Find the area of the region described.
The region bounded by yequalse Superscript x​, yequalse Superscript negative 4 x​, and xequalsln 3
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Part 1
The area of the region is

enter your response here. ​(Type an exact​ answer.)

To find the area of the region bounded by \( y = e^x \), \( y = e^{-4x} \), and \( x = \ln 3 \), we need to perform the following steps:

1. Determine the points of intersection of the curves \( y = e^x \) and \( y = e^{-4x} \).
2. Integrate the difference between the two curves from the left boundary (intersection point) to the right boundary (\( x = \ln 3 \)).

### Step 1: Find the points of intersection

To find the points of intersection, set the equations equal to each other:

\[ e^x = e^{-4x} \]

Take the natural logarithm of both sides to solve for \( x \):

\[ \ln(e^x) = \ln(e^{-4x}) \]
\[ x = -4x \]

Solving for \( x \):

\[ x + 4x = 0 \]
\[ 5x = 0 \]
\[ x = 0 \]

So, the curves intersect at \( x = 0 \).

### Step 2: Set up the integral

The region of interest is between \( x = 0 \) and \( x = \ln 3 \). We will integrate the difference of the functions \( e^x \) and \( e^{-4x} \) over this interval.

\[ \text{Area} = \int_{0}^{\ln 3} (e^x - e^{-4x}) \, dx \]

### Step 3: Evaluate the integral

Learn how to find the area of the region bounded by the curves y = e^x, y = e^{-4x}, and x = ln 3 using integral calculus. This problem involves integration of exponential functions and natural logarithms.

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