let the region r be the area enclosed by the function f(x)=e^x - 2 and g(x)=5x - 1. find the volume of the…

Explanation:

Step1: Find intersection points

Set $e^{x}-2 = 5x - 1$. Using a calculator or numerical - methods (such as Newton - Raphson), the intersection points are approximately $x_1 = 0$ and $x_2\approx2.352$.

Step2: Use the disk/washer method

The outer radius $R(x)=(5x - 1)-(-2)=5x + 1$ and the inner radius $r(x)=(e^{x}-2)-(-2)=e^{x}$. The volume $V$ of the solid of revolution about the line $y = - 2$ is given by the formula $V=\pi\int_{a}^{b}(R^{2}(x)-r^{2}(x))dx$, where $a = 0$, $b\approx2.352$, $R(x)=5x + 1$, and $r(x)=e^{x}$. So $V=\pi\int_{0}^{2.352}((5x + 1)^{2}-(e^{x})^{2})dx$.

Step3: Expand the integrand

Expand $(5x + 1)^{2}-(e^{x})^{2}=25x^{2}+10x + 1 - e^{2x}$.

Step4: Integrate term - by - term

$\int(25x^{2}+10x + 1 - e^{2x})dx=\frac{25}{3}x^{3}+5x^{2}+x-\frac{1}{2}e^{2x}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{25}{3}x^{3}+5x^{2}+x-\frac{1}{2}e^{2x}\right]_{0}^{2.352}=\pi\left(\frac{25}{3}(2.352)^{3}+5(2.352)^{2}+2.352-\frac{1}{2}e^{2\times2.352}-\left(0 + 0+0-\frac{1}{2}\right)\right)$. Using a calculator: $V\approx\pi\left(\frac{25}{3}(12.977)+5(5.532)+2.352-\frac{1}{2}(47.097)+\frac{1}{2}\right)$ $V\approx\pi\left(108.142+27.66+2.352 - 23.548+0.5\right)$ $V\approx\pi(114.106)$ $V\approx358.877$

Answer:

$358.877$