Question

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

Answer 1

${\int }_0^{2}x\sqrt{x+2}dx$
Let $x+2=t^2\Rightarrow dx=2tdt$
When $x=0,t=\sqrt{2}$ and when $x=2,t=2$
$\therefore {\int }_0^{2}x\sqrt{x+2}dx={\int }_{\sqrt{2}}^2(t^2-2)\sqrt{t^2}2tdt$
$=2{\int }_{\sqrt{2}}^2(t^2-2)t^{2}dt$
$=2{\int }_{\sqrt{2}}^2(t^4-2t^2)dt$
$=2{[\frac{t^5}{5}-\frac{2t^3}{3}]}_{\sqrt{5}}^2$
$=2[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}]$
$=2\left[\frac{96-80-12\sqrt{2}+20\sqrt{2}}{15}\right]$
$=2\left[\frac{16+8\sqrt{2}}{15}\right]$
$=\frac{16(2+\sqrt{2})}{15}$
$=\frac{16\sqrt{2}(\sqrt{2}+1)}{15}$