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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![### Finding the Average Value of a Function
To determine the average value of the function \( f(x) = 8 \sin x + 3 \cos x \) on the interval \([0, \frac{11}{3} \pi]\), use the formula for the average value of a function over an interval \([a, b]\):
\[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]
Here, the interval is \([0, \frac{11}{3} \pi]\), so \( a = 0 \) and \( b = \frac{11}{3} \pi \).
### Steps to Find the Average Value:
1. **Identify the Interval and the Function:**
- Interval: \([0, \frac{11}{3} \pi]\)
- Function: \( f(x) = 8 \sin x + 3 \cos x \)
2. **Set Up the Integral:**
\[ \int_{0}^{\frac{11}{3} \pi} (8 \sin x + 3 \cos x) \, dx \]
3. **Calculate the Integral:**
- Find the antiderivative of \( f(x) \):
\[ \int (8 \sin x + 3 \cos x) \, dx = -8 \cos x + 3 \sin x + C \]
- Evaluate the definite integral from 0 to \(\frac{11}{3} \pi\):
\[ \left[ -8 \cos x + 3 \sin x \right]_{0}^{\frac{11}{3} \pi} \]
4. **Evaluate the Antiderivative at the Bounds:**
- At \( x = \frac{11}{3} \pi \):
\[ -8 \cos \left( \frac{11}{3} \pi \right) + 3 \sin \left( \frac{11}{3} \pi \right) \]
- At \( x = 0 \):
\[ -8 \cos (0) + 3 \sin (0) \]
5. **Substitute and Simplify:**
\[ \left( -8 \cos \left( \frac{11}{3}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb2133c9-e1e5-4d56-9c72-044227328930%2F81b48c4d-c12b-4167-8120-87951f0d0af0%2Fcozc4l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Average Value of a Function
To determine the average value of the function \( f(x) = 8 \sin x + 3 \cos x \) on the interval \([0, \frac{11}{3} \pi]\), use the formula for the average value of a function over an interval \([a, b]\):
\[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]
Here, the interval is \([0, \frac{11}{3} \pi]\), so \( a = 0 \) and \( b = \frac{11}{3} \pi \).
### Steps to Find the Average Value:
1. **Identify the Interval and the Function:**
- Interval: \([0, \frac{11}{3} \pi]\)
- Function: \( f(x) = 8 \sin x + 3 \cos x \)
2. **Set Up the Integral:**
\[ \int_{0}^{\frac{11}{3} \pi} (8 \sin x + 3 \cos x) \, dx \]
3. **Calculate the Integral:**
- Find the antiderivative of \( f(x) \):
\[ \int (8 \sin x + 3 \cos x) \, dx = -8 \cos x + 3 \sin x + C \]
- Evaluate the definite integral from 0 to \(\frac{11}{3} \pi\):
\[ \left[ -8 \cos x + 3 \sin x \right]_{0}^{\frac{11}{3} \pi} \]
4. **Evaluate the Antiderivative at the Bounds:**
- At \( x = \frac{11}{3} \pi \):
\[ -8 \cos \left( \frac{11}{3} \pi \right) + 3 \sin \left( \frac{11}{3} \pi \right) \]
- At \( x = 0 \):
\[ -8 \cos (0) + 3 \sin (0) \]
5. **Substitute and Simplify:**
\[ \left( -8 \cos \left( \frac{11}{3}
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