Answer
$$ - 4$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\left( {1 - {{\left| x \right|}^3}} \right)} dx \cr
& {\text{Identify the symmetry in the function }}1 - {\left| x \right|^3},{\text{ evaluate }}f\left( { - x} \right) \cr
& f\left( { - x} \right) = 1 - {\left| { - x} \right|^3} \cr
& {\text{Where }}\left| { - x} \right| = \left| x \right| \cr
& f\left( { - x} \right) = 1 - {\left| x \right|^3} \cr
& f\left( { - x} \right) = f\left( x \right) \cr
& {\text{Therefore }}1 - {\left| x \right|^3}{\text{ is an even function}}{\text{.}} \cr
& {\text{Use the integral property }}\int_{ - a}^a {f\left( x \right)} dx = 2\int_0^2 {f\left( x \right)} dx,{\text{ }}f{\text{ is even}} \cr
& \int_{ - 2}^2 {\left( {1 - {{\left| x \right|}^3}} \right)} dx = 2\int_0^2 {\left( {1 - {{\left| x \right|}^3}} \right)} dx \cr
& {\text{Applying the absolute value definition}} \cr
& = 2\int_0^2 {\left( {1 - {x^3}} \right)} dx \cr
& {\text{Integrating}} \cr
& = 2\left[ {x - \frac{1}{4}{x^4}} \right]_0^2 \cr
& = 2\left[ {2 - \frac{1}{4}{{\left( 2 \right)}^4}} \right] - 2\left[ {0 - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr
& = - 4 \cr} $$
