Answer
$\mathop \smallint \limits_1^2 \mathop \smallint \limits_2^4 {{\rm{e}}^{3x - y}}{\rm{d}}y{\rm{d}}x \simeq 14.9529$
Work Step by Step
Write $\mathop \smallint \limits_1^2 \mathop \smallint \limits_2^4 {{\rm{e}}^{3x - y}}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 1}^2 \left( {\mathop \smallint \limits_{y = 2}^4 {{\rm{e}}^{3x - y}}{\rm{d}}y} \right){\rm{d}}x$.
Step 1. We hold $x$ constant and evaluate the inner integral with respect to $y$
$S\left( x \right) = \mathop \smallint \limits_{y = 2}^4 {{\rm{e}}^{3x - y}}{\rm{d}}y = - {{\rm{e}}^{3x}}\left( {{{\rm{e}}^{ - y}}} \right)|_2^4 = - {{\rm{e}}^{3x}}\left( {{{\rm{e}}^{ - 4}} - {{\rm{e}}^{ - 2}}} \right)$
$S\left( x \right) = \left( {{{\rm{e}}^{ - 2}} - {{\rm{e}}^{ - 4}}} \right){{\rm{e}}^{3x}}$
Step 2. Integrate $S\left( x \right)$ with respect to $x$
Thus,
$\mathop \smallint \limits_1^2 \left( {\mathop \smallint \limits_2^4 {{\rm{e}}^{3x - y}}{\rm{d}}y} \right){\rm{d}}x = \left( {{{\rm{e}}^{ - 2}} - {{\rm{e}}^{ - 4}}} \right)\mathop \smallint \limits_{x = 1}^2 {{\rm{e}}^{3x}}{\rm{d}}x$
$ = \left( {\frac{{{{\rm{e}}^{ - 2}} - {{\rm{e}}^{ - 4}}}}{3}} \right){{\rm{e}}^{3x}}|_1^2$
$ = \left( {\frac{{{{\rm{e}}^{ - 2}} - {{\rm{e}}^{ - 4}}}}{3}} \right)\left( {{{\rm{e}}^6} - {{\rm{e}}^3}} \right)$
$ = \frac{1}{3}\left( {{{\rm{e}}^4} - {{\rm{e}}^2} - {\rm{e}} + {{\rm{e}}^{ - 1}}} \right)$
$ = \frac{1}{{3{\rm{e}}}}\left( {{{\rm{e}}^5} - {{\rm{e}}^3} - {{\rm{e}}^2} + 1} \right)$
$ \simeq 14.9529$
