Evaluate the integral in Exercise: ^ 2 _ 0 x x+2 (put x+2=t^ 2 )

Question

Evaluate the integral in Exercise:$\int^{2}_{0}\text{x}\sqrt{\text{x}+2}\ (\text{put}\ \text{x}+2=\text{t}^{2})$

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Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{2}\text{x}\sqrt{\text{x}+2}\ \text{dx}$
$\text{putting}\sqrt{\text{x}+2}=\text{t}\ \Rightarrow\ \text{x}+2=\text{t}^{2}\ \Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{t}\ \Rightarrow\text{dx}=2\text{t}\ \text{dt}$
$\text{Limits of integration}\ \text{when}\ \text{x}=0,\text{t}=\sqrt{2}\ \text{and}\ \text{when}\text{x}=2,\text{t}=\sqrt{4}=2$
$\therefore\text{from}\ \text{eq}.\text{(i)},\ \text{I}=\int\limits_{\sqrt{2}}^{2}\text{t}^{2}(\text{t}^{2}-2)\text{t}.2\text{t}\ \text{dt}=2\int\limits_{\sqrt{2}}^{2}\text{t}^{2}(\text{t}^{2}-2)\text{dt}=2\int\limits_{\sqrt{2}}^{2}(\text{t}^{2}-2\text{t}^{2})\text{dt} $
$=2\Bigg[\bigg(\frac{\text{t}^{5}}{5}\bigg)^{2}_{\sqrt{2}}-2\bigg(\frac{\text{t}^{3}}{3}\bigg)^{2}_{\sqrt{2}}\Bigg]=2\bigg[\frac{1}{5}\Big(2^{5}-\big(\sqrt{2}\big)^{5}\Big)-\frac{2}{3}\Big(2^{3}-\big(\sqrt{2}\big)^{3}\Big)\bigg]$
$=2\bigg[\frac{1}{5}\big(32-4\sqrt{2}\big)-\frac{2}{3}\big(8-2\sqrt{2}\big)\bigg]=2\bigg[\frac{32}{5}-\frac{4\sqrt{2}}{5}-\frac{16}{3}+\frac{4\sqrt{2}}{3}\bigg]=2\bigg[\frac{96-12\sqrt{2}-80+20\sqrt{2}}{15}\bigg]$
$=\frac{2}{15}\big(16+8\sqrt{2}\big)=\frac{16\sqrt{2}}{15}\big(\sqrt{2}+1\big)$

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