
We give more contexts to understand integrals.

### 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.
• $s(b) -s(a)$ 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.
• $\int _a^b |v(t)| \d t$ 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: What is the displacement of the particle from $t=0$ to $t=2\pi$? That is, compute: What is the distance traveled by the particle from $t=0$ to $t=\pi$? That is, compute: What is the distance traveled by the particle from $t=0$ to $t=2\pi$? That is, compute:

### 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.

### 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 $1 = \frac {(b-a)}{(b-a)}$:

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.

Choose all the correct expressions for $\overline {v}$, the average velocity of an object moving along a straight line over the time interval $[a,b]$.

(Reminder: $s$ is the position function, and $a$ the acceleration).

$\overline {v}=\frac {1}{b-a}\int _a^b s(t)\d t$ $\overline {v}=\frac {1}{b-a}\int _a^b v(t)\d t$ $\overline {v}=\frac {1}{b-a}\int _a^b a(t)\d t$ $\overline {v}=\frac {a(b)-a(a)}{b-a}$ $\overline {v}=\frac {v(b)-v(a)}{b-a}$ $\overline {v}=\frac {s(b)-s(a)}{b-a}$

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 $g$.

### 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.

Proof
Define an accumulation (area) function, $F$, Since $F$ is continuous on the interval $[a,b]$ and differentiable on the interval $(a,b)$, we can apply the Mean Value Theorem to the function $F$ on the interval $[a,b]$. Therefore, there exist a number $c$ in $(a,b)$ such that But we know that $F'(c)=f(c)$, and that $F(b)-F(a)=\int _a^bf(t)\d t =\int _a^bf(x)\d x$. Therefore, $\blacksquare$

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