Sigma (Summation) Notation

As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some new notation here, called sigma notation (also known as summation notation). The Greek capital letter [latex]\Sigma[/latex], sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write

We could probably skip writing a couple of terms and write

which is better, but still cumbersome. With sigma notation, we write this sum as

which is much more compact.

Typically, sigma notation is presented in the form

where [latex]a_i[/latex] describes the terms to be added, and the [latex]i[/latex] is called the index. Each term is evaluated, then we sum all the values, beginning with the value when [latex]i=1[/latex] and ending with the value when [latex]i=n[/latex]. For example, an expression like [latex]\underset{i=2}{\overset{7}{\Sigma}}s_i[/latex] is interpreted as [latex]s_2+s_3+s_4+s_5+s_6+s_7[/latex]. Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter we like for the index. Typically, mathematicians use [latex]i[/latex], [latex]j[/latex], [latex]k[/latex], [latex]m[/latex], and [latex]n[/latex] for indices.

Let’s try a couple of examples of using sigma notation.

The properties associated with the summation process are given in the following rule.

Approximating Area

Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let [latex]f(x)[/latex] be a continuous, nonnegative function defined on the closed interval [latex][a,b][/latex]. We want to approximate the area [latex]A[/latex] bounded by [latex]f(x)[/latex] above, the [latex]x[/latex]-axis below, the line [latex]x=a[/latex] on the left, and the line [latex]x=b[/latex] on the right ((Figure)).

Figure 1. An area (shaded region) bounded by the curve [latex]f(x)[/latex] at top, the x-axis at bottom, the line [latex]x=a[/latex] to the left, and the line [latex]x=b[/latex] at right.

How do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many small shapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin by dividing the interval [latex][a,b][/latex] into [latex]n[/latex] subintervals of equal width, [latex]\frac{b-a}{n}[/latex]. We do this by selecting equally spaced points [latex]x_0,x_1,x_2,\cdots,x_n[/latex] with [latex]x_0=a, \, x_n=b[/latex], and

for [latex]i=1,2,3,\cdots,n[/latex].

We denote the width of each subinterval with the notation [latex]\Delta x[/latex], so [latex]\Delta x=\frac{b-a}{n}[/latex] and

for [latex]i=1,2,3,\cdots,n[/latex]. This notion of dividing an interval [latex][a,b][/latex] into subintervals by selecting points from within the interval is used quite often in approximating the area under a curve, so let’s define some relevant terminology.

We can use this regular partition as the basis of a method for estimating the area under the curve. We next examine two methods: the left-endpoint approximation and the right-endpoint approximation.

Figure 2. In the left-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the left of each subinterval.

The second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval.

Figure 3. In the right-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the right of each subinterval. Note that the right-endpoint approximation differs from the left-endpoint approximation in (Figure).

The graphs in (Figure) represent the curve [latex]f(x)=\frac{x^2}{2}[/latex]. In graph (a) we divide the region represented by the interval [latex][0,3][/latex] into six subintervals, each of width 0.5. Thus, [latex]\Delta x=0.5[/latex]. We then form six rectangles by drawing vertical lines perpendicular to [latex]x_{i-1}[/latex], the left endpoint of each subinterval. We determine the height of each rectangle by calculating [latex]f(x_{i-1})[/latex] for [latex]i=1,2,3,4,5,6[/latex]. The intervals are [latex][0,0.5], \, [0.5,1], \, [1,1.5], \, [1.5,2], \, [2,2.5], \, [2.5,3][/latex]. We find the area of each rectangle by multiplying the height by the width. Then, the sum of the rectangular areas approximates the area between [latex]f(x)[/latex] and the [latex]x[/latex]-axis. When the left endpoints are used to calculate height, we have a left-endpoint approximation. Thus,

Figure 4. Methods of approximating the area under a curve by using (a) the left endpoints and (b) the right endpoints.

In (Figure)(b), we draw vertical lines perpendicular to [latex]x_i[/latex] such that [latex]x_i[/latex] is the right endpoint of each subinterval, and calculate [latex]f(x_i)[/latex] for [latex]i=1,2,3,4,5,6[/latex]. We multiply each [latex]f(x_i)[/latex] by [latex]\Delta x[/latex] to find the rectangular areas, and then add them. This is a right-endpoint approximation of the area under [latex]f(x)[/latex]. Thus,

Approximating the Area Under a Curve

Use both left-endpoint and right-endpoint approximations to approximate the area under the curve of [latex]f(x)=x^2[/latex] on the interval [latex][0,2][/latex]; use [latex]n=4[/latex].

Show Solution

First, divide the interval [latex][0,2][/latex] into [latex]n[/latex] equal subintervals. Using [latex]n=4, \, \Delta x=\frac{(2-0)}{4}=0.5[/latex]. This is the width of each rectangle. The intervals [latex][0,0.5], \, [0.5,1], \, [1,1.5], \, [1.5,2][/latex] are shown in (Figure). Using a left-endpoint approximation, the heights are [latex]f(0)=0, \, f(0.5)=0.25, \, f(1)=1, \, f(1.5)=2.25[/latex]. Then,

[latex]\begin{array}{ll} L_4 & =f(x_0)\Delta x+f(x_1)\Delta x+f(x_2)\Delta x+f(x_3)\Delta x \\ & =0(0.5)+0.25(0.5)+1(0.5)+2.25(0.5) \\ & =1.75 \end{array}[/latex]

Figure 5. The graph shows the left-endpoint approximation of the area under [latex]f(x)=x^2[/latex] from 0 to 2.

The right-endpoint approximation is shown in (Figure). The intervals are the same, [latex]\Delta x=0.5[/latex], but now use the right endpoint to calculate the height of the rectangles. We have

[latex]\begin{array}{ll} R_4 & =f(x_1)\Delta x+f(x_2)\Delta x+f(x_3)\Delta x+f(x_4)\Delta x \\ & =0.25(0.5)+1(0.5)+2.25(0.5)+4(0.5) \\ & =3.75 \end{array}[/latex]

Figure 6. The graph shows the right-endpoint approximation of the area under [latex]f(x)=x^2[/latex] from 0 to 2.

The left-endpoint approximation is 1.75; the right-endpoint approximation is 3.75.

Looking at (Figure) and the graphs in (Figure), we can see that when we use a small number of intervals, neither the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under the curve. However, it seems logical that if we increase the number of points in our partition, our estimate of [latex]A[/latex] will improve. We will have more rectangles, but each rectangle will be thinner, so we will be able to fit the rectangles to the curve more precisely.

We can demonstrate the improved approximation obtained through smaller intervals with an example. Let’s explore the idea of increasing [latex]n[/latex], first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally 32 rectangles. Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region. (Figure) shows the area of the region under the curve [latex]f(x)=(x-1)^3+4[/latex] on the interval [latex][0,2][/latex] using a left-endpoint approximation where [latex]n=4[/latex]. The width of each rectangle is

The area is approximated by the summed areas of the rectangles, or

Figure 7. With a left-endpoint approximation and dividing the region from a to b into four equal intervals, the area under the curve is approximately equal to the sum of the areas of the rectangles.

(Figure) shows the same curve divided into eight subintervals. Comparing the graph with four rectangles in (Figure) with this graph with eight rectangles, we can see there appears to be less white space under the curve when [latex]n=8[/latex]. This white space is area under the curve we are unable to include using our approximation. The area of the rectangles is

Figure 8. The region under the curve is divided into [latex]n=8[/latex] rectangular areas of equal width for a left-endpoint approximation.

The graph in (Figure) shows the same function with 32 rectangles inscribed under the curve. There appears to be little white space left. The area occupied by the rectangles is

Figure 9. Here, 32 rectangles are inscribed under the curve for a left-endpoint approximation.

We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the same curve, using four rectangles ((Figure)), yields an area

Figure 10. Now we divide the area under the curve into four equal subintervals for a right-endpoint approximation.

Dividing the region over the interval [latex][0,2][/latex] into eight rectangles results in [latex]\Delta x=\frac{2-0}{8}=0.25[/latex]. The graph is shown in (Figure). The area is

Figure 11. Here we use right-endpoint approximation for a region divided into eight equal subintervals.

Last, the right-endpoint approximation with [latex]n=32[/latex] is close to the actual area ((Figure)). The area is approximately

Figure 12. The region is divided into 32 equal subintervals for a right-endpoint approximation.

Based on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the area under the curve better as [latex]n[/latex] gets larger. Furthermore, as [latex]n[/latex] increases, both the left-endpoint and right-endpoint approximations appear to approach an area of 8 square units. (Figure) shows a numerical comparison of the left- and right-endpoint methods. The idea that the approximations of the area under the curve get better and better as [latex]n[/latex] gets larger and larger is very important, and we now explore this idea in more detail.

Forming Riemann Sums

So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been determined by evaluating the function at either the right or left endpoints of the subinterval [latex][x_{i-1},x_i][/latex]. In reality, there is no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point [latex]x_i^*[/latex] in the subinterval [latex][x_{i-1},x_i][/latex], and use [latex]f(x_i^*)[/latex] as the height of our rectangle. This gives us an estimate for the area of the form

A sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed the idea.

Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as [latex]n[/latex] get larger and larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of [latex]n[/latex]. We are now ready to define the area under a curve in terms of Riemann sums.

Some subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limit of a function [latex]f(x)[/latex] as [latex]x[/latex] goes to infinity. Limits of sums are discussed in detail in the chapter on Sequences and Series in the second volume of this text; however, for now we can assume that the computational techniques we used to compute limits of functions can also be used to calculate limits of sums.

Second, we must consider what to do if the expression converges to different limits for different choices of [latex]\{x_i^*\}[/latex]. Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if [latex]f(x)[/latex] is continuous on the closed interval [latex][a,b][/latex], then [latex]\underset{n\to \infty }{\lim}\underset{i=1}{\overset{n}{\Sigma}}f(x_i^*)\Delta x[/latex] exists and is unique (in other words, it does not depend on the choice of [latex]\{x_i^*\}[/latex]).

We look at some examples shortly. But, before we do, let’s take a moment and talk about some specific choices for [latex]\{x_i^*\}[/latex]. Although any choice for [latex]\{x_i^*\}[/latex] gives us an estimate of the area under the curve, we don’t necessarily know whether that estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or low, we can select our value for [latex]\{x_i^*\}[/latex] to guarantee one result or the other.

If we want an overestimate, for example, we can choose [latex]\{x_i^*\}[/latex] such that for [latex]i=1,2,3,\cdots,n, \, f(x_i^*)\ge f(x)[/latex] for all [latex]x\in [x_{i-1},x_i][/latex]. In other words, we choose [latex]\{x_i^*\}[/latex] so that for [latex]i=1,2,3,\cdots,n, \, f(x_i^*)[/latex] is the maximum function value on the interval [latex][x_{i-1},x_i][/latex]. If we select [latex]\{x_i^*\}[/latex] in this way, then the Riemann sum [latex]\underset{i=1}{\overset{n}{\Sigma}}f(x_i^*)\Delta x[/latex] is called an upper sum. Similarly, if we want an underestimate, we can choose [latex]\{x_i^*\}[/latex] so that for [latex]i=1,2,3,\cdots,n, \, f(x_i^*)[/latex] is the minimum function value on the interval [latex][x_{i-1},x_i][/latex]. In this case, the associated Riemann sum is called a lower sum. Note that if [latex]f(x)[/latex] is either increasing or decreasing throughout the interval [latex][a,b][/latex], then the maximum and minimum values of the function occur at the endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations.

Finding Lower and Upper Sums

Find a lower sum for [latex]f(x)=10-x^2[/latex] on [latex][1,2][/latex]; use [latex]n=4[/latex] subintervals.

Show Solution

With [latex]n=4[/latex] over the interval [latex][1,2], \, \Delta x=\frac{1}{4}[/latex]. We can list the intervals as [latex][1,1.25], \, [1.25,1.5], \, [1.5,1.75], \, [1.75,2][/latex]. Because the function is decreasing over the interval [latex][1,2][/latex], (Figure) shows that a lower sum is obtained by using the right endpoints.

Figure 13. The graph of [latex]f(x)=10-x^2[/latex] is set up for a right-endpoint approximation of the area bounded by the curve and the x-axis on [latex][1,2][/latex], and it shows a lower sum.

The Riemann sum is

[latex]\begin{array}{ll}\underset{k=1}{\overset{4}{\Sigma}}(10-x^2)(0.25) & =0.25[10-(1.25)^2+10-(1.5)^2+10-(1.75)^2+10-(2)^2] \\ & =0.25[8.4375+7.75+6.9375+6] \\ & =7.28 \end{array}[/latex]

The area of 7.28 is a lower sum and an underestimate.

Finding Lower and Upper Sums for [latex]f(x)= \sin x[/latex]

Find a lower sum for [latex]f(x)= \sin x[/latex] over the interval [latex][a,b]=[0,\frac{\pi }{2}][/latex]; let [latex]n=6[/latex].

Show Solution

Let’s first look at the graph in (Figure) to get a better idea of the area of interest.

Figure 14. The graph of [latex]y= \sin x[/latex] is divided into six regions: [latex]\Delta x=\frac{\pi/2}{6}=\frac{\pi}{12}[/latex].

The intervals are [latex][0,\frac{\pi}{12}], \, [\frac{\pi}{12},\frac{\pi}{6}], \, [\frac{\pi}{6},\frac{\pi}{4}], \, [\frac{\pi}{4},\frac{\pi}{3}], \, [\frac{\pi}{3},\frac{5\pi}{12}][/latex], and [latex][\frac{5\pi}{12},\frac{\pi}{2}][/latex]. Note that [latex]f(x)= \sin x[/latex] is increasing on the interval [latex][0,\frac{\pi}{2}][/latex], so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum [latex]\underset{i=0}{\overset{5}{\Sigma}} \sin x_i(\frac{\pi}{12})[/latex]. We have

[latex]\begin{array}{ll}A & \approx \sin (0)(\frac{\pi }{12})+ \sin (\frac{\pi }{12})(\frac{\pi }{12})+ \sin (\frac{\pi }{6})(\frac{\pi }{12})+ \sin (\frac{\pi }{4})(\frac{\pi }{12})+ \sin (\frac{\pi }{3})(\frac{\pi }{12})+ \sin (\frac{5\pi }{12})(\frac{\pi }{12}) \\ & =0.863 \end{array}[/latex]