The area of the region enclosed between the parabolas $y^2=x+1$ and $y^2=-x+1$ is :

$\frac{8}{3}$

The correct answer is Option (1) → $\frac{8}{3}$

By symmetry

Area I = Area II = Area III = Area IV

Intersection y axis at y = 1

intersection x axis at x = 1

so Area I = ?

$=\int\limits_0^1\sqrt{1-x}dx$

total area = $4\int\limits_0^1\sqrt{1-x}dx$

alternatively,

Total area = $4\int\limits_0^11-y^2dy$

$=4[y-\frac{y^3}{3}]_0^1=\frac{8}{3}$ sq.units