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