int_{-1}^{1} [x + [x + [x]]] , dx = (where [ | vedclass

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(C) We know that $[x + n] = [x] + n$ for any integer $n$. Given the expression $[x + [x + [x]]]$,we can simplify it step by step: $[x + [x + [x]]] = [

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$-3$

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(C) We know that $[x + n] = [x] + n$ for any integer $n$. Given the expression $[x + [x + [x]]]$,we can simplify it step by step: $[x + [x + [x]]] = [x + [x] + [x]] = [x + 2[x]] = [x] + 2[x] = 3[x]$. Now,we evaluate the integral: $\int_{-1}^{1} 3[x] \, dx = 3 \int_{-1}^{1} [x] \, dx$. Splitting the integral at $x = 0$: $3 \left( \int_{-1}^{0} [x] \, dx + \int_{0}^{1} [x] \, dx \right)$. For $x \in [-1, 0)$,$[x] = -1$. For $x \in [0, 1)$,$[x] = 0$. So,$3 \left( \int_{-1}^{0} (-1) \, dx + \int_{0}^{1} (0) \, dx \right) = 3 \left( [-x]_{-1}^{0} + 0 \right) = 3(-1) = -3$.

$\int_{-1}^{1} [x + [x + [x]]] \, dx = $ (where $[\cdot]$ denotes the greatest integer function)

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