int_{-2}^{2} |[x]| , dx = | vedclass

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(D) We evaluate the integral $\int_{-2}^{2} |[x]| \, dx$ by splitting the interval at the integers where the greatest integer function $[x]$ changes v

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$4$

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(D) We evaluate the integral $\int_{-2}^{2} |[x]| \, dx$ by splitting the interval at the integers where the greatest integer function $[x]$ changes value:$\int_{-2}^{2} |[x]| \, dx = \int_{-2}^{-1} |[x]| \, dx + \int_{-1}^{0} |[x]| \, dx + \int_{0}^{1} |[x]| \, dx + \int_{1}^{2} |[x]| \, dx$For $x \in [-2, -1)$,$[x] = -2$,so $|[x]| = 2$.For $x \in [-1, 0)$,$[x] = -1$,so $|[x]| = 1$.For $x \in [0, 1)$,$[x] = 0$,so $|[x]| = 0$.For $x \in [1, 2)$,$[x] = 1$,so $|[x]| = 1$.Substituting these values:$= \int_{-2}^{-1} 2 \, dx + \int_{-1}^{0} 1 \, dx + \int_{0}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx$$= 2[x]_{-2}^{-1} + [x]_{-1}^{0} + 0 + [x]_{1}^{2}$$= 2(-1 - (-2)) + (0 - (-1)) + 0 + (2 - 1)$$= 2(1) + 1 + 1 = 4.$

$\int_{-2}^{2} |[x]| \, dx = $

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