Answer
$\frac{1-e}{6}.$
Work Step by Step
We have the integral
$$
\int_{1/2}^{0} \int_{0}^{\pi/6}e^{2y}\sin 3x d x d y=\left( \int_{0}^{\pi/6} \sin 3x d x \right)\left( \int_{1/2}^{0} e^{2y} d y\right)\\
=\left( - \frac{1}{3} \cos3x \right)_{0}^{\pi/6}\left( \frac{1}{2}e^{2y} \right)_{1/2}^{0} \\=(1/3)(1/2)(1-e)=\frac{1-e}{6}.
$$
