Weighted Average Intergral

• Sep 30th 2009, 05:45 PM
hashshashin715
Weighted Average Intergral
Hi, i spent too much time on this one problem. Any help would be helpful.

Let f(x) = x^2 for [0, 1]. The average value of f on [0, 1] is 1/3. Find a non negative weight function w such that the weighted average of f on [0, 1] is a) 1/2 b) 3/5 c) 2/3.

This is the formula that should help out.

weighted average = Integral( w(x)f(x) , x, a, b) / Integral( w(x), x, a, b)

w(x) is the weighted function.

Thanks for any help or advice you can give.
• Sep 30th 2009, 09:23 PM
mr fantastic
Quote:

Originally Posted by hashshashin715
Hi, i spent too much time on this one problem. Any help would be helpful.

Let f(x) = x^2 for [0, 1]. The average value of f on [0, 1] is 1/3. Find a non negative weight function w such that the weighted average of f on [0, 1] is a) 1/2 b) 3/5 c) 2/3.

This is the formula that should help out.

weighted average = Integral( w(x)f(x) , x, a, b) / Integral( w(x), x, a, b)

w(x) is the weighted function.

Thanks for any help or advice you can give.

Maybe you should try finding a simple weight function of the form w = x^n ....
• Oct 1st 2009, 02:42 AM
hashshashin715
Oh i see. I know the answer is x, x^2, and x^3. So if I just assume it is a simple function then all i have to do is solve for n. OK, thanks.
• Oct 1st 2009, 02:47 AM
mr fantastic
Quote:

Originally Posted by hashshashin715
Oh i see. I know the answer is x, x^2, and x^3. So if I just assume it is a simple function then all i have to do is solve for n. OK, thanks.

I haven't checked the others but I think you'll find that the
Quote:

non negative weight function w such that the weighted average of f on [0, 1] is a) 1/2
is w = 1/x, not w = x ....