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Let $[x]$ denote the greatest integer less than or equal to $x$, then the value of the integral $\int_{-1}^{1}(|x|-2[x]) d x$ is equal to
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3
Let $I=\int_{-1}^{1}(|x|-2[x]) d x$
$=\int_{-1}^{0}(|x|-2[x]) d x+\int_{0}^{1}(|x|-2[x]) d x$
$=\int_{-1}^{0}(-x-2(-1)) d x+\int_{0}^{1}(x-2(0)) d x$
$\quad=\int_{-1}^{0}(-x+2) d x+\int_{0}^{1} x d x$
$\quad=\left(-\frac{x^{2}}{2}+2 x\right)_{-1}^{0}+\left[\frac{x^{2}}{2}\right]_{0}$
$\quad=-\left(-\frac{1}{2}+2(-1)\right)+\frac{1}{2}$
$\quad=\frac{1}{2}+2+\frac{1}{2}=1+2=3$
$=\int_{-1}^{0}(|x|-2[x]) d x+\int_{0}^{1}(|x|-2[x]) d x$
$=\int_{-1}^{0}(-x-2(-1)) d x+\int_{0}^{1}(x-2(0)) d x$
$\quad=\int_{-1}^{0}(-x+2) d x+\int_{0}^{1} x d x$
$\quad=\left(-\frac{x^{2}}{2}+2 x\right)_{-1}^{0}+\left[\frac{x^{2}}{2}\right]_{0}$
$\quad=-\left(-\frac{1}{2}+2(-1)\right)+\frac{1}{2}$
$\quad=\frac{1}{2}+2+\frac{1}{2}=1+2=3$
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