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(D) Given $f(x) = \max \{|x|, |x-1|, \ldots, |x-2n|\}$.We need to evaluate $\int_0^{2n} f(x) dx$.For $x \in [0, 2n]$,the maximum value of the set $\{|

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$3n^2$

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(D) Given $f(x) = \max \{|x|, |x-1|, \ldots, |x-2n|\}$.We need to evaluate $\int_0^{2n} f(x) dx$.For $x \in [0, 2n]$,the maximum value of the set $\{|x|, |x-1|, \ldots, |x-2n|\}$ is determined by the endpoints of the interval $[0, 2n]$.Specifically,$f(x) = \max \{|x|, |x-2n|\}$.We split the integral at $x = n$:$\int_0^{2n} f(x) dx = \int_0^n f(x) dx + \int_n^{2n} f(x) dx$For $x \in [0, n]$,$|x-2n| \geq |x|$,so $f(x) = |x-2n| = 2n-x$.For $x \in [n, 2n]$,$|x| \geq |x-2n|$,so $f(x) = |x| = x$.Thus,$\int_0^{2n} f(x) dx = \int_0^n (2n-x) dx + \int_n^{2n} x dx$.Evaluating the integrals:$\int_0^n (2n-x) dx = [2nx - \frac{x^2}{2}]_0^n = 2n^2 - \frac{n^2}{2} = \frac{3n^2}{2}$.$\int_n^{2n} x dx = [\frac{x^2}{2}]_n^{2n} = \frac{4n^2}{2} - \frac{n^2}{2} = \frac{3n^2}{2}$.Adding them together: $\frac{3n^2}{2} + \frac{3n^2}{2} = 3n^2$.

Define a function $f: R \rightarrow R$ by $f(x) = \max \{|x|, |x-1|, \ldots, |x-2n|\}$,where $n$ is a fixed natural number. Then,$\int_0^{2n} f(x) dx$ is

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