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