1 Velocity and displacement, speed and distance

Some values include “direction” that is relative to some fixed point.

• $v(t)$ is the velocity of an object at time $t$. This represents the “change in position” at time $t$.

• $s(t)$ is the position of an object at time $t$. This gives location with respect to the origin.

  $$s(t) = s(a)+\int _a^t v(z) dz.$$
• $s(b) -s(a)$ is the displacement, the distance between the starting and finishing locations.

On the other hand speed and distance are values without “direction.”

• $|v(t)|$ is the speed.
• $\int _a^b |v(t)| dt$ is the distance traveled.

Consider a particle whose velocity at time $t$ is given by $v(t) = \sin (t)$.

What is the displacement of the particle from $t=0$ to $t=\pi$? That is, compute:

$$\int _0^\pi \sin (t) dt \begin{prompt}= \answer [given]{2}\end{prompt}$$

What is the displacement of the particle from $t=0$ to $t=2\pi$? That is, compute:

$$\int _0^{2\pi } \sin (t) dt \begin{prompt}= \answer [given]{0}\end{prompt}$$

What is the distance traveled by the particle from $t=0$ to $t=\pi$? That is, compute:

$$\int _0^\pi \left | \sin (t) \right | dt \begin{prompt}= \answer [given]{2}\end{prompt}$$

What is the distance traveled by the particle from $t=0$ to $t=2\pi$? That is, compute:

$$\int _0^{2\pi } \left | \sin (t) \right | dt \begin{prompt}=\answer [given]{4}\end{prompt}$$

2 Change in the amount

We can apply the Fundamental Theorems of Calculus to a variety of problems where both accumulation and rate of change play important roles. For example, we can consider a tank that is being filled with fuel at some rate. Given the rate, we can ask what is the amount of fuel in the tank at a certain time. Or, a tank that is being emptied at a given rate, or a culture of bacteria growing in a Petri dish, or a population of a city, etc. This brings us to our next theorem.

3 Average value

Conceptualizing definite integrals as “signed area” works great as long as one can actually visualize the “area.” In some cases, a better metaphor for integrals comes from the idea of average value. Looking back to your days as an even younger mathematician, you may recall the idea of an average:

If we want to know $\overline {f}$, the average value of a function on the interval $[a,b]$, a naive approach might be to introduce $n+1$ equally spaced grid points on the interval $[a,b]$

and choose a sample point $x_k^*$ in each interval $[x_k,x_{k+1}]$, $k=1,\dots ,n$.

We will approximate the average value of $f$ on the interval $[a,b]$ with the average of $f(x_1^*)$, $f(x_2^*)$, …, and $f(x_n^*)$:

Multiply this last expression by $1 = \frac {(b-a)}{(b-a)}$: $$\begin{align*} \frac 1n \sum _{k=1}^n f(x_k^*)\frac {(b-a)}{(b-a)} &= \sum _{k=1}^n f(x_k^*)\frac 1n \frac {(b-a)}{(b-a)} \\ &= \frac {1}{b-a} \sum _{k=1}^n f(x_k^*)\frac {b-a}n \\ &=\frac {1}{b-a} \sum _{k=1}^n f(x_k^*)\Delta x \end{align*}$$

where $\Delta x = (b-a)/n$. Ah! On the right we have a Riemann Sum!

What will happen as $n\to \infty$?

We take the limit as $n\to \infty$:

This leads us to our next definition:

Multiplying this equation by $(b-a)$, we obtain that

If $f$ is positive, the average value of a function gives the height of a single rectangle whose area is equal to

An application of this definition is given in the next example.

When we take the average of a finite set of values, it does not matter how we order those values. When we are taking the average value of a function, however, we need to be more careful.

For instance, there are at least two different ways to make sense of a vague phrase like “The average height of a point on the unit semi circle”

One way we can make sense of “The average height of a point on the unit semi circle” is to compute the average value of the function

on the interval $[-1,1]$.

Another way we can make sense of “The average height of a point on the unit semi circle” is the average value of the function

on $[0,\pi ]$, since $\sin (\theta )$ is the height of the point on the unit circle at the angle $\theta$.

See if you can understand intuitively why the average using $f$ should be larger than the average using $g$.

4 Mean value theorem for integrals

Just as we have a Mean Value Theorem for Derivatives, we also have a Mean Value Theorem for Integrals.

This is an existential statement. The Mean Value Theorem for Integrals tells us:

The average value of a continuous function is in the range of the function.

We demonstrate the principles involved in this version of the Mean Value Theorem in the following example.

Let $f(x)=\sin {x}$. Consider $\int _0^\pi \sin x dx$. Find a value $c$ guaranteed by the Mean Value Theorem.

We first need to evaluate $\int _0^\pi \sin x dx$.

$$\int _0^\pi \sin x\ dx =\bigg [ -\cos x \bigg ]_0^\pi = 2.$$

Thus we seek a value $c$ in $[0,\pi ]$ such that $\pi \sin c =2$.

$$\pi \sin c = 2\ \ \Rightarrow \ \ \sin c = 2/\pi \ \ \Rightarrow \ \ c_1 = \arcsin (2/\pi ) \approx 0.69.$$

There is another such point: $c_2=\pi -0.69$. Indeed,

$$f( \pi -0.69)=\sin {(\pi -0.69)}=\sin {(0.69)}=\frac {2}{\pi }=\overline {f}.$$

A graph of $\sin x$, along with a rectangle with height $\sin (0.69)$, is pictured below. The area of the rectangle is the same as the area under $\sin x$ on $[0,\pi ]$. The two points $c_1$ and $c_2$ are also shown in the figure. Recall, $f(c_1)=f(c_2)=\overline {f}$.