Answer
$$\frac{1}{2}\left( {{e^{4x + 9}} - {e^{4x}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^3 {y{e^{4x + {y^2}}}} dy \cr
& {\text{the notation }}dy{\text{ indicates integration with respect to }}y,{\text{ so we treat }}y{\text{ as the variable and}} \cr
& x{\text{ as a constant}}{\text{. then}} \cr
& \int_0^3 {y{e^{4x + {y^2}}}} dy = \frac{1}{2}\int_0^3 {{e^{4x + {y^2}}}\left( {2y} \right)} dy \cr
& {\text{using }}\int {{e^u}} du = {e^u} + C \cr
& = \frac{1}{2}\left( {{e^{4x + {y^2}}}} \right)_0^3 \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \frac{1}{2}\left( {{e^{4x + {{\left( 3 \right)}^2}}} - {e^{4x + {{\left( 0 \right)}^2}}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\left( {{e^{4x + 9}} - {e^{4x}}} \right) \cr} $$
