Mathematics

Evaluate $$\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$$


SOLUTION
$$I=\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$$. 

Let $$x+2=t^{2}$$. 

Then , $$dx=2t   \ dt$$

Also 
$$x=0\Rightarrow t^{2}=2\Rightarrow t=\sqrt{2}$$ and, $$x=2\Rightarrow t^{2}=4\Rightarrow t=2$$

$$\therefore I=\displaystyle\int_{\sqrt{2}}^{2}(t^{2}-2)\sqrt{t^{2}}2t dt$$

$$=2\displaystyle\int_{\sqrt{2}}^{2}(t^{4}-2t^{2})dt$$

Using , $$\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+c$$, we get

$$=2\left[\dfrac{t^{5}}{5}-\dfrac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$$


$$ I=2\left[\left(\dfrac{32}{5}-\dfrac{16}{3}\right)-\left(\dfrac{4\sqrt{2}}{5}-\dfrac{4\sqrt{2}}{3}\right)\right]$$

$$=2\left(\dfrac{16}{15}+\dfrac{8\sqrt{2}}{15}\right)$$

$$=\dfrac{32+16\sqrt{2}}{15}$$

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Subjective Medium Published on 17th 09, 2020
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