## Calculus (3rd Edition)

$$3$$
Given $$y=8-3 x, \quad y=6-x, \quad y=2$$ we find the intersection points $$8-3x = 6-x\to x=1$$ and $y=5$; we solve with respect to $y$, since $$6-y\geq \frac{1}{3}(8-y) \ \ \text{ for }\ \ 2\leq y\leq 5$$ Hence \begin{align*} A&= \int_{2}^{5}[(6-y)-\frac{1}{3}(8-y)]dy\\ &=\left(6y-\frac{1}{2}y^2\right)-\frac{1}{3}\left(8y-\frac{1}{2}y^2\right)\bigg|_{2}^{5}\\ &=3 \end{align*} We use the Mathematica program to plot the region.