Finding the Average Value of a Function on an Interval - AP Calculus AB Study Guide

The Core Idea: Finding the Average Value of a Function on an Interval

In many applications, we are interested in the average behavior of a quantity that is changing continuously. For instance, what is the average temperature over a 24-hour period, or the average velocity of a particle over a specific time interval? While we can find the average of a finite set of numbers by summing them and dividing by the count, this method fails for a function that takes on infinitely many values over an interval.

The concept of the average value of a function uses the definite integral to provide a solution. The integral acts as a continuous "sum" of the function's values across an interval $[a,b]$. By dividing this "sum" by the length of the interval, $(b-a)$, we find the mean, or average, value of the function. This provides a single number that represents the central tendency of the function's values over that entire continuous interval.

Key Formulas/Rules/Theorems

The calculation of the average value of a function and its theoretical underpinnings are based on a key formula and a related theorem.

The Average Value Formula

The average value of a function $f(x)$ on a closed interval $[a,b]$ is defined as:

$$f_{avg}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$$

The Mean Value Theorem for Integrals

This theorem provides a crucial link between the average value of a function and the function's actual values. It states:

If a function $f$ is continuous on the closed interval $[a,b]$, then there exists at least one number $c$ in the interval $[a,b]$ such that:

$$f(c)=f_{avg}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$$

In other words, a continuous function must achieve its average value at some point $c$ within the interval.

Understanding the Geometric Interpretation

The Mean Value Theorem for Integrals has a powerful geometric meaning. The definite integral ${\int }_a^{b}f(x)dx$ represents the area under the curve of $f(x)$ from $x=a$ to $x=b$.

The average value, $f_{avg}$, can be visualized as the height of a rectangle with the same base as the interval, $(b-a)$. The area of this rectangle is:

$${\text{Area}}_{\text{rectangle}}=\text{height}\times \text{base}=f_{avg}\cdot (b-a)$$

By substituting the formula for $f_{avg}$, we see:

$${\text{Area}}_{\text{rectangle}}=\left(\frac{1}{b-a}{\int }_a^{b}f(x)dx\right)\cdot (b-a)={\int }_a^{b}f(x)dx$$

This shows that the area of the rectangle formed by the average value is exactly equal to the area under the curve. The Mean Value Theorem for Integrals guarantees that the function's graph must intersect the top of this rectangle ($y=f_{avg}$) at least once within the interval $[a,b]$.

Core Concepts & Rules

• Calculation: The average value of a function $f$ over the interval $[a,b]$ is found by computing the definite integral of $f$ over the interval and then dividing by the length of the interval, $(b-a)$.

• The Formula: The definitive formula to memorize and apply is $f_{avg}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$.

• Existence Guarantee: The Mean Value Theorem for Integrals guarantees that for any function continuous on $[a,b]$, there is a point $c$ in that interval where the function's value $f(c)$ is precisely equal to its average value.

• Geometric Equality: The area under the curve of $f(x)$ from $a$ to $b$ is equal to the area of a rectangle with width $(b-a)$ and height equal to the average value of the function.

Step-by-Step Example 1: Calculating Average Value from a Function

Problem: Find the average value of the function $f(x)=3x^2-2x$ on the interval $[1,4]$.

Step 1: Identify the components of the formula.

• The function is $f(x)=3x^2-2x$.

• The interval is $[1,4]$, so $a=1$ and $b=4$.

• The length of the interval is $b-a=4-1=3$.

Step 2: Set up the average value integral.

Using the formula $f_{avg}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$:

$$f_{avg}=\frac{1}{3}{\int }_1^4(3x^2-2x)dx$$

Step 3: Evaluate the definite integral using the Fundamental Theorem of Calculus.

First, find the antiderivative of $f(x)$:

$$\int (3x^2-2x)dx=x^3-x^2$$

Now, evaluate this antiderivative at the bounds $a=1$ and $b=4$:

$${\left[x^3-x^2\right]}_1^4=(4^3-4^2)-(1^3-1^2)$$

$$=(64-16)-(1-1)$$

$$=48-0=48$$

Step 4: Complete the average value calculation.

Multiply the result of the integral by the $\frac{1}{b-a}$ factor:

$$f_{avg}=\frac{1}{3}\cdot (48)=16$$

Conclusion: The average value of $f(x)=3x^2-2x$ on the interval $[1,4]$ is 16.

Step-by-Step Example 2: Finding the Value 'c' Guaranteed by the MVT for Integrals

Problem: For the function $f(x)=\sqrt{x}$ on the interval $[0,9]$, find the value(s) of $c$ guaranteed by the Mean Value Theorem for Integrals.

Step 1: Calculate the average value of the function on the interval.

• $a=0$, $b=9$, so $b-a=9$.

• Set up the formula: $f_{avg}=\frac{1}{9}{\int }_0^9\sqrt{x}dx=\frac{1}{9}{\int }_0^9{x}^{1/2}dx$.

• Find the antiderivative: $\int x^{1/2}dx=\frac{x^{3/2}}{3/2}=\frac{2}{3}{x}^{3/2}$.

• Evaluate the definite integral:

  $${\left[\frac{2}{3}{x}^{3/2}\right]}_0^9=\frac{2}{3}(9{)}^{3/2}-\frac{2}{3}(0{)}^{3/2}=\frac{2}{3}(27)-0=18$$

• Calculate the average value: $f_{avg}=\frac{1}{9}\cdot (18)=2$.

Step 2: Set $f(c)$ equal to the average value.

The Mean Value Theorem for Integrals states that there is a $c$ such that $f(c)=f_{avg}$.

$$f(c)=2$$

$$\sqrt{c}=2$$

Step 3: Solve for $c$.

Square both sides of the equation:

$$(\sqrt{c}{)}^2=2^2$$

$$c=4$$

Step 4: Verify that $c$ is in the interval $[0,9]$.

The value $c=4$ is indeed between 0 and 9.

Conclusion: The value of $c$ guaranteed by the Mean Value Theorem for Integrals for $f(x)=\sqrt{x}$ on $[0,9]$ is $c=4$.

Using Your Calculator

A graphing calculator is an essential tool for finding the average value of a function, especially when the function is difficult or impossible to antidifferentiate by hand. The key is to use the calculator's numerical integration feature.

Problem: Find the average value of $f(x)=e^{-x^2}$ on the interval $[-1,1]$.

Step 1: Set up the average value formula.

$$f_{avg}=\frac{1}{1-(-1)}{\int }_{-1}^1{e}^{-x^2}dx=\frac{1}{2}{\int }_{-1}^1{e}^{-x^2}dx$$

Step 2: Use the numerical integration function (e.g., fnInt on a TI-84).

The syntax is fnInt(function, variable, lower bound, upper bound).

1. Press MATH and select 9: fnInt(.

2. Enter the expression as follows:

   fnInt(e^(-X^2), X, -1, 1)

3. Press ENTER. The calculator will return a value approximately equal to $1.4936$.

Step 3: Complete the calculation.

Remember to multiply by the $\frac{1}{b-a}$ factor.

$$f_{avg}=\frac{1}{2}\times 1.4936\approx 0.7468$$

Final Answer: The average value is approximately 0.7468. The calculator allows you to compute the average value even for functions whose antiderivatives are not elementary.

AP Exam Quick Hit

Common Question Types

• Direct Calculation (No Calculator): You will be given a function that can be antidifferentiated using basic rules (power rule, $e^x$, $\sin (x)$, etc.) and asked to find its average value on an interval.

  • Example: "Find the average value of $f(x)=\frac{1}{x}$ on the interval $[1,e]$."

• Contextual Problem (Calculator): You will be given a function representing a rate (e.g., gallons/min, people/hour) and asked to find the average rate over a time interval. This is a direct application of the average value formula.

  • Example: "The rate of water leaking from a tank is given by $R(t)=20e^{-0.1t}$ liters per hour. Find the average rate of leakage, in liters per hour, from $t=0$ to $t=10$."

• Finding $c$ from the MVT for Integrals: After finding the average value, you will be asked to find the value of $c$ where the function equals its average value.

  • Example: "Let $f_{avg}$ be the average value of $f(x)=x^2$ on $[0,3]$. Find a value $c$ in $[0,3]$ such that $f(c)=f_{avg}$."

Common Mistakes

• Forgetting the $\frac{1}{b-a}$ Factor: The most common mistake is to correctly calculate the definite integral ${\int }_a^{b}f(x)dx$ but forget to divide by the length of the interval, $(b-a)$.

• Confusing Average Value and Average Rate of Change: Students often mix up the formula for average value ($\frac{1}{b-a}{\int }_a^{b}f(x)dx$) with the formula for average rate of change ($\frac{f(b)-f(a)}{b-a}$). Read the question carefully to determine which is being asked.

• Incorrect Interval Length: A simple arithmetic error in calculating $(b-a)$, especially with negative numbers. For the interval $[-3,5]$, the length is $5-(-3)=8$, not $5-3=2$.

• Ignoring the Interval for $c$: When solving $f(c)=f_{avg}$, you may get multiple solutions for $c$. You must choose only the solution(s) that lie within the given interval $[a,b]$.