Let $f(x)=\max \{x+|x|, x-[x]\}$, where $[x]$ denotes the greatest integer less than or equal to $x$, then $\int\limits_{-2}^2 f(x) d x$ is equal to

5

Since $x+|x|>x-[x]$ for all $x>0$ and, $x+|x|=0 \leq x-[x]$ for all $x<0$.

∴  $\int\limits_{-2}^2 f(x) d x=\int\limits_{-2}^0(x-[x]) d x+\int\limits_0^2(x+|x|) d x$

$\Rightarrow \int\limits_{-2}^2 f(x) d x=2 \times \frac{1}{2}+\int\limits_0^2 2 x d x=1+4=5$