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

We have to find the area of the region $A = \left\{\left(x, y\right): \dfrac{{y}^{2}}{2} \le x \le y + 4\right\}$

Now, to find the point of intersection of the curves $x = \dfrac{{y}^{2}}{2}$ and $x = y + 4,$ we equate the value of $x,$ to get

$\dfrac{{y}^{2}}{2} = y + 4$

$⇒ {y}^{2} = 2y + 8$

$⇒ {y}^{2} - 2y - 8 = 0$

$⇒ \left(y - 4\right)\left(y + 2\right) = 0$

$⇒ y = 4$ or $y = - 2$

And, the graph of the given functions is as

The shaded area is the required area, and

Area$= {∫}_{- 2}^{4}\left({x}_{line} - {x}_{parabola}\right)dy$

$= {∫}_{- 2}^{4}\left(y + 4 - \dfrac{{y}^{2}}{2}\right)dy$

Now, using $∫ {x}^{n}dx = \dfrac{{x}^{n + 1}}{n + 1},$ we get

Area= {\left\left(\dfrac{{y}^{2}}{2} + 4y - \dfrac{{y}^{3}}{6}\right)\right|}_{- 2}^{4}

$= \left(8 + 16 - \dfrac{32}{3}\right) - \left(2 - 8 + \dfrac{4}{3}\right)$

$= 18$ sq units.