Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_{ - {x^2}}^{{x^2}} x dydx} \cr
& \int_0^1 {\left[ {\int_{ - {x^2}}^{{x^2}} x dy} \right]dx} \cr
& \int_0^1 {x\left[ {\int_{ - {x^2}}^{{x^2}} {dy} } \right]dx} \cr
& {\text{Evaluate inner integral}} \cr
& \int_{ - {x^2}}^{{x^2}} {dy} = \left[ y \right]_{y = - {x^2}}^{y = {x^2}} \cr
& = {x^2} - \left( { - {x^2}} \right) \cr
& = 2{x^2} \cr
& {\text{Therefore,}} \cr
& \int_0^1 {x\left[ {\int_{ - {x^2}}^{{x^2}} {dy} } \right]dx} = \int_0^1 {x\left( {2{x^2}} \right)dx} \cr
& = \int_0^1 {2{x^3}dx} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {\frac{1}{2}{x^4}} \right]_0^1 \cr
& = \frac{1}{2}\left[ {{{\left( 1 \right)}^4} - {{\left( 0 \right)}^4}} \right] \cr
& = \frac{1}{2} \cr} $$
