Evaluate the following integrals: _ 0 ^ 2 x x+2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits_{0}^{2}\text{x}\sqrt{\text{x}+2}\text{ dx}$

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Answer

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

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