Calculus: Early Transcendentals
8th Edition
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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Question
![**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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F777e0514-8160-4940-b57c-96884b4e2915%2F7586a6b0-f4c5-4530-9d64-cb5c4e6a9b1b%2F4yjuq29_processed.png&w=3840&q=75)
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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