Mathematics

Evaluate: $$\displaystyle\int_0^2 {x\sqrt {2 - x} } dx$$


SOLUTION
solution :-
$$ I = \int_{0}^{2} x\sqrt{2-x}dx$$

we know $$ \int_{a}^{b}f(x)dx= \int_{a}^{b}f(a+b-x)dx$$

$$ \therefore I = \int_{0}^{2}(2-x)\sqrt{2-(2-x)}dx$$

$$ = \int_{0}^{2}(2-x)\sqrt{x}dx$$

$$ = \int_{0}^{2}2\sqrt{x}dx-\int_{0}^{2}x^{3/2}dx$$

$$ I = \dfrac{4}{3}[x^{3/2}]_{0}^{2}-\dfrac{2}{5}[x^{5/2}]_{0}^{2}$$

$$ = \dfrac{4}{3}[2\sqrt{2}]-\dfrac{2}{5}[4\sqrt{2}]$$

$$ = 8\sqrt{2}[\dfrac{1}{3}-\dfrac{1}{5}]$$

$$ = 8\sqrt{2}(\dfrac{2}{15})$$

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