Answer
$$\frac{4}{5}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\int_{ - {x^2}}^{{x^2}} {\left( {{x^2} - y} \right)} dydx} \cr
& = \int_{ - 1}^1 {\left[ {\int_{ - {x^2}}^{{x^2}} {\left( {{x^2} - y} \right)} dy} \right]dx} \cr
& {\text{solve the inner integral and treat }}x{\text{ as a constant}} \cr
& = \int_{ - {x^2}}^{{x^2}} {\left( {{x^2} - y} \right)} dy \cr
& = \left( {{x^2}y - \frac{{{y^2}}}{2}} \right)_{ - {x^2}}^{{x^2}} \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \left( {{x^2}\left( {{x^2}} \right) - \frac{{{{\left( {{x^2}} \right)}^2}}}{2}} \right) - \left( {{x^2}\left( { - {x^2}} \right) - \frac{{{{\left( { - {x^2}} \right)}^2}}}{2}} \right) \cr
& {\text{simplifying}} \cr
& = \left( {{x^4} - \frac{{{x^4}}}{2}} \right) - \left( { - {x^4} - \frac{{{x^4}}}{2}} \right) \cr
& = \frac{{{x^4}}}{2} + \frac{3}{2}{x^4} \cr
& = 2{x^4} \cr
& {\text{then}} \cr
& \int_{ - 1}^1 {\left[ {\int_{ - {x^2}}^{{x^2}} {\left( {{x^2} - y} \right)} dy} \right]dx} = \int_{ - 1}^1 {2{x^4}dx} \cr
& {\text{integrating}} \cr
& = 2\left( {\frac{{{x^5}}}{5}} \right)_{ - 1}^1 \cr
& = \frac{2}{5}\left( {{{\left( 1 \right)}^3} - {{\left( { - 1} \right)}^3}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{2}{5}\left( 2 \right) \cr
& = \frac{4}{5} \cr} $$
