Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {{\rm{e}}^{3x + 4y}}{\rm{d}}A \simeq 4654.26$
Work Step by Step
We have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {{\rm{e}}^{3x + 4y}}{\rm{d}}A$ and ${\cal R} = \left[ {0,1} \right] \times \left[ {1,2} \right]$.
Write
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {{\rm{e}}^{3x + 4y}}{\rm{d}}A = \mathop \smallint \limits_{y = 1}^2 \left( {\mathop \smallint \limits_{x = 0}^1 {{\rm{e}}^{3x}}{\rm{d}}x} \right){{\rm{e}}^{4y}}{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 1}^2 \left( {\frac{1}{3}{{\rm{e}}^{3x}}|_0^1} \right){{\rm{e}}^{4y}}{\rm{d}}y$
$ = \frac{1}{3}\left( {{{\rm{e}}^3} - 1} \right)\mathop \smallint \limits_{y = 1}^2 {{\rm{e}}^{4y}}{\rm{d}}y$
$ = \frac{1}{3}\left( {{{\rm{e}}^3} - 1} \right)\left( {\frac{1}{4}{{\rm{e}}^{4y}}|_1^2} \right)$
$ = \frac{1}{{12}}\left( {{{\rm{e}}^3} - 1} \right)\left( {{{\rm{e}}^8} - {{\rm{e}}^4}} \right)$
$ = \frac{1}{{12}}\left( {{{\rm{e}}^{11}} - {{\rm{e}}^8} - {{\rm{e}}^7} + {{\rm{e}}^4}} \right)$
$ \simeq 4654.26$
