Integrals — Class 12 Maths Solution

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

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

When $x = 0,t = 2$ and when $x = 2,t = 4$

$\therefore$ $I = \int\limits_2^4 {\left( {t - 2} \right)\sqrt t dt} = \int\limits_2^4 {\left( {{t^{3/2}} - 2{t^{1/2}}} \right)dt} = \left[ {\cfrac{2}{5}{t^{5/2}} - 2 \times \cfrac{2}{3}{t^{3/2}}} \right]_2^4$

$= \left[ {\cfrac{2}{5}{t^{5/2}} - \cfrac{4}{3}{t^{3/2}}} \right]_2^4 = \left[ {\cfrac{2}{5}{{\left( 4 \right)}^{5/2}} - \cfrac{4}{3}{{\left( 4 \right)}^{3/2}}} \right] - \left[ {\cfrac{2}{5}{{\left( 2 \right)}^{5/2}} - \cfrac{4}{3}{{\left( 2 \right)}^{3/2}}} \right]$

$= \cfrac{2}{5}{\left( 2 \right)^5} - \cfrac{4}{3}{\left( 2 \right)^3} - \cfrac{2}{5} \times 4\sqrt 2 + \cfrac{4}{3} \times 2\sqrt 2$

$= \cfrac{2}{5} \times 32 - \cfrac{4}{3} \times 8 - \cfrac{8}{2}\sqrt 2 + \cfrac{8}{3}\sqrt 2$

$= \cfrac{{64}}{5} - \cfrac{{32}}{3} - \left( {\cfrac{8}{5}\sqrt 2 - \cfrac{8}{3}\sqrt 2 } \right) = \cfrac{{192 - 160}}{{15}} - \left( {\cfrac{{24\sqrt 2 - 40\sqrt 2 }}{{15}}} \right)$

$= \cfrac{{32}}{{15}} + \cfrac{{16\sqrt 2 }}{{15}} = \cfrac{{16}}{{15}}\left( {2 + \sqrt 2 } \right) = \cfrac{{16\sqrt 2 }}{{15}}\left( {\sqrt 2 + 1} \right)$