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$\int_0^3|2-x| d x \quad$ equals
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$5 / 2$
$I=\int_0^3|2-x| d x=\int_0^2(2-x) d x+\int_2^3-(2-x) d x$
$\begin{gathered}=\int_0^2(2-x) d x-\int_2^3(2-x) d x=\left[2 x-\frac{x^2}{2}\right]_0^2-\left[2 x-\frac{x^2}{2}\right]_2^3 \\ \left.\Rightarrow I=[4-2]-6-\frac{9}{2}-(4-2)\right]=2-\left[4-\frac{9}{2}\right]=\frac{5}{2} .\end{gathered}$
$\begin{gathered}=\int_0^2(2-x) d x-\int_2^3(2-x) d x=\left[2 x-\frac{x^2}{2}\right]_0^2-\left[2 x-\frac{x^2}{2}\right]_2^3 \\ \left.\Rightarrow I=[4-2]-6-\frac{9}{2}-(4-2)\right]=2-\left[4-\frac{9}{2}\right]=\frac{5}{2} .\end{gathered}$
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