## Calculus with Applications (10th Edition)

$$\frac{4}{5}$$
\eqalign{ & \iint\limits_R {\left( {{x^2} - y} \right)}dydx;\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 \leqslant x \leqslant 1,\,\,\,\,\,\, - {x^2} \leqslant y \leqslant {x^2} \cr & {\text{Replacing the limits for the region }}R \cr & \int_{ - 1}^1 {\int_{ - {x^2}}^{{x^2}} {\left( {{x^2} - y} \right)dy} dx} \cr & = \int_{ - 1}^1 {\left[ {\int_{ - {x^2}}^{{x^2}} {\left( {{x^2} - y} \right)dy} } \right]dx} \cr & {\text{solve the inner integral 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 & = \left( {{x^4} - \frac{{{x^4}}}{2}} \right) - \left( { - {x^4} - \frac{{{x^4}}}{2}} \right) \cr & = \frac{{{x^4}}}{2} + \frac{{3{x^4}}}{2} \cr & = 2{x^4} \cr & {\text{then}} \cr & \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 & = \left[ {\frac{{2{x^5}}}{5}} \right]_{ - 1}^1 \cr & {\text{evaluating}} \cr & = \frac{2}{5}\left( {{{\left( 1 \right)}^5} - {{\left( { - 1} \right)}^5}} \right) \cr & {\text{simplifying}} \cr & = \frac{2}{5}\left( 2 \right) \cr & = \frac{4}{5} \cr}