Find the area of the region enclosed by the graphs of y = e^x, y = e^(-x), and y = 6. (Use symbolic notation and fractions where needed.)

Steps to Solve

1. Find Points of Intersection
   First, we find where the graphs intersect by setting $y = e^x$ equal to $y = 6$.
   $$ e^x = 6 $$
   Taking the natural logarithm of both sides, we have
   $$ x = \ln(6) $$

Next, we find where $y = e^{-x}$ intersects $y = 6$.
$$ e^{-x} = 6 $$
Taking the natural logarithm of both sides, we get
$$ -x = \ln(6) \quad \Rightarrow \quad x = -\ln(6) $$

2. Set Up the Integral for Area Calculation
   The area enclosed by the curves can be found by integrating the difference between the top curve $y = 6$ and the bottom curve $y = e^{x}$ from $x = -\ln(6)$ to $x = \ln(6)$, and then adding the area between $y = 6$ and $y = e^{-x}$.
   The area $A$ can be expressed as:
   $$ A = \int_{-\ln(6)}^{\ln(6)} (6 - e^x) , dx + \int_{-\ln(6)}^{\ln(6)} (6 - e^{-x}) , dx $$

3. Calculate Each Integral
   Calculating the first integral:
   $$ \int (6 - e^x) , dx = 6x - e^x $$
   Now, evaluate from $-\ln(6)$ to $\ln(6)$:
   $$ \left[ 6x - e^x \right]_{-\ln(6)}^{\ln(6)} = [6\ln(6) - 6] - [(-6\ln(6)) - \frac{1}{6}] $$
   This simplifies to:
   $$ A_1 = 6\ln(6) - 6 + 6\ln(6) + \frac{1}{6} = 12\ln(6) - 5 $$

Now, calculate the second integral:
$$ \int (6 - e^{-x}) , dx = 6x + e^{-x} $$
Evaluate using the same bounds:
$$ \left[ 6x + e^{-x} \right]_{-\ln(6)}^{\ln(6)} = [6\ln(6) + \frac{1}{6}] - [6(-\ln(6)) + 6] $$
This simplifies to:
$$ A_2 = 6\ln(6) + \frac{1}{6} + 6 + 6\ln(6) = 12\ln(6) + 6 + \frac{1}{6} $$

4. Total Area Calculation
   Combine both areas:
   $$ A_{total} = A_1 + A_2 = (12\ln(6) - 5) + (12\ln(6) + 6 + \frac{1}{6}) $$
   Combine like terms:
   $$ A_{total} = 24\ln(6) + \frac{1}{6} + 1 $$
   Thus, the total area is:
   $$ A = 24\ln(6) + \frac{7}{6} $$