Why is average velocity not defined as the mean value of the velocity function?

Question

The mean value of a function is defined by $\frac{1}{b-a} \int_a^b {f(x)dx} $, but the average velocity of a moving object is defined by $\frac {\Delta x}{\Delta t}$ where $\Delta x$ is the change in the object's displacement and $\Delta t$ is the time interval in question. Since the first method could give an exact average velocity and the second only gives an approximation, why is average velocity defined this way?

Answer 1

Let's look at the first formula. The velocity is a function of time, so $f(x)$ will be written as $v(t)$, $dx$ will become $dt$, and $b$ and $a$ are the final and initial times. Then $$v_{ave}=\frac 1{t_f-t_i}\int_{t_i}^{t_f}v(t)dt$$ By definition the velocity is the derivative of the position with respect to time $$v(t)=\frac {dx}{dt}=x'(t)$$ From the fundamental theorem of calculus $$\int_{t_i}^{t_f}x'(t)dt=x(t_f)-x(t_i)$$ Introducing the notations $\Delta t=t_f-t_i$ and $\Delta x=x(t_f)-x(t_i)$, you get $$v_{ave}=\frac{\Delta x}{\Delta t}$$ So the two formulas are equivalent.

Answer 2

Note that $$\frac1{b-a}\int_a^bv(t)\,dt = \left.\frac1{b-a}s(t)\right]_a^b$$ $$=\frac{s(b)-s(a)}{b-a}$$ $$=\frac{\Delta s}{\Delta t}$$ so they're exactly the same thing.

Answer 3

This has to do with the sequencing of ideas. Typically, the idea of average velocity is introduced long before the derivative (which is used to define instantaneous velocity) or the integral. Indeed, the average velocity is often used to define the instantaneous velocity. A fairly typical exposition runs something like the following:

1. Suppose that an object is moving back and forth along the number line. At any moment in time, it is possible to measure the object's location. Denote this by $s(t)$. That is, $$ s(t) = \text{object's location at time $t$}. $$
2. If one wants to know how "fast" this object is moving, one must determine how far it has gone in some period of time (i.e. the units of velocity are $\text{distance}/\text{time}$). Thus it is possible to measure velocities over intervals, e.g. $$ v_a^b = \frac{s(b) - s(a)}{b-a} = \frac{\Delta s}{\Delta t}, $$ where $a$ and $b$ are two moments in time at which the object's position can be measured, and $s(b) - s(a)$ is the displacement of the object between the two measurements. Note that this is precisely the average velocity on the interval $[a,b]$.
3. The instantaneous velocity of the object at time $a$ is then defined to be the limit of the average velocities over intervals which "shrink to $a$": $$ v_a = \lim_{b\to a} \frac{s(b) - s(a)}{b-a}, $$ assuming that this limit exists.
4. The velocity function is then defined to be the function $v$ which takes a time as input, and returns the instantaneous velocity as output, i.e. $$v(t) = \lim_{b\to t} \frac{s(b) - s(t)}{b-t} = s'(t). $$

In this style of exposition, it would be somewhat circular to then redefine the average velocity on an interval to be mean value obtained by integrating the velocity function. Basically, it doesn't make sense to define the average velocity in terms of the velocity function, since the average velocity is used to define the velocity function.

Moreover, the average velocity (in terms of $\Delta s / \Delta t$) is defined whenever both $s(a)$ and $s(b)$ are defined. However, in order for the average of the velocity function to be defined (that is, in order for $\frac{1}{b-a} \int_{a}^{b} v(t)\,\mathrm{d}t$ to make sense as an average velocity), it must be the case that $v$ is the derivative of a differentiable function $s$. As a silly example of how things can go wrong, suppose that $s$ is the Cantor function, pictured below:

This function satisfies $s(0) = 1$ and $s(1) = 1$. Hence the average velocity over the interval $[0,1]$ is $$v_0^1 = \frac{s(1) - s(0)}{1-0} = 1. $$ On the other hand, this function is not at any point of the Cantor set, and at all other points, the derivative of this function is zero. Thus it makes little sense to attempt to understand the average velocity of an object whose position is given by the Cantor function in terms of the integral velocity (since the velocity is undefined at many points in the interval). It may be worth noting, however, that in the context of the Lebesgue integral, $$ \frac{1}{1-0} \int_{[0,1]} v(t) \,\mathrm{d}t = \int_{[0,1]\setminus\text{Cantor set}} 0 \,\mathrm{d}t + \int_{\text{Cantor set}} v(t)\,\mathrm{d}t = 0+0 = 0, $$ implying that the average velocity of this object over the interval is zero. This might even be a sensible conclusion, but it doesn't really jive with the intuition that the object has moved one unit in a period of one second.

That being said, if an object's position, $s$, is differentiable and its instantaneous velocity, $v$, is given by $v = s'$, then \begin{align} \frac{1}{b-a} \int_{a}^{b} v(t) \,\mathrm{d}t &= \frac{1}{b-a} \int_{a}^{b} s'(t) \,\mathrm{d}t \\ &= \frac{1}{b-a} \bigl( s(b) - s(a) \bigr) && \text{(Fundamental Theorem of Calculus)} \\ &= \frac{s(b) - s(a)}{b-a} \\ &= v_{a}^{b}. \end{align} Hence both notions of the average velocity agree in these cases.