Finding Where a Function Equals Its Average Value

From an Average to an Attained Value

The average value of a function over an interval summarizes its accumulated output as a single constant height. A natural question follows: does the function actually take that height at some input?

For a continuous function on a closed interval of positive length, the answer is yes. This existence result is the Mean Value Theorem for Integrals. It connects an integral, which measures accumulation across an interval, to a value the function attains at one particular point.

Central idea: If $f$ is continuous on $[a,b]$, where $a<b$, then at least one point $c$ in the interval satisfies $f(c)=f_{\text{avg}}$, where $f_{\text{avg}}$ is the average value of $f$ on $[a,b]$.

Statement of the Theorem

For a continuous function $f$ on $[a,b]$, with $a<b$, its average value on that interval is

f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.

The integral in this formula is the signed integral. The interval length $b-a$ is positive, so dividing by it gives the function’s average value over the interval.

Theorem: Mean Value Theorem for Integrals. If $f$ is continuous on $[a,b]$, where $a<b$, then there exists at least one $c\in[a,b]$ such that $$ f(c)=\frac{1}{b-a}\int_a^b f(x)\,dx. $$

Equivalently, multiplying both sides by $b-a$, the conclusion can be written as

\int_a^b f(x)\,dx=f(c)(b-a).

This says that the accumulated value of $f$ over the interval equals the accumulation of a constant function whose height is $f(c)$. The theorem asserts existence: at least one such point is guaranteed. It does not say that the point is unique or provide a formula for locating it.

Why the Theorem Is True

The proof combines two facts about continuous functions. First, on a closed interval, a continuous function has a minimum and a maximum. Second, it takes every value between its minimum and maximum. The average value must lie between those extremes, so the function must attain it.

Take the minimum and maximum.
Because $f$ is continuous on the closed interval $[a,b]$, the Extreme Value Theorem guarantees points where $f$ reaches a minimum $m$ and a maximum $M$. Thus $m\leq f(x)\leq M$ for every $x\in[a,b]$.

Integrate the bounds.
Since $a<b$, integrating $m\leq f(x)\leq M$ over $[a,b]$ gives $m(b-a)\leq\int_a^b f(x)\,dx\leq M(b-a)$.

Divide by the interval length.
Since $b-a>0$, division preserves the inequalities and gives $m\leq\frac{1}{b-a}\int_a^b f(x)\,dx\leq M$. The average value is between the minimum and maximum.

Apply the Intermediate Value Theorem.
The continuous function takes every value between $m$ and $M$. Since $f_{\text{avg}}$ lies between them, some $c\in[a,b]$ satisfies $f(c)=f_{\text{avg}}$.

That establishes the result. The continuity hypothesis matters in the final step: without continuity, a function can have values below and above its average but still skip the average value itself.

Worked Examples

Worked Example: A Quadratic on an Interval

Show that $f(x)=x^2$ takes its average value on $[0,3]$, and verify one point where it does.

The function $x^2$ is continuous on $[0,3]$, so the theorem applies. First calculate the average value:

f_{\text{avg}} =\frac{1}{3-0}\int_0^3 x^2\,dx =\frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 =\frac{1}{3}\left(\frac{27}{3}-0\right) =\frac{1}{3}(9) =3.

To verify an input, solve $x^2=3$. The solutions are $x=\sqrt{3}$ and $x=-\sqrt{3}$, but only $\sqrt{3}$ belongs to $[0,3]$. Since $0<\sqrt{3}<3$, it is in the interval, and

f(\sqrt{3})=(\sqrt{3})^2=3=f_{\text{avg}}.

Thus $c=\sqrt{3}$ verifies the theorem’s conclusion for this function and interval.

Worked Example: A Linear Function

For $f(x)=2x+1$ on $[1,4]$, calculate the average value and check that the function attains it.

The function is continuous on the closed interval. Its average value is

\begin{aligned} f_{\text{avg}} &=\frac{1}{4-1}\int_1^4(2x+1)\,dx\\ &=\frac{1}{3}[x^2+x]_1^4\\ &=\frac{1}{3}\bigl((16+4)-(1+1)\bigr)\\ &=\frac{1}{3}(20-2) =\frac{18}{3} =6. \end{aligned}

Set the function equal to this average and solve:

2x+1=6 \quad\Longrightarrow\quad 2x=5 \quad\Longrightarrow\quad x=\frac{5}{2}.

The input lies in the interval because $1<\frac{5}{2}<4$. Substitution checks the result:

f\left(\frac{5}{2}\right) =2\left(\frac{5}{2}\right)+1 =5+1 =6 =f_{\text{avg}}.

Worked Example: Two Inputs Attain the Average

For $f(x)=(x-1)^2$ on $[0,2]$, find the average value and verify points where the function takes it.

This polynomial is continuous on $[0,2]$. Calculate its average:

\begin{aligned} f_{\text{avg}} &=\frac{1}{2-0}\int_0^2(x-1)^2\,dx\\ &=\frac{1}{2}\int_0^2(x^2-2x+1)\,dx\\ &=\frac{1}{2}\left[\frac{x^3}{3}-x^2+x\right]_0^2\\ &=\frac{1}{2}\left(\frac{8}{3}-4+2\right) =\frac{1}{2}\left(\frac{2}{3}\right) =\frac{1}{3}. \end{aligned}

Solve $(x-1)^2=\frac{1}{3}$. Taking both square roots gives

x-1=\pm\frac{1}{\sqrt{3}} \quad\Longrightarrow\quad x=1\pm\frac{1}{\sqrt{3}}.

Both inputs lie in $[0,2]$, since $0<1-\frac{1}{\sqrt{3}}<1$ and $1<1+\frac{1}{\sqrt{3}}<2$. For either choice, substitution gives

f\left(1\pm\frac{1}{\sqrt{3}}\right) =\left(1\pm\frac{1}{\sqrt{3}}-1\right)^2 =\left(\pm\frac{1}{\sqrt{3}}\right)^2 =\frac{1}{3} =f_{\text{avg}}.

Here the average is attained at two inputs. The theorem guarantees at least one, not exactly one.

What the Theorem Guarantees—and What It Does Not

The theorem is useful even when the point $c$ cannot be found easily. Knowing that a continuous function attains its average can support conclusions about its behavior on an interval without requiring an exact location.

It is important to separate the theorem’s guarantee from stronger claims it does not make:

It guarantees existence: some input in the interval has function value equal to the average.
It does not guarantee uniqueness: as the quadratic example shows, more than one input may work.
It does not identify the input: an exact value of $c$ may require additional calculation.
It requires continuity on the whole closed interval: endpoint values alone or continuity on only part of the interval do not satisfy the stated hypothesis.

A Common Pitfall: Ignoring Continuity

A tempting but incorrect claim is that every function must take its average value somewhere. Consider the function on $[-1,1]$ defined by

f(x)= \begin{cases} -1,&-1\leq x<0,\\ 1,&0\leq x\leq1. \end{cases}

Its integral is the sum of the contributions from the two subintervals:

\int_{-1}^{1}f(x)\,dx =\int_{-1}^{0}(-1)\,dx+\int_0^1 1\,dx =(-1)(1)+(1)(1) =0.

The interval has length $2$, so its average value is $\frac{0}{2}=0$. But the function only takes the values $-1$ and $1$; there is no input $c$ for which $f(c)=0$. The function is discontinuous at $0$, so the theorem’s hypothesis is not met. The failed conclusion does not contradict the theorem.

Takeaway: The Mean Value Theorem for Integrals is an existence result built on continuity. Before using it, the function must be continuous on the entire closed interval, and the interval must have positive length.

Why This Result Matters

An average value is calculated from information over an entire interval, while $f(c)$ describes the function at one input. The theorem guarantees a link between these two kinds of information: for a continuous function, the interval-wide average is not merely a number produced by an integral; it is a value the function actually reaches.

This connection is especially useful in applications. If a continuous rate has an average rate over a time interval, the theorem guarantees that the instantaneous rate matches that average at some time in the interval. The theorem alone does not identify that time, but it justifies saying such a time exists.

Check Your Understanding

Use the theorem’s statement and reasoning to answer these questions.

State the hypotheses and conclusion of the Mean Value Theorem for Integrals.
Why must the average value lie between the minimum and maximum of a continuous function on a closed interval?
For $f(x)=x+2$ on $[0,4]$, calculate the average value and find an input where the function takes it.
Does the theorem guarantee exactly one input where the average is attained? Explain.
In the discontinuous example, what is the average value, and why does the theorem not apply?

Finding Where a Function Equals Its Average Value

What You'll Learn

From an Average to an Attained Value

Statement of the Theorem

Why the Theorem Is True

Worked Examples

Worked Example: A Quadratic on an Interval

Worked Example: A Linear Function

Worked Example: Two Inputs Attain the Average

What the Theorem Guarantees—and What It Does Not

A Common Pitfall: Ignoring Continuity

Why This Result Matters

Check Your Understanding

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Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Finding Where a Function Equals Its Average Value

What You'll Learn

From an Average to an Attained Value

Statement of the Theorem

Why the Theorem Is True

Worked Examples

Worked Example: A Quadratic on an Interval

Worked Example: A Linear Function

Worked Example: Two Inputs Attain the Average

What the Theorem Guarantees—and What It Does Not

A Common Pitfall: Ignoring Continuity

Why This Result Matters

Check Your Understanding

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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