Answer
$\dfrac{3}{2}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{1}_{0} \int^{2x}_{x/2} 1 \space dy \space dx+\int^{2}_{0} \int^{3-x}_{x/2}= \int^{1}_{0} [y]^{2x}_{x/2} dx+\int^{2}_{0} [y]^{3-x}_{x/2}$
or, $=\int^{1}_{0} (\dfrac{3}{2}x)dx+\int^{2}_{0} (3-\dfrac{3}{2}x)dx $
or, $=[\dfrac{3}{4} \times x^2]_0^1+[3x-\dfrac{3}{4} \times x^2]^2_1$
or, $=\dfrac{3}{2}$
