# Engineering Statics: Open and Interactive

## Section7.6Average Value of a Function

The weighted average technique discussed in (7.1.2) are a fine for averaging several discrete values, but what do we do if we need to find the average of an infinite number of values or values which change continuously?
Consider a function $$y=f(x)$$ over some interval from $$a$$ to $$b\text{.}$$ How can we find the average value $$\bar{y}$$ of the function over that interval? To understand what is meant by the average value of a function, look at the interactive below. There, the function $$f(x)$$ is the red curve, and which you can change if you like. The blue dots $$a$$ and $$b$$ mark the beginning and end of the interval and are also adjustable.
The area under this curve between $$a$$ and $$b$$ is shaded with gray, and we can find it using a definite integral
\begin{equation*} \int_a^b f(x)\ dx \end{equation*}
The blue hatched rectangle has the same area as the gray shaded region, and because the areas are the same, the height of the rectangle $$\bar{y}\text{,}$$ is the average value of $$f(x)\text{.}$$
With this in mind, we can calculate the average value of $$f(x)$$ by equating the area under the curve with the area of the rectangle and solving for $$\bar{y}\text{.}$$
\begin{align*} \int_a^b \amp f(x)\ dx = \bar{y} (b-a)\\ \bar{y} \amp = \dfrac{\int_a^b f(x)\ dx}{(b-a)} \amp \text{and since} \int_a^b dx \amp = (b-a)\\ \bar{y} \amp = \dfrac{\int_a^b f(x)\ dx}{\int_b^a dx} \end{align*}
This is a weighted average like (7.1.2) but instead of summing $$n$$ discrete values, we integrate of an infinite number of infinitesimal values. $$f(x)$$ is the value being averaged and the weighting function is $$dx\text{.}$$
This approach is true for any choice of weighting function. To find $$\bar{x}$$ for a two dimensional area, the value to be averaged is $$x$$ and the weighting function is $$dA\text{,}$$ so replacing $$dx$$ with $$dA$$ and $$x_i$$ with $$x\text{,}$$
\begin{align*} \bar{x}\amp = \frac{\sum \bar{x}_{i} A_i}{\sum A_i} \amp \amp \Rightarrow \amp \bar{x}\amp = \frac{\int x dA}{\int dA} \end{align*}
In other words, to transform a discrete summation to an equivalent continuous integral form you:
1. Replace the summation with integration, $$\Sigma \Rightarrow \smallint\text{.}$$
2. Replace the discrete weighting factor with the corresponding differential element,
\begin{equation*} \begin{cases}A_i \amp \Rightarrow dA \\ V_i \amp \Rightarrow dV \qquad \text{ etc.}\\ W_i \amp \Rightarrow dW\end{cases} \end{equation*}
3. Rename the value being averaged to eliminate the index $$i\text{.}$$ We often use $$el$$ as a subscript when referring to a differential element.
The two-dimensional centroid equations (7.5.1) become,
\begin{align*} \bar{x} \amp= \frac{ \sum {\bar{x}_i}\ A_i}{\sum A_i} \Rightarrow \frac{ \int \bar{x}_{\text{el}}\ dA}{\int dA} \amp \bar{y} \amp = \frac{ \sum {\bar{y}_i}\ A_i}{\sum A_i} \Rightarrow \frac{ \int \bar{y}_{\text{el}}\ dA}{\int dA}\text{,} \end{align*}
and in the same way the center of gravity equations become
\begin{align*} \bar{x} \amp = \frac{ \int \bar{x}_{\text{el}}\ dW}{\int dW} \amp \bar{y} \amp = \frac{ \int \bar{y}_{\text{el}}\ dW}{\int dW} \amp \bar{z} \amp = \frac{ \int \bar{z}_{\text{el}}\ dW}{\int dW}\text{.} \end{align*}

### Question7.6.2.

How far is it from the earth to the sun?