Lecture One In Integral

من طرف **Ahmed** في السبت مارس 08, 2008 9:38 pm

* partition of an interval

In mathematics, a partition of an interval [a, b] on the real line is a finite sequence of the form
a = x₀ < x₁ < x₂ < ... < x_n = b.

Such partitions are used in the theory of the Riemann integral, the Riemann-Stieltjes integral and the regulated integral.

The norm (or mesh) of the partition
x₀ < x₁ < x₂ < ... < x_n

is the length of the longest of these subintervals; it is
max{ |x_i − x_i−1| : i = 1, ..., n }.

As the mesh approaches zero, a Riemann sum based on the partition approaches the Riemann integral.

A tagged partition is a partition of an interval together with a finite sequence of numbers t₀, ..., t_n−1 subject to the conditions that for each i,
x_i

t_i

x_i+1.

In other words, it is a partition together with a distinguished
point of every subinterval. The mesh of a tagged partition is defined
the same as for an ordinary partition. We can define a partial order
on the set of all tagged partitions by saying that one tagged partition
is bigger than another if the bigger one is a refinement of the smaller
one.

Suppose that

together with

are a tagged partition of [a,b], and that

together with

are another tagged partition of [a,b]. We say that

and

together are a refinement of

together with

if for each integer i with

, there is an integer r(i) such that x_i = y_r(i) and such that t_i = s_j for some j with

.
Said more simply, a refinement of a tagged partition takes the starting
partition and adds more tags, but does not take any away.

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*Darboux integral

Definition

A partition of an interval [a,b] is a finite sequence of values x_i such that

Each interval [x_i−1,x_i] is called a subinterval of the partition. A refinement of the partition

is a partition

such that for every i with

there is an integer r(i) such that

In other words, to make a refinement, cut the subintervals into smaller pieces and do not remove any existing cuts.

Let ƒ:[a,b]→R be a bounded function, and let

be a partition of [a,b]. Let

Lower (green) and upper (green plus lavender) Darboux sums for four subintervals

The upper Darboux sum of ƒ with respect to P is

The lower Darboux sum of ƒ with respect to P is

The upper Darboux integral of ƒ is

The lower Darboux integral of ƒ is

If U_ƒ = L_ƒ, then we say that ƒ is Darboux-integrable and set

the common value of the upper and lower Darboux integrals.

Facts about the Darboux integral

When passing to a refinement, the lower sum increases and the upper sum decreases.

If

is a refinement of

then

and

If P₁, P₂ are two partitions of the same interval (one need not be a refinement of the other), then

.

It follows that

Riemann sums always lie between the corresponding lower and upper Darboux sums. Formally, if

and

together make a tagged partition

(as in the definition of the Riemann integral), and if the Riemann sum of ƒ corresponding to P and T is R, then

.

From the previous fact, Riemann integrals are at least as strong as
Darboux integrals: If the Darboux integral exists, then the upper and
lower Darboux sums corresponding to a sufficiently fine partition will
be close to the value of the integral, so any Riemann sum over the same
partition will also be close to the value of the integral. It is not
hard to see that there is a tagged partition that comes arbitrarily
close to the value of the upper Darboux integral or lower Darboux
integral, and consequently, if the Riemann integral exists, then the
Darboux integral must exist as well.
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من طرف **Ahmed** في السبت مارس 08, 2008 9:39 pm

Other referent(source) and more

The Riemann Darboux Sums

We're getting one step closer here, and we can finally define the
Riemann-Darboux Upper and Lower Sums. This is essentially the integral
only it's missing a key property which will be disclosed in the next
lecture.

As an example of the concept of what we're doing here, at least for
the upper sum, before showing the jumble of symbols, is we take a
partition's subinterval, and find the highest value of the function on
that interval, multiply it by the subinterval's length, get the
rectangle, and sum like crazy. The lower sum is similar only taking the
lowest value of the function.

The idea of this is similar to figure 5 in the textbook:

But let's reformat the image a bit:

Now, the GENERAL definition of an integral, in case your curiosity
peaks your interest, picks a RANDOM value of the function on the
subinterval, which is what is shown in figure 5.

Either way, here it is in it's glory:

U(P,f) =

L(P,f) =

For now I'll leave you a small exercise with the upper and lower sums, but this won't be the end of this lecture.

Summing Exercise

Let f: [0,1]

be f(x) = x, and let