SOLUTION 11: Compute the area of the region enclosed by the graphs of the equations $y=x$, $y=2x$, and $y=6-x$ . Begin by finding the points of intersection of the three graphs. From $y=x$ and $y=2x$ we get that $$x = 2x \ \ \longrightarrow \ \ -x = 0 \ \ \longrightarrow \ \ x = 0$$ From $y=x$ and $y=6-x$ we get that $$x = 6-x \ \ \longrightarrow \ \ 2x = 6 \ \ \longrightarrow \ \ \ x = 3$$ From $y=2x$ and $y=6-x$ we get that $$2x = 6-x \ \ \longrightarrow \ \ 3x = 6 \ \ \longrightarrow \ \ x = 2$$ Now see the given graph of the enclosed region.

Using vertical cross-sections to describe this region, which is made up of two smaller regions, we get that $$0 \le x \le 2 \ \ and \ \ x \le y \le 2x$$ in addition to $$2 \le x \le 3 \ \ and \ \ x \le y \le 6-x \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{2} (Top \ - \ Bottom) \ dx + \int_{2}^{3} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{0}^{2} (2x - x) \ dx + \int_{2}^{3} ((6-x) - x) \ dx }$$ $$= \displaystyle { \int_{0}^{2} x \ dx + \int_{2}^{3} (6-2x) \ dx }$$ $$= \displaystyle { \frac{x^{2}}{2} \Big\vert_{0}^{2} + \Big( 6x - x^{2} \Big) \Big\vert_{2}^{3} }$$ $$= \displaystyle { \Big( \frac{2^{2}}{2} - \frac{0^{2}}{2} \Big) + \Big( (6 \cdot 3 - 3^{2}) - (6 \cdot 2 - 2^{2}) \Big) }$$ $$= \displaystyle { \Big( 2 - 0 \Big) + \Big( 9 - 8 \Big) }$$ $$= \displaystyle { 2 + 1 }$$ $$= \displaystyle { 3 }$$