Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^{2}x\sqrt{2-x}dx$

Answer 1

Let $I={\int }_0^{2}x\sqrt{2-x}dx$
$I={\int }_0^2(2-x)\sqrt{x}dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$={\int }_0^2\{2x^{\frac{1}{2}}-x^{\frac{3}{2}}\}dx$
$={[2\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}]}_0^2$
$={[\frac{4}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}]}_0^2$
$=\frac{4}{3}(2{)}^{\frac{3}{2}}-\frac{2}{5}(2{)}^{\frac{5}{2}}$
$=\frac{4\times 2\sqrt{2}}{3}-\frac{2}{5}\times 4\sqrt{2}$
$=\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}$
$=\frac{40\sqrt{2}-24\sqrt{2}}{15}=\frac{16\sqrt{2}}{15}$