College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter2: Functions
Section: Chapter Questions
Problem 20P: In these problem you are asked to find a function that models a real-life situation and then use the...
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![### Calculating the Average Value of a Function
**Problem Statement:**
Find the average value of the function \( f \) over the interval \([0, 15]\).
The function \( f(x) \) is defined as:
\[ f(x) = \frac{4}{x + 1} \]
**Step-by-Step Solution:**
1. **Understand the Formula:**
The average value of a function \( f(x) \) over the interval \([a, b]\) is given by:
\[
\text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx
\]
2. **Substitute the Given Values:**
Here \( a = 0 \) and \( b = 15 \). Substitute these into the formula:
\[
\text{Average value} = \frac{1}{15 - 0} \int_{0}^{15} \frac{4}{x + 1} \, dx
\]
Simplifying the constant term, we get:
\[
\text{Average value} = \frac{1}{15} \int_{0}^{15} \frac{4}{x + 1} \, dx
\]
3. **Compute the Integral:**
To compute the integral, use the substitution method. Let \( u = x + 1 \) hence \( du = dx \).
When \( x = 0 \), \( u = 1 \).
When \( x = 15 \), \( u = 16 \).
Rewriting the integral in terms of \( u \):
\[
\int_{0}^{15} \frac{4}{x + 1} \, dx = \int_{1}^{16} \frac{4}{u} \, du
\]
The integral of \( \frac{4}{u} \) is \( 4 \ln|u| \):
\[
\int_{1}^{16} \frac{4}{u} \, du = 4 \ln|u| \Big|_{1}^{16}
\]
4. **Evaluate the Antiderivative:**
Substituting the limits into the antiderivative:
\[
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8332359c-bbef-4e4d-b367-093a92f5e054%2Faffdba01-71d9-46d3-b2c3-549dc43f7968%2Fh000x09_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Average Value of a Function
**Problem Statement:**
Find the average value of the function \( f \) over the interval \([0, 15]\).
The function \( f(x) \) is defined as:
\[ f(x) = \frac{4}{x + 1} \]
**Step-by-Step Solution:**
1. **Understand the Formula:**
The average value of a function \( f(x) \) over the interval \([a, b]\) is given by:
\[
\text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx
\]
2. **Substitute the Given Values:**
Here \( a = 0 \) and \( b = 15 \). Substitute these into the formula:
\[
\text{Average value} = \frac{1}{15 - 0} \int_{0}^{15} \frac{4}{x + 1} \, dx
\]
Simplifying the constant term, we get:
\[
\text{Average value} = \frac{1}{15} \int_{0}^{15} \frac{4}{x + 1} \, dx
\]
3. **Compute the Integral:**
To compute the integral, use the substitution method. Let \( u = x + 1 \) hence \( du = dx \).
When \( x = 0 \), \( u = 1 \).
When \( x = 15 \), \( u = 16 \).
Rewriting the integral in terms of \( u \):
\[
\int_{0}^{15} \frac{4}{x + 1} \, dx = \int_{1}^{16} \frac{4}{u} \, du
\]
The integral of \( \frac{4}{u} \) is \( 4 \ln|u| \):
\[
\int_{1}^{16} \frac{4}{u} \, du = 4 \ln|u| \Big|_{1}^{16}
\]
4. **Evaluate the Antiderivative:**
Substituting the limits into the antiderivative:
\[
4
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