Integrals

The fundamental theorem of calculus has a grandiose name. I mean, it has the word "fundamental" in its name.

In math land, the most fundamental theorems are first-class citizens. Next in the pecking order are the theorems with names, like the Mean value theorem. Then there are the theorems without names, which are referred to as "theorem X.X" in textbooks. Finally, we've got the corollaries. The corollaries are social outcasts, and they're not even awarded the most basic rights.

You might be impressed with the fundamental theorem of calculus, since it has been crowned with the epithet "fundamental". But it's a bit of a let down. Just as you might be impressed by someone with a pricey car and a watch but subsequently discover that this person wasn't all that special. You'll come to find that the fundamental theorem of calculus is a bit of a disappointment.

Ok, here's what the fundamental theorem of calculus says:

If is continuous on and:

then is differentiable and has the derivative . That's it. You can also think of as describing the area until the point , like this. So the fundamental theorem of calculus relates the area under the graph with the function value.

But there's actually an important corollary to the fundamental theorem of calculus. As it turns out, you can compute an integral as the difference in . Here's what I mean.

This corollary is used all the time. So the pecking order in math land doesn't really reflect the usefulness of each theorem.

The integral is the area under the graph. By the end of this lecture note, you'll see why. You will also know what the symbol comes from. But to learn, you have to work for it. So we'll start by doing some biking.

We're going to bike up a mountain.

We start out enthusiastically, but as we start the climb, the lactic acid builds up and we lower the speed gradually. On the top we bike slow, nice and easy. As we start rolling downhill, we let the speed increase gradually again.

The velocity-time graph for this bike trip could look like this:

Here, is the velocity and is the time.

Say we would like to know how far we have biked, and all we have is this graph, as you may be familiar with:

with the distance traveled, so for this friendly graph it's just a matter of summing areas of triangles and rectangles:

Adding these easily calculated areas together gives us the total distance we have biked.

Suppose we'd be unwilling to conform to only changing our velocity linearly. We then get some curvy velocity graph under which we need to work out the area.

Falling short of methods to calculate areas of disobedient, irregular objects, our best bet is to reduce the curve to something we can calculate easily. For example, rectangles:

In this example, we divide domain of the function into small intervals . The height of the rectangle for each interval is taken to be the function value at its midpoint.

Then, calling the first rectangle height , its area is . The next one has the area , and so on until the last one, with the area .

To get the area under the graph, take the sum of all the rectangles:

Choosing the midpoint of each interval to define the height is just an arbitrary choice. We could have chosen any point on the interval. Other ways of picking will also approximate the function.

As we sum all the rectangles, we will land at some distance from the exact answer.

However, making smaller and smaller, the approximation gets better and better:

..until we make the rectangles infinitely thin, reducing the to a tiny differential . Then, the rectangle sum becomes exactly the same as the area under the curve. What we have is no longer a sum, but an integral, denoted by a stretched-out .

Let be the value exactly at the current tiny interval. Let the end of the domain studied be . Then, the integral is:

A stringent definition will follow in the next note.

Some graphs dip under the -axis. When we want to find the area under the graph for such a curve, we subtract the area which lies under from the rest. This is analogous with saying that the heights of the infinitely thin rectangles are negative for these intervals.

This is the essence of the mean value theorem for integrals. It just says that we can find a so that the area of the rectangle is the same as the area under the graph.

The function needs to be continuous for the theorem to hold.

Formally:

Okay, seems reasonable, but why do we bother making a theorem out of something that just seems obvious? Well, it's an important part of the proof of one of the most central theorems of this course: the fundamental theorem of calculus. This theorem will be presented in the next lecture note.

Let's, for fun, see where we find the -value satisfying the theorem for:

on [0, 3].
Integrating , we get:

In our case, the interval length is . What value to put into , to get ?

We see that we need . If we let , we are finished.

We have previously covered the concept of indefinite integrals, and seen how they relate to antiderivatives. As the name hints at, there is also something called definite integrals.

In the previous section, we discussed the area under a function curve. Now we will look at how the definite integral connects the area under the curve and the indefinite integral.

Say we have a function , and the form of the indefinite integral of :

where is an antiderivative of , and C is a constant.

The definite integral is defined using the indefinite one as follows:

Here, is called the integrand, and tells us that is the variable of integration which we take the antiderivative with respect to. and are called the limits of integration, where is the lower, and the upper limit.

By this definition, we must be able to find antiderivatives of between and for it to be integrable there. It must not, however, be one and the same antiderivative over the whole interval. As we will see further down, we can split the integral into shorter sub intervals, and add the integrals of those.

Put in words, the definite integral, or simply the integral, of from to is the indefinite integral evaluated at , minus the indefinite integral evaluated at . Notice that, because of the subtraction, the constant cancels out and there is no need to include it in the calculation.

It turns out that this procedure exactly calculates the area between the -axis and the function's curve, where the area above the -axis is positive, and the area below it is considered negative.

To see what this means in practice, let's have a look at the simplest of examples:

Let , and consider the area under the graph from to . This will simply be a square with side of and an area of .

We know that the indefinite integral of is , so let's see what happens when we solve the same problem using definite integrals:

just as we expected.

There are certain properties of integrals that help us calculate them effectively:

Properties of integrals

1. Reversing the limits changes the integral's sign:

2. Integrals are linearly dependent on the integrand:
If and are constants, then:

3. The triangle inequality of extends to integrals:
If , then:

4. The integral can be divided into sub intervals:

5. The integral of an even function on an interval symmetric about 0 is equal to twice the integral of the positive side of the interval:
If , then:

6. The integral of an odd function on an interval symmetric about 0 is equal to 0:
If , then

7. Integrals preserve inequalities:
If and , then