Integrating fractional part function If $f(x) = \min(\{x\},\{-x\})$, then find $$\displaystyle\int_{-100}^{100} f(x)\, {\rm d}x,$$ where $\{x\}$ denotes fractional part of $x$.
 A: Geometrically, the plot of $f(x)$ on $[0, 1]$ is an isosceles right triangle with a hypothenuse (provided by the $x$ axis) equal to $1$. The area of such triangle is $1/4$.
$f(x)$ is periodic, and its period is $1$, so on $[-100, 100]$ the plot is just that triangle repeated 200 times. The cumulative area of those triangles is $200\cdot1/4 = 50$ which is the value of the definite integral.
A: as mniip said,there are almost 200 triangle in the [-100,100](as you can see from the graphic of function),square of one triangle equal to 1/4,so the answer will be 200*1/4=50

A: $$ x=n+p\\x=\left \lfloor x \right \rfloor  +\begin{Bmatrix} 
x
\end{Bmatrix}$$ so we can write $\begin{Bmatrix} 
x
\end{Bmatrix}=x-\left \lfloor x \right \rfloor $
now note that f(x) is even function
$$f(-x)=min(\begin{Bmatrix} 
x
\end{Bmatrix}),\begin{Bmatrix} 
-x
\end{Bmatrix})=f(x)$$so $\int_{-100}^{100}f(x)=2\int_{0}^{100}dx=2*100\int_{0}^{1}f(x)dx$ $\space $It is because f(x)   is periodic function

$$\int_{-100}^{100}f(x) dx=200\int_{0}^{1} dx=200\cdot \frac{1}{4}=\boxed{50}$$
implicit ,it is a continuous function
