# Integrals: The mathematical meaning

#### Prerequisites

The basic idea of calculus is about how changes are linked. When we look at derivatives, we pay attention to how the changes in two variables were connected when the changes were small. But we want to be able to go further than that. Suppose the changes are big? If we know the derivative, we can build up big changes by adding up the small changes. We do this with the concept of the integral. In a very real sense, the derivative and the integral are inverses of each other. On this page, we'll review the math of how the integral is constructed and what it means mathematically.

If you remember from your calculus class how the integral is built up from sums (as in the Riemann integral), you can probably skip this page and go directly to the follow-on page, How can I actually use integrals?, where we discuss the variety of ways we'll use it to make sense in this class.

If you don't remember how to construct the integral in a way that lets you think about what this means, follow our discussion on this page. Take it carefully a step at a time.

## The derivative as a ratio of small changes

Let's start with the abstract math version. We have a function, $f$, that depends on an independent variable (one whose value we can choose freely) $x$. The idea of a derivative is to define a function that represents the ratio of how much $f$ changes when $x$ changes (by a little).  So

$$g(x) = f'(x) = \frac{df}{dx}$$

The notation contains a bit of an internal contradiction. The expression on the left, $g(x)$, says that we are considering this derivative as a function of $x$ so we know it at any individual point.  The expression on the right says the derivative is the change in $f$ divided by the change in $x$.  To get a change, you have to consider two different x's.

The way we get away with this is that we consider "the change" as being from "a value of $x$ just a tiny bit smaller than $x$ to a value of $x$ just a tiny bit larger." So it's like looking at a movie film and saying "the velocity at frame 96 is the change in position from frame 95 to frame 97 divided by the time interval from frame 95 to frame 97." This is kind of cumbersome, but critical in thinking about what velocity — or any derivative — really means. People in math spend a lot of effort to get rid of this; but for us, it's better to think about this way.

## Adding up small changes: The integral

If you know $f$, you differentiate it to get $g$. But what do we do if we know the function $g$ and want to go the other way and find $f$?  This might seem strange at first. How would we know a function's derivative if we didn't know the function? But actually, this happens all the time. One of the key principles this term will be to show that an object's acceleration is determined by the totality of the forces that it feels. So we might know an object's acceleration by figuring out the forces. But the acceleration is the derivative of the velocity! So $a = dv/dt$ and we know $a$, as a function of time but want to find $v$. And then we might know $v=dx/dt$ as a function of time but want to figure out $x$.

Let's try it using $g$ and $f$ so we can focus on the general structure of the math without being distracted by the particular "what it is" for now.

If we take seriously the derivative as a ratio, we can multiply both sides of our equation by the small change "$dx$" to get the following:

$$df = g(x)dx$$

Read this as: "the change in $f$ is equal to $g$ (the derivative of $f$) times the change in $x$."  It might be clearer what this means if we used deltas instead of d's and pay attention to the ends of the interval dx.  This makes it look like the following:

$$\Delta x = x_1 - x_0$$

$$\Delta f = g(x)dx \; \rightarrow \; f(x_1) - f(x_0) = g(x_0) \Delta x$$

Since we are looking at a change, we have two values of $x$ but $g$ is only a function of $x$ once.  We pick the starting value, though it would probably be more aesthetic to choose to evaluate $g$ at the midpoint of the interval, $\frac{x_0 + x_1}{2}$. But everything would then look a lot messier and it turns out not to make a difference.

Now let's take many steps of size \Delta x:

$$x_1 - x_0 = \Delta x$$

$$x_2 - x_1 = \Delta x$$

$$x_3 - x_2 = \Delta x \\ ...$$

The ... means "imagine that this process continues".  We'll write "N" for the top value of our last step.

If we now write how $f$ changes in each step, we get

$$f(x_1) - f(x_0) = g(x_0) \Delta x$$

$$f(x_2) - f(x_1) = g(x_1) \Delta x$$

$$f(x_3) - f(x_2) = g(x_2) \Delta x \\ ...$$

where again the three dots means "keep going".

Now the trick is to add up all of the equations in the line above. This gives us

$$[f(x_N) - f(x_{N-1})] + [f(x_{N-1}) - f(x_{N-2})] +... + [f(x_2) - f(x_1)] + [f(x_1) - f(x_0)] \\ = g(x_{N-1}) \Delta x + g(x_{N-2}) \Delta x+ ... + g(x_1) \Delta x + g(x_0) \Delta x$$

Although this looks messy, something interesting happens. All the terms on the left cancel except the first and the last! (The ending point of one interval and the starting point of the next are the same but once it comes in with a + and once with a -.) But the terms on the right (with the $g$'s) just add — there is no cancellation. The result is that the change of $f$ from first to last is given by the sum of lots of $g\Delta x$ terms. We write this sum of terms using the summation notation with a big Greek "S" (= sigma = Σ) for "sum". The right hand side of the second line below just means exactly what is written above it.

$$f(x_N) - f(x_0) = [g(x_{N-1}) + g(x_{N-2}) + ... + g(x_1) + g(x_0)] \Delta x \\ = \sum_{j = 1}^{N-1}{ g(x_j) \Delta x}$$

Now we imagine making the $\Delta x$ very small and call it $dx$ again. The number of terms in the sum gets very large — but since each term is getting smaller, these compensate and the result (usually) settles down.  When we don't specify the steps in our adding up we replace our Greek Sigma by a stretched "S" or integral sign.

$$f(x_N) - f(x_0) = \int_{x_0}^{x_N}{ g(x) dx}$$

We can then write

$$f(x_f) = f(x_i) + \int_{x_0}^{x_f}{ g(x) dx}$$

changing from "0" to "i" (for initial) and from "N" to "f" (for final). This result shows that our sum (or integral) of derivatives can tell us how f changes as x moves away from the starting value. This means that our integral can give as the values of the entire function $f$ for any value of x, as long as we know a starting point and the derivative of $f$.

## Definite and indefinite integrals: the fundamental theorem of calculus

Depending on how we use that last result we can interpret it in different ways. If we change the name of our top and bottom starting points, we get an integral that goes between two specific values: a definite integral:

$$f(b) - f(a) = \int_{a}^{b}{ \frac{df}{dx}dx}$$

This says that the integral of the derivative of a function between two points gives the change in that function between those two points.

If, instead, we replace our ending point of the integral by a variable $x$, we generate the function $f(x)$ explicitly. In this case, so as not to get confused with the variable inside the integral that we are adding up, we'll call the $x$ inside the integral $x'$. This gives

$$f(x) = f(x_0) + \int_{x_0}^{x}{ \frac{df}{dx'}dx'}$$

The starting value, $f(x_0)$ is just a constant. This is often written $C$ and the above result written

$$f(x) = \int_{x_0}^{x}{ \frac{df}{dx'}dx'} + C$$

Since we are thinking of $x$ here as a variable, this is called the indefinite integral.

The expressions in this section are often referred to as the fundamental theorem of calculus, since they show that the derivative and the integral are inverses of each other.

## Some simple integrals

Since we know the derivatives of some simple algebraic expressions, we can also easily get the integrals of those expressions. Since we know

$$\frac{d}{dx}x^n = nx^{n-1}$$

it must follow by the fundamental theorem of calculus that

$$\int_{x_0}^{x} x'^n dx' = \frac{1}{n+1}x^{n+1} + C$$

And since we know (from The derivatives of trig functions) that

$$\frac{d}{dx} sin(x) = -cos(x) \quad\quad \frac{d}{dx} cos(x) = -sin(x)$$

$$\int_{x_0}^{x} cos(x') dx' = sin(x) + C$$
$$\int_{x_0}^{x} sin(x') dx' = -cos(x) + C$$