0}^{2}\left\{2 x^{\frac{1}{2}}-x^{\frac{3}{2}}\right\} d x$

$=\left[2\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_{0}^{2}$

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

$=\frac{4}{3}(2)^{\frac{3}{2}}-\frac{2}{5}(2)^{\frac{5}{2}}$

$=\frac{4 \times 2 \sqrt{2}}{3}-\frac{2}{5} \times 4 \sqrt{2}$

$=\frac{8 \sqrt{2}}{3}-\frac{8 \sqrt{2}}{5}$

$=\frac{40 \sqrt{2}-24 \sqrt{2}}{15}$

$=\frac{16 \sqrt{2}}{15}$