# Math Insight

### Averages and weighted averages

The usual notion of average of a list of $n$ numbers $x_1,\ldots,x_n$ is $$\hbox{ average of } x_1,x_2,\ldots,x_n={x_1+x_2+\ldots+x_n\over n}.$$

A continuous analogue of this can be obtained as an integral, using a notation which matches better: let $f$ be a function on an interval $[a,b]$. Then $$\hbox{ average value of f on the interval }[a,b]= {\int_a^b f(x)\;dx\over b-a}.$$

For example the average value of the function $y=x^2$ over the interval $[2,3]$ is $$\hbox{ average value of f on the interval }[a,b]= {\int_2^3 x^2\;dx\over 3-2}={[x^3/3]_2^3 \over 3-2}= {3^3-2^3 \over 3\cdot (3-2)} = 19/3.$$

A weighted average is an average in which some of the items to be averaged are ‘more important’ or ‘less important’ than some of the others. The weights are (non-negative) numbers which measure the relative importance.

For example, the weighted average of a list of numbers $x_1,\ldots,x_n$ with corresponding weights $w_1,\ldots,w_n$ is $${w_1\cdot x_1+w_2\cdot x_2+\ldots+w_n\cdot x_n\over w_1+w_2+\ldots+w_n}$$ Note that if the weights are all just $1$, then the weighted average is just a plain average.

The continuous analogue of a weighted average can be obtained as an integral, using a notation which matches better: let $f$ be a function on an interval $[a,b]$, with weight $w(x)$, a non-negative function on $[a,b]$. Then $$\hbox{ weighted average value of f on the interval }[a,b] \hbox{ with weight } w = {\int_a^b w(x)\cdot f(x)\;dx\over \int_a^b w(x) \;dx}$$ Notice that in the special case that the weight is just $1$ all the time, then the weighted average is just a plain average.

For example the average value of the function $y=x^2$ over the interval $[2,3]$ with weight $w(x)=x$ is $$\hbox{ average value of f on the interval }[a,b] \hbox{ with weight } x$$ $$={\int_2^3 x\cdot x^2\;dx\over \int_2^3 x\;dx} ={[x^4/4]_2^3 \over [x^2/2]_2^3}= {{1\over 4}(3^4-2^4) \over {1\over 2}(3^2-2^2)}$$