Let f(x) = max {x+|x|, x-[x]}, where [x] deno | vedclass

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(C) Given, $f(x) = \max \{x+|x|, x-[x]\}$.For $x \geq 0$, $x+|x| = 2x$ and $x-[x] = \{x\} \in [0, 1)$. Since $2x \geq \{x\}$ for $x \geq 0$, $f(x) = 2

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$21/2$

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(C) Given, $f(x) = \max \{x+|x|, x-[x]\}$.For $x \geq 0$, $x+|x| = 2x$ and $x-[x] = \{x\} \in [0, 1)$. Since $2x \geq \{x\}$ for $x \geq 0$, $f(x) = 2x$.For $x Thus, $\int_{-3}^{3} f(x) dx = \int_{-3}^{0} (x-[x]) dx + \int_{0}^{3} 2x dx$.Since $x-[x]$ is periodic with period $1$, $\int_{-3}^{0} (x-[x]) dx = 3 \int_{0}^{1} x dx = 3 \left[ \frac{x^2}{2} \right]_0^1 = \frac{3}{2}$.And $\int_{0}^{3} 2x dx = [x^2]_0^3 = 9$.Therefore, $\int_{-3}^{3} f(x) dx = \frac{3}{2} + 9 = \frac{21}{2}$.

Let $f(x) = \max \{x+|x|, x-[x]\}$, where $[x]$ denotes the greatest integer $\leq x$. Then, the value of $\int_{-3}^{3} f(x) dx$ is

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