The area of the region enclosed between the parabolas $y^2=x+1$ and $y^2=-x+1$ is :
Answer & explanation
Correct answer: option 1
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