JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Area Under Curves

Write both curves as $x$ in terms of $y$: $x=-3y^2$ and $x=1-4y^2$. They meet where $-3y^2=1-4y^2$, that is $y^2=1$, so $y=\pm 1$. For a fixed $y$ in $[-1,1]$ the right boundary is $x=1-4y^2$ and the left boundary is $x=-3y^2$, their horizontal gap being $(1-4y^2)-(-3y^2)=1-y^2\ge 0$. Hence the area is ${\int }_{-1}^1(1-y^2)dy=2{\int }_0^1(1-y^2)dy=2{\left[y-\frac{y^3}{3}\right]}_0^1=2\cdot \frac{2}{3}$. The required area is $\frac{4}{3}$.