let the region r be the area enclosed by the function f(x)=ln(x)+1 and g(x)=2x - 1. if the region r is the…

Explanation:

Step1: Find intersection points

Set $\ln(x)+1 = 2x - 1$. Using a calculator or software, we find the intersection points of $y=\ln(x)+1$ and $y = 2x - 1$ are $x=a$ and $x = b$ (where $a\approx0.159$ and $b\approx1.317$).

Step2: Determine side - length of square cross - section

The side - length $s$ of each square cross - section perpendicular to the $x$ - axis is $s=\vert(\ln(x)+1)-(2x - 1)\vert=\vert\ln(x)-2x + 2\vert$. Since in the interval $[a,b]$, $2x - 1\geq\ln(x)+1$, then $s=(2x - 1)-(\ln(x)+1)=2x-\ln(x)-2$.

Step3: Set up the volume integral

The volume $V$ of the solid with square cross - sections perpendicular to the $x$ - axis is given by the integral $V=\int_{a}^{b}s^{2}dx=\int_{0.159}^{1.317}(2x-\ln(x)-2)^{2}dx$. Expand $(2x-\ln(x)-2)^{2}=(2x-\ln(x)-2)(2x-\ln(x)-2)=4x^{2}-4x\ln(x)-8x+\ln^{2}(x)+4\ln(x)+4$. Then $V=\int_{0.159}^{1.317}(4x^{2}-4x\ln(x)-8x+\ln^{2}(x)+4\ln(x)+4)dx$. We know that:

• $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$
• $\int\ln(x)dx=x\ln(x)-x + C$
• $\int x\ln(x)dx=\frac{1}{2}x^{2}\ln(x)-\frac{1}{4}x^{2}+C$
• $\int\ln^{2}(x)dx=x\ln^{2}(x)-2x\ln(x)+2x + C$ $V=\left[4\times\frac{x^{3}}{3}-4\left(\frac{1}{2}x^{2}\ln(x)-\frac{1}{4}x^{2}\right)-8\times\frac{x^{2}}{2}+\left(x\ln^{2}(x)-2x\ln(x)+2x\right)+4(x\ln(x)-x)+4x\right]_{0.159}^{1.317}$ Using a calculator to evaluate the definite integral: $V\approx0.339$

Answer:

$0.339$