A) \[{{n}^{2}}/2\]
B) \[n(n-1)/2\]
C) \[n\,/\,2\]
D) \[\frac{{{n}^{2}}}{2}-n\]
Correct Answer: C
Solution :
\[x-[x]\] is a periodic function with period 1. \[\therefore \int_{0}^{n}{\left\{ x-[x] \right\}\,dx=n\int_{0}^{1}{(x-[x])\,\,dx}}\] \[=n\left[ \int_{0}^{1}{x\,\,dx-\int_{0}^{1}{[x]\,dx}} \right]\]\[=n\left[ \left( \frac{{{x}^{2}}}{2} \right)_{0}^{1}-0 \right]=\frac{n}{2}\].
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