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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

If you want the average y value (height) of a

function, then you can get the area under the

curve and divide by the interval width.

Well I was hoping to find a way to do this a

little differently, but I may need some help.

Instead of integrating first and then dividing,

I would like to do something in preparation

before the integration, so after the definite

integral, there will be no division.

If anyone can help. Thank you.

If this can be done, I think it would help me

to grasp underlying principles better for

maybe other applications.

**igloo** **myrtilles** **fourmis**

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

You can divide by the class width before you integrate instead of afterwards, but that's pretty much the same thing. Other than that, I don't think there's a way.

Why did the vector cross the road?

It wanted to be normal.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I don't understand.

What about y = x from 10 to 20, so the average should be 15?

So

Oh you mean divide by 10 and factor out the

before you do the integral?

*Last edited by John E. Franklin (2005-12-12 06:52:31)*

**igloo** **myrtilles** **fourmis**

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

Yes. So, in your example, that would be ∫ (x/10)dx. As I said, they're really just the same thing in a different order.

Why did the vector cross the road?

It wanted to be normal.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Thanks, I got it.

But if you don't know the class width, you probably can't do it ahead of

time. Like divide by

But that only works for y=constant?

What I mean is

so the 1/x is like dividing by the class width.

But it doesn't work if it is another function because

and then divide by class width.

But this is different than the following where I divide by first:

And clearly is not the same as

.

Sorry I left out the dx's in the equations; I don't understand what they are yet. I know it is an incremental piece of x, but I don't know where it is

coming from. Does it appear when you decide to take the integral with

respect to x?

*Last edited by John E. Franklin (2005-12-12 07:39:59)*

**igloo** **myrtilles** **fourmis**

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

You are getting a little confused with the notation. f(x)=x^2 is the same thing as y=x^2. f'(x) is the same as dy/dx=2x. Taking an integral is the exact opposite of taking an integral. An integral is also called an antiderivative. ∫dy = ∫2x dx is the same as y=x^2.

Too many people get hung up on the different notation. In my opinion, you should become more comfortable with the dy/dx type than the f'(x) kind. It makes the higher maths a little easier to grasp.

Those increments are the most important piece of calculus to understand. Without those increments you really can't understand limits. And limits are used as the basis for almost all of calculus' proofs.

By the way if you want to remember to divide by the "class width" in a general way, just divide your function by (b-a). They are constants and will not affect your integral. But that is more general notation. a being the upper and b the lower bound of a definite integral.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,618

["Taking an integral is the exact opposite of taking an integral" - you intended to say derivative for one of those, I think. If you want to, just edit your own post and I will delete this comment and it will all look really neat ]

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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

Yes, but it will still have the *last edited by...* bit so people will know that something's up. Plus, my post is here now.

Why did the vector cross the road?

It wanted to be normal.

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