In a calculus class integration is introduced as 'finding the
area under a curve'. While this interpretation is certainly useful,
we instead want to think of 'integration' as more sophisticated form of
summation. Geometric considerations, in our situation, will not
be so fruitful, whereas the summation interpretation of integration
will make many of its properties easy to remember.
First, as usual, we need to define integration before we can discuss
its properties. We will start with defining the Riemann integral,
and we will move to the Riemann-Stieltjes and the Lebesgue integral
later.
- A partition P of the closed interval [a, b] is a finite set
of points
P = { x0, x1, x2, ..., xn}
such that
- a = x0 < x1 < x2 <
... < xn-1 < xn = b
- The maximum difference between any two consecutive points
of the partition is called the norm or mesh of the partition
and denoted as | P |, i.e.
- | P | = max { xj - xj-1, j = 1 ... n }
- A refinement of the partition P is another partition P' that
contains all the points from P and some additional points, again
sorted in order of magnitude.
Examples:
What is the norm of a partition of 10 equally spaced points
in the interval [0, 2] ?
What is the norm of a partition of n equally spaced
points in the interval [a, b] ?
Show that if P' is a refinement of P then
| P' |
| P |.
Using these partitions, we can define the following finite sum:
If
P = { x0, x1, x2, ..., xn}
is a partition of the closed interval [a, b] and f is
a function defined on that interval, then the Riemann Sum of f
with respect to the partition P is defined as:
- R(f, P) =
f(tj) (xj - xj-1)
where tj is an arbitray number in the interval
[xj-1, xj].
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Note: If the function f is positive, a Riemann Sum
geometrically corresponds to a summation of areas of rectangles with
length xj - xj-1 and height
f(tj).
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Examples
Suppose f(x) = x2 on [0, 2]. Find
- the fifth Riemann sum for an equally spaced partition, taking always the
left endpoint of each subinterval
- the fifth Riemann sum for an equally spaced partition, taking always the
right endpoint of each subinterval
- the fifth Riemann sum for an equally spaced partition, taking always the
midpoint of each subinterval.
Riemann sums have the practical disadvantage that we do not know which point to take
inside each subinterval. To remedy that one could agree to always take the left
endpoint (resulting in what is called the left Riemann sum) or always
the right one (resulting in the right Riemann sum). However, it will turn
out to be more useful to single out two other close cousins of Riemann sums:
Let
P = { x0, x1, x2, ..., xn}
be a partition of the closed interval [a, b] and f a bounded
function defined on that interval. Then:
- the upper sum of f with respect to the
partition P is defined as:
- U(f, P) =
cj (xj - xj-1)
where cj is the supremum of f(x) in
the interval [xj-1, xj].
- the lower sum of f with respect to the
partition P is defined as
- L(f, P) =
dj (xj - xj-1)
where dj is the infimum of f(x) in the
interval [xj-1, xj].
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Here is an example where the upper sum in displayed in dark brown
and the lower sum in light brown.
The partition P = {0.5, 1, 1.5, 2}, and the numbers
for the sums are chosen:
- for the upper sum:
c1 = f(1),
c2 = f(2), and
c3 = f(1)
- for the lower sum:
d1 = f(0.5),
d2 = f(1), and
d3 = f(2)
|
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Examples:
Suppose f(x) = x2-1 for x in the interval [-1, 1].
Find:
- The left and right sums where the interval [-1, 1] is
subdivided into 10 equally spaced subintervals.
- The upper and lower sums where the interval [-1, 1]
is subdivided into 10 equally spaced subintervals.
- The upper and lower sums where the interval [-1,1] is subdivided
into n equally spaced subintervals.
Why is, in general, an upper (or lower) sum not a special case of a Riemann sum ?
Find a condition for a function f so that the upper and lower sums
are actually special cases of Riemann sums.
Find conditions for a function so that the upper sum can be computed by always
taking the left endpoint of each subinterval of the partition, or conditions for
always being able to take the right endpoints.
Suppose f is the Dirichlet function, i.e. the function that is
equal to 1 for every rational number and 0 for every irrational number.
Find the upper and lower sums over the interval [0, 1] for an
arbitrary partition.
These various sums are related via a basic inequality, and they
are also related to a refinement of the partition in the following
theorem:
Suppose
P = { x0, x1, x2, ..., xn}
is a partition of the closed interval [a, b], f a bounded
function defined on that interval. Then we have:
- The lower sum is increasing with respect to refinements of
partitions, i.e.
L(f, P)
L(f, P') for every refinement
P' of the partition P
- The upper sum is decreasing with respect to refinements of
partitions, i.e.
U(f, P)
U(f,P') for every refinement
P' of the partition P
- L(f, P)
R(f, P)
U(f, P) for every partition P
In other words, the lower sum is always less than or equal to the upper sum, and
the upper sum is decreasing with respect to a refinement of the partition while
the lower sum is increasing with respect to a refinement of the partition. Hence,
a natural question is: will the two quantities ever coincide ?
Suppose f is a bounded function defined on a closed, bounded interval
[a, b]. Define
- the upper integral
I*(f) = inf{ U(f,P): P a partition of [a, b]}
- the lower integral
I*(f) = sup{ L(f,P): P a partition of [a, b]}
Then if I*(f) = I*(f) the function f is called
Riemann integrable and the Riemann integral of f over the interval
[a, b] is denoted by
f(x) dx
Note that upper and lower sums depend on the particular partition chosen, while
the upper and lower integrals are independent of partitions. However, this
definition is very difficult for practical applications, since we need to find
the sup and inf over any partition.
Examples
Show that the constant function
f(x) = c is Riemann integrable on any interval [a, b] and find the
value of the integral.
Is the function f(x) = x2
Riemann integrable on the interval [0,1] ? If so, find the value of the Riemann
integral. Do the same for the interval [-1, 1].
Is the Dirichlet function Riemann
integrable on the interval [0, 1] ?
The third example shows that not every function is Riemann integrable, and
the second one shows that we need an easier condition to determine
integrability of a given function. The next lemma provides such a condition for
integrability.
Suppose f is a bounded function defined on the closed, bounded interval
[a, b]. Then f is Riemann integrable if and only if for every
> 0 there exists at least one partition
P such that
| U(f,P) - L(f,P) | <
Examples
Is the function f(x) = x2
Riemann integrable on the interval [0,1] ? If so, find the value of the Riemann
integral. Do the same for the interval [-1, 1] (since this is the same example
as before, using Riemann's Lemma will hopefully simplify the solution).
Let f(x) = 0 if x is irrational and f(p/q) = 1/q if x = p/q
is rational, p, q relatively prime and q > 0, and assume
f is restricted to [0, 1]. Is f Riemann integrable ? If so, what is
the value of the integral ?
Now we can state some easy conditions that the Riemann integral satisfies. All
of them are easy to memorize if one thinks of the Riemann integral as a somewhat
glorified summation.
Suppose f is a Riemann integrable function defined on [a, b]. Then
c f(x) + d g(x) dx =
c
f(x) dx +
d
g(x) dx
- If a then
f(x) dx =
f(x) dx +
f(x) dx
- |
f(x) dx |
| f(x) | dx
- If g is another function defined on [a, b] such that
g(x) on [a, b], then
g(x) dx
f(x) dx
- If g is another Riemann integrable function on [a, b] then
f(x) . g(x) is integrable on [a, b]
Examples
Find an upper and lower estimate for
x sin(x) dx over the interval [0, 4].
Suppose f(x) = x2 if x
1
and f(x) = 3 if x > 1. Find
f(x) dx over the interval
[-1, 2].
If f is an integrable function defined on [a, b] which is bounded by
M on that interval, prove that
M (a - b)
f(x) dx
M (b - a)
Now we can illustrate the relation between Riemann integrable and
continuous functions.
Every continuous function on a closed, bounded interval is Riemann integrable.
The converse is false.
Note that this theorem does not say anything about the actual value of the Riemann
integral. Also, we can drop the explicit condition that f is bounded, since
every continuous function on a compact set is automatically bounded.
Since differentiable functions are continuous, this result therefore implies that
-
{ integrable functions }
{ continuous functions }
{ differentiable functions}
and each set is in fact a proper subset of the next.
Examples
Find a function that is not integrable, a function that is integrable but not
continuous, and a function that is continuous but not differentiable.
To finalize the relation between integrable and continuous functions, the
following theorem can be proved:
- If f is a bounded function defined on a closed, bounded interval
[a, b] and f is continuous except at countably many points, then f is
Riemann integrable.
The converse is also true:
- If f is a bounded function defined on a closed, bounded interval
[a, b] and f is Riemann integrable, then f is continuous on [a, b]
except possibly at countably many points.
Examples
Show that every monotone function defined on [a, b] is Riemann integrable.
Now that we have easy conditions to determine which functions are integrable,
it would also be convenient to have a nice shortcut to easily compute the actual
value of an integral.
Suppose f is an integrable function defined on the closed, bounded interval
[a, b], define a new function:
- F(x) =
f(t) dt
Then F is continuous in [a, b]. Moreover, if f is also continous,
then F is differentiable in (a, b) and
- F'(x) = f(x) for all x in [a, b].
Note that this theorem has an easy corollary that enables us to quickly find the
value of an integral in many situations.
Suppose f is an continuous function defined on the closed, bounded interval
[a, b], and F is a function on [a, b] such that F'(x) = f(x) for
all x in (a, b). Then
f(x) dx = F(b) - F(a)
Before we look at several examples, we should rephrase these results in less
rigorous notation. The first theorem says, basically:
f(x) dx = f(x)
while the corollary states, basically, that:
f(x) dx = f(b) - f(a)
Hence, loosely speaking, integration and differentiation are inverse
operations of each other.
Examples
Define a function
F(x) =
x2 sin(x) dx
for x in the interval [a, a + 10].
- Find F(a)
- Find F'(x)
- Find F''(x)
- Find all critical points of F(x) in [a, a + 10]
Find the value of the following integrals:
x5 - 4 x2 dx
on the interval [0, 2].
1/x2 + cos(x) dx
on the interval [1, 4].
(1 + x2)-1 dx
on the interval [-1, 1].
Show that if one starts with an integrable function f in the Fundamental
Theorem of Calculus that is not continuous, the corresponding function F
may not be differentiable.
In the next chapter we will introduce the Riemann-Stieltjes integral, and also
learn two more shortcuts to compute the value of an integral called substitution
method and integration by parts.
(bgw)