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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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