JEE Main 2023 (Online) 30th January Evening ShiftQuestion 1
Let $A$ be the area of the region $\left\{(x,y):y\ge x^{2},\ y\ge(1-x)^{2},\ y\le2x(1-x)\right\}$. Then $540A$ is equal to
Application of Integrals
Answer$25$
Solution
Curves meet at $x=\frac13$ ($(1-x)^2$ and $2x(1-x)$), $x=\frac12$ ($x^2$ and $(1-x)^2$), $x=\frac23$. By symmetry $A=2\int_{1/2}^{2/3}(2x-3x^2)dx=2\left[x^2-x^3\right]_{1/2}^{2/3}=\frac5{108}$, so $540A=25$.