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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**Average Value of a Function**
**Problem Statement:**
Find the average value of the function \( f(x) = x^2 - 9 \) on the interval \([0, 3]\).
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**Solution:**
The average value of the function \( f(x) = x^2 - 9 \) on \([0, 3]\) is given by
\[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
where \( a = 0 \) and \( b = 3 \).
\[ \text{Average value} = \frac{1}{3-0} \int_0^3 (x^2 - 9) \, dx \]
\[ = \frac{1}{3} \left[ \int_0^3 x^2 \, dx - \int_0^3 9 \, dx \right] \]
\[ = \frac{1}{3} \left[ \left. \frac{x^3}{3} \right|_0^3 - 9(x) \bigg|_0^3 \right] \]
\[ = \frac{1}{3} \left[ \frac{3^3}{3} - \frac{0^3}{3} - (9 \times 3 - 9 \times 0) \right] \]
\[ = \frac{1}{3} \left[ \frac{27}{3} - 27 \right] \]
\[ = \frac{1}{3} \left[ 9 - 27 \right] \]
\[ = \frac{1}{3} \left[ -18 \right] \]
\[ = -6 \]
---
The average value of the function \( f(x) = x^2 - 9 \) on \([0, 3]\) is \(-6\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe97fa865-4410-453a-83a1-c6f2a5dd62aa%2F02f7c349-8642-4e89-90a0-126151941465%2Furoyhv6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
**Average Value of a Function**
**Problem Statement:**
Find the average value of the function \( f(x) = x^2 - 9 \) on the interval \([0, 3]\).
---
**Solution:**
The average value of the function \( f(x) = x^2 - 9 \) on \([0, 3]\) is given by
\[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
where \( a = 0 \) and \( b = 3 \).
\[ \text{Average value} = \frac{1}{3-0} \int_0^3 (x^2 - 9) \, dx \]
\[ = \frac{1}{3} \left[ \int_0^3 x^2 \, dx - \int_0^3 9 \, dx \right] \]
\[ = \frac{1}{3} \left[ \left. \frac{x^3}{3} \right|_0^3 - 9(x) \bigg|_0^3 \right] \]
\[ = \frac{1}{3} \left[ \frac{3^3}{3} - \frac{0^3}{3} - (9 \times 3 - 9 \times 0) \right] \]
\[ = \frac{1}{3} \left[ \frac{27}{3} - 27 \right] \]
\[ = \frac{1}{3} \left[ 9 - 27 \right] \]
\[ = \frac{1}{3} \left[ -18 \right] \]
\[ = -6 \]
---
The average value of the function \( f(x) = x^2 - 9 \) on \([0, 3]\) is \(-6\).
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