Mathematics

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


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

Let $$x+2=t^{2}$$. 

Differentiating with respect to $$x$$, we get

 $$dx=2t dt$$


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

So, lower limit is $$\sqrt 2$$ and upper limit is $$2$$

$$\therefore I=\displaystyle\int_{\sqrt{2}}^{2}(t^{2}-2)\sqrt{t^{2}} \ \cdot 2t dt=2\displaystyle\int_{\sqrt{2}}^{2}(t^{4}-2t^{2})dt=2\left[\dfrac{t^{5}}{5}-\dfrac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$$

$$\Rightarrow I=2\left[\left(\dfrac{32}{5}-\dfrac{16}{3}\right)-\left(\dfrac{4\sqrt{2}}{5}-\dfrac{4\sqrt{2}}{3}\right)\right]=2\left(\dfrac{16}{15}+\dfrac{8\sqrt{2}}{15}\right)=\dfrac{32+16\sqrt{2}}{15}$$
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Subjective Medium Published on 17th 09, 2020
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