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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![### Calculating the Length of a Curve by Integration
To find the length of a curve defined by a given integral, we use the formula for arc length in calculus. The provided problem is detailed below with intermediate steps and explanations.
#### Step-by-Step Calculation:
1. **Original Integral Setup:**
\[
s = \int_{1}^{8} \sqrt{4 + \left(\frac{1}{t}\right)^2 + 4t} \, dt
\]
2. **Simplification of the Integrand:**
The integrand is simplified to:
\[
s = \int_{1}^{8} (2t + \frac{1}{t}) \, dt
\]
3. **Integration Process:**
The integral of \(2t + \frac{1}{t}\) is evaluated:
\[
\int (2t + \frac{1}{t}) \, dt = t^2 + t \ln(t)
\]
4. **Evaluating at the Boundaries:**
The result is then evaluated from \(t = 1\) to \(t = 8\):
However, there are errors indicated at these steps in the provided image (marked with red crosses):
- The integration result is:
\[
\left[ t^2 + t \ln (t) \right]_1^8
\]
5. **Simplification:**
The proper evaluation at the boundaries should yield:
\[
t^2 + t \ln (t) \bigg|_1^8
\]
This indicates that we need to substitute \(t = 8\) and \(t = 1\) into the expression and find the final result.
6. **Substitution and Calculation:**
After substituting the values:
\[
\left(8^2 + 8 \ln (8)\right) - \left(1^2 + 1 \ln(1)\right)
\]
Since \(\ln(1) = 0\), this simplifies further:
\[
64 + 8 \ln (8) - 1
\]
The final simplified result is:
\[
63 + 8 \ln (8)
\]
### Conclusion
Thus, the length of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee7991ba-658a-4880-acb7-04815ad6e6cd%2Ffe4f5dd8-a23d-4bf2-8ddf-f08f0140d903%2F7ldajfm.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Length of a Curve by Integration
To find the length of a curve defined by a given integral, we use the formula for arc length in calculus. The provided problem is detailed below with intermediate steps and explanations.
#### Step-by-Step Calculation:
1. **Original Integral Setup:**
\[
s = \int_{1}^{8} \sqrt{4 + \left(\frac{1}{t}\right)^2 + 4t} \, dt
\]
2. **Simplification of the Integrand:**
The integrand is simplified to:
\[
s = \int_{1}^{8} (2t + \frac{1}{t}) \, dt
\]
3. **Integration Process:**
The integral of \(2t + \frac{1}{t}\) is evaluated:
\[
\int (2t + \frac{1}{t}) \, dt = t^2 + t \ln(t)
\]
4. **Evaluating at the Boundaries:**
The result is then evaluated from \(t = 1\) to \(t = 8\):
However, there are errors indicated at these steps in the provided image (marked with red crosses):
- The integration result is:
\[
\left[ t^2 + t \ln (t) \right]_1^8
\]
5. **Simplification:**
The proper evaluation at the boundaries should yield:
\[
t^2 + t \ln (t) \bigg|_1^8
\]
This indicates that we need to substitute \(t = 8\) and \(t = 1\) into the expression and find the final result.
6. **Substitution and Calculation:**
After substituting the values:
\[
\left(8^2 + 8 \ln (8)\right) - \left(1^2 + 1 \ln(1)\right)
\]
Since \(\ln(1) = 0\), this simplifies further:
\[
64 + 8 \ln (8) - 1
\]
The final simplified result is:
\[
63 + 8 \ln (8)
\]
### Conclusion
Thus, the length of
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