4.3 Area

One basic problem we have been exploring in this chapter is how to compute the distance traveled from a given velocity function. We examined this in terms of antiderivatives in section 4.1 and reworked this as an area problem in section 4.2. In this section, we will develop the general problem of calculation of areas in some detail.

You are all familiar with the formulas for computing the area of a rectangle, a circle and a triangle. From endless use of these formulas over the years, you probably have good intuition about what area is: one measure of the size of a two-dimensional region. But, how can we compute the area of a region that's not a rectangle, circle or triangle? Notice that we've used the word compute. For most regions, areas are not measured directly, but rather are computed using some one-dimensional measurements and a formula.

The problem of computing areas is more important than you might think. There are fairly obvious examples such as the area of your front yard (how much grass seed is needed?) and the area of the living room floor (how much carpet is needed?). We saw in section 4.2 that areas are also computed to determine the distance traveled. We see later how to interpret area to give the probability of a random event or the volume of oil flowing through a pipeline.

What we need, then, is a more general description of area, one that can be used to find the area of almost any two-dimensional region imaginable. In this section, we develop a general process for computing area. It turns out that this process (which we generalize to the notion of the definite integral in section 4.4) has significance far beyond the calculation of area. In fact, this powerful and flexible tool is one of the central ideas of calculus, with applications in a wide variety of fields.

Let's start our investigation of area by looking at a fairly simple example: we'll try to compute the area bounded between the graph of y = 2x-2x2 and the x - axis. Since 2x-2x2 = 0 for x = 0 and x = 1 , the region we are interested in extends from x = 0 to x = 1 (see Figure 4.5).

Figure 4.5
Area under 2x-2x2.

The region is clearly not a rectangle, circle or triangle, so we have no handy formula available for computing the area. Since we don't know how to compute the area exactly, we first try to approximate it, using what we already know about area. We'll then suggest a means for systematically improving the approximation.

Although we often think of graphing calculators and computers as drawing curves, what they actually do is plot points, usually by coloring in small squares on the screen called pixels. If you look closely enough at a calculator display of the graph of y = 2x-2x2 , you will probably see a picture similar to Figure 4.6.

Figure 4.6
Calculator display of y = 2x-2x2.

(Computer screens look the same way, but their resolution is so fine as to make individual pixels indistinguishable to the naked eye.) In the case of Figure 4.6, each pixel represents a square of side 0.1. To estimate the area, then, we simply count the number of pixels in the region of interest and multiply the total by 0.01, the area of 1 pixel. Try this now: count the highlighted pixels forming the curve and the pixels underneath, but not the pixels forming the x - axis. You should find 33 pixels, yielding an approximate area of 0.33. We can simplify the counting and gain some valuable insight by rethinking the problem somewhat. In Figure 4.7, we have organized the pixels we want to count into columns: the first column is 2 pixels high, the second column is 3 pixels high and so on.

Figure 4.7
Counting pixels to approximate area.

Notice that the area indicated in Figure 4.5 can be approximated by the sum of the areas of the rectangles indicated in Figure 4.7. Each rectangle has width 0.1 (1 pixel wide) but the heights differ. What determines the height of each rectangle? Since the top pixel is supposed to represent the graph of y = f (x) , the height is determined by the function values. For example, we have f (0.1) = 0.18 and 0.18 rounds off to 0.2 or two pixels, so that the first column is 2 pixels high. Similarly, f (0.2) = 0.32 and so, the second column is 3 pixels high. Ignoring the rounding off, each rectangle has width 0.1 and height f (x) , for some x. The total area is then
A = 0.1f (0.1)+0.1f (0.2)+0.1f (0.3)+ +0.1f (0.9)+0.1f (1.0).
After factoring out the common width of 0.1 and evaluating the function, we get
A = (0.18+0.32+0.42+0.48+0.5+0.48+0.42+0.32+0.18+0)(0.1) = 0.33.
We will see later that the exact area in Figure 4.5 is 1/3 , so we have found a fairly good approximation.

How can we improve on this approximation? It should be clear graphically that the smaller the pixels are, the higher the resolution of the picture is. The same is true computationally. In general, the more rectangles we use, the better the area approximation. Since more rectangles means a better approximation, it should make sense that the actual area is the limit of the approximations as the number of rectangles tends to infinity. The following definitions make this idea more precise.

The general problem is to estimate the area below the graph of y = f (x) and above the x - axis for a x b. For now, we assume that f (x) 0 and f  is continuous on the interval [a, b] , as in Figure 4.8.

Figure 4.8
Area under y = f (x).

We start by dividing the interval [a, b] into n equal pieces. (This is called a regular partition of [a, b] and is equivalent to marking off the boundaries of the pixels.) The width of each subinterval in the partition is then , which we denote by x (meaning a small change in x). The points in the partition are denoted by x0 = a , x1 = x0+ x , x2 = x1+ x , and so on. In general,

(Think about this some.) See Figure 4.9 for an illustration of a regular partition for the case where n = 6. On each subinterval [xi-1, xi] ( i = 1, 2, , n ), construct a rectangle of

Figure 4.9
Regular partition of [a, b].

height f (xi) (the value of the function at the right endpoint of the subinterval), as illustrated in Figure 4.10, for the case where n = 4. It should be clear from Figure 4.10 that the area under the curve A is approximately the same as the sum of the areas of the four rectangles,
A f (x1) x+f (x2) x+f (x3) x+f (x4) x = A4.

Figure 4.10
A A4.

In particular, notice that although two of these rectangles enclose more area than that under the curve and two enclose less area, on the whole the sum of the areas of the four rectangles provides an approximation to the total area under the curve. More generally, if we construct n rectangles of equal width on the interval [a, b] , we have

Approximating an Area with Rectangles
Approximate the area under the curve y = f (x) = 2x-2x2 on the interval [0, 1] , using 10 rectangles.
The partition divides the interval into 10 subintervals, each of length x = 0.1 , namely [0, 0.1], [0.1, 0.2], , [0.9, 1.0]. In Figure 4.11, we have drawn in rectangles of height f (xi) on each subinterval [xi-1, xi] for i = 1, 2, , n. Notice that the sum of the areas of the 10 rectangles indicated provides an approximation to the area under the curve.

Figure 4.11
A A10.

That is,
This should look familiar to you as the same estimate of the area in Figure 4.5 that we had arrived at earlier.

A Better Approximation Using More Rectangles
Repeat example 3.1, with n = 20.
Here, we partition the interval [0, 1] into 20 subintervals, each of width
We then have x0 = 0, x1 = 0+ x = 0.05, x2 = x1+ x = 2(0.05) and so on, so that xi = (0.05)i, for i = 0, 1, 2, , n. From (3.1), the area is then approximately
where the details of the calculation are left for the reader. Figure 4.12a shows an approximation using 20 rectangles and in Figure 4.12b, we see 40 rectangles.

Figure 4.12a
A A20.

Figure 4.12b
A A40.

Based on Figures 4.11, 4.12a and 4.12b, you should expect that the larger we make n, the better An will approximate the actual area, A. The obvious drawback to this idea is the length of time it would take to compute An, for n large. However, your CAS or programmable calculator can compute these sums for you, with ease. The table shown in the margin indicates approximate values of An for various values of n.

n An
10 0.33
20 0.3325
30 0.332963
40 0.333125
50 0.3332
60 0.333241
70 0.333265
80 0.333281
90 0.333292
100 0.3333

Notice that as n gets larger and larger, An seems to be approaching .

Examples 3.1 and 3.2 give strong evidence that the larger the number of rectangles we use, the better our approximation of the area is. Thinking this through, we arrive at the following definition of the area under a curve.

For a function f  defined on the interval [ a, b], if f  is continuous on [ a, b] and f (x) 0 on [a, b], the area A under the curve y = f (x) on [ a, b] is given by

In the following example, we use the limit defined in (3.2) to find the exact area under the curve from examples 3.1 and 3.2.

Computing the Area Exactly
Find the area under the curve y = f (x) = 2x-2x2 on the interval [0, 1].
Here, using n subintervals, we have
and so, x0 = 0, and so on. Then, for i = 0, 1, 2, , n. From (3.1), the area is approximately
A   An
  From Theorem 2.1
(ii) and (iii).
You should provide the algebraic details for the last step.) Note that since we have a formula for An, for any n, we can compute various values with ease. We have
and so on. Finally, we can compute the limiting value of An explicitly. We have
Therefore, the exact area in Figure 4.5 is 1/3.

Estimating the Area Under a Curve
Find the area under the curve on the interval [1, 3].
Here, we have
and x0 = 1 , so that
and so on, so that
Thus, we have from (3.1) that
A An
You should quickly recognize that we have no formulas like those in Theorem 2.1 for simplifying this last sum (unlike the sum in example 3.3). Our only choice, then, is to compute An for a number of values of n. Of course, this is easy to do using a CAS or programmable calculator. The accompanying table lists values of An displayed to six digits. Although we can't compute the area exactly (as yet), you should get the sense that the area is approximately 3.4478.

n An
10 3.50595
50 3.45942
100 3.45357
500 3.44888
1000 3.44830
5000 3.44783


We pause now to define some of the mathematical objects we have been examining.

Let { x0, x1, , xn} be a regular partition of the interval [a, b], with xi-xi-1 = x = for all i. Pick points c1, c2, , cn, where ci is any point in the subinterval [ xi-1, xi], for i = 1, 2, , n. (These are called evaluation points.) The Riemann sum for this partition and set of evaluation points is


So far in our development, we have shown that for a continuous, nonnegative function f, the area under the curve y = f (x) is the limit of the Riemann sums:
where ci = xi, for i = 1, 2, , n. Surprisingly, for any continuous function f, the limit in (3.3) is the same for any choice of the evaluation points ci [ xi-1, xi] (although the proof is beyond the level of this course). In examples 3.3 and 3.4, we used the evaluation points ci = xi, for each i (the right endpoint of each subinterval). This is usually the most convenient choice when working by hand, but usually does not produce particularly accurate approximations).


Figure 4.13a
ci = xi.

Figure 4.13b
ci = xi-1.

Figure 4.13c
ci = (xi-1+xi).

Computing Riemann Sums with Different Evaluation Points
Compute Riemann sums for on the interval [1, 3] , for n = 10, 50, 100, 500, 1000 and 5000, using the left endpoint, right endpoint and midpoint of each subinterval as the evaluation points.
The numbers given in the following table are from a program written for a programmable calculator. We suggest that you test your own program or one built into your CAS against these values (rounded off to six digits).

n Left Endpoint Midpoint Right Endpoint
10 3.38879 3.44789 3.50595
50 3.43598 3.44772 3.45942
100 3.44185 3.44772 3.45357
500 3.44654 3.44772 3.44888
1000 3.44713 3.44772 3.44830
5000 3.44760 3.44772 3.44783

There are several conclusions to be drawn from these numbers. First, there is good evidence that all three sets of numbers are converging to a common limit of approximately 3.4477. You should notice that the limit is independent of the particular evaluation point used. Second, even though the limits are the same, the different rules approach the limit at different rates. You should try computing left and right endpoint sums for larger values of n, to see that these eventually approach 3.44772, also.

Riemann sums using midpoint evaluation usually approach the limit far faster than left or right endpoint rules. If you think about the rectangles being drawn, you may be able to explain why. Finally, notice that the left and right endpoint sums in example 3.5 approach the limit from opposite directions and at about the same rate. We take advantage of this in an approximation technique called the Trapezoidal Rule, to be discussed in section 4.7. If your CAS or graphics calculator does not have a command for calculating Riemann sums, we suggest that you write a program for computing them yourself.

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