The minimum of two curves

Let this Integral be $I$

now as shown in The graph By symmetry This Integral is equal to

$Inetegral\quad =\quad 4\quad \times \quad { Area }_{ shaded\quad tringle }\\ \quad \quad \quad \quad \quad \quad \quad \quad =\quad 4\quad \times \quad (\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times 1)\\ \quad \quad \quad \quad \quad \quad \quad \quad =\quad 1\quad$ .

graph

The integrand is

$\large min\{\{x\},\{-x\}\}$

Well if we analyze its graph between $0$ to $1$ , it will sort of like be an isosceles triangle with height $=\frac{1}{2}$ & base equal to $1$ unit. Since it's a periodic function, there will be a total of four such triangles in $[-2,2]$ .

Thus the required definite integral equal to the total area of these triangles which is obviously $4(\frac{1}{2}) (\frac{1}{2})=\boxed 1$