int_{-2}^2 |[x]| , dx is equal to | vedclass

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(D) We have $\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} |[x]| \, dx + \int_{-1}^0 |[x]| \, dx + \int_0^1 |[x]| \, dx + \int_1^2 |[x]| \, dx$. Since $[x]

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$4$

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(D) We have $\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} |[x]| \, dx + \int_{-1}^0 |[x]| \, dx + \int_0^1 |[x]| \, dx + \int_1^2 |[x]| \, dx$. Since $[x]$ is the greatest integer function: For $x \in [-2, -1)$, $[x] = -2$, so $|[x]| = |-2| = 2$. For $x \in [-1, 0)$, $[x] = -1$, so $|[x]| = |-1| = 1$. For $x \in [0, 1)$, $[x] = 0$, so $|[x]| = |0| = 0$. For $x \in [1, 2)$, $[x] = 1$, so $|[x]| = |1| = 1$. Substituting these values: $\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} 2 \, dx + \int_{-1}^0 1 \, dx + \int_0^1 0 \, dx + \int_1^2 1 \, dx$. $= 2[x]_{-2}^{-1} + [x]_{-1}^0 + 0 + [x]_1^2$. $= 2(-1 - (-2)) + (0 - (-1)) + (2 - 1)$. $= 2(1) + 1 + 1 = 4$.

$\int_{-2}^2 |[x]| \, dx$ is equal to

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