The area (in sq. units) of the region bounded by the curves x ^ 2 +...

The given curves are ${x}^{2} + 2y - 1 = 0, {y}^{2} + 4x - 4 = 0$ and ${y}^{2} - 4x - 4 = 0 .$

Drawing their graphs, we get

The required area is shown by the shaded part and since the curves ${y}^{2} + 4x - 4 = 0$ and ${y}^{2} - 4x - 4 = 0$ are symmetric about the $y$-axis, hence they have the same area.

Hence, the required area$= 2\left[{∫}_{0}^{2}\left(\dfrac{4 - {y}^{2}}{4}\right)dy - {∫}_{0}^{1}\left(\dfrac{1 - {x}^{2}}{2}\right)dx\right]$

$= 2\left[{∫}_{0}^{2}\dfrac{1}{4}\left(4 - {y}^{2}\right)dy - {∫}_{0}^{1}\dfrac{1}{2}\left(1 - {x}^{2}\right)dx\right]$

$= 2\left[\dfrac{1}{4}{\left(4y - \dfrac{{y}^{3}}{3}\right)}_{0}^{2} - \dfrac{1}{2}{\left(x - \dfrac{{x}^{3}}{3}\right)}_{0}^{1}\right]$

$= 2\left[\dfrac{1}{4}\left(8 - \dfrac{8}{3} - 0\right) - \dfrac{1}{2}\left(1 - \dfrac{1}{3} - 0\right)\right]$

$= 2\left[\dfrac{4}{3} - \dfrac{1}{3}\right] = 2sq . units$ .