Integral of floor function

Evaluate the integral:

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

Solution



 We note that:

$$\left.\begin{matrix}
\displaystyle K_1= \int_{-n}^{-n+1}\left \lfloor x \right \rfloor\, {\rm d}x\\
\displaystyle K_2 = \int_{-n+1}^{-n+2}\left \lfloor x \right \rfloor\, {\rm d}x\\
\displaystyle K_3 =\int_{-n+2}^{-n+3}\left \lfloor x \right \rfloor \, {\rm d}x\\
\vdots \\
\displaystyle K_{n-4}= \int_{n-4}^{n-3}\left \lfloor x \right \rfloor\, {\rm d}x\\
\displaystyle K_{n-3}= \int_{n-3}^{n-2} \left \lfloor x \right \rfloor\, {\rm d}x\\
\displaystyle K_{n-2}= \int_{n-2}^{n-1}\left \lfloor x \right \rfloor\, {\rm d}x\\
\displaystyle K_{n-1} = \int_{n-1}^{n}\left \lfloor x \right \rfloor\, {\rm d}x
\end{matrix}\right\}\overset{(+)}{\Rightarrow } \sum_{k=1}^{n-1}K_n = K \Rightarrow $$

$$\Rightarrow \sum_{k=0}^{n}\int_{k-n}^{k+1-n}\left \lfloor x \right \rfloor\, {\rm d}x =K\Rightarrow $$

$$\Rightarrow -n + \left ( 1- n  \right )+ \left ( 2-n \right )+\cdots + \left ( n-4 \right )+\cdots + \left ( n-1 \right )= K \Rightarrow  $$

$$\Rightarrow K=-n $$

which completes the exercise.