SOLUTION 14: Compute the area of the
region enclosed by the graphs of the equations $ \displaystyle{ y= \frac{1}{4}x^4-x^2 } $ and $ y=x(x-2)(x+2) $ . Begin by finding the points of
intersection of the two graphs. From $ y= \frac{1}{4}x^4-x^2 $ and $ y=x(x-2)(x+2) $ we get that
$$ \frac{1}{4}x^4-x^2 = x(x-2)(x+2) \ \ \longrightarrow $$
$$ x^4-4x^2 = 4x(x-2)(x+2) \ \ \longrightarrow $$
$$ x^2(x-2)(x+2) - 4x(x-2)(x+2) = 0 \ \ \longrightarrow $$
$$ x(x-2)(x+2)[x-4] = 0 \ \ \longrightarrow \ \ x=0, x=2, x=-2, \ \ or \ \ x=4$$
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
$$ -2 \le x \le 0 \ \ and \ \ \frac{1}{4}x^4-x^2 \le y \le x(x-2)(x+2) $$
in addition to
$$ 0 \le x \le 2 \ \ and \ \ x(x-2)(x+2) \le y \le \frac{1}{4}x^4-x^2 $$
and
$$ 2 \le x \le 4 \ \ and \ \ \frac{1}{4}x^4-x^2 \le y \le x(x-2)(x+2) \ , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{-2}^{0} (Top \ - \ Bottom) \ dx +
\int_{0}^{2} (Top \ - \ Bottom) \ dx +
\int_{2}^{4} (Top \ - \ Bottom) \ dx } $$
$$ = \displaystyle { \int_{-2}^{0} \Big(x(x-2)(x+2) - \Big( \frac{1}{4}x^4-x^2 \Big) \Big) \ dx
+ \int_{0}^{2} \Big( \Big( \frac{1}{4}x^4-x^2 \Big) - x(x-2)(x+2) \Big) \ dx
+ \int_{2}^{4} \Big( x(x-2)(x+2) - \Big( \frac{1}{4}x^4-x^2 \Big) \Big) \ dx } $$
$$ = \displaystyle { \int_{-2}^{0} \Big( - \frac{x^{4}}{4} +x^3+x^2-4x \Big) \ dx
+ \int_{0}^{2} \Big( \frac{x^{4}}{4} -x^3-x^2+4x \Big) \ dx
+ \int_{2}^{4} \Big( - \frac{x^{4}}{4} +x^3+x^2-4x \Big) \ dx } $$
$$ = \displaystyle { \Big( - \frac{x^{5}}{20} + \frac{x^4}{4} + \frac{x^3}{3} - 2x^2 \Big) \Big\vert_{-2}^{0}
+ \Big( \frac{x^{5}}{20} - \frac{x^4}{4} - \frac{x^3}{3} + 2x^2 \Big) \Big\vert_{0}^{2}
+ \Big( - \frac{x^{5}}{20} + \frac{x^4}{4} + \frac{x^3}{3} - 2x^2 \Big) \Big\vert_{2}^{4} } $$
$$ = \displaystyle { \Big( \Big( - \frac{(0)^{5}}{20} + \frac{(0)^4}{4} + \frac{(0)^3}{3} - 2(0)^2 \Big)
- \Big( - \frac{(-2)^{5}}{20} + \frac{(-2)^4}{4} + \frac{(-2)^3}{3} - 2(-2)^2 \Big) \Big)
+ \Big( \Big( \frac{(2)^{5}}{20} - \frac{(2)^4}{4} - \frac{(2)^3}{3} + 2(2)^2 \Big)
- \Big( \frac{(0)^{5}}{20} - \frac{(0)^4}{4} - \frac{(0)^3}{3} + 2(0)^2 \Big) \Big) } $$
$$ \displaystyle { + \Big( \Big( - \frac{(4)^{5}}{20} + \frac{(4)^4}{4} + \frac{(4)^3}{3} - 2(4)^2 \Big)
- \Big( - \frac{(2)^{5}}{20} + \frac{(2)^4}{4} + \frac{(2)^3}{3} - 2(2)^2 \Big) \Big) } $$
$$ = \displaystyle { \Big( \Big( 0 \Big)
- \Big( \frac{8}{5} + 4 - \frac{8}{3} - 8 \Big) \Big)
+ \Big( \Big( \frac{8}{5} - 4 - \frac{8}{3} + 8 \Big)
- \Big( 0 \Big) \Big)
+ \Big( \Big( - \frac{256}{5} + 64 + \frac{64}{3} - 32 \Big)
- \Big( - \frac{8}{5} + 4 + \frac{8}{3} - 8 \Big) \Big) } $$
$$ = \displaystyle { \Big(
- \frac{8}{5} - 4 + \frac{8}{3} + 8
+ \frac{8}{5} - 4 - \frac{8}{3} + 8 \Big)
+ \Big( - \frac{256}{5} + 64 + \frac{64}{3} - 32
+ \frac{8}{5} - 4 - \frac{8}{3} + 8 \Big) \Big) } $$
$$ = \Big( 8 \Big) + \Big( - \frac{248}{5} + 36 + \frac{56}{3} \Big) $$
$$ = 44 - \frac{248}{5} + \frac{56}{3} $$
$$ = \frac{660}{15} - \frac{744}{15} + \frac{280}{15} $$
$$ = \frac{196}{15} $$
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