Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\left( {1 - \left| x \right|} \right)} dx \cr
& {\text{Let }}f\left( x \right) = 1 - \left| x \right|{\text{ and evaluate }}f\left( { - x} \right) \cr
& f\left( { - x} \right) = 1 - \left| { - x} \right| \cr
& f\left( { - x} \right) = 1 - \left| x \right| \cr
& f\left( { - x} \right) = f\left( x \right),{\text{ then the function is even}} \cr
& \cr
& {\text{Use the property }}\int_{ - a}^a {f\left( x \right)} dx = 2\int_0^a {f\left( x \right)dx} ,\,\,\,{\text{ If }}f{\text{ is even}} \cr
& \int_{ - 1}^1 {\left( {1 - \left| x \right|} \right)} dx = 2\int_0^1 {\left( {1 - \left| x \right|} \right)} dx \cr
& = 2\int_0^1 {dx} - 2\int_0^1 {\left| x \right|} dx \cr
& {\text{Integrating}} \cr
& = 2\int_0^1 {dx} - 2\int_0^1 x dx \cr
& = \left[ {2x - {x^2}} \right]_0^1 \cr
& = 2\left( 1 \right) - {\left( 1 \right)^2} \cr
& = 1 \cr} $$
