JEE Main 2020 (Online) 3rd September Morning ShiftQuestion 1
The area (in sq. units) of the region $\{(x,y):0\le y\le x^2+1,\ 0\le y\le x+1,\ \frac12\le x\le2\}$ is :
Application of Integrals
Solution
$x^2+1\le x+1$ for $x\in[0,1]$, and $x+1\le x^2+1$ for $x\ge1$. So $y\le\min$ gives $A=\int_{1/2}^1(x^2+1)dx+\int_1^2(x+1)dx=\left(\frac7{24}+\frac12\right)+\frac52=\frac{19}{24}+\frac{60}{24}=\frac{79}{24}$. Option (D).