Integral Question

1. Dec 21, 2011

Miike012

I am confused on what the paragraph is saying which is copied onto the paint document...
I think it is saying that the graph, which I added to the paint document, has two
Integrable function s(x) and t(x) in which s(x) <= f(x) <= t(x). If the rectangles that I highlighted in green and orange have equal areas then f(x) whose area is surrounded by these rectanels must be integrable because ∫s(x)dx = ∫t(x)dx on [a,b] . how far off am I?

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Last edited: Dec 22, 2011
2. Dec 22, 2011

CompuChip

You're close, but not entirely there.

Remember that we used to calculate the area under a graph by drawing little rectangles below it? This is the generalisation of that procedure.

Note that it is only talking about step functions, i.e. functions of the form
$$s(x) = \begin{cases} c_0 & x \in [a_0, b_0], \\ c_1 & x \in [a_1, b_1], \\ \vdots & \vdots \\ c_n & x \in [a_n, b_n] \end{cases}$$

The idea is, we define the integral of such a function, as
$$\int_{a_0}^{b_n} s(x) \, dx = \sum_{i = 0}^n c_i (b_i - a_i)$$
Then we define the lower and upper Riemann sums of f as
$$I_{a}^{b}(f) = \sup_{s \le f} \int_a^b s(x) \, dx$$
and
$$J_{a}^{b}(f) = \inf_{s \ge f} \int_a^b s(x) \, dx$$
respectively. This is bascially a approximation with rectangles of the function f.

Now you have a sequence of inequalities like $a \le I \le J \le d$, and the second part of the theorem basically says that if a = d, then I = J.
In that case, we call I = J the integral of f, denoted by the familiar notation, and we call it integrable.

So we basically extend the "simple" intuitive definition to (almost) arbitrary functions.

3. Dec 23, 2011

LCKurtz

Is that what you meant to write?

4. Dec 24, 2011

CompuChip

Oops, of course I meant them to be continuous and open intervals, good catch!

$$s(x) = \begin{cases} c_0 & x \in )a_0, a_1), \\ c_1 & x \in (a_1, a_2), \\ \vdots & \vdots \\ c_n & x \in (a_n, a_{n + 1}) \end{cases}$$