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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: 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 $ m$ and $ M$ be the minimum and maximum values of $ f$ on the interval $ [a,b]$.
These exist and are values of $ f$ somewhere on the interval, by the Extreme Value Theorem.
So we have $ 0 \le m \le f(t) \le M$ for all $ t$ in $ [a,b]$. So now assume that $ M > 0$. So now assume that $ m < M$. Our first corollary is the first version of the Mean Value Theorem for integrals: Proof: Our second corollary is the complete Mean Value Theorem for integrals: Proof: So now we assume that $ a \ne b$ We can see geometrically why this theorem holds in the case that $ a < b$ and $ f(t) > 0$.
In that case, the integral is the area under the curve from $ t = a$ to $ t = b$ and the right-hand side is the area of a rectangle of base $ b - a$ and height $ f(\xi)$.
Denote the this area by $ J$.
If we consider a $ LRS$ for the area with just one interval, then its value is $ m(b - a)$ where $ m$ is the minimum of $ f$ on $ [a,b]$.
If we consider a $ URS$ for the area with just one interval, then its value is $ M(b - a)$ where $ M$ is the maximum of $ f$ on $ [a,b]$.
Since $ LRS \le J\le URS$, we have $ m(b -a)\le J\le M(b - a)$.
Now make a rectangle with height $ h$ with base the interval $ [a,b]$.
As $ h$ increases from $ m$ to $ M$, its area increases continuously from $ m(b - a)$ to $ M(b - a)$.
So there is a particular height at $ h$, between $ m$ and $ M$, when its area exactly matches the area under the curve $ J$ (of course $ h =
\frac{J}{b - a}$).
But on the interval $ [a,b]$, $ f$ attains all values between $ m$ and $ M$, so some $ \xi$ in the interval $ [a,b]$ exists with $ f(\xi) = h$ and then $ f(\xi)(b - a) = J$, as required.
Note that one way of visualizing the level $ h = f(\xi)$ is that it is the level such that there is an equal amount of area under the curve $ f$ and above the level $ h$ as there is below, since $ \int_a^b (f(t) - h) dt = \int_a^b f(t) dt - h(b - a) = 0$ and geometrically $ \int_a^b (f(t) -
h) dt $ is the area under the curve $ y = f(t)$ above the level $ y = h$ minus the area above the curve $ y = f(t)$ and below the level $ y = h$.
next up previous
Next: Summary of the Mean Up: Internet Calculus II Previous: Internet Calculus II
George A. J. Sparling 2002-01-27