AP Calculus AB and BC

When the MVT for Integrals Fails

Q: Is the Mean Value Theorem for Integrals related to the ordinary Mean Value Theorem?

Yes. Apply the ordinary Mean Value Theorem to the accumulation function F(x) = \int_a^x f, whose derivative is f by the Fundamental Theorem, and you get exactly this statement.

Q: Does the point c have to be strictly inside the interval?

The usual statement allows c to be an endpoint. A slightly stronger version puts c strictly inside, which follows from the same argument with a little more care.

The Mean Value Theorem for Integrals promises that a continuous function actually attains its own average value somewhere on the interval. Drop continuity and the average value can be a number the function never takes at any point, as the sign function shows.

Mean Value Theorem for Integrals

f(c) = \frac{1}{b-a}\int_a^b f(x)\, dx \quad \text{for some } c \in [a,b]

If f is continuous on the closed interval, then somewhere on that interval f takes a value exactly equal to its own average value across it.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
f is continuous on the closed interval [a, b]
$f \in C[a, b]$
The proof is the Intermediate Value Theorem applied to the average value, so it inherits exactly the same dependence on continuity. The average always exists as a number; what continuity adds is the promise that the function reaches it.
Drop it and the theorem fails
$f(x) = \frac{x}{|x|} \ \ (x \ne 0), \qquad f(0) = 5$
The sign function on [-1, 1], with f(0) = 5
This function is odd away from the origin, so its integral over $[-1,1]$ is 0 and its average value is 0. Changing the value at the single point $x = 0$ does not affect the integral at all. But the function only ever takes the values $-1$ , 5 and $1$ . It never takes its own average value of 0 anywhere on the interval.
2
The interval has positive length, so a is strictly less than b
$a < b$
The average value divides by $b - a$ , so a degenerate interval makes the statement meaningless rather than false. There is nothing to break here and no counterexample to display, but it is worth noticing that the theorem quietly assumes you are averaging over something.
No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Since $f$ is continuous on a closed bounded interval, the Extreme Value Theorem gives a minimum $m$ and a maximum $M$ that are actually attained.
Bounding the integral between the two constant functions gives $m(b-a) \le \int_a^b f \le M(b-a)$ , so the average value lies between $m$ and $M$ .
The average value is therefore an intermediate value between two values that $f$ genuinely attains.
The Intermediate Value Theorem, which again needs continuity, supplies a point $c$ where $f(c)$ equals it.

What it does not say

The average value is the same as the average rate of change.

They are different quantities. The average value of $f$ is $\frac{1}{b-a}\int_a^b f$ ; the average rate of change is $\frac{f(b)-f(a)}{b-a}$ , which is the average value of $f'$ . Confusing them is the most common error in Unit 8.

It tells you where the average value is attained.

Like the Mean Value Theorem for derivatives, it is pure existence. Finding $c$ means solving $f(c) =$ average, which is a separate computation.

A discontinuous function has no average value.

It usually does. The average value only needs the integral to exist, which is a much weaker requirement than continuity. What fails is the promise that the function reaches that average.

Frequently asked questions

Is the Mean Value Theorem for Integrals related to the ordinary Mean Value Theorem?

Yes. Apply the ordinary Mean Value Theorem to the accumulation function $F(x) = \int_a^x f$ , whose derivative is $f$ by the Fundamental Theorem, and you get exactly this statement.

Does the point c have to be strictly inside the interval?

The usual statement allows $c$ to be an endpoint. A slightly stronger version puts $c$ strictly inside, which follows from the same argument with a little more care.