Integrals — Class 12 Maths Solution

Let$I = \int\limits_0^2 {x\sqrt {2 - x} } dx$

Put $2 - x = t$ $\Rightarrow$ $- dx = dt$

When $x = 0,t = 2$ and when $x = 2,t = 0$

$\therefore$ $I = - \int\limits_2^0 {\left( {2 - t} \right)\sqrt t dt} = \int\limits_0^2 {\left( {2{t^{1/2}} - {t^{3/2}}} \right)dt} = \left[ {\cfrac{{2{t^{3/2}}}}{{3/2}} - \cfrac{{{t^{5/2}}}}{{5/2}}} \right]_0^2$

$= \left[ {\cfrac{4}{3}{t^{3/2}} - \cfrac{2}{5}{t^{5/2}}} \right]_0^2 = \cfrac{4}{3}{\left( 2 \right)^{3/2}} - \cfrac{2}{5}{\left( 2 \right)^{5/2}}$

$= \cfrac{4}{3} \times 2\sqrt 2 - \cfrac{2}{5} \times 4\sqrt 2 = \cfrac{{8\sqrt 2 }}{3} - \cfrac{{8\sqrt 2 }}{5} = \cfrac{{16\sqrt 2 }}{{15}}$