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The function $f(x) = x - [x]$ is a periodic function. In fact, $f(x) = x\mod1,$ and its period is $1.$

Considering only the first period $0 \leq x < 1,$ since $x$ is less than the modulus, $f(x) = x.$ The integral in this period is therefore $\int_0^1 f(x) = \frac{1^2}{2} = 0.5$ Since it is a periodic function, the integral will be the same in every period. We are finding the integral in $2$ periods, so the answer is $2\times0.5=\boxed{1}$