## Calculus with Applications (10th Edition)

$$\frac{{52}}{5}$$
\eqalign{ & \int_1^2 {\int_2^{2{x^2}} {ydy} dx} \cr & = \int_1^2 {\left[ {\int_2^{2{x^2}} {ydy} } \right]dx} \cr & {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr & = \int_2^{2{x^2}} {ydy} = \left[ {\frac{{{y^2}}}{2}} \right]_2^{2{x^2}} \cr & {\text{evaluating the limits in the variable }}y \cr & = \frac{{{{\left( {2{x^2}} \right)}^2}}}{2} - \frac{{{{\left( 2 \right)}^2}}}{2} \cr & {\text{simplifying}} \cr & = \frac{{4{x^4}}}{2} - \frac{4}{2} \cr & = 2{x^4} - 2 \cr & \cr & \int_1^2 {\left[ {\int_2^{2{x^2}} {ydy} } \right]dx} = \int_1^2 {\left( {2{x^4} - 2} \right)} dx \cr & {\text{integrating}} \cr & = \left[ {\frac{{2{x^5}}}{5} - 2x} \right]_1^2 \cr & = \left( {\frac{{2{{\left( 2 \right)}^5}}}{5} - 2\left( 2 \right)} \right) - \left( {\frac{{2{{\left( 1 \right)}^5}}}{5} - 2\left( 1 \right)} \right) \cr & = \frac{{44}}{5} - \left( { - \frac{8}{5}} \right) \cr & = \frac{{44 + 8}}{5} \cr & = \frac{{52}}{5} \cr}