We will begin our discussions by considering the simple geometric problem of determining the area underneath the graph of a function. In fact, we will later see how to use the same ideas to compute the volume of many different solids. These issues were of great importance to ancient mathematicians and led to the discovery of many of the ideas we'll talk about.

The area of a triangle

Of course, we all know that the area of a triangle is equal to half the product of the base and height of the triangle. However, to illustrate our methods, we will consider this simple example from another perspective.

To be more specific, we will compute the area underneath the graph as

xvaries from0to1.Our method for computing this area will be first to approximate this area by the area of some rectangles which lie underneath the graph. Then we will increase the number of these rectangles so that our approximation becomes better and better. If you drag the red dot in the following figure, you can vary how many rectangles will be used in the approximation. As more rectangles are added, the area of the triangle is filled up by the rectangles. This means that the area of all the rectangles will be a good approximation to the area of the triangle.

Let's get started by building the rectangles. We will use

nto denote the number of rectangles we are going to use. Since the rectangles will span the interval fromx = 0tox = 1,we will break this interval intonpieces of width as in the figure.The points are

Let us now consider one of the rectangles as in the figure. We know that the width of the rectangle is So that the rectangle fits underneath the graph, its upper left corner should lie on the graph which is

This means that the height of the rectangle is and hence its area is since

To put everything together now, the area of all the rectangles is

since the first rectangle we consider has left endpoint while the last rectangle has left endpoint

Notice that each term has a common factor so that we may write

If you remember, the sum we need to evaluate is simply that of an arithmetic sequence: in particular, the sum

In our case,

k = n - 1so thatNow consider what happens as the number of rectangles

nbecomes very large. Then the term becomes very small and we see that which is as we expect since we know that the area of the triangle is equal toTo summarize, we have computed the area of a triangle in a roundabout way. We approximated it by the area of some rectangles and observed that as the number of rectangles gets very large, the appproximation becomes more precise. We can, and will, use this method to find some areas that we do not already know.

The triangle revisited

Now we would like to go back and revisit our triangle. Instead of computing the are under the graph asxvaries from0to1,we would like to allow the right endpoint of the interval to vary. It might not be immediately clear to you why we want to do this, but we will see a very interesting relationship if we do.In the following figure, you can vary the right endpoing by dragging the ball around.

We will use the notation A(x)to denote the function which measures the area under the curve between0andx.We could, of course, determineA(x)as we did above. However, let's use the fact that this is measuring the area of a right isosceles triangle where the length of the legs isx.This means thatNow we have something very interesting: we are considering the area under the graph and we have found that this area is Notice that the derivative

In other words, the derivative of the area function is the original function which defines the graph. Is this merely a coincidence? Sorry, but you'll have to wait and see.

The area under a parabola

In the previous example, we did not really need a new technique since we know the area of a triangle. However, in this example, we will find the area of a region which cannot be computed through conventional means. Namely, we will compute the area under the parabola asxvaries from0to1.As in the example with the triangle, we will approximate this area with the area of many rectangles. Then we will let the number of rectangles increase so that we have a better approximation.

As before, we divide the interval from

0to1intonpieces of width The endpoints of these sub-intervals are, as before,In this case, the height of the rectangles are given by and so the area of a rectangle is We then find that the sum of the areas of all the rectangles is

In order to evaluate this, we need to be able to sum a sequence consisting of squares. We have discussed this sum earlier: recall that we found

In the case we are now looking at, the sum ends at

This says that

Notice that as

nbecomes very large, the two terms on the right become very small and Since the area of the rectangles provide a better and better approximation asnbecomes large, we can say that the area under the parabola between and is equal toHere we were able to compute an area which we could not have found through standard geometry. This will be a very useful technique for us throughout the course. If you are uncomfortable with any of the ways in which we have used the sigma notation above, you should write out the sums represented by the notation and verify the ways in which the notation has been manipulated.

More generally:Let us now explore an area function similar to the one we considered for the area of a triangle. In particular, will represent the area under the parabola between0andx.In the following diagram, you may explore this function by dragging the ball around.As before, we will divide the interval from

0toxintonpieces each of width The corresponding points will be denoted by where The height of the rectangles will again be so that the area of a rectangle will beNow if we sum the area of the rectangles, we obtain

As is now familiar, as

nbecomes very large, we find that the rectangles provide a better and better approximation for the area and henceOnce again, we see a remarkable relationship: measures the area under the graph of the function and we see that

The area under an exponential

Finally, we will consider the area under the graph of over the interval from

0tox.The following diagram will illustrate this function.A large part of this computation proceeds just as in the previous example. We devide the interval into

npieces of width The points defined by these sub-intervals are Now the height of the rectangles will be given by the value of he function at so that the height is This means that the area is If we now sum all the areas of the rectangles, we haveYou may recognize this last sum: it is a geometric series (meaning that the ratio of one term to its predecessor is constant) and it may be easily evaluated. For instance, suppose we want to add

The trick is to multiply the sum by the ratio

r:If we subtract the second expression from the first, we find that

which means that the sum

If we apply this to our example, we have

Now let's consider what happens when we have a lot of rectangles. This means that

nis very large or rather that is very close to0.To study this situation, we will consider the linear approximation of the functoin at the point Since we have it follows that both and This means that the linear approximation is and that this approximation is good when is close to

0.In other words, whentis close to0.Now when

nbecomes very large, we have that is very close to0.This means that In other words,and so Once again, we have found that the area function has the original function as its derivative. This is still more evidence for the Fundamental Theorem of Calculus which we will soon meet.

Summary

Let's summarize what we have done. To find the area under the graph of a function between and , we divide the interval into

npieces. These pieces each have width and we call the left endpoint of these subintervalsNow we form a rectangle whose height is and whose area is If we sum all the areas of these rectangles, we have

Finally, as we consider more and more rectangles, the quantity gives a better and better approximation so that we may write the area as

We will see many expressions like this throughout the term.