Question
By using property of definite integrals, evaluate $\int_0^2 x \sqrt{2-x} d x$
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Answer
$\begin{array}{l}\int_0^2 x \sqrt{2-x} d x=\int_0^2(2-x)(2-(2-x))^{\frac{1}{2}} d x \text { (by property } P _4 \text { ) } \\ =\int_0^2(2-x) x^{\frac{1}{2}} d x=\int_0^2\left(2 x^{\frac{1}{2}}-x^{\frac{3}{2}}\right) d x \\ =\left[\frac{4}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}\right]_0^2=\frac{4}{3} \cdot 2 \sqrt{2}-\frac{2}{5} \cdot 4 \sqrt{2}=\frac{16}{15} \sqrt{2}\end{array}$
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