Why the Hypotheses Matter
A formula for an average value is not, by itself, a guarantee that the function actually takes that value. The Mean Value Theorem for Integrals supplies such a guarantee, but only when its hypotheses are satisfied.
The key condition is continuity on the entire closed interval. Checking that condition means considering not only the interior of the interval, but also the behavior at both endpoints. It also means distinguishing a theorem that guarantees existence from stronger claims—such as uniqueness—that it does not make.
The Mean Value Theorem for Integrals
The theorem applies when \(f\) is continuous on \([a,b]\). A continuous function on a closed, bounded interval is integrable, so its average value is defined. The conclusion is that the function takes that average value at some point strictly inside the interval.
The hypothesis is continuity on the whole closed interval, including \(a\) and \(b\). Continuity at an endpoint is understood one-sidedly within the interval: for example, at \(a\), the values of \(f(x)\) must approach \(f(a)\) as \(x\) approaches \(a\) from the right.
This is the Mean Value Theorem for Integrals. It is different from the differential Mean Value Theorem, which has differentiability as a hypothesis. For the integral theorem, continuity is sufficient; differentiability is not required.
Checking Continuity on the Closed Interval
For familiar formulas, continuity can often be checked from standard facts. Polynomials are continuous everywhere, and absolute value is continuous everywhere. Sums and compositions of continuous functions are continuous wherever they are defined. A formula involving a denominator requires checking that the denominator is nonzero throughout the interval.
Piecewise-defined functions need special care. Check each formula on the portion where it applies, then check every point where the formula changes. At a joining point, the left-hand and right-hand limits must agree with each other and with the function’s value. At the endpoints of the full interval, check the corresponding one-sided limit against the endpoint value.
Record the full closed interval \([a,b]\), with \(a<b\).
Determine whether the function is continuous on each part of the interval where its formula applies.
Test every piecewise join, possible denominator zero, or other point where continuity could fail. Include both endpoints.
If continuity holds on all of \([a,b]\), the theorem guarantees at least one interior point with function value equal to the average.
Worked Examples
Worked Example: A Polynomial on a Closed Interval
Consider \(f(x)=x^2+1\) on \([0,2]\). First check the hypothesis: \(f\) is a polynomial plus a constant, so it is continuous at every real number, including \(0\) and \(2\). The theorem applies.
Compute the integral using the antiderivative \(F(x)=\frac{x^3}{3}+x\):
The interval length is \(2-0=2\), so the average value is
Therefore, the theorem guarantees at least one \(c\in(0,2)\) for which \(f(c)=\frac{7}{3}\). As a check on the average calculation, differentiating \(F(x)=\frac{x^3}{3}+x\) gives \(F'(x)=x^2+1\), the original integrand; and the computed average lies between the function’s minimum \(1\) and maximum \(5\) on this interval. The theorem guarantees existence; this application does not require finding the value of \(c\).
Worked Example: Continuity Without Differentiability
Consider \(f(x)=|x|\) on \([-1,1]\). Absolute value is continuous everywhere, including at \(x=0\) and at both endpoints, so the theorem applies. The function is not differentiable at \(0\), but differentiability is not a hypothesis of the Mean Value Theorem for Integrals.
Split the integral at \(0\). On \([-1,0]\), \(|x|=-x\); on \([0,1]\), \(|x|=x\):
The interval length is \(1-(-1)=2\), so
The theorem guarantees at least one \(c\in(-1,1)\) such that \(|c|=\frac{1}{2}\). As a check, the two pieces of the integral sum to \(1\), and the antiderivatives used above differentiate to \(-x\) and \(x\), respectively. The corner at \(0\) does not prevent the theorem from applying because continuity, not differentiability, is required.
Worked Example: A Discontinuity Can Defeat the Conclusion
Define \(g\) on \([0,1]\) by
This bounded piecewise-constant function is integrable, but it is discontinuous at \(x=\frac{1}{2}\): its left-hand values are \(0\), while its value and right-hand values are \(2\). It is therefore not continuous on \([0,1]\), and the Mean Value Theorem for Integrals cannot be applied.
Its integral and average can still be computed:
But \(g(x)\) is always either \(0\) or \(2\), so there is no \(c\in(0,1)\) with \(g(c)=1\). This verifies that the theorem’s continuity hypothesis matters: without it, the stated existence conclusion need not hold. The integral calculation is checked directly by the two constant contributions, whose sum is \(1\).
Why Continuity Guarantees a Point
The existence conclusion follows from two facts about continuous functions. First, the Extreme Value Theorem says that a continuous function on \([a,b]\) attains a minimum \(m\) and a maximum \(M\). Second, the Intermediate Value Theorem says that it takes every value between any two of its values.
Let
Because \(m\le f(x)\le M\) throughout the interval, integrating these inequalities and dividing by the positive number \(b-a\) gives \(m\le A\le M\). If \(f\) is constant, its value equals \(A\), and any interior point works.
If \(f\) is not constant, the inequalities are strict. In fact, \(f\) is greater than \(m\) somewhere, and by continuity it remains greater than \(m\) on an interval of positive length. Thus its integral is greater than \(m(b-a)\), giving \(A>m\). Similarly, \(f\) is less than \(M\) on an interval of positive length, so \(A<M\). Continuity also ensures these conclusions when an extremum occurs at an endpoint: the relevant one-sided neighborhood has positive length.
Since \(m<A<M\), continuity provides interior points where \(f\) is below \(A\) and above \(A\). To see why the points can be interior, if a point with value below \(A\) is an endpoint, continuity gives nearby interior points that are still below \(A\); the same reasoning applies to a point above \(A\). Applying the Intermediate Value Theorem between such interior points gives a point \(c\in(a,b)\) with \(f(c)=A\). This proves the theorem’s existence conclusion.
What the Theorem Does—and Does Not—Say
The conclusion is deliberately limited: at least one interior point attains the average value. The theorem does not say that the point is unique. A function may attain its average at multiple interior points, and a constant function attains it at every interior point.
The theorem also does not say that every integrable function takes its average value. The discontinuous example shows that an average can be well-defined even when there is no input at which the function equals that average. If continuity fails, the theorem’s guarantee is unavailable; this does not by itself prove that equality is impossible. A discontinuous function may happen to take its average value, but that must be established by other reasoning.
Check Your Understanding
For each question, distinguish carefully between checking the hypotheses and asserting the conclusion.
- What continuity condition must hold before the Mean Value Theorem for Integrals guarantees a point attaining the average?
- Does the theorem require \(f\) to be differentiable on the interval? Explain.
- A piecewise formula changes at an interior point. What must be checked there before applying the theorem?
- If the theorem applies, does it guarantee exactly one point where \(f\) equals its average?
- Why does the step function in the third worked example show that integrability alone is not enough for the theorem’s conclusion?
