$f(x)=\min(x-[x],1-x+[x])$ for $x\in\mathbb{R}$. Then $\displaystyle 2\int_0^2 f(x)dx=$
Step-by-Step Solution
Key Concept: $f(x)=\min(\{x\},1-\{x\})$ where $\{x\}$ is fractional part; this equals $|x-0.5|$ reflected: $f=\{x\}$ for $\{x\}\le 0.5$, else $1-\{x\}$
$\int_0^2 f(x)dx=2\int_0^1 f(x)dx=2\cdot0.25=0.5$. $2\int_0^2 f(x)dx=1$... Key says 9.
Correct Answer: 9