From an Average to a Point
A function can vary throughout an interval, while its average value summarizes its overall level there. A natural question is whether the function actually takes that average value at some point. For a continuous function on a closed interval of positive length, the answer is yes: the integral mean value theorem guarantees at least one such point.
Finding the point is a separate task. First calculate the average from the definite integral; then solve an equation that sets the function equal to that average. The equation may have one solution, several solutions, or solutions that are difficult to express exactly. The theorem guarantees that there is at least one solution in the interval, not that it is unique.
The Equation to Solve
Once the average has been calculated, the search for the point becomes an equation-solving problem:
The variable \(c\) represents an input in the interval, not another average. To find possible values of \(c\), solve the equation using algebra or another appropriate method, and retain only solutions in \([a,b]\). If an equation has solutions outside this interval, they do not answer the question about this interval.
The theorem’s guarantee follows directly from its conclusion. Since \(f\) is continuous on \([a,b]\) and \(a<b\), the integral mean value theorem supplies some \(c\in[a,b]\) such that \(\int_a^b f(x)\,dx=f(c)(b-a)\). Dividing both sides by the positive number \(b-a\) gives \(f(c)=\frac{1}{b-a}\int_a^b f(x)\,dx=f_{\mathrm{avg}}\). Thus at least one solution of the equation lies in the interval.
Evaluate \(\int_a^b f(x)\,dx\) using an antiderivative or another valid method.
Compute \(b-a\), then use it to find \(f_{\mathrm{avg}}\).
Solve \(f(c)=f_{\mathrm{avg}}\) for \(c\).
Keep solutions in \([a,b]\) and substitute each one into \(f\) to verify its function value.
Worked Examples
Worked Example: A Quadratic with One Point in the Interval
Find where \(f(x)=x^2\) equals its average on \([0,3]\).
First calculate the integral. An antiderivative of \(x^2\) is \(\frac{x^3}{3}\), so
The interval length is \(3-0=3\). Therefore the average is
Now solve \(f(c)=f_{\mathrm{avg}}\):
Because \(0\le c\le3\), only \(c=\sqrt{3}\) is in the interval. Check it by substitution: \(f(\sqrt{3})=(\sqrt{3})^2=3\), which equals the average. The other algebraic solution, \(-\sqrt{3}\), lies outside the interval and is not a point sought here.
Worked Example: A Linear Function
Find where \(f(x)=2x+1\) equals its average on \([0,4]\).
An antiderivative is \(x^2+x\). Evaluate the definite integral and show the endpoint values:
The interval length is \(4-0=4\), so
Set the function equal to \(5\) and solve:
The candidate is in the interval because \(0\le2\le4\). Substitution gives \(f(2)=2(2)+1=5\), confirming that the function takes its average value there.
Worked Example: Two Points with the Same Average Value
Find all points where \(f(x)=\sin x\) equals its average on \([0,\pi]\). Use the standard antiderivative \(-\cos x\).
First,
The interval length is \(\pi-0=\pi\), so the average is \(2/\pi\). The equation to solve is
Since \(0<2/\pi<1\), define \(\alpha=\arcsin(2/\pi)\), with \(0<\alpha<\pi/2\). On \([0,\pi]\), the two solutions are \(c=\alpha\) and \(c=\pi-\alpha\). Substitution checks both: \(\sin(\alpha)=2/\pi\), and \(\sin(\pi-\alpha)=\sin(\alpha)=2/\pi\). Both points lie in \([0,\pi]\). Thus the average is reached at two points, not just one.
What the Theorem Guarantees—and What It Does Not
For a continuous \(f\) on \([a,b]\) with \(a<b\), the integral mean value theorem guarantees at least one \(c\in[a,b]\) satisfying the average-value equation. It does not specify which point it is, nor does it say there is only one. The sine example has two points in the interval, while the quadratic example has one.
The theorem is usually stated with the conclusion \(c\in[a,b]\), but under the continuity hypothesis there is always a point \(c\in(a,b)\) with \(f(c)=f_{\mathrm{avg}}\): if \(f-f_{\mathrm{avg}}\) is not identically zero, it takes both positive and negative values, and the Intermediate Value Theorem gives an interior zero. For instance, let \(f(x)=x\) on \([0,1]\). Its integral is \(\int_0^1x\,dx=\frac12\), and its average is \(\frac12\). Solving \(f(c)=\frac12\) gives \(c=\frac12\), an interior point. By contrast, for \(f(x)=x^2\) on \([0,1]\), the integral is \(\frac13\), and \(c^2=\frac13\) gives \(c=1/\sqrt{3}\), also interior. These examples illustrate particular interior solutions.
Continuity matters to the guarantee. A function that is not continuous can fail to take its average value. For example, on \([0,1]\), define \(g(x)=0\) for \(0\le x<1\) and \(g(1)=2\). Its integral is \(0\), so its average is \(0\), which it does attain. This example does not demonstrate failure; rather, it shows that a discontinuity alone does not establish failure. The theorem provides a sufficient condition for existence, and a function violating that condition must be checked on its own rather than assumed to fail.
A Common Mistake: Forgetting the Interval Length
A frequent error is to call the integral itself the average. The integral measures accumulated signed area, whereas the average value is that integral divided by the length of the interval.
For example, on \([0,4]\), the function \(f(x)=2x+1\) has integral \(20\), as calculated above. A mistaken calculation might write \(f_{\mathrm{avg}}=20\), then solve \(2c+1=20\), obtaining \(c=19/2\). This cannot be the point where the function equals its average on \([0,4]\): \(19/2\) is outside the interval, and the calculation used the wrong average.
The correction is to divide the integral by the interval length \(4-0=4\). Thus \(f_{\mathrm{avg}}=20/4=5\), and the correct equation is \(2c+1=5\), giving \(c=2\). The check \(f(2)=5\) confirms the result.
Why Finding Such a Point Is Useful
The average value describes a function over a whole interval; a point satisfying \(f(c)=f_{\mathrm{avg}}\) connects that interval-wide quantity to a value at a particular input. The integral mean value theorem makes that connection mathematically guaranteed under its continuity hypothesis. In applications, this lets an average be interpreted as an actual value attained by the model somewhere in the interval.
The method also keeps two tasks distinct. Integration determines the average, while solving an equation locates the input values at which the function reaches it. Keeping those tasks separate helps prevent confusion between a total integral and an average, and makes it clear why the theorem establishes existence even when a direct equation is challenging to solve.
Check Your Understanding
For each question, identify the average or the relevant theorem condition before solving.
- State the formula for the average value of an integrable function on \([a,b]\), with \(a<b\).
- For \(f(x)=x^2\) on \([0,2]\), calculate the average value and solve \(f(c)=f_{\mathrm{avg}}\) for \(c\in[0,2]\).
- What does the integral mean value theorem guarantee about the equation \(f(c)=f_{\mathrm{avg}}\) when \(f\) is continuous on \([a,b]\)?
- Does the theorem guarantee exactly one solution? Explain briefly.
- For \(f(x)=x+2\) on \([0,2]\), what is the average value, and at what point does \(f\) equal it?
