## First Application of Definite Integral

Read this section to see how some applied problems can be reformulated as integration problems. Work through practice problems 1-4.

### Average Value of a Function

We know the average of n numbers, $a_{1}, a_{2}, \ldots, a_{n}$, is their sum divided by $n: average (mean) = \frac{1}{n} \sum_{k=1}^{n} a_{k}$.

Finding the average of a function on an interval, an infinite number of values, requires an integral.

To find a Riemann sum approximation of the average value of $f$ on the interval $[a,b]$, we can partition $[a,b]$ into $n$ equally long subintervals of length $\Delta x=(b-a) / n$, pick a value $\mathrm{c}_{\mathrm{i}}$ of $x$ in each subinterval, and find the average of the numbers $\mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right)$. Then

$\text { average of } \mathrm{f} \approx \frac{\mathrm{f}\left(\mathrm{c}_{1}\right)+\mathrm{f}\left(\mathrm{c}_{2}\right)+\ldots+\mathrm{f}\left(\mathrm{c}_{\mathrm{n}}\right)}{\mathrm{n}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right)=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right) \cdot \frac{1}{\mathrm{n}}$

This last sum is not a Riemann sum since it does not have the form $\sum \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right) \cdot \Delta \mathrm{x}_{\mathrm{i}}$ , but it can be manipulated into one:

$\sum_{i=1}^{n} f\left(c_{i}\right) \cdot \frac{1}{n}=\sum_{i=1}^{n} f\left(c_{i}\right) \cdot \frac{b-a}{n} \cdot \frac{1}{b-a}=\frac{1}{b-a} \sum_{i=1}^{n} f\left(c_{i}\right) \cdot \frac{b-a}{n}=\frac{1}{b-a} \sum_{i=1}^{n} f\left(c_{i}\right) \cdot \Delta x$

Then $\{\text { average of } f\} \approx \frac{1}{b-a} \sum_{k=1}^{n} f\left(c_{i}\right) \cdot \Delta x \longrightarrow \frac{1}{b-a} \int_{a}^{b} f(x) d x=\{\text { average of } f\}$

as the number of points $n$ gets larger and the mesh , $(\mathrm{b}-\mathrm{a}) / \mathrm{n}$, approaches 0.

Definition: Average (Mean) Value of a Function

For an integrable function $f$ on the interval $[a,b]$,

the average value of $f$ on $[a,b]$ is $\frac{1}{b-a} \int{a}^{b} f(x) d x$

The average value of a positive $f$ has a nice geometric interpretation. Imagine that the area under $f$ (Fig.6a) is a liquid that can "leak" through the graph to form a rectangle with the same area (Fig. 6b). If the height of the rectangle is $H$, then the area of the rectangle is$\mathrm{H} \cdot(\mathrm{b}-\mathrm{a})$. We know the area of the rectangle is the same as the area under $f$ so

$\mathrm{H} \cdot(\mathrm{b}-\mathrm{a})=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{dx}$ then

$\mathrm{H}=\frac{1}{\mathrm{~b}-\mathrm{a}} \int{a}^{b} \mathrm{f}(x) \mathrm{d} \mathrm{x}$

The average value of positive $f$ is the height $H$ of the rectangle whose area is the same as the area under $f$.

Example 4: Find the average value of $\mathrm{f}(x)=\sin (x)$ on the interval $[0, \pi]$. (Fig. 7)

Solution: Average value = $\frac{1}{\pi-0} \int_{0}^{\pi} \sin (x) \mathrm{dx}$

$=\left.\frac{1}{\pi}(-\cos (x))\right|_{0} ^{\pi}=\frac{1}{\pi}\{-(-1)-(-1)\}=\frac{2}{\pi} \approx 0.6366$

A rectangle with height $2 / \pi \approx 0.64$ on the interval $[0, \pi]$ encloses the same area as one arch of the sine curve. The average value of $\sin (x)$ on the interval $[0,2 \pi]$ is 0 since $\frac{1}{2 \pi} \int_{0}^{2 \pi} \sin (x) \mathrm{dx}=0$.

Practice 2: During a 9 hour work day, the production rate at time $t$ hours was $\mathrm{r}(t)=5+\sqrt{t}$ cars per hour. Find the average hourly production rate.

Function averages, involving means and more complicated averages, are used to "smooth" data so that underlying patterns are more obvious and to remove high frequency "noise" from signals. In these situations, the original function $f$ is replaced by some "average of $f$". If f is rather jagged time data, then the ten year average of $f$ is the integral $g(x)=\frac{1}{10} \int_{x-5}^{x+5} f(t) d t$. An average of $f$ over 5 units on each side of $x$. For example, Fig. 9 shows the graphs of a Monthly Average (rather “noisy” data) of surface temperature data, an Annual Average (still rather “jagged), and a Five Year Average (a much smoother function). Typically the average function reveals the pattern much more clearly than the original data. This use of a “moving average” value of “noisy” data (weather information, stock prices) is a very common.