Answer
$$\frac{1}{{14}}\left( {{e^{ - 8}} - {e^{ - 4}} - {e^{ - 1}} + {e^3}} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\int_3^5 {{e^{2x - 7y}}dx} dy} \cr
& = \int_1^2 {\left[ {\int_3^5 {{e^{2x - 7y}}dx} } \right]dy} \cr
& {\text{solve the inner integral}} \cr
& \int_3^5 {{e^{2x - 7y}}dx} = \frac{1}{2}\int_3^5 {{e^{2x - 7y}}\left( 2 \right)dx} \cr
& = \frac{1}{2}\left[ {{e^{2x - 7y}}} \right]_3^5 \cr
& {\text{evaluate limits for }}x \cr
& = \frac{1}{2}\left[ {{e^{2\left( 5 \right) - 7y}} - {e^{2\left( 3 \right) - 7y}}} \right] \cr
& = \frac{1}{2}\left[ {{e^{10 - 7y}} - {e^{6 - 7y}}} \right] \cr
& {\text{then}} \cr
& \int_1^2 {\int_3^5 {{e^{2x - 7y}}dx} dy} = \int_1^2 {\frac{1}{2}\left[ {{e^{10 - 7y}} - {e^{6 - 7y}}} \right]dy} \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left[ {\frac{{{e^{10 - 7y}}}}{{ - 7}} - \frac{{{e^{6 - 7y}}}}{{ - 7}}} \right]_1^2 \cr
& = \frac{1}{{14}}\left[ {{e^{6 - 7y}} - {e^{10 - 7y}}} \right]_1^2 \cr
& {\text{evaluating}} \cr
& = \frac{1}{{14}}\left[ {{e^{6 - 7\left( 2 \right)}} - {e^{10 - 7\left( 2 \right)}}} \right] - \frac{1}{{14}}\left[ {{e^{6 - 7\left( 1 \right)}} - {e^{10 - 7\left( 1 \right)}}} \right] \cr
& = \frac{1}{{14}}\left( {{e^{ - 8}} - {e^{ - 4}}} \right) - \frac{1}{{14}}\left( {{e^{ - 1}} - {e^3}} \right) \cr
& = \frac{1}{{14}}\left( {{e^{ - 8}} - {e^{ - 4}} - {e^{ - 1}} + {e^3}} \right) \cr} $$
