JEE Main 2019 (Online) 11th January Morning ShiftQuestion 1
The area (in sq. units) of the region bounded by the curve $x^{2}=4y$ and the straight line $x=4y-2$ is :
Application of Integrals
Solution
$y=\frac{x^2}{4}$ and $y=\frac{x+2}{4}$ meet at $x=-1,2$. Area $=\frac14\int_{-1}^{2}(x+2-x^2)dx=\frac98$.