JEE Main 2017 (Offline)Question 1
The area (in sq. units) of the region $\{(x,y):x\ge0,\ x+y\le3,\ x^2\le4y\text{ and }y\le1+\sqrt x\}$ is :
Application of Integrals
Solution
Lower boundary $y=\frac{x^2}4$; upper is $1+\sqrt x$ on $[0,1]$ and $3-x$ on $[1,2]$ (the line meets $y=1+\sqrt x$ at $x=1$ and the parabola at $x=2$). Area $=\int_0^1\left(1+\sqrt x-\frac{x^2}4\right)dx+\int_1^2\left(3-x-\frac{x^2}4\right)dx=\frac{19}{12}+\frac{11}{12}=\frac52$.