Question

By using the properties of the definite integral, evaluate the integral:

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

Answer 1

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

Put 2 - x = t

⇒ dx = dt

When x = 0, t = 2

and x = 2, t =  0

∵ ${\displaystyle I=-{\int }_2^0(2-t)\sqrt{t} dt}$

${\displaystyle ={\int }_0^2(2t^{\frac{1}{2}}-t^{\frac{3}{2}})dt}$

${\displaystyle ={[\frac{2t^{\frac{3}{2}}}{\frac{3}{2}}-\frac{t^{\frac{5}{2}}}{\frac{5}{2}}]}_0^2}$     ${\displaystyle ...[\because -{\int }_a^{0}f\left(x\right)dx={\int }_0^{a}f\left(x\right)dx]}$

${\displaystyle ={[\frac{4}{3}{t}^{\frac{3}{2}}-\frac{2}{5}{t}^{\frac{5}{2}}]}_0^2}$

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

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

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

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