Question

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

Answer 1

Let ${\int }_0^{2}x\sqrt{2-x}dx.$
Also, $I={\int }_0^2(2-x)\sqrt{2-(2-x)}dx={\int }_0^2(2-x)\sqrt{x}dx={\int }_0^2(2x^{\frac{1}{2}}-x^{\frac{3}{2}})dx={[\frac{2x^{\left(\frac{1}{2}\right)+1}}{\left(\frac{1}{2}\right)+1}-\frac{x^{\left(\frac{3}{2}\right)+1}}{\left(\frac{3}{2}\right)+1}]}_0^2={[\frac{4}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}]}_0^2=\frac{4}{3}.2^{\frac{3}{2}}-\frac{2}{5}.2^{\frac{5}{2}}-0=\frac{4}{3}2\sqrt{2}-\frac{2}{5}4\sqrt{2}=(\frac{8}{3}-\frac{8}{5})\sqrt{2}=\left(\frac{40-24}{15}\right)\sqrt{2}=\frac{16\sqrt{2}}{15}$