JEE Main 2025 (Online) 22nd January Evening ShiftQuestion 1
The area of the region enclosed by the curves $y=x^{2}-4x+4$ and $y^{2}=16-8x$ is :
Application of Integrals
Solution
The curves meet at $(0,4)$ and $(2,0)$. In terms of $y$: $\int_0^4\left(\frac{16-y^2}8-(2-\sqrt y)\right)dy=-\frac83+\frac{16}3=\frac83$.