- Consider the following hypotheses:
- and are given real numbers, with .
- is continuous on the closed interval .
- on .

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.

- If , then for all and and we are done.

- 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:

- Consider the following hypotheses:
- and are given real numbers, with .
- is continuous on the closed interval joining and .
- .

- 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.

- Consider the following hypotheses:
- and are given real numbers.
- is continuous on the closed interval joining and .

- Some exists in the closed interval joining and , such that:
- Further, if , then may be taken to lie in the open interval joining and .

- If , then both sides of the desired relation are zero, so the result holds with .

- Make the following definitions:
- Put .
- Put .

- is continuous on the interval joining and .
- The integral of on the interval joining and vanishes:

Therefore by the previous corollary, vanishes at some in the open interval joining and .

Then , so and we are done.

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 .