Find the area of the region described. The region bounded by y = ex, y=e -2x, and x = In6

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Finding the Area of the Bounded Region**

To determine the area of the region described, consider the following boundaries:

- \( y = e^x \)
- \( y = e^{-2x} \)
- \( x = \ln{6} \)

**Goal:**
Calculate the exact area of the region enclosed by these boundaries.

**Steps to Follow:**

1. **Identify the Points of Intersection:**
   - Solve \( e^x = e^{-2x} \) to find the \( x \)-coordinates where the functions intersect.
   - Solution:
     \[
     e^x = e^{-2x}
     \]
     \( x = -2x \)
     \[
     3x = 0
     \]
     \[
     x = 0
     \]

2. **Determine the Limits of Integration:**
   - The bounds for \( x \) are from \( x = 0 \) to \( x = \ln{6} \).
   
3. **Set Up the Integral:**
   - The area \( A \) can be found by integrating the difference between the two functions from \( 0 \) to \( \ln{6} \):
     \[
     A = \int_{0}^{\ln{6}} \left( e^x - e^{-2x} \right) \, dx
     \]

4. **Evaluate the Integral:**
   - Break the integral into two parts:
     \[
     A = \int_{0}^{\ln{6}} e^x \, dx - \int_{0}^{\ln{6}} e^{-2x} \, dx
     \]
   - Compute each integral:
     \[
     \int e^x \, dx = e^x
     \]
     \[
     \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x}
     \]
   - Apply the limits of integration:
     \[
     A = \left[ e^x \right]_0^{\ln{6}} - \left[ -\frac{1}{2} e^{-2x} \right]_0^{\ln{6}}
     \]
     \[
     A = \left( e^{\ln{6}} - e^0 \right) - \left( -\frac
Transcribed Image Text:**Finding the Area of the Bounded Region** To determine the area of the region described, consider the following boundaries: - \( y = e^x \) - \( y = e^{-2x} \) - \( x = \ln{6} \) **Goal:** Calculate the exact area of the region enclosed by these boundaries. **Steps to Follow:** 1. **Identify the Points of Intersection:** - Solve \( e^x = e^{-2x} \) to find the \( x \)-coordinates where the functions intersect. - Solution: \[ e^x = e^{-2x} \] \( x = -2x \) \[ 3x = 0 \] \[ x = 0 \] 2. **Determine the Limits of Integration:** - The bounds for \( x \) are from \( x = 0 \) to \( x = \ln{6} \). 3. **Set Up the Integral:** - The area \( A \) can be found by integrating the difference between the two functions from \( 0 \) to \( \ln{6} \): \[ A = \int_{0}^{\ln{6}} \left( e^x - e^{-2x} \right) \, dx \] 4. **Evaluate the Integral:** - Break the integral into two parts: \[ A = \int_{0}^{\ln{6}} e^x \, dx - \int_{0}^{\ln{6}} e^{-2x} \, dx \] - Compute each integral: \[ \int e^x \, dx = e^x \] \[ \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} \] - Apply the limits of integration: \[ A = \left[ e^x \right]_0^{\ln{6}} - \left[ -\frac{1}{2} e^{-2x} \right]_0^{\ln{6}} \] \[ A = \left( e^{\ln{6}} - e^0 \right) - \left( -\frac
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