JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Area Under Curves

For $|x|\le 2$ the outer boundary is $y=4-x^2$ and the inner boundary is $y=x^2$, and the two parabolas meet where $x^2=4-x^2$, that is at $x=\pm \sqrt{2}$. The line $y=1$ meets $y=x^2$ at $x=\pm 1$, so the lower boundary is $y=1$ for $|x|\le 1$ and $y=x^2$ for $1\le |x|\le \sqrt{2}$. By symmetry about the $y$-axis, $A=2\left[{\int }_0^1((4-x^2)-1)dx+{\int }_1^{\sqrt{2}}((4-x^2)-x^2)dx\right]$. The first integral is ${\int }_0^1(3-x^2)dx=\frac{8}{3}$, and the second is ${\int }_1^{\sqrt{2}}(4-2x^2)dx=\frac{8\sqrt{2}-10}{3}$. Hence $A=2\cdot \frac{8\sqrt{2}-2}{3}=\frac{4}{3}(4\sqrt{2}-1)$.