The value of $\int\limits_0^{[x]}(-x-[x]) d x$, is

[x] 2[x] (1/2)[x] none of these

(1/2)[x]

We have, $I=\int\limits_0^{[x]}(x-[x]) d x$ $\Rightarrow I=[x] \int\limits_0^1(x-[x]) d x$                  [∵ x - [x] is periodic] $\Rightarrow I=\frac{1}{2}[x]$