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Let \( \left \lfloor x \right \rfloor \) denote the floor function.



a) Prove that it is integrable on \( [-n, n ] , \;\; n \in \mathbb{N} \).

b) Evaluate the integral:

$$\mathcal{J}=\int_{-n}^{n}\left \lfloor x \right \rfloor\, {\rm d}x$$

Hello Tolaso.

a) If \(\displaystyle{n\in\mathbb{N}}\), then :

\(\displaystyle{\left[-n,n\right]=\bigcup_{k=0}^{2\,n-1}\left[k-n,k+1-n\right]}\) and

\(\displaystyle{\forall\,k\in\left\{0,1,...,2\,n-1\right\}: [x]=k-n}\) if \(\displaystyle{x\in\left[k-n,k+1-n\right]}\) .

So, the floor function is integrable on \(\displaystyle{\left[-n,n\right]}\) since it is bounded on \(\displaystyle{\left[-n,n\right]}\)

and it has only \(\displaystyle{2\,n-1}\) points of discontinuity.

b) Since

\(\displaystyle{\left(k-n,k+1-n\right)\cap \left(m-n,m+1-n\right)=\varnothing\,\,,k\,,m\in\left\{0,1,...,2\,n-1\right\}\,,k\neq m}\)

we have that

\(\displaystyle{\begin{aligned} J&=\sum_{k=0}^{2\,n-1}\int_{k-n}^{k+1-n}[x]\,\mathrm{d}x\\&=\sum_{k=0}^{2\,n-1}\int_{k-n}^{k+1-n}(k-n)\,\mathrm{d}x\\&=\sum_{k=0}^{2\,n-1}(k-n)\\&=\sum_{k=0}^{2\,n-1}k-\sum_{k=0}^{2\,n-1}n\\&=\sum_{k=1}^{2\,n-1}k-2\,n^2\\&=\dfrac{\left(2\,n-1\right)\cdot 2\,n}{2}-2\,n^2\\&=n\cdot \left(2\,n-1\right)-2\,n^2\\&=2\,n^2-n-2\,n^2\\&=-n\end{aligned}}\)