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
$$ AREA = \displaystyle{ \int_{0}^{1} (Top \ - \ Bottom) \ dx +
\int_{1}^{8} (Top \ - \ Bottom) \ dx +
\int_{8}^{12} ( Top \ - \ Bottom ) \ dx } $$
$$ = \displaystyle{ \int_{0}^{1} (2x - 0) \ dx +
\int_{1}^{8} (2 - 0) \ dx +
\int_{8}^{12} \Big( 2 - \Big( \frac{1}{2}x-4 \Big) \Big) \ dx } $$
$$ = \displaystyle{ \int_{0}^{1} 2x \ dx +
\int_{1}^{8} 2 \ dx +
\int_{8}^{12} \Big( 6 - \frac{1}{2}x \Big) \ dx } $$
$$ = \displaystyle { ( x^2 ) \Big\vert_{0}^{1}
+ (2x) \Big\vert_{1}^{8}
+ \Big(6x - \frac{1}{4}x^2 \Big) \Big\vert_{8}^{12} } $$
$$ = \displaystyle { ( (1)^2-(0)^2)
+ (2(8)-2(1))
+ \Big( \Big( 6(12) - \frac{1}{4}(12)^2 \Big) - \Big( 6(8) - \frac{1}{4}(8)^2 \Big) \Big) } $$
$$ = \displaystyle { (1)
+ (14)
+ (36-32) } $$
$$ = 19 $$
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