let the region r be the area enclosed by the ...

# Explanation: ## Step1: Find intersection points Set $e^{x}=5x + 1$. Using a calculator or numerical - methods (such as Newton - Raphson), the intersection points are approximately $x_1 = 0$ and $x_2\approx2.542$. ## Step2: Use the washer method formula The formula for the volume $V$ of the solid of revolution about the line $y = k$ using the washer method is $V=\pi\int_{a}^{b}([R(x)]^{2}-[r(x)]^{2})dx$, where $R(x)$ is the outer - radius and $r(x)$ is the inner - radius. Here, $R(x)=e^{x}+2$ and $r(x)=5x + 1+2=5x + 3$, and $a = 0$, $b\approx2.542$. So, $V=\pi\int_{0}^{2.542}((e^{x}+2)^{2}-(5x + 3)^{2})dx$. ## Step3: Expand the integrand Expand $(e^{x}+2)^{2}=e^{2x}+4e^{x}+4$ and $(5x + 3)^{2}=25x^{2}+30x + 9$. The integrand becomes $e^{2x}+4e^{x}+4-(25x^{2}+30x + 9)=e^{2x}+4e^{x}-25x^{2}-30x - 5$. ## Step4: Integrate term - by - term $\int e^{2x}dx=\frac{1}{2}e^{2x}+C$, $\int4e^{x}dx = 4e^{x}+C$, $\int25x^{2}dx=\frac{25}{3}x^{3}+C$, $\int30x dx = 15x^{2}+C$, $\int5dx = 5x+C$. So, $\int_{0}^{2.542}(e^{2x}+4e^{x}-25x^{2}-30x - 5)dx=\left[\frac{1}{2}e^{2x}+4e^{x}-\frac{25}{3}x^{3}-15x^{2}-5x\right]_{0}^{2.542}$. ## Step5: Evaluate the definite integral $\left(\frac{1}{2}e^{2\times2.542}+4e^{2.542}-\frac{25}{3}(2.542)^{3}-15(2.542)^{2}-5(2.542)\right)-\left(\frac{1}{2}e^{0}+4e^{0}-0 - 0 - 0\right)$. Using a calculator: $\frac{1}{2}e^{5.084}+4e^{2.542}-\frac{25}{3}(2.542)^{3}-15(2.542)^{2}-5(2.542)-\left(\frac{1}{2}+4\right)$ $V=\pi\times$ (the result of the above calculation) $\approx104.797$. # Answer: $104.797$