Question

Evaluate the integrals using substitution.
${\int }_0^{2}x\sqrt{x+2}$ (Put $x+2=t^2$).

Answer 1

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

Let $x+2=t^2\Rightarrow dx=2tdt$

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

$=2[(\frac{32}{5}-\frac{2\times 8}{3})-(\frac{4\sqrt{2}}{5}-\frac{2.2\sqrt{2}}{3})]=2(\frac{16}{15}+\frac{8\sqrt{2}}{15})=\frac{32+16\sqrt{2}}{15}$