JEE Main 2025 (Online) 29th January Morning ShiftQuestion 1
Let the area of the region $\{(x,y):2y\le x^2+3,\ y+|x|\le3,\ y\ge|x-1|\}$ be $A$. Then $6A$ is equal to :
Application of Integrals
Solution
Vertices $(-1,2),(1,2),(2,1)$. Area $=\int_{-1}^1\left(\frac{x^2+3}2-(1-x)\right)dx+\int_1^2(4-2x)dx=\frac43+1=\frac73$. So $6A=14$.