Area Solution 16

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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.

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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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