JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Area Under Curves

In the upper half plane the curves are $y=\frac{1-x^2}{2}$, meeting the $x$-axis at $(\pm 1,0)$, together with $y^2=4(1-x)$ and $y^2=4(x+1)$, whose vertices are $(1,0)$ and $(-1,0)$. The two sideways parabolas meet where $1-\frac{y^2}{4}=\frac{y^2}{4}-1$, that is at $(0,2)$, so the enclosed region has corners $(-1,0)$, $(1,0)$ and $(0,2)$, with the downward parabola below and the two sideways parabolas above. The figure is symmetric about the $y$-axis, and for $x\ge 0$ the upper boundary is $y=2\sqrt{1-x}$, so

The region has area $2$ square units.