The area (in sq. units) of the region x ∈ R: x ≥ 0, y ≥ 0, y ≥ x -...

$y = \sqrt{x}$

$\sqrt{x} = x - 2$

$( \sqrt{x}{)}^{2} - ( \sqrt{x} ) - 2 = 0$

$( \sqrt{x} - 2 ) ( \sqrt{x} + 1 ) = 0$

$\sqrt{x} = 2 \to x = 4$
Therefore, the area of the region is
{∫}_{0}^{4}\sqrt{x}dx - Areaof ∆ ABC

$= \dfrac{2}{3}{x}^{3 / 2}{}_{0}^{4} - \dfrac{1}{2} × 2 × 2$

$= \dfrac{2}{3} × {4}^{3 / 2} - 2$
$= \dfrac{16}{3} - 2 = \dfrac{10}{3}sq . unit$