# 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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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!

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