Answer
$\dfrac{9}{2}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{3}_{0} \int^{2x-x^2}_{-x} \space dy \space dx = \int^{3}_{0}(3x-x^2)dx $
or, $=[\dfrac{3x^2}{2}-\dfrac{x^3}{3}]^3_0$
or, $=\dfrac{27}{2}-9$
or, $=\dfrac{9}{2}$
