The definite integral generalizes and formalizes a simple and intuitive concept: that of area. So far, you've been solving indefinite integrals, and it may be difficult to imagine how all those calculations could be remotely related to area.

We'll discover how that relationship works with the fundamental theorem of calculus. What I want to do now is to define the definite integral and give you some intuition.

As we did in the case of derivatives, we'll use a physical example to explain the concept of the definite integral. This is natural, as calculus was originally inspired by physics.

Let's say the distance traveled by a particle is a function of time:

We use the letter "s" to denote distance. As we saw in definition of the derivative, if we derive this function we'll get velocity.

Let's graph these two functions. First distance:

It is a parabola. Now, if we take its derivative, we get:

And we can graph this derivative, as it also a function of time:

It is a straight line. When we take the derivative of distance we get velocity. Now, let's consider the reverse problem.

Let's say that given velocity, we want to find the distance traveled over a period of time. We know from physics that distance equals velocity times time.

So, let's go back to our graph of velocity. Let's say we want to find the distance traveled over a very small time interval. We'll call this time interval "dt":

I drew it large, but this interval is very very small. So small, in fact, that we can consider that the speed is constant during this time interval.

What does it mean that speed is constant? It means that its graph is an horizontal line. Because the time interval is so small, the velocity didn't have enough time to change. This is an approximation. And it is a good approximation if the time interval, dt, is really small. The smaller the interval, the better. We can see this in this graph:

So, if we have a constant velocity during this time interval, it's easy to calculate the distance traveled. We just have to multiply velocity and time. We'll call this small distance "ds", and it equals:

Don't feel overwhelmed by the fancy notation. This is just velocity times time, which equals distance.

Now, if we look at the graph we'll see that the formula we've got for distance is equal to the colored area:

What a coincidence! We have a rectangle with base "dt", and height "V". So, multiplying these we get the rectangle's area. We're arriving somewhere here...

What if we wanted to find the distance traveled during a longer time interval? Using the method in the previous section we're limited to a very small time intervals. This is because we used the approximation that velocity is constant during that interval.

During longer intervals, velocity definitely isn't constant. We can see that in its graph. But there's hope, and here is when things start to get interesting.

Let's say we want to find the distance traveled between time 0 and a later time we'll call "t subzero". This is the time interval:

What we can do is to divide this long interval into smaller chunks. We'll get something like this:

Each of these smaller intervals has width "dt". And now we transformed this complex problem into a series of very simple problems. For each of these small time intervals we can do what we did in the previous section.

We can say that for each interval velocity is constant. This means that each interval has only one velocity, and it is constant. So, just using common sense, to get the distance traveled during the total time:

- We calculate the distance traveled over each small time interval. This is done multiplying velocity and "dt".
- We sum these distances, and get total distance.

Also, remember that each distance is equal to the area of each rectangle. So, basically we just have to find the area of each rectangle and sum them up. Quite simple.

However, remember that this is just an approximation. There is an error for each small interval.

How can we make this approximation more exact? We've already seen that. We can take smaller and smaller intervals to get more exact results. You can see this in this animation:

Now, the really important question... What will be the real distance we're looking for? You've probably guessed that by now. We get the real value when we take the limit as each interval approaches zero. And this limit is also equal to the area under the curve.

This is a really intuitive concept, and that's all definite integrals are about.

Now we'll introduce notation. We saw that the distance traveled during each small time interval is:

And that if we want to approximate the distance over a longer interval we need to sum all the small intervals it is made up of. If we take the limit of this sum as the width of each interval approaches zero, we get the exact distance. We write this limit as:

The curvy integral sign in reality represents a sum. In fact, it represents the limit of a sum. And we also have the upper and lower limits:

These two little numbers are called upper limit and lower limit. Don't confuse them with limits of functions. They just specify the interval over which we're adding little rectangles. In this particular case, the lower limit is zero, and the upper limit is "t subzero".

We've seen what a definite integral is and how to approximate it using small rectangles. But how do we calculate its exact value? You may think that we need to solve a limit for each definite integral.

Fortunately, there is a simple rule which we can use. Here all your skills for solving indefinite integrals will show its real value. We'll learn this when we talk about the fundamental theorem of calculus.

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