JEE Main 2019 (Online) 9th January Morning ShiftQuestion 1
The area (in sq. units) bounded by the parabola $y=x^2-1$, the tangent at the point $(2,3)$ to it and the $y$-axis is
Application of Integrals
Solution
Tangent at $(2,3)$: $y=4x-5$. Area $=\int_0^2(x^2-1-4x+5)dx=\int_0^2(x-2)^2dx=\frac83$.