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

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

The concept of an "average" is typically associated with a finite set of numbers. However, calculus provides a tool to determine the average value of a function, which has infinitely many values over a continuous interval. The core idea is to find a single representative y-value, or height, that encapsulates the function's behavior across the entire interval.

Geometrically, the average value of a function $f(x)$ on an interval $[a,b]$ is the height of a rectangle with base $(b-a)$ that has the exact same area as the area under the curve of $f(x)$ from $x=a$ to $x=b$. This provides a way to "flatten" the curve into a single, constant value that represents the function's mean height over that domain. This concept is formalized through a definite integral, which measures the total accumulation under the curve, divided by the length of the interval.

Key Formulas

The average value of a function is defined by a specific formula derived from the relationship between the area under a curve and the area of a rectangle.

The Average Value of a Function

The average value of a function $f$ over a closed interval $[a,b]$ is given by the formula:

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

Where:

• $f(x)$ is the function being averaged.

• $[a,b]$ is the closed interval over which the average is being calculated.

• ${\int }_a^{b}f(x)dx$ represents the definite integral of $f$ from $a$ to $b$, which calculates the net area under the curve.

• $(b-a)$ is the length of the interval.

Understanding the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals provides a crucial link between the average value of a function and the function's actual values. It guarantees the existence of a point within the interval where the function achieves its average value.

The Theorem

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

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

In other words, $f(c)$ is equal to the average value of the function on that interval.

The key condition for this theorem is the continuity of the function $f$ on $[a,b]$. If the function is continuous, it does not have any jumps, holes, or vertical asymptotes within the interval. This continuity ensures that the function must pass through its average value at some point. Geometrically, this means that if you draw a horizontal line at the height of the average value ($y=f_{avg}$), that line is guaranteed to intersect the graph of $f(x)$ at least once within the interval $[a,b]$.

Core Concepts & Rules

• Definition of Average Value: The average value of a function $f$ on an interval $[a,b]$ is calculated by finding the definite integral of the function over that interval and then dividing by the length of the interval, $(b-a)$.

• Geometric Interpretation: The average value represents the height of a rectangle with a base spanning the interval $[a,b]$ whose area is identical to the net area under the curve of $f(x)$ on that same interval.

• The Mean Value Theorem for Integrals (MVT for Integrals): This theorem states that for any function $f$ that is continuous on a closed interval $[a,b]$, there is a point $c$ within that interval where the function's value, $f(c)$, is exactly equal to its average value on $[a,b]$.

• Condition for MVT for Integrals: The guarantee provided by the MVT for Integrals only applies if the function is continuous over the entire closed interval $[a,b]$.

Step-by-Step Example 1: Calculating Average Value

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

Step 1: Identify the function and interval.

The function is $f(x)=3x^2-2x$.

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

Step 2: Set up the average value formula.

Substitute the function and interval bounds into the formula $f_{avg}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$.

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

Step 3: Evaluate the definite integral.

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

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

Next, apply the Fundamental Theorem of Calculus using the bounds $[1,4]$.

$${\int }_1^4(3x^2-2x)dx=[x^3-x^2{]}_1^4$$

$$=(4^3-4^2)-(1^3-1^2)$$

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

$$=48-0=48$$

Step 4: Complete the calculation.

Multiply the result of the integral by the $\frac{1}{b-a}$ term.

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

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

Step-by-Step Example 2: Applying 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: Verify the condition of the theorem.

The function $f(x)=\sqrt{x}$ is continuous on the closed interval $[0,9]$. Therefore, the MVT for Integrals applies.

Step 2: Calculate the average value of the function.

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

$$f_{avg}=\frac{1}{9-0}{\int }_0^9\sqrt{x}dx=\frac{1}{9}{\int }_0^9{x}^{1/2}dx$$

Evaluate the integral:

$$\frac{1}{9}{\left[\frac{x^{3/2}}{3/2}\right]}_0^9=\frac{1}{9}{\left[\frac{2}{3}{x}^{3/2}\right]}_0^9$$

$$=\frac{1}{9}\left(\frac{2}{3}(9{)}^{3/2}-\frac{2}{3}(0{)}^{3/2}\right)$$

$$=\frac{1}{9}\left(\frac{2}{3}(27)-0\right)=\frac{1}{9}(18)=2$$

The average value of the function on $[0,9]$ is 2.

Step 3: Set $f(c)$ equal to the average value.

The theorem guarantees a $c$ such that $f(c)=f_{avg}$.

$$\sqrt{c}=2$$

Step 4: Solve for $c$.

Square both sides of the equation.

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

$$c=4$$

Step 5: Verify that $c$ is in the interval.

The value $c=4$ is in the interval $[0,9]$.

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 integral is difficult or impossible to compute by hand. The calculator is primarily used to evaluate the definite integral part of the formula.

To find the average value of $f(x)$ on $[a,b]$:

1. Access the numerical integration function. On a TI-84 style calculator, this is often $math$ -> 9: fnInt(.

2. Enter the arguments. The syntax is typically $fnInt(function,variable,lowerbound,upperbound)$.

   • For example, to find ${\int }_1^4(3x^2-2x)dx$, you would enter fnInt(3X^2-2X, X, 1, 4).

3. Execute the command. The calculator will return the value of the definite integral (in this case, $48$).

4. Divide by the interval length. Manually divide the calculator's result by $(b-a)$.

   • $48/(4-1)=48/3=16$.

To find the value of $c$ from the MVT for Integrals:

1. Calculate the average value as described above. Let this value be $k$.

2. Graph the functions. In your calculator's graphing menu, set Y1 = f(x) and Y2 = k.

3. Find the intersection. Use the calculator's "intersect" feature (2nd -> TRACE -> $5:intersect$) to find the point(s) where the two graphs cross within the window $x_{m}in=a$ and $x_{m}ax=b$. The x-coordinate of any such intersection is a valid value for $c$.

AP Exam Quick Hit

Common Question Types

• Direct Calculation: A multiple-choice or free-response question may ask directly for the average value of a given function on an interval.

  • Example: "What is the average value of $f(x)=e^x$ on the interval $[0,2]$?"

• Application of MVT for Integrals: A question may ask for the value $c$ that satisfies the conclusion of the Mean Value Theorem for Integrals.

  • Example: "For what value of $c$ on the interval $[0,3]$ is $f(c)$ equal to the average value of $f(x)=x^2$?"

• Contextual Problem (FRQ): A free-response question may present a scenario where you must recognize the need to calculate an average value.

  • Example: "The velocity of a particle is given by $v(t)=t^2\sin (t^3)$. Find the average velocity of the particle from $t=0$ to $t=3$."

Common Mistakes

• Forgetting to Divide by $(b-a)$: The most common mistake is to correctly calculate the definite integral ${\int }_a^{b}f(x)dx$ but forget to multiply by the $\frac{1}{b-a}$ factor. This finds the total accumulation, not the average value.

• 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, $\frac{f(b)-f(a)}{b-a}$. Remember that average value involves an integral, while average rate of change does not.

• Incorrect Interval Length: Making an arithmetic error when calculating $b-a$, especially with negative numbers. For the interval $[-3,2]$, the length is $2-(-3)=5$, not 2-3 = -1`. - **MVT for Integrals vs. MVT for Derivatives:** Confusing the conclusion of the MVT for Integrals ($f(c) = f_{avg}) with the conclusion of the MVT for Derivatives ($f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$).

• Forgetting the Interval for $c$: When solving for $c$ in an MVT for Integrals problem, you may find multiple solutions. A common error is failing to discard solutions that fall outside the specified interval $[a,b]$.