Find the area of the region enclosed by the graphs of y = e^x, y = e^{-x}, and y = 6.

Steps to Solve

1. Determine the intersection points

To find where the curves intersect, we need to set the equations equal to each other.

First, we find the points of intersection between the curves $y = e^x$ and $y = 6$: $$ e^x = 6 \implies x = \ln(6) $$

Next, we find the intersection between $y = e^{-x}$ and $y = 6$: $$ e^{-x} = 6 \implies -x = \ln(6) \implies x = -\ln(6) $$

Thus, the points of intersection are $x = -\ln(6)$ and $x = \ln(6)$.

2. Set up the integral for the area

The area between the curves from $x = -\ln(6)$ to $x = \ln(6)$ is given by the integral of the top curve minus the bottom curve.

The top curve in this region is $y = 6$ and the bottom is $y = e^{-x}$ (for $x < 0$, since $e^x$ is greater than 6 for positive x), and for positive x, $y = e^x$ dominates. So, we can express the area $A$ as:

$$ A = \int_{-\ln(6)}^{0} (6 - e^{-x}) , dx + \int_{0}^{\ln(6)} (e^x - 6) , dx $$

3. Calculate the first integral

Calculate the first part of the integral:

$$ \int (6 - e^{-x}) , dx = 6x - (-e^{-x}) = 6x + e^{-x} $$

Evaluate from $x = -\ln(6)$ to $x = 0$:

$$ [6(0) + e^0] - [6(-\ln(6)) + e^{\ln(6)}] = 1 - [-6\ln(6) + 6] = 7 - 6\ln(6) $$

4. Calculate the second integral

Now calculate the second integral:

$$ \int (e^x - 6) , dx = e^x - 6x $$

Evaluate from $x = 0$ to $x = \ln(6)$:

$$ [e^{\ln(6)} - 6(\ln(6))] - [e^0 - 6(0)] = [6 - 6\ln(6)] - [1] = 5 - 6\ln(6) $$

5. Combine both integrals

Now, we can find the total area by adding both parts:

$$ A = (7 - 6\ln(6)) + (5 - 6\ln(6)) = 12 - 12\ln(6) $$