let the region r be the area enclosed by the function f(x)=x³ + 1 and g(x)=4x + 1. if the region r is the…

Explanation:

Step1: Find intersection points

Set $x^{3}+1 = 4x + 1$. Then $x^{3}-4x=0$, factoring gives $x(x - 2)(x + 2)=0$. So the intersection - points are $x=-2,0,2$.

Step2: Determine side - length of square cross - section

The side - length of each square cross - section perpendicular to the $x$ - axis is $s=(4x + 1)-(x^{3}+1)=4x - x^{3}$.

Step3: Set up volume integral

The volume $V$ of the solid with square cross - sections is given by the integral $V=\int_{a}^{b}s^{2}dx$, where $a=-2$, $b = 2$, and $s = 4x - x^{3}$. So $V=\int_{-2}^{2}(4x - x^{3})^{2}dx=\int_{-2}^{2}(16x^{2}-8x^{4}+x^{6})dx$.

Step4: Evaluate the integral

Since $\int_{-2}^{2}(16x^{2}-8x^{4}+x^{6})dx = 2\int_{0}^{2}(16x^{2}-8x^{4}+x^{6})dx$ (because the integrand is an even function). $\int(16x^{2}-8x^{4}+x^{6})dx=\frac{16}{3}x^{3}-\frac{8}{5}x^{5}+\frac{1}{7}x^{7}+C$. $2\left[\frac{16}{3}x^{3}-\frac{8}{5}x^{5}+\frac{1}{7}x^{7}\right]_{0}^{2}=2\left(\frac{16}{3}\times2^{3}-\frac{8}{5}\times2^{5}+\frac{1}{7}\times2^{7}\right)$. $=2\left(\frac{128}{3}-\frac{256}{5}+\frac{128}{7}\right)$. $=2\times\frac{128\times35 - 256\times21+128\times15}{105}$. $=2\times\frac{4480 - 5376+1920}{105}$. $=2\times\frac{1024}{105}\approx19.505$.

Answer:

$19.505$