What the Mean Value Theorem for Integrals Says
The Mean Value Theorem for Integrals connects a definite integral to the average value of a function. It says that if a function is continuous on a closed interval, then there is at least one point where the function's instantaneous value equals its average value over the entire interval.
Suppose is continuous on . Then there is at least one number such that:
This is an existence theorem: it guarantees that such a value of exists. It does not necessarily tell you that there is only one such point.
The Average Value Formula
The quantity on the left side of the theorem is the average value of the function on the interval. Therefore:
Notice the structure. First find the total accumulated value using the definite integral. Then divide by the length of the interval, .
This is closely related to an ordinary arithmetic average. If you average several numbers, you add them and divide by how many numbers there are. For a continuous function, the integral plays the role of adding up infinitely many values, and dividing by the interval length converts that total into an average.
Why the Formula Makes Sense
Think about a graph of a continuous function over an interval. The definite integral gives the signed area between the graph and the -axis. If you divide that accumulated area by the width of the interval, you get the height of a rectangle with the same width and the same area.
The Mean Value Theorem for Integrals says that, because the function is continuous, the graph must actually reach that average height somewhere in the interval.
So the theorem gives a geometric interpretation:
- Integral: total accumulated signed area.
- Divide by interval length: convert total accumulation into an average height.
- Continuity: guarantees that the function actually reaches that average height.
How to Find the Average Value
- Identify the interval .
- Find the definite integral .
- Divide by the interval length .
- Interpret the result as the average value of over the interval.
Worked Examples
Find the average value of on the interval .
Let . Find the value of guaranteed by the Mean Value Theorem for Integrals on .
A continuous function is represented by the table below.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| f(x) | 2 | 4 | 5 | 3 | 1 |
Suppose you are also given . Find the average value of on .
Let and define . Find the average value of on .
Mean Value Theorem vs. Average Value
These two ideas are closely connected, but they are not exactly the same thing.
- The average value formula tells you what the average value is.
- The Mean Value Theorem for Integrals guarantees that the function reaches that average value at some point in the interval.
For example, if the average value of on is , then the theorem guarantees that there is at least one in the interval for which:
What Continuity Has to Do With It
The function must be continuous on the interval for the theorem to apply. Continuity is what guarantees that the graph cannot jump over its average value.
This is the same basic idea behind the Intermediate Value Theorem. If a continuous function takes values below and above a target value, it must pass through that target value somewhere in between.
For the Mean Value Theorem for Integrals, the target value is specifically the function's average value.
The average value of a function is generally not . You must use the integral formula. The endpoint average only happens to work in special situations, such as when the function is linear.
The definite integral gives total accumulated value, not average value. Always divide by . Omitting this step produces the total integral instead of the average.
The theorem says there is at least one point . There may be several points where the function equals its average value.
The Mean Value Theorem for Integrals requires continuity on the entire closed interval. If the function has a discontinuity inside the interval, you cannot automatically invoke the theorem.
A Useful Connection to the Fundamental Theorem of Calculus
The Mean Value Theorem for Integrals becomes especially useful when you connect it to the Fundamental Theorem of Calculus.
If , then:
Therefore, the average value of can often be found by evaluating an antiderivative at the endpoints and dividing by the interval length:
This is one reason the average value formula is so efficient on the AP exam. You usually do not need to calculate the integral from scratch using numerical methods if an antiderivative is available.
Practice Problems
Find the average value of on .
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Find the average value of on .
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Let . Find the value of guaranteed by the Mean Value Theorem for Integrals on .
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A continuous function satisfies . Explain what the Mean Value Theorem for Integrals guarantees.
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