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Riemann Integration · Tutorial 485 of 1000

Proof of the Integral Mean Value Theorem

See how continuity forces a function to attain its integral average, and why a nonconstant continuous function cannot have its average equal to its minimum or maximum.

Advanced 9 min read

What You'll Learn

  • Prove that a continuous function with zero integral must vanish at some point
  • Apply the zero-integral principle to prove the Integral Mean Value Theorem
  • Track how linearity and the integral of a constant enter the proof
  • Determine when the average of a continuous function can equal its minimum or maximum
  • Verify the theorem in examples with corners, sign changes, and a radical function

Turn the Average into a Zero-Integral Problem

The Integral Mean Value Theorem says that the average value of a continuous function on a nondegenerate closed interval is attained at some point of that interval. The previous tutorial stated the theorem and located the average between the function’s minimum and maximum. Here we prove the existence claim.

The key step is to subtract the average from the function. The resulting continuous function has integral zero. If it never vanished, the Intermediate Value Theorem would force it to have one strict sign throughout the interval, which is incompatible with a zero integral. This argument isolates the role of continuity: it prevents a function from changing sign without taking the value zero.

Recall that for \(a<b\), the average value is

$$ \operatorname{Avg}_{[a,b]}f =\frac{1}{b-a}\int_a^b f(x)\,dx. $$

The interval length is positive, so this quantity is defined. We first formalize the zero-integral step.

Lemma (Zero-Integral Crossing Principle): Let \(h:[a,b]\to\mathbb{R}\) be continuous, where \(a<b\). If $$ \int_a^b h(x)\,dx=0, $$ then there exists \(c\in[a,b]\) such that \(h(c)=0\).

Proof. Suppose, to the contrary, that \(h(x)\ne0\) for every \(x\in[a,b]\). Choose any \(x_0\in[a,b]\). Either \(h(x_0)>0\) or \(h(x_0)<0\).

First suppose \(h(x_0)>0\). If there were some \(x_1\in[a,b]\) with \(h(x_1)<0\), the Intermediate Value Theorem applied between \(x_0\) and \(x_1\) would give a point where \(h\) equals zero. That contradicts our supposition. Since \(h\) has no zeros, it follows that \(h(x)>0\) for every \(x\in[a,b]\). In particular, \(0\leq h(x)\) everywhere, with strict inequality at every point. The Strict Integral Comparison for Continuous Functions, established earlier in the course, then gives

$$ \int_a^b 0\,dx < \int_a^b h(x)\,dx. $$

The left side is \(0\), so this implies \(\int_a^b h(x)\,dx>0\), contrary to the assumed zero integral.

Now suppose \(h(x_0)<0\). The same Intermediate Value Theorem argument shows that \(h(x)<0\) everywhere: a positive value anywhere would force a zero between that point and \(x_0\). Thus \(h(x)\leq0\) everywhere and is strictly less than zero at every point. Strict Integral Comparison gives

$$ \int_a^b h(x)\,dx < \int_a^b 0\,dx =0, $$

again contradicting the assumed zero integral. Both possible signs at \(x_0\) lead to a contradiction. Therefore \(h\) must vanish at some \(c\in[a,b]\). \(\square\)

Proof of the Integral Mean Value Theorem

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, \(f(c)=\operatorname{Avg}_{[a,b]}f\).

Proof. Because \(f\) is continuous on \([a,b]\), it is Riemann integrable there by the Integrability of Continuous Functions theorem. Define its average \(A\) by

$$ A=\frac{1}{b-a}\int_a^b f(x)\,dx, $$

and define \(h(x)=f(x)-A\) for \(x\in[a,b]\). The function \(h\) is continuous, since it is the difference of the continuous function \(f\) and the constant function with value \(A\). By linearity of the Riemann integral and the integral of a constant,

$$ \begin{aligned} \int_a^b h(x)\,dx &=\int_a^b \bigl(f(x)-A\bigr)\,dx\\ &=\int_a^b f(x)\,dx-\int_a^b A\,dx\\ &=\int_a^b f(x)\,dx-A(b-a)\\ &=\int_a^b f(x)\,dx -\frac{1}{b-a}\left(\int_a^b f(x)\,dx\right)(b-a)\\ &=0. \end{aligned} $$

The Zero-Integral Crossing Principle now applies to \(h\). It gives a \(c\in[a,b]\) for which \(h(c)=0\). By the definition of \(h\), this means \(f(c)-A=0\), so \(f(c)=A\). Multiplying by \(b-a\) yields

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

as required. \(\square\)

Every hypothesis has a role in this proof. Continuity supplies the crossing principle, while \(a<b\) ensures that the average is defined and that \(b-a\) is positive. No derivative of \(f\) is needed. The argument also does not require finding an antiderivative: it uses only integrability, linearity, the integral of a constant, and the Intermediate Value Theorem.

Worked Applications of the Proof

Worked Example: A Continuous Function with a Corner

Let \(f(x)=|x-1|\) on \([0,3]\). This function is continuous, although its graph has a corner at \(x=1\). Split the integral at that point:

$$ \begin{aligned} \int_0^3 |x-1|\,dx &=\int_0^1(1-x)\,dx+\int_1^3(x-1)\,dx\\ &=\left[x-\frac{x^2}{2}\right]_0^1 +\left[\frac{x^2}{2}-x\right]_1^3\\ &=\frac{1}{2}+2 =\frac{5}{2}. \end{aligned} $$

The interval has length \(3\), so the average is \((5/2)/3=5/6\). To find points that attain it, solve \(|c-1|=5/6\). The two solutions are

$$ c=1-\frac{5}{6}=\frac{1}{6} \qquad\text{and}\qquad c=1+\frac{5}{6}=\frac{11}{6}. $$

Both points belong to \([0,3]\). Direct substitution verifies \(|1/6-1|=5/6\) and \(|11/6-1|=5/6\). This example shows that differentiability is not needed and that the point attaining the average need not be unique.

Worked Example: Cancellation in a Signed Integral

Consider \(f(x)=x^3-3x\) on \([-2,2]\). An antiderivative is \(x^4/4-3x^2/2\), so

$$ \begin{aligned} \int_{-2}^{2}(x^3-3x)\,dx &=\left[\frac{x^4}{4}-\frac{3x^2}{2}\right]_{-2}^{2}\\ &=(4-6)-(4-6)\\ &=0. \end{aligned} $$

The interval length is \(4\), so the average is \(0/4=0\). The equation \(f(c)=0\) factors as

$$ c^3-3c=c(c^2-3)=0. $$

Its solutions are \(c=0\), \(c=\sqrt{3}\), and \(c=-\sqrt{3}\). They all lie in \([-2,2]\), since \(0\leq\sqrt{3}<2\). Substitution gives \(f(0)=0\) and, for either \(c=\sqrt{3}\) or \(c=-\sqrt{3}\), \(c^2=3\) implies \(c^3-3c=c(3)-3c=0\). The average is attained at three points. Here the zero average also reflects cancellation between positive and negative parts of the function.

Worked Example: A Radical Function

Let \(f(x)=\sqrt{x}\) on \([0,4]\). It is continuous on the closed interval, including at \(0\). Its integral is

$$ \int_0^4\sqrt{x}\,dx =\left[\frac{2}{3}x^{3/2}\right]_0^4 =\frac{2}{3}\cdot 8 =\frac{16}{3}. $$

The interval length is \(4\), giving the average

$$ \operatorname{Avg}_{[0,4]}f =\frac{16/3}{4} =\frac{4}{3}. $$

Solving \(\sqrt{c}=4/3\) gives \(c=16/9\). This lies in the interval because \(0\leq16/9\leq4\), and direct substitution verifies \(\sqrt{16/9}=4/3\). Thus the theorem’s point can be located explicitly in this case.

What the Proof Says About Extreme Averages

The average lies between the minimum and maximum of a continuous function, as established in the previous tutorial. The proof above gives an additional fact about the boundary cases: for a continuous function on a nondegenerate interval, the average can equal the minimum or maximum only when the function is constant. This rules out equality at an extreme for every nonconstant continuous function.

Theorem (Rigidity at an Extreme Average): Let \(f:[a,b]\to\mathbb{R}\) be continuous, where \(a<b\). If the average of \(f\) equals its minimum on \([a,b]\), or equals its maximum on \([a,b]\), then \(f\) is constant on \([a,b]\). Conversely, a constant function has average equal to both its minimum and its maximum.

Proof. Suppose first that the average equals the minimum \(m=\min_{x\in[a,b]}f(x)\). Then \(f(x)-m\geq0\) for all \(x\in[a,b]\). Using linearity and the integral of a constant,

$$ \int_a^b\bigl(f(x)-m\bigr)\,dx =\int_a^b f(x)\,dx-m(b-a) =0, $$

because the average is \(m\). If there were a point \(x_0\) with \(f(x_0)-m>0\), then the continuous function \(f-m\) would be nonnegative everywhere and strictly positive at \(x_0\). Strict Integral Comparison, applied to the zero function and \(f-m\), would imply

$$ 0=\int_a^b0\,dx < \int_a^b\bigl(f(x)-m\bigr)\,dx, $$

contradicting the zero integral just obtained. Hence \(f(x)-m=0\) for every \(x\in[a,b]\), and \(f\) is constant.

Now suppose the average equals the maximum \(M=\max_{x\in[a,b]}f(x)\). The function \(M-f\) is continuous and nonnegative, and

$$ \int_a^b\bigl(M-f(x)\bigr)\,dx =M(b-a)-\int_a^b f(x)\,dx =0. $$

If \(M-f\) were positive at any point, strict comparison with the zero function would make this integral positive, a contradiction. Thus \(M-f\) is zero everywhere, so \(f\) is constant. Conversely, if \(f(x)=K\) for all \(x\in[a,b]\), then its integral is \(K(b-a)\), its average is \(K\), and both its minimum and maximum are \(K\). \(\square\)

A Common Pitfall: Integrability Is Not Enough

The crossing principle depends on continuity, not merely on the existence of the integral. For example, define \(g\) on \([0,1]\) by \(g(x)=0\) for \(0\leq x<1/2\) and \(g(x)=1\) for \(1/2\leq x\leq1\). This step function is Riemann integrable and its integral, hence its average on this interval, is \(1/2\). But \(g\) takes only the values \(0\) and \(1\), so it never attains its average. The function can jump from one value to the other without taking the values in between.

This also pinpoints why the zero-integral proof cannot be applied to every integrable function: without continuity, a function with no zero can change sign by a jump. For continuous functions, the Intermediate Value Theorem rules out that possibility. The proof therefore establishes an existence statement, not a general recipe for calculating \(c\); explicit calculations require additional information about the particular function.

Check Your Understanding

Use the proof and examples to answer the following questions.

  1. Why must a continuous function with zero integral have a zero somewhere on a nondegenerate closed interval?
  2. In the proof of the Integral Mean Value Theorem, why is the function \(h(x)=f(x)-\operatorname{Avg}_{[a,b]}f\) continuous and why does it have integral zero?
  3. Which theorem rules out a strict sign for a continuous function whose integral is zero?
  4. Can a continuous nonconstant function have its average equal its minimum? Explain using the rigidity theorem.
  5. Why does the step function in the final section show that integrability alone does not guarantee that the average is attained?

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