Answer
$$\frac{1}{2}\left( {{e^{12 + 3y}} - {e^{4 + 3y}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_2^6 {{e^{2x + 3y}}} dx \cr
& {\text{the notation }}dx{\text{ indicates integration with respect to }}x,{\text{ so we treat }}x{\text{ as the variable and}} \cr
& y{\text{ as a constant}}{\text{. then}} \cr
& \int_2^6 {{e^{2x + 3y}}} dx = \frac{1}{2}\int_2^6 {{e^{2x + 3y}}} \left( 2 \right)dx \cr
& {\text{using }}\int {{e^u}} du = {e^u} + C \cr
& = \frac{1}{2}\left( {{e^{2x + 3y}}} \right)_2^6 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \frac{1}{2}\left( {{e^{2\left( 6 \right) + 3y}} - {e^{2\left( 2 \right) + 3y}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\left( {{e^{12 + 3y}} - {e^{4 + 3y}}} \right) \cr} $$
