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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,930

Hi;

I came across this link: Please have a look.

http://betterexplained.com/articles/a-c … plication/

Does everyone agree with this? What's wrong with thinking about definite integration as an area?

*Last edited by bobbym (2009-08-30 05:55:20)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**random_fruit****Member**- Registered: 2008-12-25
- Posts: 39

I had a look and I liked it. Whilst the "area under the curve" is good, the idea of "multiplication for things that vary" could help some who can't apply concepts away from the way they are taught. I also suspect that it does provide the "ah" moment.

Thanks for showing me that.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Thinking of integration as an area is certainly a useful technique when it works, but the problem is that there are some integrable functions where that doesn't make sense.

One example would be

f(x) = {1, if x is rational.

{0, otherwise.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,930

Hi random_fruit;

Your welcome, can you think of a specifc example where that helped you over the area way?

Hi mathsyperson;

A piecewise function like that is strange. Wouldn't it be difficult to evaluate the definite integral?

Using your f(x):

What is the value of that integral?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The integral of that function on any interval is 0.

My lecturer only mentioned it in passing though, saying the proof was far beyond the scope of the course.

That said, it's kind of understandable since the rationals are countable and the irrationals are not.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,930

Hi mathsyperson;

I figured that. If you think about it as an area you will have an infinite number of rectangles of 0 width ( just a vertical line ) for every rational x.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

My lecturer only mentioned it in passing though, saying the proof was far beyond the scope of the course.

That integral does not exist as a Riemann integral. We can take the limit as Δx goes to zero by choosing all rationals for Δx, in which case each term in the sequence is 1, or we can take the limit by choosing Δx to be irrationals, in which case each term in the sequence is 0. This prove the limit does not exist.

Now that is using the calculus definition of the integral, rather than the analysis definition. The analysis definition will also not exist.

The integral is however zero when you consider it as a Lebesgue integral. Lebesgue integration works like this: come up with a lower approximation for the integral using functions you can compute the integral of, and then call the highest one of those the value of the integral. We make these approximations using simple functions which are functions which take on finitely many values. The integral of a simple function f is

Where a_i is a value the simple function takes, E_i = f^{-1}(a_i), and m(E_i) is the measure of E_i. This uses a bit of a fancy definition, but in this case you're helped a bit by the theorem that the measure of any countable set is 0, and m(A U B) = m(A) + m(B) when A and B are disjoint. This means the measure of the rationals in [0,1] is 0, and so your function has integral 0.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**random_fruit****Member**- Registered: 2008-12-25
- Posts: 39

An example where area is not directly relevant is the work done in charging a capacitor. See http://en.wikipedia.org/wiki/Capacitance for more information.

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