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Definite Integrals in Context

Sarah Miller

Sarah Miller

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Study Guide Overview

This study guide covers the definition of the average value of a function over an interval, using the formula involving a definite integral. It introduces the Mean Value Theorem for Integrals, which relates the average value to a constant function with equivalent total change. The guide also includes a geometrical interpretation and worked examples. Key terms like average value and continuous function are explained, and practice questions are provided for reinforcement.

#Average Value of a Function

#Table of Contents

  1. Introduction
  2. Definition
  3. Mean Value Theorem for Integrals
  4. Geometrical Interpretation
  5. Exam Tip
  6. Worked Example
  7. Practice Questions
  8. Glossary
  9. Summary and Key Takeaways

#Introduction

In this section, we will discuss the concept of the average value of a function, its mathematical formulation, and its applications. Understanding this concept is crucial for solving various problems in calculus.

#Definition

Key Concept

The average value of a function fff over the interval [a,b][a, b][a,b] provides a single number that represents the average output of the function over that interval.

If fff is a continuous function, the average value of fff over the interval [a,b][a, b][a,b] is given by:

Average value of f on [a,b]=1b−a∫abf(x),dx\text{Average value of } f \text{ on } [a, b] = \frac{1}{b-a} \int_{a}^{b} f(x) , dxAverage value of f on [a,b]=b−a1​∫ab​f(x),dx

#Mean Value Theorem for Integrals

The average value of a function will be a number kkk such that:

k⋅(b−a)=∫abf(x),dxk \cdot (b - a) = \int_{a}^{b} f(x) , dxk⋅(b−a)=∫ab​f(x),dx

This result is referred to as the Mean Value Theorem for Integrals.

Key Concept

The constant function ggg defined by g(x)=kg(x) = kg(x)=k will represent the same accumulation of change as fff between x=ax = ax=a and x=bx = bx=b because:

∫abk,dx=k[x]ab=k(b−a)\int_{a}^{b} k , dx = k \left[ x \right]_{a}^{b} = k (b - a)∫ab​k,dx=k[x]ab​=k(b−a)

#Geometrical Interpretation

While we have removed the diagram, the geometric interpretation of the Mean Value Theorem for Integrals can be described textually:

Imagine a curve y=f(x)y = f(x)y=f(x) over the interval [a,b][a, b][a,b]. The average value is the height of a rectangle with the same base [a,b][a, b][a,b] whose area is equal to the area under the curve y=f(x)y = f(x)y=f(x). This height represents the average value of the function over that interval.

#Exam Tip

Exam Tip

Remember that you can't talk about the 'average value of a function' in general terms. The average value is only defined for a specific interval [a,b][a, b][a,b], and it will usually be different for different intervals.

#Worked Example

Let fff be the function defined by f(x)=sin⁡xf(x) = \sin xf(x)=sinx. Calculate the average value of fff over the interval [0,π][0, \pi][0,π].

#Solution:

Use the formula for the average value of fff on [a,b][a, b][a,b]:

Average value of f on [0,π]=1π−0∫0πsin⁡x,dx\text{Average value of } f \text{ on } [0, \pi] = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin x , dxAverage value of f on [0,π]=π−01​∫0π​sinx,dx

Calculate the integral:

Average value=1π∫0πsin⁡x,dx=1π[−cos⁡x]0π=1π(−cos⁡(π)−(−cos⁡(0)))=1π(−(−1)−(−1))=1π(1+1)=2π\begin{aligned} \text{Average value} &= \frac{1}{\pi} \int_{0}^{\pi} \sin x , dx \\ &= \frac{1}{\pi} \left[ -\cos x \right]_{0}^{\pi} \\ &= \frac{1}{\pi} \left( -\cos(\pi) - (-\cos(0)) \right) \\ &= \frac{1}{\pi} \left( -(-1) - (-1) \right) \\ &= \frac{1}{\pi} \left( 1 + 1 \right) \\ &= \frac{2}{\pi} \end{aligned}Average value​=π1​∫0π​sinx,dx=π1​[−cosx]0π​=π1​(−cos(π)−(−cos(0)))=π1​(−(−1)−(−1))=π1​(1+1)=π2​​

Therefore, the average value is:

Average value=2π\text{Average value} = \frac{2}{\pi}Average value=π2​

#Practice Questions

Practice Question
  1. Calculate the average value of the function f(x)=x2f(x) = x^2f(x)=x2 over the interval [1,3][1, 3][1,3].
Practice Question
  1. Find the average value of f(x)=exf(x) = e^xf(x)=ex over the interval [0,1][0, 1][0,1].

#Glossary

  • Average Value: A single number representing the average output of a function over a specific interval.
  • Mean Value Theorem for Integrals: A theorem providing a way to find the average value of a function over an interval.
  • Continuous Function: A function that is unbroken or uninterrupted over a given interval.

#Summary and Key Takeaways

  • The average value of a function fff over an interval [a,b][a, b][a,b] is given by 1b−a∫abf(x),dx\frac{1}{b-a} \int_{a}^{b} f(x) , dxb−a1​∫ab​f(x),dx.
  • The Mean Value Theorem for Integrals helps in finding a constant value that represents the same accumulation of change as the function over the interval.
  • The average value of a function is specific to the interval chosen and can vary for different intervals.

Key Takeaways:

  1. The formula for the average value of a function is essential for solving related problems.
  2. Understanding the geometrical interpretation can help visualize the concept.
  3. Always specify the interval when discussing the average value of a function.

Use these notes to reinforce your understanding of the average value of a function and practice with the provided questions to ensure mastery of the concept.

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Question 1 of 7

Alright, let's start with something you know! 🚀 What is the formula for finding the average value of a continuous function f(x)f(x)f(x) over the interval [a,b][a, b][a,b]?

∫abf(x),dx\int_{a}^{b} f(x) , dx∫ab​f(x),dx

ddxf(x)\frac{d}{dx}f(x)dxd​f(x)

1b−a∫abf(x),dx\frac{1}{b-a} \int_{a}^{b} f(x) , dxb−a1​∫ab​f(x),dx

f(b)−f(a)f(b) - f(a)f(b)−f(a)