# What is an integral

1. Oct 31, 2004

### Alkatran

I feel embarassed for asking this, but:

An integral is just dividing a curve into smaller sections and adding up their areas, right? The smaller the section the closer the estimate?

How does an integral differ from an anti-derivative?

What do the numbers at the top and bottom of the integral long s sign mean? From (bottom value) to (top value)?

2. Oct 31, 2004

### Hurkyl

Staff Emeritus
Very loosely speaking, yes.

Antidifferentiation is the inverse operation to differentiation.

Boundary values. For 1-dimensional integrals, this notation is shorthand for saying the domain of integration is the (directed) interval (a, b).

3. Oct 31, 2004

### T@P

as hurkly said, thats an integral is really a riemann sum. ul learn it later so dont rush :)

4. Oct 31, 2004

### Alkatran

But doesn't anti-differentiation give the area under the curve?

If I go from position to position 4 in 4 seconds, I was moving at 1m/s:

x = x^1 -> 1*x^0 = 1

So if my speed is 1 I can do the inverse to get the area under the curve

1 = 1*x^0 -> x^1/1 = x (+ R)

5. Oct 31, 2004

### Alkatran

Well I'm getting physics questions now asking me to find the integral. It's all via hand but I like to know the theory behind things.

6. Oct 31, 2004

### Hurkyl

Staff Emeritus
Yes; it can be proven that anti-differentiation very nicely corresponds to integration. The theorem is aptly named the Fundamental Theorem of Calculus! It has two parts:

$$\int_a^b f'(x) \, dx = f(b) - f(a)$$

and

$$\frac{d}{dx} \int_a^x f(t) \, dt = \lim_{t \rightarrow x} f(x)$$

The proof, IIRC, is fairly straightforward: just write down the riemann sum definition of the integral and the limit definition of the derivative, and chug your way through it.

7. Oct 31, 2004

### Alkatran

Let's say we want to count the average number of times you have to flip a coin and get a head.

The chance of 1 is 1/2, the chance of 2 is 1/4. The chance of x is 1/2^x

So for the average, we take each x, multiply it by it's chance, and sum them all up.
t(x) = x/2^x

I know this isn't exactly an 'area under the curve' because you actually only count integers and not all numbers, but anyways:

What is the sum of all positive integers x in t(x)? (I know the answer is 2)

8. Oct 31, 2004

### Hurkyl

Staff Emeritus
Actually, in the general theory of integration, sums are integrals too.

I'll loosely introduce the idea of a Lesbegue integral, applying it to your coin flipping example.

Here, you have a topological space, X, which is simply the positive integers with the discrete topology.

You have a measure &mu; -- a function which takes subsets of X to real numbers that satisfies some axioms that encapsulate the idea of "size".

Here, &mu; can be defined by its values on singleton sets: &mu;({n}) = 2^-n. &mu; is actually a special sort of measure called a probability measure. (I think simply because &mu;(X) = 1)

To define an integral with respect to this measure, you first start with characteristic functions: for any subset S of X, define

$$\chi_S(x) := \left\{ \begin{tabular}{ll} 1 & x \in S \\ 0 & x \notin S \end{tabular}$$

It's easy to define the integral of these things:

$$\int_T \chi_S \, d\mu := \mu(S \cap T)$$

You then extend the integral in the obvious way to any linear combination of character functions. (for technical reasons, you do it only for nonnegative ones) Call these step functions.

You can then extend the integral to any positive "measurable" function. (I'll omit the definition) If f is a nonnegative measurable function, consider all possible step functions that are less than or equal to f. Then, you take the integral of all of them, and the sup is the integral of f.

That is:

$$\int_T f \, d\mu := \sup_{\phi <= f} \int_T \phi \, d\mu$$

In this particular example, you can prove that it is equal to:

$$\int_T f \, d\mu := \sum_{t \in T} f(t) \mu({t})$$

whether T is finite or infinite.

In the case where d&mu; is the usual measure on the real numbers, and f is a nonnegative Riemann integrable function, you can show that the integral defined above has the same value as the Riemann integral.

Then, you continue extending the definition of the integral to other cases (such as allowing functions to be negative), and you get the Lesbegue integral.

9. Nov 1, 2004

### HallsofIvy

You don't really need the Lebesque integral to treat sums as integrals, just the "Riemann-Stieljes" integral.

The difference between the "Riemann" integral and "Riemann-Stieljes" integral is that, with Riemann-Stieljes, instead of measuring the "base" by Δ x= xn- xn-1, you measure it by α(xn)- &alpha(xn-1) where α can be any increasing function.

As long as α is differentiable, the Riemann-Stieljes integral, $\int f(x)d\alpha$, is the same as the Rieman integral, $\int f(x)\frac{d\alpha}{dx} dx$.

But if &alpha; is the step function [x], The Riemann-Stieljes integral is just the sum
$\Sigma f(n)$.