Path:  physics insights > basics > Calculus >

## Integrals:  Definite and Indefinite, and Defined as Limits

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 xk 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 xk in this section.

First, I defined xk 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 xk 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 well-defined, and is the value of that limit.
If, instead, I had used the expression in (3) to set the value of xk, I would have had to say something like this:
Whenever the limit exists, and the function being integrated is sufficiently well-behaved, then the Riemann integral is well-defined 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 xk is to be chosen; all we've said is that xk 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 xk 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 xk such that f(xk) is as large as possible within each panel, we call the result the upper sum.  Conversely, if we choose xk such that f(xk) 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 b2 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/(2k) around the kth 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