The area (in sq. units) of the region A = \ (x, y): x ^ 2 ≤ y ≤ x +...

The two given curves are $y = x + 2$ and $y = {x}^{2}$

To find their points of intersection, we equate the value of $y$ to get

${x}^{2} = x + 2$

$⇒ {x}^{2} - x - 2 = 0$

$⇒ \left(x - 2\right)\left(x + 1\right) = 0$

$⇒ x = 2$ or $x = - 1 .$

The graphs of the curves are given below.

Required area is the area of the shaded portion.

Area $= {∫}_{- 1}^{2}\left({y}_{line} - {y}_{parabola}\right)dx$
$= {∫}_{- 1}^{2}\left(x + 2 - {x}^{2}\right)dx$

Using $∫ {x}^{n}dx = \dfrac{{x}^{n + 1}}{n + 1}$

$= {\left[\dfrac{{x}^{2}}{2} + 2x - \dfrac{{x}^{3}}{3}\right]}_{- 1}^{2}$

$= \left(2 + 4 - \dfrac{8}{3}\right) - \left(\dfrac{1}{2} - 2 + \dfrac{1}{3}\right)$

$= 6 - \dfrac{8}{3} - \dfrac{1}{2} + 2 - \dfrac{1}{3}$

$= 8 - 3 - \dfrac{1}{2}$

$= \dfrac{9}{2}$ sq units.