SOLUTION 13: Compute the area of the
region enclosed by the graphs of the equations $ y=x^3 $ and $ y=4x $ . Begin by finding the points of
intersection of the two graphs. From $ y=x^3 $ and $ y=4x $ we get that
$$ x^3 = 4x \ \ \longrightarrow $$
$$ x^3-4x = 0 \ \ \longrightarrow $$
$$ x(x^2-4) = 0 \ \ \longrightarrow $$
$$ x(x-2)(x+2) = 0 \ \ \longrightarrow \ \ x=0, x=2, \ \ or \ \ x=-2 $$
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
$$ -2 \le x \le 0 \ \ and \ \ 4x \le y \le x^3 $$
in addition to
$$ 0 \le x \le 2 \ \ and \ \ x^3 \le y \le 4x \ ,$$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{-2}^{0} (Top \ - \ Bottom) \ dx +
\int_{0}^{2} (Top \ - \ Bottom) \ dx } $$
$$ = \displaystyle { \int_{-2}^{0} (x^3 - 4x) \ dx + \int_{0}^{2}
(4x - x^3) \ dx } $$
$$ = \displaystyle { \Big( \frac{x^{4}}{4} - 4 \cdot \frac{x^2}{2} \Big) \Big\vert_{-2}^{0} + \Big( 4 \cdot \frac{x^2}{2} -
\frac{x^{4}}{4} \Big) \Big\vert_{0}^{2} } $$
$$ = \displaystyle { \Big( \ \Big( \frac{(0)^{4}}{4} - 4 \cdot \frac{(0)^2}{2} \Big)
- \Big( \frac{(-2)^{4}}{4} - 4 \cdot \frac{(-2)^2}{2} \Big) \ \Big)
+ \Big( \ \Big( 4 \cdot \frac{(2)^2}{2} - \frac{(2)^{4}}{4} \Big)
- \Big( 4 \cdot \frac{(0)^2}{2} - \frac{(0)^{4}}{4} \Big) \ \Big) } $$
$$ = \displaystyle { \Big( \ \Big( 0 - 0 \Big)
- \Big( 4 - 8 \Big) \ \Big)
+ \Big( \ \Big( 8 - 4 \Big)
- \Big( 0 - 0 \Big) \ \Big) } $$
$$ = \displaystyle { 4 + 4 } $$
$$ = \displaystyle { 8 } $$
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