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(C) The expression ${x - [x]}$ represents the fractional part of $x$,denoted as $\{x\}$.Since the fractional part function $f(x) = \{x\} = x - [x]$ is

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$n/2$

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(C) The expression ${x - [x]}$ represents the fractional part of $x$,denoted as $\{x\}$.Since the fractional part function $f(x) = \{x\} = x - [x]$ is a periodic function with period $T = 1$,we can use the property $\int_0^{nT} f(x) \,dx = n \int_0^T f(x) \,dx$.Therefore,$\int_0^n \{x\} \,dx = n \int_0^1 \{x\} \,dx$.In the interval $[0, 1)$,$[x] = 0$,so $\{x\} = x - 0 = x$.Thus,$\int_0^n \{x\} \,dx = n \int_0^1 x \,dx$.Evaluating the integral: $n \left[ \frac{x^2}{2} \right]_0^1 = n \left( \frac{1}{2} - 0 \right) = \frac{n}{2}$.

If $n$ is a positive integer and $[x]$ is the greatest integer not exceeding $x$,then $\int_0^n {\{x - [x]\} \,dx}$ equals

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