Question

Evaluate: ${\int }_0^{2}x\sqrt{x+2}dx$

Answer 1

Let $t={(x+2)}^2\Rightarrow dt=2(x+2)dx$

When $x=0\Rightarrow t=4$ and when $x=2\Rightarrow t=16$

${\int }_0^2(x+2-2)\sqrt{x+2}dx$

$={\int }_0^2(x+2)\sqrt{x+2}dx-2{\int }_0^2\sqrt{x+2}dx$

$=\frac{1}{2}{\int }_4^{16}\sqrt{t}dt-2{\int }_0^2\sqrt{x+2}dx$

$=\frac{1}{2}{\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_4^{16}-2{\left[\frac{{(x+2)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^2$

$=\frac{1}{3}{\left[t^{\frac{3}{2}}\right]}_4^{16}-2\times \frac{2}{3}{\left[{(x+2)}^{\frac{3}{2}}\right]}_0^2$

$=\frac{1}{2}[{16}^{\frac{3}{2}}-4^{\frac{3}{2}}]-\frac{4}{3}[{(2+2)}^{\frac{3}{2}}-{(0+2)}^{\frac{3}{2}}]$

$=\frac{1}{2}[64-8]-\frac{4}{3}[8-2\sqrt{2}]$

$=28-\frac{32}{3}+\frac{8\sqrt{2}}{3}$

$=\frac{52+8\sqrt{2}}{3}$