Mathematics

Evaluate the following integral:

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


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

Let $$x+2=t^{2}$$. Then , $$dx=2t dt$$

Also $$x=0\Rightarrow t^{2}=2$$ 

$$\Rightarrow t=\sqrt{2}$$ 

$$x=2\Rightarrow t^{2}=4$$ 

$$\Rightarrow t=2$$

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

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

$$=2\left[\dfrac{t^{5}}{5}-\dfrac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$$   [$$\because\int x^n=\dfrac{x^{n+1}}{n+1}$$]


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

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

$$=\dfrac{32+16\sqrt{2}}{15}$$
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Subjective Medium Published on 17th 09, 2020
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