SOLUTION 5: Compute the area of the region enclosed by the graphs of the equations $y=x^3+x^2$ and $y=3x^2+3x$ . Begin by finding the points of intersection of the two graphs. From $y=x^3+x^2$ and $y=3x^2+3x$ we get that $$x^{3} + x^{2} = 3x^{2} + 3x \ \ \longrightarrow$$ $$x^{3} - 2x^{2} - 3x = 0 \ \ \longrightarrow$$ $$x(x^{2} - 2x - 3) = 0 \ \ \longrightarrow$$ $$x(x-3)(x+1) = 0 \ \ \longrightarrow \ \ x=0, x=3, \ or \ x=-1$$ 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 $$-1 \le x \le 0 \ \ and \ \ 3x^{2} + 3x \le y \le x^{3} + x^{2}$$ in addition to $$0 \le x \le 3 \ \ and \ \ x^{3} + x^{2} \le y \le 3x^{2} + 3x ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{-1}^{0} (Top \ - \ Bottom) \ dx + \int_{0}^{3} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{-1}^{0} ((x^{3}+x^{2})-(3x^{2}+3x)) \ dx + \int_{0}^{3} ((3x^{2}+3x) -(x^{3}+x^{2})) \ dx }$$ $$= \displaystyle { \int_{-1}^{0} (x^{3}-2x^{2}-3x) \ dx + \int_{0}^{3} (-x^{3}+2x^{2}+3x) \ dx }$$ $$= \displaystyle { \Big( \frac{x^{4}}{4} - \frac{2x^{3}}{3} - \frac{3x^{2}}{2} \Big) \Big\vert_{-1}^{0} + \Big( -\frac{x^{4}}{4} + \frac{2x^{3}}{3} + \frac{3x^{2}}{2} \Big) \Big\vert_{0}^{3}}$$ $$= \displaystyle { \Big( \frac{0^{4}}{4} - \frac{2(0)^{3}}{3} - \frac{3(0)^{2}}{2} \Big) - \Big( \frac{(-1)^{4}}{4} - \frac{2(-1)^{3}}{3} - \frac{3(-1)^{2}}{2} \Big) + \Big( -\frac{3^{4}}{4} + \frac{2(3)^{3}}{3} + \frac{3(3)^{2}}{2} \Big) }$$ $$\displaystyle { \ \ \ \ - \Big( -\frac{0^{4}}{4} + \frac{2(0)^{3}}{3} + \frac{3(0)^{2}}{2} \Big) }$$ $$= \displaystyle { \Big( 0 \Big) - \Big( \frac{1}{4} + \frac{2}{3} - \frac{3}{2} \Big) + \Big( -\frac{81}{4} + 18 + \frac{27}{2} \Big) - \Big( 0 \Big) }$$ $$= \displaystyle { - \Big( \frac{3}{12} + \frac{8}{12} - \frac{18}{12} \Big) + \Big( -\frac{81}{4} + \frac{72}{4} + \frac{54}{4} \Big) }$$ $$= \displaystyle { - \Big( -\frac{7}{12} \Big) + \Big( \frac{45}{4} \Big) }$$ $$= \displaystyle { \frac{7}{12} + \frac{45}{4} }$$ $$= \displaystyle { \frac{7}{12} + \frac{135}{12} }$$ $$= \displaystyle { \frac{142}{12} }$$ $$= \displaystyle { \frac{71}{6} }$$