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## The Mean Value Theorem for integrals of continuous functions

To get to the mean value theorem for integrals of continuous functions, we first prove the following preliminary, but basic and intuitively clear result:
• Consider the following hypotheses:
• and are given real numbers, with .
• is continuous on the closed interval .
• on .
Then either on , or .
Intuitively this is saying that that area under the graph of a non-negative continuous function can only be zero if the function is everywhere zero.
This is intuitively correct, because if the function is ever non-zero, then it must be strictly positive in some region and then the area under the graph should be also strictly positive.

Proof:

Let and be the minimum and maximum values of on the interval .
These exist and are values of somewhere on the interval, by the Extreme Value Theorem.
So we have for all in .
• If , then identically and we are done.
So now assume that .
• If , then for all and and we are done.
So now assume that .
• Then the number is positive and strictly smaller than .
Let , with in .
Since is continuous on , we can make as close as we like to , if we get sufficiently close to , so in particular we can make larger than if we are close enough to , so there is a neighbourhood containing , such that on that interval, where we have: and .
Then, since on , we have and , so we get:

So and we are done.
Our first corollary is the first version of the Mean Value Theorem for integrals:
• Consider the following hypotheses:
• and are given real numbers, with .
• is continuous on the closed interval joining and .
• .
Then must vanish somewhere on the open interval joining and .
Proof:
• By interchanging and if necessary, we may assume that .
• Now suppose that never vanishes on .
Then since is continuous, cannot change sign on the interval , since otherwise if and exist in with and of opposite signs, then by the Intermediate Value Theorem, vanishes somewhere on the interval joining and .
By replacing by if necessary, we can assume that on the interval .
But now we can apply the previous theorem and we conclude that the integral , contradicting the hypothesis that .
So must vanish somewhere on and we are done.
Our second corollary is the complete Mean Value Theorem for integrals:
• Consider the following hypotheses:
• and are given real numbers.
• is continuous on the closed interval joining and .
Then we conclude:
• Some exists in the closed interval joining and , such that:

• Further, if , then may be taken to lie in the open interval joining and .
Proof:
• If , then both sides of the desired relation are zero, so the result holds with .
So now we assume that
• Make the following definitions:
• Put .
• Put .
Then we have:
• is continuous on the interval joining and .
• The integral of on the interval joining and vanishes:

So the previous corollary applies.
Therefore by the previous corollary, vanishes at some in the open interval joining and .
Then , so and we are done.
We can see geometrically why this theorem holds in the case that and .
In that case, the integral is the area under the curve from to and the right-hand side is the area of a rectangle of base and height .
Denote the this area by .
If we consider a for the area with just one interval, then its value is where is the minimum of on .
If we consider a for the area with just one interval, then its value is where is the maximum of on .
Since , we have .
Now make a rectangle with height with base the interval .
As increases from to , its area increases continuously from to .
So there is a particular height at , between and , when its area exactly matches the area under the curve (of course ).
But on the interval , attains all values between and , so some in the interval exists with and then , as required.
Note that one way of visualizing the level is that it is the level such that there is an equal amount of area under the curve and above the level as there is below, since and geometrically is the area under the curve above the level minus the area above the curve and below the level .

Next: Summary of the Mean Up: Internet Calculus II Previous: Internet Calculus II
George A. J. Sparling 2002-01-27