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By using the properties of definite integrals, evaluate the integrals
$\int_0^2 x \sqrt{2-x} d x=I($ say $)$
$\int_0^2 x \sqrt{2-x} d x=I($ say $)$
Solution:
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Verified Answer
Let $2-x=t \Rightarrow-d x=d t$
When $x=0, t=2$ and when $x=2, t=0$
$\begin{aligned}
&\therefore \quad I=-\int_2^0(2-t) \sqrt{t} d t=\int_0^2\left(2 t^{1 / 2}-t^{3 / 2}\right) d t \\
&\left.=\frac{4}{3} t^{3 / 2}-\frac{2}{5} t^{5 / 2}\right]_0^2=\frac{8 \sqrt{2}}{3}-\frac{8 \sqrt{2}}{5}=\frac{16 \sqrt{2}}{15}
\end{aligned}$
When $x=0, t=2$ and when $x=2, t=0$
$\begin{aligned}
&\therefore \quad I=-\int_2^0(2-t) \sqrt{t} d t=\int_0^2\left(2 t^{1 / 2}-t^{3 / 2}\right) d t \\
&\left.=\frac{4}{3} t^{3 / 2}-\frac{2}{5} t^{5 / 2}\right]_0^2=\frac{8 \sqrt{2}}{3}-\frac{8 \sqrt{2}}{5}=\frac{16 \sqrt{2}}{15}
\end{aligned}$
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