let the region r be the area enclosed by the function f(x)=e^x and g(x)=5x + 1. if the region r is the base…

Explanation:

Step1: Find intersection points

Set $e^{x}=5x + 1$. Using a calculator or numerical - methods, the intersection points are $x = 0$ and $x\approx2.542$.

Step2: Determine base of cross - section

The base of each cross - section perpendicular to the $x$ - axis is $b=e^{x}-(5x + 1)$.

Step3: Determine height of cross - section

The height of each cross - section is $h=\frac{1}{2}(e^{x}-(5x + 1))$.

Step4: Find area of cross - section

The area of a rectangle cross - section $A(x)=\text{base}\times\text{height}=\frac{1}{2}(e^{x}-(5x + 1))^{2}=\frac{1}{2}(e^{2x}-2(5x + 1)e^{x}+(5x + 1)^{2})$.

Step5: Calculate the volume

The volume $V=\int_{a}^{b}A(x)dx=\int_{0}^{2.542}\frac{1}{2}(e^{2x}-2(5x + 1)e^{x}+(25x^{2}+10x + 1))dx$. We know that:

• $\int e^{2x}dx=\frac{1}{2}e^{2x}+C$;
• Using integration by parts $\int(5x + 1)e^{x}dx=(5x + 1)e^{x}-5\int e^{x}dx=(5x + 1)e^{x}-5e^{x}=(5x - 4)e^{x}+C$;
• $\int(25x^{2}+10x + 1)dx=\frac{25}{3}x^{3}+5x^{2}+x+C$.

$V=\frac{1}{2}\left[\frac{1}{2}e^{2x}-2(5x - 4)e^{x}+\frac{25}{3}x^{3}+5x^{2}+x\right]_{0}^{2.542}$

$V=\frac{1}{2}\left[\left(\frac{1}{2}e^{2\times2.542}-2(5\times2.542 - 4)e^{2.542}+\frac{25}{3}(2.542)^{3}+5(2.542)^{2}+2.542\right)-\left(\frac{1}{2}e^{0}-2(0 - 4)e^{0}+0 + 0+0\right)\right]$

Using a calculator: $V\approx10.364$

Answer:

$10.364$