Find weight functions so that the weighted average has given values

by RoRi

August 17, 2015

Let $f(x) = x^2$ for $0 \leq x \leq 1$ . We know the average of $f$ on $[0,1]$ is $\frac{1}{3}$ . Find a nonnegative function $w$ such that the weighted average of $f$ on $[0,1]$ is

$\frac{1}{2}$ ,
$\frac{3}{5}$ ,
$\frac{2}{3}$ ,

First, we recall that given functions $f$ and $w$ (with $w$ nonnegative) defined on an interval $[a,b]$ , the weighted average is given by

$A(f) = \frac{\int_a^b w(x) f(x) \, dx }{\int_a^b w(x) \, dx}.$

Let $w(x) = x$ , then
$A(f) = \frac{\int_0^1 x \cdot x^2 \, dx}{\int_0^1 x \, dx} = \frac{1/4}{1/2} = \frac{1}{2}.$
Let $w(x) = x^2$ , then
$A(f) = \frac{\int_0^1 x^2 \cdot x^2 \, dx}{\int_0^1 x^2 \, dx} = \frac{1/5}{1/3} = \frac{3}{5}.$
Let $w(x) = x^3$ , then
$A(f) = \frac{\int_0^1 x^3 \cdot x^2 \, dx}{\int_0^1 x^3 \, dx} = \frac{1/6}{1/4} = \frac{2}{3}.$