Subdivisions for Definite Integrals based on Uniform Areas

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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  e1 ]  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 
 2e
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 
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