If $n \in N$, then $\int\limits_0^n(x-[x]) d x$ is equal to

n n/2 2n none of these

n/2

Since x - [x] is a periodic function with period one unit. ∴  $\int\limits_0^n(x-[x]) d x=n \int\limits_0^1(x-[x]) d x=n \int\limits_0^n x d x=\frac{n}{2}$