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

Explanation:

Step1: Find intersection points

Set $e^{x}=4x + 1$. Using a calculator or numerical - methods (e.g., Newton - Raphson), the intersection points are $x = 0$ and $x\approx1.512$.

Step2: Determine side - length of square cross - section

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

Step3: Find the area of the square cross - section

The area of a square $A(x)=s^{2}=(e^{x}-4x - 1)^{2}=e^{2x}-8xe^{x}-2e^{x}+16x^{2}+8x + 1$.

Step4: Calculate the volume using the integral

The volume $V$ of the solid with square cross - sections is given by the integral $V=\int_{a}^{b}A(x)dx$, where $a = 0$ and $b\approx1.512$. [ \begin{align*} V&=\int_{0}^{1.512}(e^{2x}-8xe^{x}-2e^{x}+16x^{2}+8x + 1)dx\ &=\left[\frac{1}{2}e^{2x}-8(xe^{x}-e^{x})-2e^{x}+\frac{16}{3}x^{3}+4x^{2}+x\right]_{0}^{1.512}\ \end{align*} ] First, substitute $x = 1.512$: [ \begin{align*} &\frac{1}{2}e^{2\times1.512}-8(1.512e^{1.512}-e^{1.512})-2e^{1.512}+\frac{16}{3}(1.512)^{3}+4(1.512)^{2}+1.512\ \end{align*} ] Then substitute $x = 0$: $\frac{1}{2}-8(0 - 1)-2+0 + 0+0=\frac{1}{2}+8 - 2=\frac{1 + 16-4}{2}=\frac{13}{2}=6.5$ Subtract the two results: [ \begin{align*} V&\approx\frac{1}{2}e^{3.024}-8\times0.512e^{1.512}-2e^{1.512}+\frac{16}{3}(1.512)^{3}+4(1.512)^{2}+1.512-6.5\ &\approx\frac{1}{2}\times20.67-8\times0.512\times4.54-2\times4.54+\frac{16}{3}\times3.46+4\times2.29+1.512 - 6.5\ &\approx10.335-18.67+(-9.08)+18.45+9.16+1.512 - 6.5\ &\approx4.207 \end{align*} ]

Answer:

$4.207$