# What do I have to know about integrals?

#### Prerequisites

The lead-in article to this, Integrals: The mathematical meaning, was pretty long, tedious, and mathematical.  In this class, we are not going to ask you to make extensive mathematical derivations like that. So what is it we want you to know?

## The basics

Here are a few of the main points you need to understand about an integral.

1. An integral is a summing up of products of a function ($g$) times a small change ($dx$).
2. By summing up the products of $g$ times $dx$ we are creating the function (call it $f$) whose derivative is $g$.
3. Since the integral is a sum of products, the dimensionality of the integral will be a product of the dimensionality of $g$ times the dimensionality of $dx$.
4. Typically, the integral only gives you the shape of the function whose derivative is $g$.  Since you can add a constant to $f$ and the new function will have the same derivative (since the derivative of a constant is 0), to get the function that is represented by an integral, you have to know the value of the function you are creating at one point ($f(x_0)$).
5. You need to know a few "anti-derivatives" — that is, what function do you get if you integrate a few simple functions.  At most you will need to know these:
• The integral of $x^n$ is $x^{n+1}/(n+1)$
• The integral of $e^x$ is $e^x$.
• The integral of $\cos{x}$ is $\sin{x}$ and the integral of $\sin{x}$ is $-\cos{x}$

There are a number of useful ways of looking at integrals:

• Integral as a product
• Integral as a stepping rule
• Integral on a graph

## Integral as a product

The key equations for understanding the integral are multiplying up from the definition of the derivative (remember, $g = \frac{df}{dx}$) and then opening up the change in $f$ and solving for the final value. These are written as three equations that follow easily one from the next.

$$g=\frac{\Delta f}{\Delta x} \rightarrow \Delta f = g\Delta x$$

$$f_f - f_i \approx g(x_i)\Delta x$$

$$f_f \approx f_i + g(x_i)\Delta x$$

where we are using a subscript "i" to indicate "initial value" and "f" to mean "final value". This isn't too bad.  It's really just algebra (with lots of symbols, as is usual in science!).

The equation above says the small change in $f$ = the small change, $g\Delta x$. The integral is just added up these small changes to get a big change.

## The integral as a stepping rule

After we had the product, in the above equations we just opened $\Delta f$ up to show it as an explicit change. We then moved the initial value to the right. So the right side only depends on the starting place and gives us the final value at a nearby point. If, for example, our "$g$" stands for an acceleration, since the acceleration is the derivative of the velocity, then the velocity should be the integral of the acceleration.  If we know the acceleration as a function of time (for any time), and we know a starting value of the velocity, we can generate velocities at later times. The triplet of equations above would now look like this:

$$a=\frac{\Delta v}{\Delta t} \rightarrow \Delta v = a\Delta t$$

$$v_f - v_i \approx a(t_i)\Delta t$$

$$v_f \approx v_i + a(t_i)\Delta t$$

We can get a velocity at a later time from the velocity and acceleration at an earlier time — predicting the future by integrating! This method turns out to be a very valuable way to solve for the motion of objects using our fundamental principle of motion, Newton's 2nd law. (See Newton 2 as a stepping rule.)

## The integral as a graph and the relation to the derivative

We can also look at the integral as something on a graph, just as we could a derivative. For example, the velocity ($v$) is the derivative of the position ($x$) with respect to time, $v = dx/dt$.  This means that the position is the integral of the velocity with respect to time, $dx = v dt$.  Note that the integral only gives us the change in the function it creates. To get the whole thing, you have to know a starting value.

These ideas are represented in the following graphs with their associated equations.

The first set represents how to get the graph of the derivative (v) from the graph of the function (x) by taking the slope ("little triangles" or "rise over run")

These equations show what is going on:

$$v(t) = \frac{dx}{dt}$$

$$v(t) = \frac{x(t+\frac{\Delta t}{2}) - x(t-\frac{\Delta t}{2}) }{\Delta t}$$

We start with the x-t graph.  At each time on the x-t graph (5 are shown by dotted lines) we make small triangles showing what the change is in x during a small time interval around each selected instant. The ratio of those two changes is the velocity at that instant.

To go backwards — from velocity to position — we have to integrate. That process is represented by these graphs.

These equations show what is going on:

$$dx=v(t)dt$$

$$\Delta x=\sum{dx} = \int{v(t)dt}$$

Matching the equations to the graph, we start on the v graph at a particular time and choose a small time interval, $dt$. The product of the average velocity in that interval times $dt$ is the area of the little rectangle shown. That area has dimensions [$v dt$] = (L/T) (T) = L so it's the change in the position. If we have a starting value of $x$ at the initial time, we can use the area on the v-t graph to figure out the change in $x$ and therefore what the new value of $x$ will be at the end of the time interval.

Using the position and velocity example, the pair of equations

$$v =\frac{\Delta x}{\Delta t} \quad \quad \Delta x = v \Delta t$$

shows how to think about going from a function graph to a derivative graph and back. The derivative is the ratio (so... "rise over run" or "little triangles") and integral is the product (so area equals change).

Joe Redish 9/4/11

Article 283