If $n \in N$, then $\int\limits_0^n(x-[x]) d x$ is equal to
Answer & explanation
Correct answer: option 2
Since x - [x] is a periodic function with period one unit.
∴ $\int\limits_0^n(x-[x]) d x=n \int\limits_0^1(x-[x]) d x=n \int\limits_0^n x d x=\frac{n}{2}$