Average Value and Average Rate of Change

One connection I like to make in my class is that, on any interval, the average rate of change of a differentiable function is equal the the average value of its derivative. This follows directly from the Fundamental Theorem, since the average value of \(\scriptsize f’\) on \(\scriptsize[a,b]\) is $$\frac{\int_a^bf'(x){\rm d}x}{b-a}=\frac{f(b)-f(a)}{b-a} .$$From my point of view, this connection is what justifies the use of the word “average” in describing a rate of change. What are we averaging? The derivative, which is to say the rate of change, of f. What are we averaging over? The entire interval \(\scriptsize[a,b]\).

In my experience, students are often confused by this use of the word “average.” And no wonder: we start using the phrase “average rate of change” long before defining the word “average” in the context of a continuous variable. I have had students who think that the average rate of change on \(\scriptsize[a,b]\) should be \(\frac{f'(a)+f'(b)}{2}\), and I don’t blame them: this is a totally reasonable thing to think if your concept of average is “add a list of n numbers and then divide by n.” Students have this confusion because we start using “average” in this different way, before explaining what we mean by it.

.

Comments

Leave a Reply

Your email address will not be published. Required fields are marked *