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On first blush it seems that we have proved the Fundamental Theorem of Calculus outright in the main text. But in reality we have made a hidden assumption, and without proving that assumption, there is no guarantee that a Riemann sum even has a limit as the number of rectangles goes to infinity.

The hidden assumption (and it is subtle) was that we could look at what happens
with just one rectangle and then generalize to what happens with an infinite
sum of them. It turns out that the condition of `f(x)` being continuous
is enough to guarantee what we need for any rectangle in particular, but not enough
for the infinite crowd of them that we end up with when we take the limit.
There is a stronger form of continuity called *uniform continuity* that
is required for the Riemann sum to be guaranteed a limit.

Look at what's happening in the diagram. The `e`'s indicate how far
the top of the rectangle is from the function at a distance, `d`, from
`x`. If, for example, you chose the width of the rectangle to be
`d _{1}`, then the height of the rectangle
is within

0 £ xYou can see that if you demand a smaller_{1}£ |e_{1}d_{1}|

The problem lies in that the recipe is only guaranteed to work for *this*
rectangle. Other rectangles in the Riemann sum might require a different recipe.
So a `d` that guarantees a desired
`e` for this rectangle might not give you
as small an `e` at some other rectangle.

This is where the concept of *uniform continuity* comes in. Uniform continuity
of a function over some domain of `x` says not only that the function,
`f(x)`, is continuous at *every* `x` in that
domain, but additionally that for any
`e > 0``d > 0`*every* `x` in the domain. That is this *same*
`d` will assure that

|f(x+d) - f(x)| £ enot just for a particular

With the Riemann sums we are approximating the area under the curve, `f(x)`,
within some interval of `x`. That interval is the domain over which we would
like `f(x)` to be uniformly continous. Why? Suppose you name
an `e` that must be the worst error for the height
of *any* rectangle in the Riemann sum. Uniform continuity of `f(x)`
says that we can find a *single* `d` to use
as the width of *every* rectangle to ensure this.

Now what happens to the inaccuracy for the area of each rectangle as a result?

0 £ x £ |ed|where

But what is `d`? If the length of the interval
in which we are finding the area is `L` and we are using `n` rectangles,
then

d = L/nSo substitute that into the inequality and you get

0 £ x £ |eL/n|And that is true for every one of the

0 £ xUniform continuity ensures that we can make_{total}£ |eL|

But what about this uniform continuity requirement we are forcing onto `f(x)`?
It's a lot tougher than just plain continuity. How do we guarantee that the function
we have chosen to find the area under is uniformly continuous over the entire interval
we are interested in?

There is a theorem (whose proof is well beyond first year students) that when
a domain interval is *closed and bounded* (that is it includes the endpoints
and has endpoints at *both* ends),
then any function that is continuous on that entire interval is also
uniformly continuous on that entire interval.
If you are dying of curiosity for more information on this theorem,
search the web for "Heine-Borel theorem," "Lebesgue covering lemma" and
"Lebesgue number." But you do need to have had either a course in real analysis
or a course in topology to be equipped for these ideas you will find there.

The great thing about all this is that closed and bounded intervals are exactly the domains we are interested in taking Riemann sums and their limits over. So as long as the domain interval is closed and bounded, the function you are finding the area under needs only to be continuous to qualify.

email me at
*hahn@netsrq.com*