Average Value of a Function

Today I want to consider a way of developing the expression for finding the average value of a function, f (x), on an interval [a, b].

Ask students how to find the average of a bunch of numbers and they will say, “add them up and divide by the number of numbers.” Then ask if they can average an infinite number of numbers. Most will say no since you cannot divide by infinity.

Well, what if the numbers were all the same?

Such as all the y-values of f (x) = 2 between x =1 and x = 5. Isn’t the average 2? So, apparently, you can sometimes average an infinite number of numbers.

Next suggest something like f (x) = x between x = 0 and x = 3. Since half the values are obviously above 1.5 and half below, can’t 1.5 be their average? Sketch this situation and draw the segment at y = 1.5 to help them see this.

Suggest another situation, say f (x) = 2 + sin(x) on the interval \left[ 0,2\pi  \right] and some others. The average appears to be 2, again since half the values are above and half below 2.

When you draw the line that is the apparent average lead the students to see that the rectangle formed by this line, the x-axis and the ends of the interval has the same area as between the function and the x-axis.

Continue with more difficult examples until someone hits on the idea of finding the “area” with an integral and then dividing the result by the width of the interval to find the height of the rectangle that is also the average value:

\displaystyle \overline{y}=\frac{\text{''area''}}{\text{length of interval}}=\frac{\int_{a}^{b}{f\left( x \right)dx}}{b-a}

Click here for an activity that you can use to develop this idea from scratch.

The next post: A fun application of average value – Most Triangles Are Obtuse.

Advertisements

One thought on “Average Value of a Function

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s