Answer
$$4$$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{2}_{0} \int^{3y}_{y} (1) \space dx \space dy = \int^{2}_{0}[x]^{3y}_y \space dy $
or, $=\int^{2}_{0}(2y)dy $
or, $=[y^2]^2_0$
So, $ Area=4$
