## Calculus with Applications (10th Edition)

$$\frac{{32544}}{{35}}$$
\eqalign{ & \int_2^4 {\int_2^{{x^2}} {\left( {{x^2} + {y^2}} \right)} dydx} \cr & = \int_2^4 {\left[ {\int_2^{{x^2}} {\left( {{x^2} + {y^2}} \right)} dy} \right]dx} \cr & {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr & = \int_2^{{x^2}} {\left( {{x^2} + {y^2}} \right)} dy \cr & {\text{integrate by using the power rule }} \cr & = \left[ {{x^2}y + \frac{{{y^3}}}{3}} \right]_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)}^3}}}{3}} \right) - \left( {{x^2}\left( 2 \right) + \frac{{{{\left( 2 \right)}^3}}}{3}} \right) \cr & {\text{simplifying}} \cr & = \left( {{x^4} + \frac{{{x^6}}}{3}} \right) - \left( {2{x^2} + \frac{8}{3}} \right) \cr & = {x^4} + \frac{{{x^6}}}{3} - 2{x^2} - \frac{8}{3} \cr & \cr & \int_2^4 {\left[ {\int_2^{{x^2}} {\left( {{x^2} + {y^2}} \right)} dy} \right]dx} = \int_2^4 {\left( {{x^4} + \frac{{{x^6}}}{3} - 2{x^2} - \frac{8}{3}} \right)dx} \cr & {\text{integrating by the power rule}} \cr & = \left( {\frac{{{x^5}}}{5} + \frac{{{x^7}}}{{21}} - \frac{{2{x^3}}}{3} - \frac{8}{3}x} \right)_2^4 \cr & {\text{evaluate}} \cr & = \left( {\frac{{{4^5}}}{5} + \frac{{{4^7}}}{{21}} - \frac{{2{{\left( 4 \right)}^3}}}{3} - \frac{8}{3}\left( 4 \right)} \right) - \left( {\frac{{{2^5}}}{5} + \frac{{{2^7}}}{{21}} - \frac{{2{{\left( 2 \right)}^3}}}{3} - \frac{8}{3}\left( 2 \right)} \right) \cr & {\text{simplifying by using a calculator}} \cr & = \frac{{32608}}{{35}} - \frac{{64}}{{35}} \cr & = \frac{{32544}}{{35}} \cr}