Subdivisions for Definite Integrals Based on Uniform Areas All of the standard numerical integration techniques, in practice, typically begin with the idea of partitioning an interval, say [a, b], into n uniformly spaced subintervals. 2 In this article, we look at what is more or less the reverse issue: Can we partition the region under the curve for some function f on an Area = 0.5 interval into a series of subregions having the same area? That is, we seek to partition the interval of definition into a collection of subintervals at x0, x1, x2, x3, …, where the area x2 of each subregion under the curve is the same as the area between x0 and x1; we call this the baseline. Area = 0.5 We begin by considering the linear 0 0 2 x1 = 1 x2 function y = x on the interval [0, 1], so the baseline area is simply ½. We next seek a point x2 such that the area under the line from x1 = 1 to x2 is also ½. We do this geometrically, as illustrated in Figure 1. The area of the larger triangle is ½x22, so that we require that ½x22 - ½ = ½ or ½x22 = 1. Consequently, x22 = 2 and x2 = 2. We continue this argument, as illustrated in Figure 2, to determine x3 so that the area under 2 the line from x2 to x3 is also ½ Now, the area Area = 0.5 of the larger triangle is ½x32 and the area of the mid-size triangle extending to x2 = 2 is Area = 0.5 1, so that ½x32 - 1 = ½ and so x32 = 3. x3 That is, x3 = 3, and in general, it is apparent that xn = n, for any positive integer n. We note that this rather remarkable result about the sequence 2, 3, 4, 5, … Area = 0.5 0 does not appear to be widely known. x1 = 1 x2 x3 2 0 Furthermore, we note that the identical result holds for the linear function y = mx, for any positive slope m. We leave that to the interested readers or their students to either derive or prove in general. However, as we discuss later, things get considerably more complicated if the line does not pass through the origin. Before considering this, though, we first look at a variety of other fundamental functions. Exponential Growth and Decay Functions We start with the exponential function f(x) = ex on the interval x > 0. For a baseline, we use the interval [0, 1], so that 1 e dx e 1. x 0 We now attempt to find the interval [1, x2] such that the region under the exponential curve has the same area. That is, x2 1 e x dx e x2 e1 e x2 e e 1. We therefore find that e x2 2e 1, and so x2 = ln (2e – 1). We note that this is numerically equal to roughly 1.48988, so that the same area is traced out in just under half the distance needed for the interval [0, 1]. We continue to determine the number x3 so that the region under the curve on the interval [x2, x3] also has the same area. Thus, x3 x2 e x dx e x3 e x2 e 1. Using the fact that e x2 2e 1, we therefore find that e x3 2e 1 + e – 1 = 3e - 2 and so x3 = ln (3e – 2). We note that this is approximately x3= 1.81724 and, as expected, the length of the interval is shorter than the previous interval. Continuing this process, we next seek the number x4 so that the region under the curve on the interval [x3, x4] also has the same area. Thus, x4 x3 e x dx e x4 e x3 e 1. Since e x3 3e 2 , we therefore find that e x4 (3e – 2) + e – 1 = 4e - 3 and so x4 = ln (4e – 3). Numerically, this is roughly 2.06346. By this point, the pattern is evident: the nth interval, for any positive integer n, extends from xn – 1 = ln ((n – 1)e – (n-2)) to xn = ln (ne – (n – 1)). We show the pattern 4 xn in the sequence of these values in Figure 3 3 to demonstrate how the length of each successive interval decreases slowly as the 2 successive xn values become ever closer 1 together. n We next consider the exponential 0 -x 0 5 10 15 20 decay function f(x) = e . As before, let’s consider the interval [0, 1] as our baseline, so that 1 e 0 x 1 dx 1 . e We now attempt to find the interval [1, x2] on which the region under the curve has the same area. That is, x2 1 e x dx [e x2 e1 ] 1e e x2 1 1e . Therefore, e x2 1 2e . However, when we try to solve this equation for x2 using logs, we find that 2 e x2 ln 1 ln , e 2e and the latter expression is undefined because 2 – e < 0. Let’s see what went wrong. The total area under the exponential decay curve for x > 0 is 1. Of this total, the area that corresponds to the interval [0, 1] is 1 – 1/e 0.6321, so that almost ⅔ of the total has already been accounted for. There simply is not enough area remaining beyond x = 1 to match the area from 0 to 1. The function simply dies out much too quickly. However, if a smaller baseline interval were used, say [0, 0.1], then we should expect that it would be possible to find a number of intervals in which the same area recurs. We leave it to the interested readers or their students to experiment with this idea and to determine the relationship between the length of the baseline interval and the number of partitions over which one can obtain sections of the areas under the curve that have the same area as the baseline region. The Quadratic Function f(x) = x2 We next consider the quadratic function f(x) = x2 and choose as our baseline interval [0, 1]. The corresponding baseline area is clearly ⅓. As a result, we find that, for the next interval, x2 x23 1 1 2 x dx 1 3 3 3 and so x23 2 and x2 21/ 3. Continuing, we find that x3 x33 x23 1 2 x dx . x2 3 3 3 Since x23 2, we have x33 3 and so x3 31/ 3. Doing this once more, we find that x4 x43 x33 1 2 x dx . x3 3 3 3 Since x33 3, we have x43 4 and so x4 41/ 3. The pattern is clearly evident for all subsequent points of the subdivision. If one desires, it can be proved formally. We leave it to interested readers to see what comparable patterns exist for higher degree power functions with integer powers. We also suggest looking at some power functions with non-integer powers, such as f(x) = x; at least with this case, some reasonably simple patterns arise for the points of the subdivision starting with [0, 1] as the baseline interval. The Power Function f(x) = 1/x As our next function, we look at the decaying power function f(x) = 1/x and select the baseline interval to be [1, 2]. Clearly, the baseline area is simply ln 2. For the next interval [x1, x2], x2 1 2 xdx ln x2 ln 2 ln 2, and so ln x2 = 2 ln 2 = ln 4, so that x2 = 4. Continuing, we find that x3 1 x2 xdx ln x3 ln x2 ln x3 ln 4 ln 2, and therefore ln x3 = ln 4 + ln 2 = 3 ln 2 = ln 8, so that x3 = 8. In a comparable way, we find that ln x4 = ln 16 and so x4 = 16. The pattern is obvious. Interestingly, for this function, each interval must be twice as long as the preceding one to accommodate the same amount of area under the curve. The Logarithmic Function We next consider the logarithmic function f(x) = ln x and select the baseline interval to be [1, 2]. The corresponding baseline area B0 is then 2 B0 = ln xdx ( x ln x x) 21 (2ln 2 2) (ln1 1) 2ln 2 1. 1 This is approximately 0.38629. Consequently, for the next interval [x1, x2], x2 2 ln xdx ( x ln x x) x2 2 ( x2 ln x2 x2 ) (2ln 2 2). Since this is to be equal to the area B0 of the baseline region, we have x2 2 ln xdx ( x2 ln x2 x2 ) (2ln 2 2) 2ln 2 1, and so x2 ln x2 x2 4ln 2 3. It is not possible to solve this transcendental equation for x2 in closed form, although we can find it approximately using either numerical or geometrical approaches; the result is x2 = 2.48012, correct to 5 decimal places. See Figure 4. We next determine the number x3 so that the region under the logarithmic curve on the interval [x2, x3] also has the same area. However, we will approach this in a somewhat different manner to simplify the resulting algebra, particularly 2 1 0 0 -1 1 2 3 4 if the process is to be continued farther. Thus, x3 x2 x3 x2 x3 ln xdx ln xdx ln xdx ln xdx 2 B0 B0 1 1 1 and therefore x3 1 ln xdx 3B0 3(2ln 2 1). When we expand the definite integral, we get the transcendental equation x3 ln x3 x3 6ln 2 4, whose solution turns out to be, approximately, x = 2.87285. In a comparable way, we find that, for the next interval [x3, x4], x4 x3 x4 x3 x4 ln xdx ln xdx ln xdx ln xdx 3B0 B0 1 1 1 and therefore x4 1 ln xdx 4B0 4(2ln 2 1). When we expand the definite integral, we obtain x4 ln x4 x4 8ln 2 5, whose solution is, approximately, x = 3.21980. Moreover, the pattern in the successive transcendental equations is now apparent. Despite the fact that the logarithmic function grows incredibly slowly, we know that the total area under its curve becomes infinite and this process can be continued indefinitely. The slow growth in the function suggests that we should expect the lengths of the successive intervals needed to encompass the same area should be fairly comparable. Let’s examine precisely what happens in practice. The baseline area, 0.38629, occurs from x = 1 to x = 2; the same area then recurs from x = 2 to x = 2.48012, an interval of length 0.48012. The identical area occurs again from x = 2.48012 to x = 2.87285, an interval of length 0.39273 and then once more from x = 2.87285 to x = 3.21980, whose length is 0.34695. So, the lengths of the successive intervals are decreasing slightly in a concave up pattern, as we might have expected. The General Linear Function Finally, as promised, we come back to consider the general linear function f(x) = ax + b and use the interval [0, 1] as our baseline interval. It turns out that the presence of the two parameters here makes things considerably more complicated. To start, we have the baseline area B0 = (ax b)dx (a x2 bx) 01 ( a 2 b). 0 To find the interval [1, x2] over which the same area is encompassed under the line, we write 1 x2 0 2 2 (ax b)dx (a x2 bx) x2 0 (a x22 2 bx2 ) 2B0 a 2b. This leads to the quadratic equation a x 2 bx a 2b 0, 2 2 2 whose solutions are given by the quadratic formula as x2 b b2 2a 2 4ab , a after some simplification. We note that the solution with the positive sign typically produces a value for x2 that is larger than 1, as required, while the other solution leads to a negative value for x2, which is not realistic in this context. When we extend this argument to find the next interval [x2, x3] over which the area is again equal to B0, we similarly obtain x3 0 2 (ax b)dx (a x2 bx) x3 0 (a x23 2 bx3 ) 3B0 32 a 3b. This leads to the quadratic equation a x 2 bx 3 a 3b 0, 3 2 2 3 whose solutions are b b2 3a 2 6ab . a When we repeat this argument to find the next interval [x3, x4] over which the area is again equal to B0, we get x3 x4 0 2 (ax b)dx (a x2 bx) The resulting quadratic equation x4 0 (a x24 2 bx4 ) 4B0 2a 4b. a x 2 bx 2a 4b 0 3 2 3 has the solutions b b2 4a 2 8ab . a By this point, the patterns in both the successive quadratic equations and the corresponding solutions are apparent. We leave it to the interested reader to investigate what happens with different values for the parameters a and b. x4