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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

I've tried a number of equations for the above statement, and from the answers I get it seems like it's true?

If it is true, is there any theory that I can use to explain the statement?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,206

hi Shelled,

I believe it is the direct result of the definition of 'integral'.

An integral is a summation process. In the area example in my diagram proceed as follows:

Divide the space into vertical strips with width Δx. The height of each strip is f(x) which will be negative in your case.

Calculate the area of one strip = f(x).Δx

Add up all the areas for strips starting at 'a' and ending at 'b' = ∑f(x).Δx

Now let the widths tend to zero and find the limit as Δx -> zero of ∑f(x).Δx

This is denoted by ∫ f(x).dx

As all the f(x) are negative, the sums will be too, and so the limit will be too.

It is possible to extend this idea to summate things other then areas. But it is also possible to 'convert' such problems into area problems, so the result will generalise.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

Thanks!

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Bob bundy's reasoning is probably the most intuitive way to look at it. For a more rigorous proof, it may help to know some more details about f -- for instance, is f continuous? Whilst it is true that continuity => Riemann integrability, the converse is not true in general -- but f being continuous does make the proof a tad easier, without having to resort to measure theory. A good starting point might be how to apply Riemann's Criterion for Integrability here, i.e. f is Riemann integrable if, and only if;

where P is a partition of [a,b], and U(f,P) and L(f,P) are the lower Darboux sums of f with respect to the partition P, defined by

where

and follow their usual definitions:It is probably doable with a careful choice of P. The fact that f is never positive might be of some use here!

This is also part of a more general property called the *order property* of integrals, i.e. if f,g are Riemann integrable on [a,b], then

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

Thanks for posting this up

Can you exaplin in a little more depth what this means?:

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Epsilon is just some (arbitrarily small) positive real number. We claim that, if f is Riemann integrable on [a,b], then we can find some partition, P, of [a,b], such that the difference between the upper and lower Darboux sums is arbitrarily small. A partition P of [a,b] is a finite sequence of real numbers

such that .Essentially, you choose how you want to split up the interval. For instance, P = {0, 1/2, 1} and Q = {0, 1/2, 3/4, 1} are both partitions of [0,1]. (In fact, for this particular example, we say Q is a *refinement* of P.)

A perfectly good question is: **Why is this useful?** It turns out that many functions which aren't seemingly 'nice' can actually be integrated -- in particular, functions whose set of points of discontinuity is everywhere dense in their domain can be Riemann integrable, which can be quite a non-intuitive result.

*Last edited by zetafunc (2014-05-21 22:01:39)*

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

Thanks!

*Last edited by Shelled (2014-05-22 12:26:37)*

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