Answer
$\dfrac{4}{3}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{2}_{0} \int^{y-y^2}_{-y} \space dx \space dy =\int^{2}_{0}[ x]^{y-y^2}_{-y} \space dy $
or, $=\int^{2}_{0}(2y-y^2)\space dy $
or, $=[y^2-\dfrac{y^3}{3}]^2_0$
or, $=\dfrac{12-8}{3}$
or, $=\dfrac{4}{3}$
