Mathematics

 Evaluate: $$\displaystyle\int _{ 0 }^{ 2 }{ x\sqrt { x+2 } dx } $$


SOLUTION
Let $$t={\left(x+2\right)}^{2}\Rightarrow dt=2\left(x+2\right)dx$$

When $$x=0\Rightarrow t=4$$ and when $$x=2\Rightarrow t=16$$

$$\displaystyle\int_{0}^{2}{\left(x+2-2\right)\sqrt{x+2}dx}$$

$$=\displaystyle\int_{0}^{2}{\left(x+2\right)\sqrt{x+2}dx}-2\displaystyle\int_{0}^{2}{\sqrt{x+2}dx}$$

$$=\dfrac{1}{2}\displaystyle\int_{4}^{16}{\sqrt{t}dt}-2\displaystyle\int_{0}^{2}{\sqrt{x+2}dx}$$

$$=\dfrac{1}{2}\left[\dfrac{{t}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{4}^{16}-2\left[\dfrac{{\left(x+2\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{0}^{2}$$

$$=\dfrac{1}{3}\left[{t}^{\frac{3}{2}}\right]_{4}^{16}-2\times\dfrac{2}{3}\left[{\left(x+2\right)}^{\frac{3}{2}}\right]_{0}^{2}$$

$$=\dfrac{1}{2}\left[{16}^{\frac{3}{2}}-{4}^{\frac{3}{2}}\right]-\dfrac{4}{3}\left[{\left(2+2\right)}^{\frac{3}{2}}-{\left(0+2\right)}^{\frac{3}{2}}\right]$$

$$=\dfrac{1}{2}\left[64-8\right]-\dfrac{4}{3}\left[8-2\sqrt{2}\right]$$

$$=28-\dfrac{32}{3}+\dfrac{8\sqrt{2}}{3}$$

$$=\dfrac{52+8\sqrt{2}}{3}$$
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Subjective Medium Published on 17th 09, 2020
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