I spent a lot of time yesterday trying to confirm to myself that

$$\frac{f(b) - f(a)}{b - a}$$

is the average rate of change of a function over $[a, b]$. I did this by first calculating this value, and then calculating $\dfrac{\Delta y}{\Delta x}$ by discretizing $\Delta x$ to a very small time step, and then taking the arithmetic mean of the results for every $\Delta y$.

Indeed, I proved to myself that given a small enough discretization they are equal within some acceptable tolerance.

This lead me to read more on the difference quotient and reminding me of my numerical analysis classes. This wikipedia article on the Mean of a Function shows an alleged equivalence between the difference quotient and the arithmetic mean but it uses an integral.

The arithmetic mean is defined:

$$\mu = \frac{1}{n}\sum \limits_{i = 0}^n x_i$$

How can I analytically prove to myself that

$$\mu = \frac{f(b) - f(a)}{b - a}?$$

I believe it intuitively due to my results above, but I would like to see it analytically because I'm legitimately curious at this result.


1 Answer 1


Actually, average rate of change is simply defined to be $\mu=\frac{f(b)-f(a)}{b-a}$. It's a natural way of answering the question "If I traveled 100 km in 4 hours, how fast was I going on average?" with only the information given.

Now I'll turn to your investigation, where you quite literally average the rates of change over small intervals. This is good thinking--you are very close in spirit to the construction of integrals and to the fundamental theorem of calculus.

Let's first suppose $\Delta x$ is constant, meaning you divide $[a,b]$ into $n$ intervals of equal length, and $\Delta x=\frac{b-a}{n}$. I am going to put an index on the $\Delta y$'s, since I don't want to assume that these are constant as well.

Then $$\mu=\frac{1}{n}\sum_{i=0}^{n-1} \frac{\Delta y_i}{\Delta x}.$$ (Notice that your indicies are slightly off. Since you are adding up $n$ summands, you should either start with $i=1$ or end with $i=n-1$.)

$$\mu=\frac{1}{n}\sum_{i=0}^{n-1} \frac{\Delta y_i}{\Delta x}=\frac{1}{n}\sum_{i=0}^{n-1} \frac{\Delta y_i}{\left(\frac{b-a}{n}\right)}=\frac{1}{n}\sum_{i=0}^{n-1} \frac{n\Delta y_i}{b-a}=\frac{1}{b-a}\sum_{i=0}^{n-1}\Delta y_i.$$ Notice that the $\Delta y_i$ add up to the total change in $y$, which is $f(b)-f(a)$, and you have your proof.

What if the intervals aren't constant? Let's take a simple example: $f(x)=x^2$ on $[0,3]$. If we consider two subintervals, $[0,2]$ and $[2,3]$, then our arithemetic mean of the two difference quotients is


while the average rate of change is $\frac{9-0}{3-0}=3$, so these notions are not the same.

As you yourself pointed out, you do get something close with a small enough discretization. In fact, when $f$ is differentiable you can approximate the average rate of change with as little error as you like, just by picking small enough intervals.

The key results here is the the fundamental theorem of calculus. But developing those ideas from a precalculus background would go far beyond what I can write in this space.

  • $\begingroup$ Ah! I think the key insight for me there was that $sum(\Delta Y_i) = Y$ and $sum(\Delta x) = n$. If we take the sum of the $\Delta x$ we get our $n$ observations, and the sum of the $\Delta y_i$ is exactly equal to the total change over the interval. Fascinating. Thank you! $\endgroup$
    – CL40
    Aug 14, 2018 at 20:17

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