Answer
$\dfrac{13}{3}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{1}_{0} \int^{\sqrt{x}}_{-x} (1) \space dy \space dx +\int^{4}_{0} \int^{\sqrt{x}}_{x-2} (1) \space dy \space dx $
or, $=\int^{1}_{0} [y]^{\sqrt{x}}_{-x}dx+\int^{4}_{0} [y]^{\sqrt{x}}_{x-2}dx $
or, $=\int^{1}_{0} (\sqrt{x}+x)dx+\int^{4}_{0} (\sqrt{x}-x+2)dx $
or, $=[\dfrac{2x^{3/2}}{3}+\dfrac{x^2}{2}]^1_0+[\dfrac{2x^{3/2}}{3}-\dfrac{x^2}{2}+2x]^4_1$
or, $ Area=\dfrac{13}{3}$
