question let the region r be the area enclosed by the function f(x)=ln(x)-2 and g(x)=x - 4. if the region r…

Explanation:

Step1: Find intersection points

Set $\ln(x)-2=x - 4$. Let $y_1=\ln(x)-2$ and $y_2=x - 4$. By using a graph - ing utility or numerical methods (e.g., Newton - Raphson method), we find the intersection points of $y_1$ and $y_2$. The intersection points are $x = 1$ and $x = 4$.

Step2: Determine the length of the leg of the isosceles right - triangle

The length of the leg $L$ of the isosceles right - triangle cross - section perpendicular to the $x$ - axis is $L=(x - 4)-(\ln(x)-2)=x-\ln(x)-2$.

Step3: Find the area formula of the isosceles right - triangle

The area formula of an isosceles right - triangle with leg length $L$ is $A=\frac{1}{2}L^{2}$. Substituting $L=x-\ln(x)-2$ into the formula, we get $A(x)=\frac{1}{2}(x - \ln(x)-2)^{2}=\frac{1}{2}(x^{2}-2x\ln(x)-4x + \ln^{2}(x)+4\ln(x)+4)$.

Step4: Calculate the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = a$ to $x = b$ is given by $V=\int_{a}^{b}A(x)dx$. Here, $a = 1$, $b = 4$, so $V=\frac{1}{2}\int_{1}^{4}(x^{2}-2x\ln(x)-4x+\ln^{2}(x)+4\ln(x)+4)dx$. We use the following integral formulas:

• $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$
• $\int\ln(x)dx=x\ln(x)-x + C$
• $\int x\ln(x)dx=\frac{1}{2}x^{2}\ln(x)-\frac{1}{4}x^{2}+C$
• $\int\ln^{2}(x)dx=x\ln^{2}(x)-2x\ln(x)+2x + C$

[ \begin{align*} V&=\frac{1}{2}\left[\frac{x^{3}}{3}-2\left(\frac{1}{2}x^{2}\ln(x)-\frac{1}{4}x^{2}\right)-4\times\frac{x^{2}}{2}+\left(x\ln^{2}(x)-2x\ln(x)+2x\right)+4(x\ln(x)-x)+4x\right]{1}^{4}\ &=\frac{1}{2}\left[\frac{x^{3}}{3}-x^{2}\ln(x)+\frac{1}{2}x^{2}-2x^{2}+x\ln^{2}(x)-2x\ln(x)+2x + 4x\ln(x)-4x + 4x\right]{1}^{4}\ &=\frac{1}{2}\left[\frac{x^{3}}{3}-x^{2}\ln(x)+\frac{1}{2}x^{2}-2x^{2}+x\ln^{2}(x)+2x\ln(x)+2x\right]_{1}^{4} \end{align*} ]

[ \begin{align*} &\text{When }x = 4:\ &\frac{4^{3}}{3}-4^{2}\ln(4)+\frac{1}{2}\times4^{2}-2\times4^{2}+4\ln^{2}(4)+2\times4\ln(4)+2\times4\ &=\frac{64}{3}-16\ln(4)+8 - 32+4\ln^{2}(4)+8\ln(4)+8\ &=\frac{64}{3}-16\ln(4)+4\ln^{2}(4)+8\ln(4)-16 \end{align*} ]

[ \begin{align*} &\text{When }x = 1:\ &\frac{1^{3}}{3}-1^{2}\ln(1)+\frac{1}{2}\times1^{2}-2\times1^{2}+1\times\ln^{2}(1)+2\times1\times\ln(1)+2\times1\ &=\frac{1}{3}-0+\frac{1}{2}-2 - 0+0 + 2\ &=\frac{1}{3}+\frac{1}{2}=\frac{2 + 3}{6}=\frac{5}{6} \end{align*} ]

[ \begin{align*} V&=\frac{1}{2}\left[\left(\frac{64}{3}-16\ln(4)+4\ln^{2}(4)+8\ln(4)-16\right)-\frac{5}{6}\right]\ &=\frac{1}{2}\left[\frac{128 - 5}{6}-16\ln(4)+4\ln^{2}(4)+8\ln(4)-16\right]\ &=\frac{1}{2}\left[\frac{123}{6}-16\ln(4)+4\ln^{2}(4)+8\ln(4)-16\right]\ &=\frac{1}{2}\left[\frac{41}{2}-16\ln(4)+4\ln^{2}(4)+8\ln(4)-16\right]\ &=\frac{1}{2}\left[\frac{41 - 32}{2}-8\ln(4)+4\ln^{2}(4)\right]\ &=\frac{1}{2}\left[\frac{9}{2}-8\ln(4)+4\ln^{2}(4)\right]\ &\approx\frac{1}{2}\left[\frac{9}{2}-8\times1.3863+4\times(1.3863)^{2}\right]\ &=\frac{1}{2}\left[4.5-11.0904 + 4\times1.9218\right]\ &=\frac{1}{2}\left[4.5-11.0904+7.6872\right]\ &=\frac{1}{2}\times1.0968\ &=0.5484\approx0.548 \end{align*} ]

Answer:

$0.548$