JEE Main 2025 (Online) 7th April Evening ShiftQuestion 1
If the area of the region $\{(x,y):1+x^{2}\le y\le\min\{x+7,11-3x\}\}$ is $A$, then $3A$ is equal to
Application of Integrals
Solution
$1+x^2=x+7\Rightarrow x=-2,3$; $1+x^2=11-3x\Rightarrow x=-5,2$; $x+7=11-3x$ at $x=1$. Region: $x\in[-2,2]$, upper curve $x+7$ for $x\le1$, $11-3x$ for $x\ge1$. $A=\int_{-2}^1(6+x-x^2)dx+\int_1^2(10-3x-x^2)dx=\frac{27}{2}+\frac{19}{6}=\frac{50}{3}$. $3A=50$. Option (A).