# Integral in terms of summation?

1. Jul 20, 2013

### jaydnul

So i realize that the integral of [f(x)dx] is pretty much the height of the rectangle f(x), multiplied by the width dx. But that is the area of 1 infinitesimally skinny rectangle. How does the integral sign add up an infinite amount of rectangles? I've taken cal 2 so if you could show what the integral does in terms of sigma that would be nice. Thanks!

2. Jul 20, 2013

### Staff: Mentor

Suppose you want to find the integral of f(x) from x = a to x = b. Suppose you divide the interval between a and b into n equal increments, so that, for each increment $Δx=\frac{(b-a)}{n}$. The center of the i'th interval is located at $x = (i - \frac{1}{2})Δx$. Suppose that you evaluate the function f(x) at the center of each interval, and form a rectangle of width Δx and height $f\left((i - \frac{1}{2})Δx\right)$. The area of each rectangle should be a pretty good approximation to the area under f(x) over that interval. If you add the areas of all the rectangles together, you will get:

$$Total Area = \sum_1^n{f\left((i - \frac{1}{2})\frac{(b-a)}{n}\right)}\frac{(b-a)}{n}$$

Now, if you consider the value of this sum as the number of intervals n is made very large, the width and area of each rectangle (and its area) becomes smaller, but their number increases. As a result, the sum approaches a limit as the number of intervals is made very large. This limit is equal to the integral of the function between a and b.

3. Jul 20, 2013

### skiller

Almost!

This should read $x = a + (i - \frac{1}{2})Δx$

This should read $f\left(a + (i - \frac{1}{2})Δx\right)$

$$Total Area = \sum_1^n{f\left(a + (i - \frac{1}{2})\frac{(b-a)}{n}\right)}\frac{(b-a)}{n}$$

I'm either being pedantic or I'm incorrect because it's late and I'm tired. Either way, sorry!

4. Jul 20, 2013

### SteamKing

Staff Emeritus
5. Jul 21, 2013

### verty

This is the easiest way I know to give an intuition about a limit, or how a limit relates to the things that are being added.

The series 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... has the following property: term n (counting from 1) is half the gap between the sum of the first n-1 terms and 2. The first term is 1, half the gap between 0 and 2, and so on for all the other terms. Let $S_n$ be the sum of the first n terms. For any ε > 0, there is an n such that $2 - ε < S_n < 2$. We just keep halving the gap until we get close enough to 2. The number 2 is the only number that this sum gets ε-close to (if you disagree, try to prove it).

In essentially the same way, the limit of the sum of the rectangles is the unique number that the sum, as we reduce the width of the rectangles, gets ε-close to.

6. Jul 21, 2013

### Staff: Mentor

Oops. You're right. Thanks very much for the correction.

7. Jul 21, 2013

### Mandelbroth

It's not in terms of sigma, but my fellow forumer micromass wrote an excellent blog entry on integration. You can find it here.

Isn't this assuming that $f$ is Riemann integrable? :tongue: