question let the region r be the area enclosed by the function f(x)=x³ and g(x)=4x. if the region r is the…

Explanation:

Step1: Find intersection points

Set $x^{3}=4x$. Rearrange to $x^{3}-4x = 0$, factor out $x$: $x(x^{2}-4)=0$, then $x(x - 2)(x + 2)=0$. The intersection - points are $x=-2,0,2$. Since the region is in the first - quadrant (from the graph), we consider $x$ from $0$ to $2$.

Step2: Determine the base of the rectangle

The base of the rectangle $b$ for a given $x$ is $b = 4x-x^{3}$ (since $g(x)\geq f(x)$ for $0\leq x\leq2$).

Step3: Determine the height of the rectangle

The height of the rectangle $h$ is $h = 2(4x - x^{3})$ (height is twice the base).

Step4: Set up the volume integral

The volume $V$ of the solid with cross - sectional area $A(x)$ and integrating over the interval $[a,b]$ is $V=\int_{a}^{b}A(x)dx$. Here, $A(x)=b\times h=(4x - x^{3})\times2(4x - x^{3})=2(4x - x^{3})^{2}$. So $V = \int_{0}^{2}2(4x - x^{3})^{2}dx$.

Step5: Expand the integrand

Expand $(4x - x^{3})^{2}=(4x)^{2}-2\times4x\times x^{3}+(x^{3})^{2}=16x^{2}-8x^{4}+x^{6}$. Then $2(4x - x^{3})^{2}=32x^{2}-16x^{4}+2x^{6}$.

Step6: Integrate term - by - term

$\int(32x^{2}-16x^{4}+2x^{6})dx=32\times\frac{x^{3}}{3}-16\times\frac{x^{5}}{5}+2\times\frac{x^{7}}{7}+C$.

Step7: Evaluate the definite integral

$V=\left[\frac{32x^{3}}{3}-\frac{16x^{5}}{5}+\frac{2x^{7}}{7}\right]_{0}^{2}=\frac{32\times2^{3}}{3}-\frac{16\times2^{5}}{5}+\frac{2\times2^{7}}{7}$. $V=\frac{32\times8}{3}-\frac{16\times32}{5}+\frac{2\times128}{7}=\frac{256}{3}-\frac{512}{5}+\frac{256}{7}$. Find a common denominator, which is $105$. $V=\frac{256\times35}{105}-\frac{512\times21}{105}+\frac{256\times15}{105}=\frac{8960 - 10752+3840}{105}=\frac{2048}{105}\approx19.50476\approx20$ (rounded to the nearest thousandth).

Answer:

$20$