## Calculus with Applications (10th Edition)

$$\frac{1}{2}\left( {{e^7} - {e^6} - {e^3} + {e^2}} \right)$$
\eqalign{ & \iint\limits_R {y{e^{x + {y^2}}}}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \leqslant x \leqslant 3,\,\,\,\,\,\,\,0 \leqslant y \leqslant 2 \cr & {\text{replacing the limits for the region }}R \cr & \int_0^2 {\int_2^3 {y{e^{x + {y^2}}}dx} dy} \cr & = \int_0^2 {\left[ {\int_2^3 {y{e^{x + {y^2}}}dx} } \right]dy} \cr & {\text{solve the inner integral, treating }}y{\text{ as a constant}} \cr & = y\int_2^3 {{e^{x + {y^2}}}dx} \cr & {\text{integrate by using the rule }}\int {{e^u}} du = {e^u} + C \cr & = y\left[ {{e^{x + {y^2}}}} \right]_2^3 \cr & {\text{evaluating the limits in the variable }}x \cr & = y\left[ {{e^{3 + {y^2}}} - {e^{2 + {y^2}}}} \right] \cr & {\text{simplifying}} \cr & = y{e^{3 + {y^2}}} - y{e^{2 + {y^2}}} \cr & \cr & \int_0^2 {\left[ {\int_2^3 {y{e^{x + {y^2}}}dx} } \right]dy} = \int_0^2 {\left( {y{e^{3 + {y^2}}} - y{e^{2 + {y^2}}}} \right)dy} \cr & = \int_0^2 {y{e^{3 + {y^2}}}dy} - \int_0^2 {y{e^{2 + {y^2}}}dy} \cr & = \frac{1}{2}\int_0^2 {2y{e^{3 + {y^2}}}dy} - \frac{1}{2}\int_0^2 {2y{e^{2 + {y^2}}}dy} \cr & {\text{integrating}} \cr & = \frac{1}{2}\left( {{e^{3 + {y^2}}} - {e^{2 + {y^2}}}} \right)_0^2 \cr & {\text{evaluate}} \cr & = \frac{1}{2}\left( {{e^{3 + {2^2}}} - {e^{2 + {2^2}}}} \right) - \frac{1}{2}\left( {{e^{3 + {0^2}}} - {e^{2 + {0^2}}}} \right) \cr & {\text{simplifying}} \cr & = \frac{1}{2}\left( {{e^7} - {e^6}} \right) - \frac{1}{2}\left( {{e^3} - {e^2}} \right) \cr & = \frac{1}{2}\left( {{e^7} - {e^6} - {e^3} + {e^2}} \right) \cr}