Integrate. Transitive verb. To
indicate the whole of; to give the sum or total of; as, an integrating
anemometer, one that indicates or registers the entire action of the
wind in a given time. [From the 1913 Webster's dictionary]
When we
integrate
a function we find the area under its curve. As the notation
implies, in ordinary Riemann integration, we perform the operation by
slicing the region up into many pieces and summing the area of all of
them.
A Little More Background: Definite versus Indefinite Integrals
Figure 1a: Definite integral

Figure 1b: Integral evaluated

In our
definition of an integral we didn't say much about the end points of the region over which we're finding the area. The
region of integration is only well defined for a
definite integral, which is one for which the bounds are specified (
figure 1a).
As we said previously, the notation provides us with a recipe for computing the value. As shown in
figure 1b, we divide the region into
panels, each
dx units wide, find the area of each by multiplying
f(x) for the panel by
dx, and add them up.
Of course the value of
f(x)
typically varies across the width of each panel; we'll say more about
that later on the page. But for the purposes of understanding the
process, we can just pretend that
dx is so small that
f doesn't vary across each panel  or varies by such a tiny amount we can ignore it.
Figure
2: Indefinite integral

If the bounds are not specified, then the integral is
indefinite, and it no longer corresponds to a particular numeric value (
figure 2). In this case, while we can't evaluate the integral to an actual number, we can still ask what
function the integral represents, if we take the argument of the function to be the
end value of the region of integration. For example, the indefinite integral of
1 is the function
x. However, since we never said where we
started
the integration, the total area "to the left" is undefined; this is
shown in figure 2 by the "fade out" of region whose area we're
measuring.
To make indefinite integrals useful in equations,
it's necessary to allow for the indefinite extension "off to the left",
and this is done by adding an indefinite
constant of integration to the result. Thus, to take our trivial example above a little farther,
where "
k" is undetermined. In a physics problem, the value
k might typically be determined by the constraints on the problem.
More on the Integral as a Limit
For completeness, we should say exactly what is meant by the formal definition we gave for an integral,
definition (1b). The main statement of the definition is (hopefully!) clear enough; we'll repeat it here:
(
1)
but the condition on
x_{k}
deserves some comment. I also haven't defined limits anywhere,
and while the reader is probably familiar with the concept, I should
still say exactly what I mean by it. I'll define the limit and
say more about the condition on
x_{k} in this section.
First, I defined
x_{k} by saying:
(2)
rather than by picking a particular point in each panel. I could, for instance, have said,
(
3)
That would have defined the area of one panel as its width, times the value of
f in the
middle of the panel. But with the statement I actually used, we don't know how to pick
x_{k} in each panel  the choice of exact point is left unspecified.
The reason for using (2) rather than (3) is just
convenience. Because of the way definition 1b is stated, I can say this:
Whenever the limit in definition 1b exists, the Riemann integral is welldefined, and is the value of that limit.
If, instead, I had used the expression in (
3) to set the value of
x_{k}, I would have had to say something like this:
Whenever the limit exists, and
the function being integrated is sufficiently wellbehaved, then the
Riemann integral is welldefined and is the value of that limit.
And
then I'd need to go on to define what I mean by "well behaved".
By using the slightly less obvious statement of the definition, I
avoid the need to put extra conditions on the definition to avoid
finding "false integrals"; for badly behaved functions, the limit
simply won't exist, and the issue doesn't come up. We'll give an
example of a badly behaved function later on this page. But
first, let's deal with the limit.
The Limit Defined
The limit in (
1) has the values
S,
(4)
if and only if we can force the value of the sum to be
arbitrarily close to
S just by choosing
n large enough. In English, given any ε>0, we can find some value
N such that, for
any n>
N, the value of the sum will be within ε of
S. Symbolically,
(
5)
There is a catch here, however. We haven't said exactly how the value
x_{k} is to be chosen; all we've said is that
x_{k}
must lie in the bounds of the k'th panel. So, the implication 
and the intended meaning  is that (5) must be true no matter how the
values
x_{k} are chosen.
The Upper and Lower Sums
If the limit in (
5) is to equal
S for
any choice of
x values within each panel, then, in particular, it must be true for the
x values corresponding to both the
largest and
smallest values for
f within each panel. Conversely, if it's true when we select
x such that
f is at a maximum in each panel, and it's true when we select
x such that
f is at a minimum within each panel, then it seems clear that it must be true for
any selection of
x values, for all other sums must lie between these two extremes.
When we choose
x_{k} such that
f(x_{k}) is as large as possible within each panel, we call the result the
upper sum. Conversely, if we choose
x_{k} such that
f(x_{k}) is as small as possible within each panel, we call that the
lower sum.
Now we have specified how
x is to be chosen, and we can consequently obtain a specific value for both the upper and lower sums for a given value of
n. If the limits of both the upper and lower sums are defined,
and they are
equal, then the integral is equal to their (mutual) limit.
When the limits don't exist: A Counterexample
A simple counterexample is the function which is 1 for each irrational value, and zero for each rational value:
Consider the integral of this function between zero and 1. No matter how small we make the panels, the minimum value for
f(x)
on each panel is 0, and the maximum value is 1  so, the upper and
lower sums, as defined in the previous section, never approach each
other.
In fact, if we use any simple approach to picking the value for
f in each panel, we'll obtain a value of 0 for the sum of the areas of the panels. For example, consider the sum:
No matter how large we make
n, the value of the sum remains zero, as the points at which we evaluate
f are always rational, and so every term of the sum is zero. Yet it's not hard to show that
nearly every point
is irrational, and that, in fact, the "length" occupied by all the
rational numbers together is zero  so we are somehow missing nearly
every point when we attempt to integrate this function.
The
rationals are countable. (Write each rational value between 0 and
1 as a fraction, a/b. Now, sort the fractions by
denominator.
It's easy to see that every fraction we can name is on this
sorted list, at some finite location; in fact an arbitrary fraction a/b
must appear no later than location b
^{2} on the list.)
So, take a list of all the rational numbers between 0 and 1.
Put an interval of length
n/2 around the first number on the list.
Put an interval of length
n/4 around the second.
Continue, putting an interval of length
n/(2
^{k}) around the k
^{th} member of the list.
The
total length of all the intervals is
Consequently, if choose
n
small enough, we can make the total length as small as we like  yet
these intervals, taken together, contain all the rational numbers, plus
some "fuzz" around each one.
The "proper" result of integrating
f from 0 to 1 would thus seem to be
one, not zero.
Page first posted on 11/04/2007