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#1 2005-12-12 04:31:22

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

average within interval

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.


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#2 2005-12-12 04:37:13

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

Re: average within interval

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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#3 2005-12-12 06:49:36

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: average within interval

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)


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#4 2005-12-12 06:58:38

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

Re: average within interval

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.


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It wanted to be normal.

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#5 2005-12-12 07:32:41

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: average within interval

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)


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#6 2005-12-13 15:52:34

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: average within interval

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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#7 2005-12-13 20:05:01

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,552

Re: average within interval

["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 smile]


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

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#8 2005-12-14 04:48:19

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

Re: average within interval

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. smile


Why did the vector cross the road?
It wanted to be normal.

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