Mean Value Theorem for Integrals

From Integral Bounds to an Attained Value

The previous tutorial, Bounds for Integrals, showed how pointwise bounds on an integrable function give bounds on its integral. For a continuous function on an interval, this idea has a useful converse flavor: the integral, divided by the interval length, is not merely trapped between values of the function. It is itself one of the values the function attains.

This statement is the Integral Mean Value Theorem. It resembles the Mean Value Theorem for derivatives in that it identifies a point inside an interval, but it concerns an average over the whole interval rather than a derivative or a difference quotient. Continuity is essential to the existence claim; differentiability is not required.

Definition: Let $f$ be Riemann integrable on $[a,b]$, where $a<b$. The average value of $f$ on $[a,b]$ is $$ \operatorname{Avg}_{[a,b]} f =\frac{1}{b-a}\int_a^b f(x)\,dx. $$

The factor $1/(b-a)$ normalizes the integral by the length of the interval. For a constant function with value $K$, the integral is $K(b-a)$, so its average value is $K$. For a nonconstant function, the average is a single number that summarizes the signed integral over the interval. It need not be a value taken at the midpoint, and the point where it is attained need not be unique.

Theorem (Integral Mean Value Theorem): Let $f:[a,b]\to\mathbb{R}$ be continuous, where $a<b$. There exists $c\in[a,b]$ such that $$ \int_a^b f(x)\,dx=f(c)(b-a). $$ Equivalently, $\operatorname{Avg}_{[a,b]}f=f(c)$. The existence proof is developed in the next tutorial.

The theorem has two parts that should be kept distinct. The average value is determined by the integral and the interval length. The theorem then asserts that at least one point of the interval has that function value. It does not usually specify that point in advance, nor does it promise that there is only one such point.

The Average Lies Between the Extreme Values

Before using the existence statement, it is useful to locate the average numerically. A continuous function on a closed bounded interval attains a minimum $m$ and a maximum $M$. Since $m\leq f(x)\leq M$, the established Bounds for the Integral theorem gives an immediate bound on the average.

Theorem (Bounds for the Average Value): If $f:[a,b]\to\mathbb{R}$ is continuous and $a<b$, then $$ \min_{x\in[a,b]}f(x) \leq \operatorname{Avg}_{[a,b]}f \leq \max_{x\in[a,b]}f(x). $$

Proof. By the Extreme Value Theorem, there are points in $[a,b]$ where $f$ attains its minimum $m$ and maximum $M$. Thus $m\leq f(x)\leq M$ for every $x\in[a,b]$. The Bounds for the Integral theorem gives

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

Because $b-a>0$, division by $b-a$ preserves the inequalities. The result is $m\leq \operatorname{Avg}_{[a,b]}f\leq M$, as claimed. $\square$

This bound is a useful check on a calculation. If a proposed average lies outside the range of a continuous function, then either the integral, the interval length, or the arithmetic has been mishandled. The bound alone, however, does not establish the Integral Mean Value Theorem: it locates the average between extreme values, while the theorem’s existence conclusion uses continuity to ensure that intermediate values are attained.

Worked Example: Averages and More Than One Attainment Point

Let $f(x)=x^2$ on $[-1,2]$. The integral and interval length are

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

Therefore the average value is $3/3=1$. The equation $f(c)=1$ becomes $c^2=1$, whose solutions are $c=-1$ and $c=1$. Both belong to $[-1,2]$, and direct substitution verifies $f(-1)=(-1)^2=1$ and $f(1)=1^2=1$. Thus the average value is attained at two points in this example.

The minimum of $x^2$ on this interval is $0$, and the maximum is $4$. The bound $0\leq1\leq4$ is consistent with the calculation, but the equation $x^2=1$ identifies the actual points where the average is attained.

Finding the Point in Concrete Examples

When a function is one-to-one, its average value equation may determine the point $c$ directly. The calculation has two stages: first evaluate and normalize the integral; then solve $f(c)=\operatorname{Avg}_{[a,b]}f$. The theorem guarantees a solution for continuous functions, while algebra or inverse functions can sometimes locate it explicitly.

Worked Example: The Average Value of Sine

Take $f(x)=\sin x$ on $[0,\pi/2]$. Its integral is

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

The interval length is $\pi/2$, so

\operatorname{Avg}_{[0,\pi/2]}\sin =\frac{1}{\pi/2} =\frac{2}{\pi}.

A point $c$ that attains this average must satisfy $\sin c=2/\pi$, so $c=\arcsin(2/\pi)$. Since $0<2/\pi<1$, this value lies in $(0,\pi/2)$, where sine takes values between $0$ and $1$. Substitution gives $\sin(\arcsin(2/\pi))=2/\pi$, as required.

Worked Example: An Exponential Average

Let $f(x)=e^x$ on $[0,1]$. Since an antiderivative is $e^x$,

\int_0^1 e^x\,dx =[e^x]_0^1 =e^1-e^0 =e-1.

The interval has length $1$, so the average is $e-1$. Solving $e^c=e-1$ gives $c=\ln(e-1)$. To check that this point belongs to the interval, note that $e>2$, so $e-1>1$, and also $e-1<e$. Since the logarithm is strictly increasing, these inequalities imply $0<\ln(e-1)<1$. Finally, $e^c=e^{\ln(e-1)}=e-1$, verifying that this interior point attains the average.

How Subinterval Averages Combine

An interval can be divided into pieces, and the average on the whole interval is related to the averages on those pieces. The relevant weights are the subinterval lengths: longer pieces contribute more to the overall average. This is a useful way to organize a calculation and to interpret what happens when a function has different typical values in different regions.

Theorem (Averages Across a Partition): Let $f$ be Riemann integrable on $[a,b]$, and let $a=x_0<x_1<\cdots<x_n=b$ be a partition. For each $i$, define $$ A_i=\frac{1}{x_i-x_{i-1}}\int_{x_{i-1}}^{x_i}f(x)\,dx. $$ Then $$ \operatorname{Avg}_{[a,b]}f =\sum_{i=1}^n\frac{x_i-x_{i-1}}{b-a}A_i. $$

Proof. Each subinterval has positive length, so each $A_i$ is defined. By the definition of $A_i$,

(x_i-x_{i-1})A_i =\int_{x_{i-1}}^{x_i}f(x)\,dx.

Add these equalities over $i=1,\ldots,n$. Additivity of the Riemann integral across a partition gives

\sum_{i=1}^n(x_i-x_{i-1})A_i =\sum_{i=1}^n\int_{x_{i-1}}^{x_i}f(x)\,dx =\int_a^b f(x)\,dx.

Divide by $b-a$, which is positive. The resulting equality is precisely the asserted formula. $\square$

The coefficients $(x_i-x_{i-1})/(b-a)$ are positive and sum to $1$, because the subinterval lengths add to $b-a$. Thus the full-interval average is a weighted average of the subinterval averages. In particular, if every $A_i$ lies between numbers $L$ and $U$, multiplying $L\leq A_i\leq U$ by the corresponding positive weight and summing gives $L\leq\operatorname{Avg}_{[a,b]}f\leq U$.

Worked Example: Combining Two Subinterval Averages

Suppose $f(x)=x+1$ on $[0,3]$, divided at $x=1$. On $[0,1]$, the integral is

\int_0^1(x+1)\,dx =\left[\frac{x^2}{2}+x\right]_0^1 =\frac{1}{2}+1 =\frac{3}{2},

so its average there is $A_1=(3/2)/1=3/2$. On $[1,3]$,

\int_1^3(x+1)\,dx =\left[\frac{x^2}{2}+x\right]_1^3 =\left(\frac{9}{2}+3\right)-\left(\frac{1}{2}+1\right) =6,

so $A_2=6/(3-1)=3$. The partition formula gives the full average as

\operatorname{Avg}_{[0,3]}f =\frac{1}{3}\left(1\cdot\frac{3}{2}+2\cdot3\right) =\frac{1}{3}\left(\frac{3}{2}+6\right) =\frac{5}{2}.

A direct calculation confirms this: $\int_0^3(x+1)\,dx=15/2$, and $(15/2)/3=5/2$. The point where $f$ attains the average solves $c+1=5/2$, giving $c=3/2$, which lies in $[0,3]$.

When Is the Attainment Point Unique?

The Integral Mean Value Theorem guarantees at least one suitable point, not exactly one. The example $x^2$ on $[-1,2]$ showed that multiple points can attain the average. Strict monotonicity gives a simple sufficient condition for uniqueness: a strictly monotone function can take a given value at most once.

Proposition (Uniqueness for Strictly Monotone Functions): Suppose $f$ is continuous and strictly monotone on $[a,b]$, where $a<b$. Then there is exactly one $c\in[a,b]$ such that $f(c)=\operatorname{Avg}_{[a,b]}f$.

Proof. The Integral Mean Value Theorem gives at least one such $c$. If there were two distinct points $c_1$ and $c_2$ with $f(c_1)=f(c_2)=\operatorname{Avg}_{[a,b]}f$, strict monotonicity would imply $f(c_1)\ne f(c_2)$, a contradiction. Hence the point is unique. $\square$

This proposition identifies a situation in which the theorem’s point can be discussed unambiguously, even if finding its exact value requires solving an equation. Without strict monotonicity, uniqueness can fail; conversely, failure of strict monotonicity does not by itself imply that there are multiple attainment points.

Interpretation and a Common Pitfall

The average value is an average with respect to interval length. It should not be confused with the value at the midpoint, or with the value at a point selected for some unrelated reason. The Integral Mean Value Theorem says there is some $c$ whose function value matches the average; it generally does not say which $c$ to choose unless additional information is available.

The hypotheses also matter. Continuity on the closed interval ensures that the theorem applies and that the function is Riemann integrable, by the established Integrability of Continuous Functions theorem. Integrability alone gives bounds for the integral, but does not ensure that the average is attained. For example, a step function can have an average strictly between its two values without taking that intermediate value anywhere. Thus it is unsafe to replace continuity by integrability in the theorem’s existence claim.

Finally, the theorem concerns the signed integral. If $f$ changes sign, positive and negative contributions may cancel before the integral is divided by $b-a$. The resulting average is still attained for continuous $f$, but it need not describe the typical size of $|f|$. For a question about magnitude, the integral of $|f|$ is a different quantity.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

How is the average value of a Riemann integrable function on $[a,b]$ defined, and why must $a<b$?
What does the Integral Mean Value Theorem guarantee for a continuous function, and does it guarantee a unique point?
If the minimum and maximum of a continuous function are $m$ and $M$, what bounds apply to its average value?
Why does a strictly monotone continuous function have exactly one point where it attains its average?
In the partition formula, why are the subinterval averages weighted by their lengths?
Why can integrability alone not replace continuity in the existence statement?