If [x] represents the greatest integer functi | vedclass

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(B) Let $I = \int_{-2}^2 [2-x] \, dx$. We split the integral based on the intervals where $[2-x]$ is constant: $I = \int_{-2}^{-1} [2-x] \, dx + \int_

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$6$

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(B) Let $I = \int_{-2}^2 [2-x] \, dx$. We split the integral based on the intervals where $[2-x]$ is constant: $I = \int_{-2}^{-1} [2-x] \, dx + \int_{-1}^0 [2-x] \, dx + \int_0^1 [2-x] \, dx + \int_1^2 [2-x] \, dx$. For $x \in [-2, -1)$, $2-x \in (3, 4]$, so $[2-x] = 3$. For $x \in [-1, 0)$, $2-x \in (2, 3]$, so $[2-x] = 2$. For $x \in [0, 1)$, $2-x \in (1, 2]$, so $[2-x] = 1$. For $x \in [1, 2)$, $2-x \in (0, 1]$, so $[2-x] = 0$. Thus, $I = \int_{-2}^{-1} 3 \, dx + \int_{-1}^0 2 \, dx + \int_0^1 1 \, dx + \int_1^2 0 \, dx$. $I = 3[x]_{-2}^{-1} + 2[x]_{-1}^0 + 1[x]_0^1 + 0$. $I = 3(-1 - (-2)) + 2(0 - (-1)) + 1(1 - 0) = 3(1) + 2(1) + 1(1) = 3 + 2 + 1 = 6$.

If $[x]$ represents the greatest integer function, then $\int_{-2}^2 [2-x] \, dx = $

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