let the region r be the area enclosed by the function f(x)=ln(x) - 2 and g(x)=x - 4. write an integral in…

Explanation:

Step1: Find intersection points

Set $\ln(x)-2=x - 4$. Let $h(x)=\ln(x)-x + 2$. By using a numerical - method (such as Newton - Raphson method: $x_{n + 1}=x_{n}-\frac{h(x_{n})}{h'(x_{n})}$, where $h'(x)=\frac{1}{x}-1$), or by using a graphing utility, we find the intersection points. The intersection points of $y = \ln(x)-2$ and $y=x - 4$ are approximately $x_1\approx1.110$ and $x_2\approx3.146$.

Step2: Determine the integrand in terms of $x$

For $x\in[x_1,x_2]$, the upper - curve is $y_1=\ln(x)-2$ and the lower - curve is $y_2=x - 4$. The area $A_x$ in terms of $x$ is given by the integral $\int_{x_1}^{x_2}[(\ln(x)-2)-(x - 4)]dx=\int_{x_1}^{x_2}(\ln(x)-x + 2)dx$, where $x_1\approx1.110$ and $x_2\approx3.146$.

Step3: Rewrite the functions in terms of $y$

From $y=\ln(x)-2$, we can solve for $x$: $x = e^{y + 2}$. From $y=x - 4$, we have $x=y + 4$.

Step4: Find the $y$ - limits

When $x=x_1\approx1.110$, $y_1=\ln(1.110)-2\approx - 1.905$. When $x=x_2\approx3.146$, $y_2=\ln(3.146)-2\approx - 0.900$.

Step5: Determine the integrand in terms of $y$

For $y\in[y_1,y_2]$, the right - curve is $x_1=y + 4$ and the left - curve is $x_2=e^{y + 2}$. The area $A_y$ in terms of $y$ is given by the integral $\int_{y_1}^{y_2}[(y + 4)-e^{y + 2}]dy$, where $y_1\approx - 1.905$ and $y_2\approx - 0.900$.

Answer:

$x_1\approx1.110$, $x_2\approx3.146$, $\int_{1.110}^{3.146}(\ln(x)-x + 2)dx$ $y_1\approx - 1.905$, $y_2\approx - 0.900$, $\int_{-1.905}^{-0.900}[(y + 4)-e^{y + 2}]dy$