Thread: average value of a function for 'advanced calc'

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

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

Asked and answered here: http://www.mathhelpforum.com/math-he...intergral.html

Thread closed.