To find the area of the region $R$, we need to set up the integral that represents the area bounded by the curve $y = 4\ln(3 - x)$, the line $y = 6$, and the vertical line $x = 2$.

Steps:

1. Determine the bounds:

   • The curve $y = 4\ln(3 - x)$ intersects the horizontal line $y = 6$ at a certain point. We need to solve for the intersection to find the upper bound for the region.

2. Solve for the intersection: Set $4\ln(3 - x) = 6$. $\ln(3 - x) = \frac{6}{4} = 1.5$ Now exponentiate both sides to eliminate the logarithm: $3 - x = e^{1.5}$ Solving for $x$: $x = 3 - e^{1.5}$

3. Set up the integral: The area $A$ is given by the integral of the difference between the upper curve $y = 6$ and the lower curve $y = 4\ln(3 - x)$, over the interval from $x = 0$ to $x = 2$: $A = \int_0^2 \left( 6 - 4\ln(3 - x) \right) \, dx$

4. Evaluate the integral:

Let's compute the integral.

$A = \int_0^2 6 \, dx - \int_0^2 4 \ln(3 - x) \, dx$

We'll evaluate these two integrals separately.

First integral:

$\int_0^2 6 \, dx = 6x \bigg|_0^2 = 6(2) - 6(0) = 12$

Second integral:

We use substitution for the second integral. Let $u = 3 - x$, so $du = -dx$, and when $x = 0$, $u = 3$, and when $x = 2$, $u = 1$.

Now the integral becomes: $-4 \int_3^1 \ln(u) \, du$ This simplifies to: $4 \int_1^3 \ln(u) \, du$ The integral of $\ln(u)$ is $u \ln(u) - u$, so: $4 \left[ u \ln(u) - u \right]_1^3$ Evaluating at the limits: $4 \left[ 3 \ln(3) - 3 - (1 \ln(1) - 1) \right]$ Since $\ln(1) = 0$, we get: [ 4 \left[ 3 \ln(3) - 3 + 1 \right] = 4 \left[ 3 \ln(3) - 2 \right] ] So, the second integral is: $4(3 \ln(3) - 2) = 12 \ln(3) - 8$

Final area calculation:

Now, combine the results: $A = 12 - (12 \ln(3) - 8)$ $A = 12 - 12 \ln(3) + 8$ $A = 20 - 12 \ln(3)$

Thus, the area of the region $R$ is: $A = 20 - 12 \ln(3)$

Let me know if you need any further details or clarifications!

Related questions:

1. How do logarithmic functions affect the shape of a graph?
2. What are other common methods for solving integrals involving logarithms?
3. Can you explain the substitution method in more detail?
4. How does the area under a curve relate to real-world applications like physics or economics?
5. How would you compute the area between two curves instead of just one?

Tip: Always check the limits of integration carefully, especially when solving for intersections or using substitution.