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 $$AREA = \displaystyle{ \int_{-2}^{0} (Top \ - \ Bottom) \ dx + \int_{0}^{2} (Top \ - \ Bottom) \ dx + \int_{2}^{4} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{-2}^{0} \Big(x(x-2)(x+2) - \Big( \frac{1}{4}x^4-x^2 \Big) \Big) \ dx + \int_{0}^{2} \Big( \Big( \frac{1}{4}x^4-x^2 \Big) - x(x-2)(x+2) \Big) \ dx + \int_{2}^{4} \Big( x(x-2)(x+2) - \Big( \frac{1}{4}x^4-x^2 \Big) \Big) \ dx }$$ $$= \displaystyle { \int_{-2}^{0} \Big( - \frac{x^{4}}{4} +x^3+x^2-4x \Big) \ dx + \int_{0}^{2} \Big( \frac{x^{4}}{4} -x^3-x^2+4x \Big) \ dx + \int_{2}^{4} \Big( - \frac{x^{4}}{4} +x^3+x^2-4x \Big) \ dx }$$ $$= \displaystyle { \Big( - \frac{x^{5}}{20} + \frac{x^4}{4} + \frac{x^3}{3} - 2x^2 \Big) \Big\vert_{-2}^{0} + \Big( \frac{x^{5}}{20} - \frac{x^4}{4} - \frac{x^3}{3} + 2x^2 \Big) \Big\vert_{0}^{2} + \Big( - \frac{x^{5}}{20} + \frac{x^4}{4} + \frac{x^3}{3} - 2x^2 \Big) \Big\vert_{2}^{4} }$$ $$= \displaystyle { \Big( \Big( - \frac{(0)^{5}}{20} + \frac{(0)^4}{4} + \frac{(0)^3}{3} - 2(0)^2 \Big) - \Big( - \frac{(-2)^{5}}{20} + \frac{(-2)^4}{4} + \frac{(-2)^3}{3} - 2(-2)^2 \Big) \Big) + \Big( \Big( \frac{(2)^{5}}{20} - \frac{(2)^4}{4} - \frac{(2)^3}{3} + 2(2)^2 \Big) - \Big( \frac{(0)^{5}}{20} - \frac{(0)^4}{4} - \frac{(0)^3}{3} + 2(0)^2 \Big) \Big) }$$ $$\displaystyle { + \Big( \Big( - \frac{(4)^{5}}{20} + \frac{(4)^4}{4} + \frac{(4)^3}{3} - 2(4)^2 \Big) - \Big( - \frac{(2)^{5}}{20} + \frac{(2)^4}{4} + \frac{(2)^3}{3} - 2(2)^2 \Big) \Big) }$$ $$= \displaystyle { \Big( \Big( 0 \Big) - \Big( \frac{8}{5} + 4 - \frac{8}{3} - 8 \Big) \Big) + \Big( \Big( \frac{8}{5} - 4 - \frac{8}{3} + 8 \Big) - \Big( 0 \Big) \Big) + \Big( \Big( - \frac{256}{5} + 64 + \frac{64}{3} - 32 \Big) - \Big( - \frac{8}{5} + 4 + \frac{8}{3} - 8 \Big) \Big) }$$ $$= \displaystyle { \Big( - \frac{8}{5} - 4 + \frac{8}{3} + 8 + \frac{8}{5} - 4 - \frac{8}{3} + 8 \Big) + \Big( - \frac{256}{5} + 64 + \frac{64}{3} - 32 + \frac{8}{5} - 4 - \frac{8}{3} + 8 \Big) \Big) }$$ $$= \Big( 8 \Big) + \Big( - \frac{248}{5} + 36 + \frac{56}{3} \Big)$$ $$= 44 - \frac{248}{5} + \frac{56}{3}$$ $$= \frac{660}{15} - \frac{744}{15} + \frac{280}{15}$$ $$= \frac{196}{15}$$