Question

Evaluate the integral ${\int }_0^{2}x\sqrt{x+2}dx$.

Answer 1

$I={\int }_0^{2}x\sqrt{x+2}dx$. Let $x+2=t^2$. Then , $dx=2tdt$

Also $x=0\Rightarrow t^2=2\Rightarrow t=\sqrt{2}$ and, $x=2\Rightarrow t^2=4\Rightarrow t=2$

$\therefore I={\int }_{\sqrt{2}}^2(t^2-2)\sqrt{t^2}2tdt=2{\int }_{\sqrt{2}}^2(t^4-2t^2)dt=2{[\frac{t^5}{5}-\frac{2t^3}{3}]}_{\sqrt{2}}^2$

$\Rightarrow I=2[(\frac{32}{5}-\frac{16}{3})-(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3})]=2(\frac{16}{15}+\frac{8\sqrt{2}}{15})=\frac{32+16\sqrt{2}}{15}$