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(A) Let $I = \int_{-1}^{1}(|x|-2[x]) \, dx$. We split the integral at $x=0$: $I = \int_{-1}^{0}(|x|-2[x]) \, dx + \int_{0}^{1}(|x|-2[x]) \, dx$. For $

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$3$

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(A) Let $I = \int_{-1}^{1}(|x|-2[x]) \, dx$. We split the integral at $x=0$: $I = \int_{-1}^{0}(|x|-2[x]) \, dx + \int_{0}^{1}(|x|-2[x]) \, dx$. For $x \in [-1, 0)$, $|x| = -x$ and $[x] = -1$. For $x \in [0, 1)$, $|x| = x$ and $[x] = 0$. Substituting these values: $I = \int_{-1}^{0}(-x - 2(-1)) \, dx + \int_{0}^{1}(x - 2(0)) \, dx$. $I = \int_{-1}^{0}(-x + 2) \, dx + \int_{0}^{1} x \, dx$. Evaluating the integrals: $I = \left[ -\frac{x^2}{2} + 2x \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{1}$. $I = (0 - 0) - \left( -\frac{(-1)^2}{2} + 2(-1) \right) + \left( \frac{1^2}{2} - 0 \right)$. $I = - \left( -\frac{1}{2} - 2 \right) + \frac{1}{2}$. $I = \frac{1}{2} + 2 + \frac{1}{2} = 1 + 2 = 3$.

Let $[x]$ denote the greatest integer less than or equal to $x$, then the value of the integral $\int_{-1}^{1}(|x|-2[x]) \, dx$ is equal to

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