Originally Posted by

**twostep08** ok, i tried everything i could think of for this, but no luck with anything

let f(x)=x^2 for x's between 0 and 1. The average value of f on [0,1] is 1/3. Find a non-negative weight function such that the weighted average of f on [0,1] as defined by

A(f)=[ integral from a to b of w(x)f(x) dx]/ [integral from a to b of w(x) dx] where w is a nonneggative weight function where its integral isnt 0

is:

a)1/2

b)3/5

c)2/3

i imagine id be able to get b and c if someone gave me a push in the right direction with a, so do what you can..by the way the answers are x, x^2, and x^3 respectively, if that helps you guys...

thanks in advance