Area Under Curves Question 1
Question 1 - 2024 (01 Feb Shift 1)
The area enclosed by the curves $x y+4 y=16$ and $x+y=6$ is equal to :
(1) $28-30 \log _{e} 2$
(2) $30-28 \log _{e} 2$
(3) $30-32 \log _{e} 2$
(4) $32-30 \log _{e} 2$
Show Answer
Answer (3)
Solution
$x y+4 y=16 \quad, \quad x+y=6$
$y(x+4)=16$
$\mathrm{x}+\mathrm{y}=6$
on solving, (1) $\backslash &$ (2)
we get $x=4, x=-2$
$$ \begin{gathered} \text { Area }=\int_{-2}^{4}\left((6-x)-\left(\frac{16}{x+4}\right)\right) d x \ =30-32 \ln 2 \end{gathered} $$
