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### Riemann Sums and the Integral

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Date: 01/18/99 at 03:34:41
From: Brian J.
Subject: Fund. Calc Theorem/Riemann Sums

Dear Dr. Math,

I have always wondered why the limit of Riemann sums just happened to
be the integral? How is that derived? I guess a related question is
the proof of the fundamental theorem of calculus:

/b
|
|    =    F(b) - F(a)
/a

One source I had said that the proof involves advanced calculus, so it
wasn't included, but my textbook has a proof printed. Is it that
```

```
Date: 01/18/99 at 11:03:48
From: Doctor Anthony
Subject: Re: Fund. Calc Theorem/Riemann Sums

If you draw the curve and consider an element of area dA formed by
verticals through the points x and x+dx along the x axis, then by
elementary geometry (and for a very short interval dx)

dA = f(x) dx    where f(x) is the ordinate at point x

So   dA/dx = f(x)

Therefore f(x) is the derivative of the area A. Since by definition,
integration is the inverse of differentiation, the integral of f(x)
will be the area.

That means

F'(x) = f(x)

F(x) = INT[f(x) dx]

and on the interval [a,b] this gives

INT[f(x) dx] = F(b) - F(a)

so that F(b) - F(a) is the area under the curve between x = a and
x = b.

Riemann sums, which calculate the areas algebraically (with n ->
infinity), will therefore approach the value of the integral on the
closed interval [a,b].

The advanced calculus (analysis) is mostly in the rather long-winded
justification for algebraic manipulation of infinitesmals. You can
probably leave that to analysts and agree that the basic concept of
Riemann sums is very reasonable.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus

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