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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. ∴ $\int\limits_0^n(x-[x]) d x=n \int\limits_0^1(x-[x]) d x=n \int\limits_0^n x d x=\frac{n}{2}$ |
