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Finding the Average Value of a Function on an Interval

Abigail Young

Abigail Young

5 min read

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Study Guide Overview

This study guide covers the average value of a function on an interval using definite integrals. It presents the formula for calculating the average value, provides a step-by-step walkthrough with an example, and offers practice problems with solutions. The guide emphasizes the application of this concept in solving accumulation of change problems and its relevance to AP Calculus FRQs.

8.1 Finding the Average Value of a Function on an Interval

Welcome back to AP Calculus with Fiveable! This topic focuses on finding the average value of a continuous function using definite integrals.

🔢 Average Value of a Function

The average value of a function will allow us to solve problems that involve the accumulation of change over an interval, which will later be used to understand more difficult topics of integration.

For questions that require the average value of a function, we are never given a finite number of data points. Therefore, we must use integration to determine what the average value is.

This idea is fairly simple once you memorize a key piece of information: if f is continuous on [a,b],[a,b], then the average value of f on [a,b[a,b] is the following.

Average Value=1b−a∫abf(x),dx\text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) , dx

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Image of average value equation and corresponding graph.

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Previous Topic - Applications of IntegrationNext Topic - Connecting Position, Velocity, and Acceleration of Functions Using Integrals

Question 1 of 6

What is the correct setup to find the average value of the function f(x)=2x+1f(x) = 2x + 1 on the interval [0,3][0, 3]? 🤔

13∫03(2x+1),dx\frac{1}{3} \int_{0}^{3} (2x + 1) , dx

12∫03(2x+1),dx\frac{1}{2} \int_{0}^{3} (2x + 1) , dx

∫03(2x+1),dx\int_{0}^{3} (2x + 1) , dx

13∫14(2x+1),dx\frac{1}{3} \int_{1}^{4} (2x + 1) , dx