Let $$f\left( x \right) = $$   maximum $$\left\{ {x + \left| x \right|,\,x - \left[ x \right]} \right\},$$     where $$\left[ x \right] = $$  the greatest integer $$ \leqslant x.$$  Then $$\int_{ - 2}^2 {f\left( x \right)dx} $$    is equal to :

Solution :
$$\left| x \right| \geqslant - \left[ x \right]$$    for all positive $$x,$$ and $$\left| x \right| \leqslant - \left[ x \right]$$    for all negative $$x.$$
$$\eqalign{ & \therefore \int_{ - 2}^2 {f\left( x \right)dx} = \int_{ - 2}^0 {\left( {x - \left[ x \right]} \right)dx} + \int_0^2 {\left( {x + \left| x \right|} \right)dx} \cr & = \int_{ - 2}^0 {x\,dx} - \int_{ - 2}^0 {\left[ x \right]dx} + \int_0^2 {2x\,dx} \cr & = \left[ {\frac{{{x^2}}}{2}} \right]_{ - 2}^0 - \left\{ {\int_{ - 2}^{ - 1} {\left[ x \right]dx} + \int_{ - 1}^0 {\left[ x \right]dx} } \right\} + \left[ {{x^2}} \right]_0^2 \cr & = 0 - 2 - \int_{ - 2}^{ - 1} { - 2\,dx} - \int_{ - 1}^0 { - 1\,dx + 4} \cr & = - 2 + 2\left[ x \right]_{ - 2}^{ - 1} + \left[ x \right]_{ - 1}^0 + 4 \cr & = - 2 + 2\left( { - 1 + 2} \right) + \left( {0 + 1} \right) + 4 \cr & = 5 \cr} $$