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$\int_0^2[x] d x+\int_0^2|x-1| d x=$
(where $[x]$ denotes the greatest integer function.)
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(where $[x]$ denotes the greatest integer function.)
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The correct answer is:
2
$\begin{aligned} & \int_0^2[x] d x+\int_0^2[x-1] d x \\ & =\int_0^1 0 . d x+\int_1^2 1 . d x+\int_0^1(1-x) d x+\int_1^2(x-1) d x \\ & =[0]_0^1+[x]_1^2+\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]_1^2 \\ & =0+1+\frac{1}{2}+\frac{1}{2}=2\end{aligned}$
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