JEE Main 2024 (Online) 1st February Morning ShiftQuestion 1
The area enclosed by the curves $xy+4y=16$ and $x+y=6$ is equal to :
Application of Integrals
Solution
$y=\tfrac{16}{x+4}$ and $y=6-x$: $(6-x)(x+4)=16\Rightarrow x^2-2x-8=0\Rightarrow x=-2,4$ (line above curve between). Area $=\int_{-2}^4\left[6-x-\tfrac{16}{x+4}\right]dx=\left[6x-\tfrac{x^2}2\right]_{-2}^4-16\ln\tfrac82=30-16\ln4=30-32\ln2$. Answer: (B).