let the region r be the area enclosed by the function f(x)=2x³ and g(x)=2x. find the volume of the solid…

Explanation:

Step1: Find intersection points

Set $2x^{3}=2x$, then $2x^{3}-2x = 0$, $2x(x^{2}-1)=0$, $2x(x - 1)(x + 1)=0$. The solutions are $x=-1,0,1$.

Step2: Use the washer - method formula

The outer - radius $R(x)=g(x)+1=2x + 1$ and the inner - radius $r(x)=f(x)+1=2x^{3}+1$. The volume formula for the washer method when rotating about $y =-1$ is $V=\pi\int_{a}^{b}[(R(x))^{2}-(r(x))^{2}]dx$, where $a=-1$ and $b = 1$. [ \begin{align*} V&=\pi\int_{-1}^{1}[(2x + 1)^{2}-(2x^{3}+1)^{2}]dx\ &=\pi\int_{-1}^{1}[(4x^{2}+4x + 1)-(4x^{6}+4x^{3}+1)]dx\ &=\pi\int_{-1}^{1}(4x^{2}+4x + 1 - 4x^{6}-4x^{3}-1)dx\ &=\pi\int_{-1}^{1}(4x^{2}-4x^{6}+4x-4x^{3})dx \end{align*} ] Since $\int_{-1}^{1}4x dx = 0$ and $\int_{-1}^{1}-4x^{3}dx = 0$ (because they are odd functions), we have: [ \begin{align*} V&=\pi\left(2\int_{0}^{1}(4x^{2}-4x^{6})dx\right)\ &=2\pi\left[\frac{4x^{3}}{3}-\frac{4x^{7}}{7}\right]_{0}^{1}\ &=2\pi\left(\frac{4}{3}-\frac{4}{7}\right)\ &=2\pi\times\frac{28 - 12}{21}\ &=2\pi\times\frac{16}{21}\ &\approx4.789 \end{align*} ]

Answer:

$4.789$