Which of these is the correct average rate of change? The question is to find the average rate of change of $2000e^{0.21x}$ between $2$ and $5$?
I found the derivative of the equation ($2000e^{0.21x}$) and averaged them, I found the average rate of change. i.e.
$$\frac{\mathrm{d}}{\mathrm{d}x} 2000e^{0.21x} = 4200e^{0.21x}$$
so
$$\frac{f'(2) + f'(5)}{2} = 9197.$$
But you can also just use the original equation to find it i.e.
$\frac{f(5) - f(2)}{3} = 8904.6$, since there is $3$ intervals between them.
My question is why is there two different answers because the first method is the average rate of change between the two points. But the second one is too?
 A: They are both averages of the derivative, but over different sets. The first is the average of the derivative over the endpoints of the interval, i.e $\{2, 5\}$. The latter is the average of the derivative over the whole interval $[2, 5]$.
How can we define the average value of a function over an interval? We use integrals. Essentially, the average of a function $g$ over an interval $[a, b]$ is
$$\frac{1}{b - a}\int_a^b g(x) \,\mathrm{d}x.$$
Just think of the integral as a kind of "sum" here, and the $b - a$ factor as sort of the number of points on the interval $[a, b]$, and it starts to resemble the average of finitely many points.
What happens if we plug in $f'$? We get
$$\frac{1}{b - a}\int_a^b f'(x) \, \mathrm{d}x = \frac{1}{b - a}(f(b) - f(a)),$$
which is precisely the formula for the average rate of change.
So yeah, both are averages of the derivative, one over the endpoints of the interval, and one over the whole interval. Normally, when we talk about the average rate of change, we mean the latter.
A: The first equation you calculated is the average of the rates of change. The second equation you calculated is the average rate of change. Notice the small, unfortunate, semantic difference between them.
