$$\int_{ - 2}^2 {\left| {1 - {x^2}} \right|dx} $$    is :

Solution :
$$\eqalign{ & \left| {1 - {x^2}} \right|{\text{ is an even function}}{\text{.}} \cr & \therefore I = 2\int_0^2 {\left| {1 - {x^2}} \right|dx} \cr & \therefore I = 2\int_0^1 {\left( {1 - {x^2}} \right)dx} + 2\int_1^2 {\left( {{x^2} - 1} \right)dx} \cr & = 2.\left[ {x - \frac{{{x^3}}}{3}} \right]_0^1 + 2.\left[ {\frac{{{x^3}}}{3} - x} \right]_1^2 \cr & = 2\left\{ {\left( {1 - \frac{1}{3}} \right) - 0} \right\} + 2\left\{ {\left( {\frac{8}{3} - 2} \right) - \left( {\frac{1}{3} - 1} \right)} \right\} \cr & = \frac{4}{3} + \frac{8}{3} \cr & \,\,\,\,\,\,\,\,\,\, = 4 \cr} $$