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If $f(x)=x-[x]$, for every real number $x$, where $[x]$, is the integral part of $x$. Then, $\int_{-1}^{1} f(x) d x$ is equal to
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$\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}],-1 \leq \mathrm{x} < 0$
$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}+1$
When $0 \leq \mathrm{x} < 1$
$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}$
$\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{-1}^{0} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
$=\int_{-1}^{0}(\mathrm{x}+1) \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{x} \mathrm{dx}$
$=\left[\frac{\mathrm{x}^{2}}{2}+\mathrm{x}\right]_{-1}^{0}+\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1}$
$=0-\left[\frac{(-1)^{2}}{2}-1\right]+\frac{1}{2}=1$
$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}+1$
When $0 \leq \mathrm{x} < 1$
$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}$
$\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{-1}^{0} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
$=\int_{-1}^{0}(\mathrm{x}+1) \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{x} \mathrm{dx}$
$=\left[\frac{\mathrm{x}^{2}}{2}+\mathrm{x}\right]_{-1}^{0}+\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1}$
$=0-\left[\frac{(-1)^{2}}{2}-1\right]+\frac{1}{2}=1$
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