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I have the common misconception in my business calculus classes that the Average Rate of Change, say from $x=1$ to $x=5$, is the statistical average of the rates on the four unit intervals $1$ to $2$, $2$ to $3$, etc. So when introducing/reviewing Average Rate of Change, I'm tempted to say "This is not an Average like you learn about in your stats class." Except, in reality, it is. From the perspective at the end of the course, the difference quotient is the result of finding the average value of the derivative. And the idea of the average value of a continuous function is very much like the averages they compute in a stats class.

So my question is: How can I tell them not to compute a statistical average without lying to them about the relation between average rate of change and continuous statistics?

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  • $\begingroup$ I am not sure I understand the question: if the intermediate intervals are all unitary (this is critical), then the ARC (of a function of $x$ I guess) is exactly equal to the arithmetic average of the ARC's in the intermediate intervals. What am I missing? $\endgroup$ – Alecos Papadopoulos Aug 26 '14 at 19:41
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    $\begingroup$ @AlecosPapadopoulos I am not sure what you mean by "unitary", but the fundamental problem which Aeryk is asking is that understanding the average rate of change as an actual average, you need the fundamental theorem of calculus. Personally, I think the fundamental theorem is such a basic idea that we should be teaching it to students in the first couple of weeks. $\endgroup$ – Steven Gubkin Aug 26 '14 at 20:44
  • $\begingroup$ @StevenGubkin By "unitary" I just meant what the OP writes, in the 2nd line of the post about "unit intervals, 1 to 2, 2 to 3 etc", i.e. having length equal to unity. $\endgroup$ – Alecos Papadopoulos Aug 26 '14 at 21:52
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    $\begingroup$ This property cannot be used to define the ARC, it does not seem, since it is circular. But since it reduces the problem to smaller subintervals, perhaps the apparent circularity can be removed. It should end up being the Riemann sum definition, and then you need essentially fundamental theorem to prove the equivalence. $\endgroup$ – Steven Gubkin Aug 26 '14 at 22:41
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$$\frac{1}{4}\int_1^5 f' = \frac{1}{4}\left( \int_1^2 f' + \int_2^3 f' + \int_3^4 f' + \int_4^5 f' \right)$$ The LHS is an average rate of change.

The RHS is a statistical average of four average rates of change.

This requires some differentiability, but I would not treat this as a misconception!

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From the point of view of an outsider like me, what the OP describes is the following: there is a function $f()$ and we define the Average Rate of Change of this function between points $a<b$ as

$$ARC_f = \frac {f(b) - f(a)}{b-a} \tag{1}$$

Then the OP considers as an example the interval $[1,5]$, for which we have

$$ARC_f(1,5) = \frac {f(5) - f(1)}{5-1} = \frac 14 [f(5) - f(1)]\tag{2}$$

He then speaks of the "misconception" that students believe that they can recover $ARC_f(1,5)$ by calculating the arithmetic average of the Average Rates of Change in the intermediate unit-length intervals, namely

$$ARC_f(1,5) = ?? \frac 14 \left(\frac {f(2) - f(1)}{2-1}+\frac {f(3) - f(2)}{3-2}+\frac {f(4) - f(3)}{4-3}+\frac {f(5) - f(4)}{5-4}\right) \tag{3}$$

It is obvious that the right-hand-side of $(3)$ is identical to the right-hand side of $(2)$, and so there is no misconception at all, the students are right, as long as we are talking about an interval that can be exactly decomposed into unit-length intervals, and that we take the arithmetic average of the ARCs related to these unit-length intervals.

Rigor? I am more than certain that all involved have in their minds a function that is defined everywhere in $[a,b]$, and any other regularity condition that might be needed -after all, this is business calculus, as the OP informs us.

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