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 $$AREA = \displaystyle{ \int_{-2}^{0} (Top \ - \ Bottom) \ dx + \int_{0}^{2} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{-2}^{0} (x^3 - 4x) \ dx + \int_{0}^{2} (4x - x^3) \ dx }$$ $$= \displaystyle { \Big( \frac{x^{4}}{4} - 4 \cdot \frac{x^2}{2} \Big) \Big\vert_{-2}^{0} + \Big( 4 \cdot \frac{x^2}{2} - \frac{x^{4}}{4} \Big) \Big\vert_{0}^{2} }$$ $$= \displaystyle { \Big( \ \Big( \frac{(0)^{4}}{4} - 4 \cdot \frac{(0)^2}{2} \Big) - \Big( \frac{(-2)^{4}}{4} - 4 \cdot \frac{(-2)^2}{2} \Big) \ \Big) + \Big( \ \Big( 4 \cdot \frac{(2)^2}{2} - \frac{(2)^{4}}{4} \Big) - \Big( 4 \cdot \frac{(0)^2}{2} - \frac{(0)^{4}}{4} \Big) \ \Big) }$$ $$= \displaystyle { \Big( \ \Big( 0 - 0 \Big) - \Big( 4 - 8 \Big) \ \Big) + \Big( \ \Big( 8 - 4 \Big) - \Big( 0 - 0 \Big) \ \Big) }$$ $$= \displaystyle { 4 + 4 }$$ $$= \displaystyle { 8 }$$