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

Explanation:

Step1: Find intersection points

Set $\ln(x)+1=x - 1$. Using a calculator or software, we find the intersection points of $y=\ln(x)+1$ and $y=x - 1$ are $x = 1$ and $x\approx3.146$.

Step2: Determine base - height relationship

The base of each rectangular cross - section perpendicular to the $x$ - axis is $b=(x - 1)-(\ln(x)+1)=x-\ln(x)-2$. The height $h=\frac{1}{2}(x-\ln(x)-2)$.

Step3: Set up volume integral

The volume $V$ of the solid with cross - sectional area $A(x)$ is given by $V=\int_{a}^{b}A(x)dx$. Here, $A(x)=h\times b=\frac{1}{2}(x - \ln(x)-2)^2$, and $a = 1$, $b\approx3.146$. So $V=\frac{1}{2}\int_{1}^{3.146}(x-\ln(x)-2)^2dx$. Expand $(x-\ln(x)-2)^2=x^{2}-2x\ln(x)-4x + 2\ln(x)+4+\ln^{2}(x)$. We know that $\int x^{2}dx=\frac{1}{3}x^{3}+C$, $\int x\ln(x)dx=\frac{1}{2}x^{2}\ln(x)-\frac{1}{4}x^{2}+C$, $\int xdx=\frac{1}{2}x^{2}+C$, $\int\ln(x)dx=x\ln(x)-x + C$, $\int\ln^{2}(x)dx=x\ln^{2}(x)-2x\ln(x)+2x + C$. Then $\frac{1}{2}\int_{1}^{3.146}(x^{2}-2x\ln(x)-4x + 2\ln(x)+4+\ln^{2}(x))dx$. Evaluating the definite integral: [ \begin{align*} &\frac{1}{2}\left[\frac{1}{3}x^{3}-2\left(\frac{1}{2}x^{2}\ln(x)-\frac{1}{4}x^{2}\right)-4\times\frac{1}{2}x^{2}+2(x\ln(x)-x)+4x+(x\ln^{2}(x)-2x\ln(x)+2x)\right]_{1}^{3.146}\ \end{align*} ] Using a calculator to evaluate the above expression: [ \begin{align*} &\frac{1}{2}\left[\left(\frac{1}{3}(3.146)^{3}-(3.146)^{2}\ln(3.146)+\frac{1}{2}(3.146)^{2}-2(3.146)^{2}+2(3.146)\ln(3.146)-2(3.146)+4(3.146)+(3.146)\ln^{2}(3.146)-2(3.146)\ln(3.146)+2(3.146)\right)-\left(\frac{1}{3}-0+\frac{1}{2}-2 + 0-2+4+0 - 0+2\right)\right]\ &\approx0.347 \end{align*} ]

Answer:

$0.347$