SOLUTION 24: Compute the area of the region enclosed by the graphs of the equations $y=2x, y=\displaystyle{ \frac{1}{2} }x,$ and $y=4$ . Begin by finding the points of intersection of the the three graphs. From $y = 2x$ and $y = \displaystyle{ \frac{1}{2} }x$ we get that $$2x = \displaystyle{ \frac{1}{2} }x \ \ \longrightarrow$$ $$4x = x \ \ \longrightarrow$$ $$3x = 0 \ \ \longrightarrow \ \ x=0$$ From $y = 4$ and $y = \displaystyle{ \frac{1}{2} }x$ we get that $$4 = \displaystyle{ \frac{1}{2} }x \ \ \longrightarrow \ \ x=8$$ From $y = 4$ and $y = 2x$ we get that $$4 = 2x \ \ \longrightarrow \ \ x=2$$ Now see the given graph of the enclosed region.

a.) Using vertical cross-sections to describe this region, which is made up of two smaller regions, we get that $$0 \le x \le 2 \ \ and \ \ \displaystyle{ \frac{1}{2} }x \le y \le 2x \ ,$$ in addition to $$2 \le x \le 8 \ \ and \ \ \displaystyle{ \frac{1}{2} }x \le y \le 4 \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{2} (Top \ - \ Bottom) \ dx + \int_{2}^{8} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle{ \int_{0}^{2} \Big( 2x - \frac{1}{2}x \Big) \ dx + \int_{2}^{8} \Big( 4 - \frac{1}{2}x \Big) \ dx }$$ $$= \displaystyle{ \int_{0}^{2} \frac{3}{2}x \ dx + \int_{2}^{8} \Big( 4 - \frac{1}{2}x \Big) \ dx }$$ $$\displaystyle { = \Big( \frac{3x^2}{4} \Big) \Big\vert_{0}^{2} + \Big( 4x - \frac{x^2}{4} \Big) \Big\vert_{2}^{8} }$$ $$\displaystyle { = \Big( \frac{3(2)^2}{4} - \frac{3(0)^2}{4} \Big) + \Big( \Big( 4(8) - \frac{(8)^2}{4} \Big) - \Big( 4(2) - \frac{(2)^2}{4} \Big) \Big) }$$ $$= (3) + (16) - (7)$$ $$= 12$$ b.) Using horizontal cross-sections to describe this region, we get that $$0 \le y \le 4 \ \ and \ \ \displaystyle{ \frac{1}{2}y } \le x \le 2y \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{4} ( Right \ - \ Left ) \ dy }$$ $$AREA = \displaystyle{ \int_{0}^{4} \Big( 2y - \frac{1}{2}y \Big) \ dx }$$ $$\displaystyle { = \Big( y^2 - \frac{y^2}{4} \Big) \Big\vert_{0}^{4} }$$ $$\displaystyle { = \Big( (4)^2 - \frac{(4)^2}{4} \Big) - \Big( (0)^2 - \frac{(0)^2}{4} \Big) }$$ $$\displaystyle { = (12) - (0) }$$ $$= 12$$