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

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(B) Given $f(x) = \max \{x + |x|, x - [x]\}$.We know that $x - [x] = \{x\}$, where $\{x\}$ is the fractional part of $x$.For $x \in [-3, 0)$, $x + |x|

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$\frac{21}{2}$

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(B) Given $f(x) = \max \{x + |x|, x - [x]\}$.We know that $x - [x] = \{x\}$, where $\{x\}$ is the fractional part of $x$.For $x \in [-3, 0)$, $x + |x| = x - x = 0$. Since $\{x\} \ge 0$, $f(x) = \max \{0, \{x\}\} = \{x\}$.For $x \in [0, 3]$, $x + |x| = x + x = 2x$. Since $2x \ge \{x\}$ for $x \ge 0$, $f(x) = 2x$.Now, calculate the integral:$\int_{-3}^3 f(x) \, dx = \int_{-3}^0 \{x\} \, dx + \int_0^3 2x \, dx$$\int_{-3}^0 \{x\} \, dx = \int_{-3}^0 (x - [x]) \, dx$. Since the integral of the fractional part over an interval of length $n$ is $\frac{n}{2}$, $\int_{-3}^0 \{x\} \, dx = 3 	imes \frac{1}{2} = \frac{3}{2}$.$\int_0^3 2x \, dx = [x^2]_0^3 = 9 - 0 = 9$.Thus, $\int_{-3}^3 f(x) \, dx = \frac{3}{2} + 9 = \frac{21}{2}$.

Let $f(x) = \max \{x + |x|, x - [x]\}$, where $[x]$ stands for the greatest integer not greater than $x$. Then $\int_{-3}^3 f(x) \, dx$ has the value:

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