Let [x] be the greatest integer function. The | vedclass

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(C) Given the integral $I = \int_{-1}^{1} [x+2[x+2[x]]] dx$. Using the property $[x+n] = [x]+n$ for any integer $n$, we simplify the expression inside

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$-7$

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(C) Given the integral $I = \int_{-1}^{1} [x+2[x+2[x]]] dx$. Using the property $[x+n] = [x]+n$ for any integer $n$, we simplify the expression inside the integral: $[x+2[x+2[x]]] = [x] + 2[x+2[x]] = [x] + 2([x] + 2[x]) = [x] + 2[x] + 4[x] = 7[x]$. Now, substitute this back into the integral: $I = \int_{-1}^{1} 7[x] dx = 7 \int_{-1}^{1} [x] dx$. Split the integral into intervals: $I = 7 \left( \int_{-1}^{0} [x] dx + \int_{0}^{1} [x] dx \right)$. For $x \in [-1, 0)$, $[x] = -1$. For $x \in [0, 1)$, $[x] = 0$. $I = 7 \left( \int_{-1}^{0} (-1) dx + \int_{0}^{1} (0) dx \right) = 7 ([-x]_{-1}^{0} + 0) = 7(-1) = -7$.

Let $[x]$ be the greatest integer function. Then, $\int_{-1}^{1} [x+2[x+2[x]]] dx = $

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