The integral


The integral $\int_{0}^{2}|| x-1|-x| d x$ is equal to


$\int_{0}^{2}|| x-1|-x| d x=\int_{0}^{1}|1-x-x| d x+\int_{1}^{2}|| x-1-x \mid d x$

$=\int_{0}^{1}(1-2 x) d x+\int_{1 / 2}^{1}(2 x-1) d x+\int_{1}^{2} d x$



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