If n is a positive integer and [x] is the gre | vedclass

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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: $n \left[ \frac{x^2}{2} \right]_0^1 = n \left( \frac{1}{2} - 0 \right) = \frac{n}{2}$.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

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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