Why an Average Lies Between the Minimum and Maximum

Why an Average Has a Natural Bound

An average summarizes values across an interval. It would be surprising for that summary to be smaller than every value being summarized, or larger than every one. For a continuous function on a closed interval, this intuition is a precise theorem: the average value lies between the function’s minimum and maximum.

The result is useful both as a fact about averages and as a quick check on calculations. If a computed average falls outside the function’s range, at least one part of the calculation or setup must be wrong.

Definition and Assumptions

Let $f$ be continuous on a closed interval $[a,b]$, where $a<b$. Its average value on that interval is the integral of the function divided by the interval’s length:

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

The interval length $b-a$ is positive, so dividing by it preserves inequalities. Continuity on the closed, bounded interval also guarantees that $f$ attains both a minimum and a maximum there. Write those values as

m=\min_{x\in[a,b]}f(x) \qquad\text{and}\qquad M=\max_{x\in[a,b]}f(x).

In particular, $m\le f(x)\le M$ for every $x$ in the interval. The continuity and closed-interval assumptions matter: they ensure that these minimum and maximum values exist and that the integral is defined.

Theorem: If $f$ is continuous on $[a,b]$ with $a<b$, then its average value satisfies $$ m\le f_{\mathrm{avg}}\le M, $$ where $m$ and $M$ are the minimum and maximum of $f$ on $[a,b]$.

Why the Bounds Hold

The proof uses two basic facts about definite integrals: inequalities between continuous functions are preserved when integrated over the same interval, and division by a positive number preserves an inequality.

Bound the function pointwise.
By the definitions of minimum and maximum, $m\le f(x)\le M$ throughout $[a,b]$.

Integrate the bounds.
Integrating the pointwise inequalities over $[a,b]$ gives $m(b-a)\le\int_a^b f(x)\,dx\le M(b-a)$.

Divide by the interval length.
Since $b-a>0$, division by $b-a$ yields $m\le f_{\mathrm{avg}}\le M$.

To see explicitly why the integration step preserves order, $f(x)-m\ge0$ and $M-f(x)\ge0$ at every point. The integral of a nonnegative continuous function is nonnegative. Therefore,

\int_a^b (f(x)-m)\,dx\ge0 \qquad\text{and}\qquad \int_a^b (M-f(x))\,dx\ge0.

Expanding each integral gives

\int_a^b f(x)\,dx-m(b-a)\ge0, \qquad M(b-a)-\int_a^b f(x)\,dx\ge0.

Rearranging these two inequalities gives the integrated bounds in Step 2. Dividing by the positive interval length and using the definition of $f_{\mathrm{avg}}$ completes the proof. The argument does not depend on whether $f$ is positive or negative; it depends only on the function’s lower and upper bounds.

Worked Examples

Worked Example: A Quadratic on an Interval

Find the average value of $f(x)=x^2$ on $[-1,2]$, then check it against the minimum and maximum.

The square is nonnegative, so the minimum is $m=0$, attained at $x=0$. On this interval, the largest square is $4$, attained at $x=2$, so $M=4$. The interval length is $2-(-1)=3$.

Compute the integral:

\int_{-1}^{2}x^2\,dx =\left[\frac{x^3}{3}\right]_{-1}^{2} =\frac{2^3}{3}-\frac{(-1)^3}{3} =\frac{8}{3}+\frac{1}{3} =3.

Divide by the interval length:

f_{\mathrm{avg}}=\frac{1}{3}\cdot 3=1.

The bound predicts $0\le f_{\mathrm{avg}}\le4$, and the computed value satisfies it: $0\le1\le4$. As a direct check of the average calculation, $3\cdot1=3$, which is the integral over the interval.

Worked Example: A Trigonometric Function

Find the average value of $f(x)=2+\sin x$ on $[0,\pi/2]$, and verify that it lies within the function’s range.

On this interval, $0\le\sin x\le1$. Hence $2\le f(x)\le3$, so the minimum is $2$ and the maximum is $3$. The interval length is $\pi/2$.

Integrate and evaluate at the endpoints:

\int_0^{\pi/2}(2+\sin x)\,dx =\left[2x-\cos x\right]_0^{\pi/2} =\left(\pi-0\right)-\left(0-1\right) =\pi+1.

Now divide by the interval length:

f_{\mathrm{avg}} =\frac{\pi+1}{\pi/2} =2+\frac{2}{\pi}.

Because $\pi>2$, we have $0<2/\pi<1$. Adding $2$ gives $2<2+2/\pi<3$, as required by the bounds. Multiplying the result by the interval length also recovers the integral:

\frac{\pi}{2}\left(2+\frac{2}{\pi}\right)=\pi+1.

Worked Example: Checking an Average Rate

A rate is given by $r(t)=3t+1$ for $1\le t\le3$. Find its average value and use the range bound to check the calculation.

Since $r$ increases on this interval, its minimum is $r(1)=3(1)+1=4$, and its maximum is $r(3)=3(3)+1=10$. Thus any average value must be between $4$ and $10$. The interval length is $3-1=2$.

The accumulated total over the interval is

\int_1^3(3t+1)\,dt =\left[\frac{3}{2}t^2+t\right]_1^3 =\left(\frac{27}{2}+3\right)-\left(\frac{3}{2}+1\right) =\frac{33}{2}-\frac{5}{2} =14.

Divide the total by the interval length:

r_{\mathrm{avg}}=\frac{14}{2}=7.

The result passes the bound check, $4\le7\le10$. It also passes the total check: $7\cdot2=14$, the integral just calculated. If the rate is measured in units per second and $t$ in seconds, the integral has units of the measured quantity, while the average rate has units of units per second.

Using the Bound as a Calculation Check

Before or after evaluating an average, find the function’s range on the interval whenever it is reasonably straightforward. The theorem then supplies a necessary test for the result. This test does not determine the average by itself: many values may lie between the minimum and maximum. But it can rule out an impossible answer immediately.

For example, in the rate calculation above, the integral $14$ is a total accumulated quantity, not the average rate. Calling $14$ the average would fail the range test because $14>10$. Dividing the total by the interval length gives $7$, which is within the permitted range.

The check is also meaningful when the function takes negative values. The minimum and maximum are still the least and greatest function values, and the same inequality proof applies. A negative average is not inherently a problem; it is a problem only if it falls below the minimum or above the maximum.

A Common Mistake: Forgetting to Divide by the Length

The integral $\int_a^b f(x)\,dx$ is the accumulated total, while the average value is that total divided by $b-a$. Confusing the two can produce a result with the wrong size and, when units are involved, the wrong units.

For $r(t)=3t+1$ on $[1,3]$, a mistaken calculation might label the integral itself as the average:

r_{\mathrm{avg}}\mathrel{\color{red}{\not=}}\int_1^3(3t+1)\,dt=14.

This answer is impossible: every value of the rate is between $4$ and $10$, so its average cannot be $14$. The correction is to divide the integral by the interval length $2$:

r_{\mathrm{avg}}=\frac{1}{3-1}\int_1^3(3t+1)\,dt=\frac{14}{2}=7, \qquad 4\le7\le10.

A bound check is a useful safeguard, but it does not replace checking the setup, integration, and interval length. A wrong answer can still lie between the minimum and maximum.

What the Bound Tells You—and What It Does Not

The theorem gives a necessary range for the average and provides a justification for rejecting values outside it. If the minimum and maximum are close, the bound can be especially informative. If they are far apart, it may only provide a broad check.

The inequalities allow equality. For instance, if $f$ is the constant function $f(x)=c$, then its minimum, maximum, and average are all $c$. In general, the bound alone does not tell you where, or how often, the function takes its average value. Its purpose here is to locate the average within the function’s range and to check calculations against that range.

Key takeaway: For a continuous function on $[a,b]$, first identify its minimum $m$ and maximum $M$. The computed average must satisfy $m\le\frac{1}{b-a}\int_a^b f(x)\,dx\le M$.

Check Your Understanding

Use the definition and the range bound to answer each question.

What assumptions ensure that the minimum and maximum used in the theorem exist?
If $f$ is continuous on $[2,5]$ and $1\le f(x)\le6$ throughout the interval, what bounds must its average satisfy?
For $f(x)=x^2$ on $[0,2]$, identify the minimum and maximum. Without calculating the integral, give the range in which the average must lie.
Why does dividing the integrated inequalities by $b-a$ preserve their direction?
A function has values between $3$ and $8$ on an interval, but a calculation gives an average of $9$. What does the theorem imply about the calculation?

Why an Average Lies Between the Minimum and Maximum

What You'll Learn

Why an Average Has a Natural Bound

Definition and Assumptions

Why the Bounds Hold

Worked Examples