The area enclosed by the curves xy+4y=16 and x+y=6 is equal to:

Question

The area enclosed by the curves $xy+4y=16$ and $x+y=6$ is equal to:

Accepted Answer

$y=\dfrac{16}{x+4}$ and $y=6-x$. Setting equal: $16=(6-x)(x+4)\Rightarrow x^2-2x-8=0\Rightarrow x=4,-2$.

$$\text{Area}=\int_{-2}^{4}\left[(6-x)-\frac{16}{x+4}\right]dx=\left[6x-\frac{x^2}{2}-16\ln|x+4|\right]_{-2}^{4}$$

$$=(16-16\ln8)-(-14-16\ln2)=30-16(\ln8-\ln2)=30-16\ln4=30-32\ln2.$$

Solutions · 1