SOLUTION 16: Compute the area of the region enclosed by the graphs of the equations $y= 2x, y= \displaystyle \frac{1}{2}x-4, y=0,$ and $y=2$. Begin by finding the points of intersection of the four graphs. From $y=2$ and $y = \displaystyle \frac{1}{2}x-4$ we get that $$2 = \displaystyle \frac{1}{2}x-4 \ \ \longrightarrow$$ $$4 = x-8 \ \ \longrightarrow \ \ x = 12$$ From $y = 2x$ and $y = 2$ we get that $$2x = 2 \ \ \longrightarrow \ \ x = 1$$ From $y=0$ and $y = \displaystyle \frac{1}{2}x-4$ we get that $$0 = \displaystyle \frac{1}{2}x-4 \ \ \longrightarrow$$ $$0 = x-8 \ \ \longrightarrow \ \ x = 8$$ From $y=0$ and $y = 2x$ we get that $$0 = 2x \ \ \longrightarrow \ \ x = 0$$ Now see the given graph of the enclosed region.

Using vertical cross-sections to describe this region, which is made up of three smaller regions, we get that $$0 \le x \le 1 \ \ and \ \ 0 \le y \le 2x$$ in addition to $$1 \le x \le 8 \ \ and \ \ 0 \le y \le 2 \ ,$$ and $$8 \le x \le 12 \ \ and \ \ \displaystyle \frac{1}{2}x-4 \le y \le 2 \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{1} (Top \ - \ Bottom) \ dx + \int_{1}^{8} (Top \ - \ Bottom) \ dx + \int_{8}^{12} ( Top \ - \ Bottom ) \ dx }$$ $$= \displaystyle{ \int_{0}^{1} (2x - 0) \ dx + \int_{1}^{8} (2 - 0) \ dx + \int_{8}^{12} \Big( 2 - \Big( \frac{1}{2}x-4 \Big) \Big) \ dx }$$ $$= \displaystyle{ \int_{0}^{1} 2x \ dx + \int_{1}^{8} 2 \ dx + \int_{8}^{12} \Big( 6 - \frac{1}{2}x \Big) \ dx }$$ $$= \displaystyle { ( x^2 ) \Big\vert_{0}^{1} + (2x) \Big\vert_{1}^{8} + \Big(6x - \frac{1}{4}x^2 \Big) \Big\vert_{8}^{12} }$$ $$= \displaystyle { ( (1)^2-(0)^2) + (2(8)-2(1)) + \Big( \Big( 6(12) - \frac{1}{4}(12)^2 \Big) - \Big( 6(8) - \frac{1}{4}(8)^2 \Big) \Big) }$$ $$= \displaystyle { (1) + (14) + (36-32) }$$ $$= 19$$