Integration of $\min \{x-[x] , -x-[-x]\} $where $[x]$ is floor function

Question

How to calculate integration of $ \min \{x-[x] , -x-[-x]\}$ (where $[x]$ is floor function) without drawing graph

Answer 1

Graphically it can be calculated as below. Without graph i'm unable to solve this problem.

Answer 2

Note that $x-[x] = \{x\}$ where $\{x\}$ is the fraction part of $x$. Also note that $\{x\}$ is periodic with period $1$ because:

$$\color{red}{\{x+1\}}= x+1-[x+1] = x -[x] = \color{red}{\{x\}} $$

Thus your integrand converts to $\min\{ \{x\}, \{-x\}\}$. This function is periodic too again with period $1$, so we just have to analyse in integer interval of length $1$.

Therefore value of integral is $4$ times the value on $(0,1)$

$$I = \int_{-2}^{2} \min\{ \{x\}, \{-x\}\} = 4\int_{0}^{1} \min\{ \{x\}, \{-x\}\}$$

For $0\lt x \lt 1$ we have that $\{x\} = x$ and $\{-x\} = 1-x$. Thus for $x \in (0, 0.5)$ integrand is $x$ and from $[0.5, 1)$ integrand is $1-x$. Area of this one unit is found to be $\boxed {1/4}$.

Hence $I = 1.$

Note: For fraction part function, refer Graham et. al definition here: Fractional Part - Wolfram Mathworld