Integral in terms of summation?

In summary, the integral of [f(x)dx] is the height of the rectangle f(x), multiplied by the width dx. If you divide the interval between a and b into n equal increments, the center of the i'th interval is located at x = (i - \frac{1}{2})Δx. The sum of the areas of all the rectangles formed by evaluating the function at the center of each interval is equal to the integral of the function between a and b.
  • #1
jaydnul
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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!
 
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  • #2
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 [itex]Δx=\frac{(b-a)}{n}[/itex]. The center of the i'th interval is located at [itex]x = (i - \frac{1}{2})Δx[/itex]. Suppose that you evaluate the function f(x) at the center of each interval, and form a rectangle of width Δx and height [itex]f\left((i - \frac{1}{2})Δx\right)[/itex]. 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:

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

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
Almost! :smile:

Chestermiller said:
The center of the i'th interval is located at [itex]x = (i - \frac{1}{2})Δx[/itex].
This should read [itex]x = a + (i - \frac{1}{2})Δx[/itex]

Suppose that you evaluate the function f(x) at the center of each interval, and form a rectangle of width Δx and height [itex]f\left((i - \frac{1}{2})Δx\right)[/itex].
This should read [itex]f\left(a + (i - \frac{1}{2})Δx\right)[/itex]

If you add the areas of all the rectangles together, you will get:

[tex]Total Area = \sum_1^n{f\left((i - \frac{1}{2})\frac{(b-a)}{n}\right)}\frac{(b-a)}{n}[/tex]
This should read:

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

I'm either being pedantic or I'm incorrect because it's late and I'm tired. Either way, sorry! :redface:
 
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  • #4
  • #5
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
oay said:
Almost! :smile:


This should read [itex]x = a + (i - \frac{1}{2})Δx[/itex]


This should read [itex]f\left(a + (i - \frac{1}{2})Δx\right)[/itex]


This should read:

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

I'm either being pedantic or I'm incorrect because it's late and I'm tired. Either way, sorry! :redface:
Oops. You're right. Thanks very much for the correction.
 
  • #7
lundyjb said:
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!
It's not in terms of sigma, but my fellow forumer micromass wrote an excellent blog entry on integration. You can find it here.

oay said:
[tex]Total Area = \sum_1^n{f\left(a + (i - \frac{1}{2})\frac{(b-a)}{n}\right)}\frac{(b-a)}{n}[/tex]
Isn't this assuming that ##f## is Riemann integrable? :tongue:
 

1. What is the definition of an integral in terms of summation?

The integral in terms of summation is a mathematical concept used to find the area under a curve by approximating it with rectangles. This is done by dividing the area into smaller rectangles and summing the areas of these rectangles. As the number of rectangles increases, the approximation of the area under the curve becomes more accurate.

2. How is an integral in terms of summation calculated?

An integral in terms of summation is calculated by dividing the area under the curve into smaller rectangles, with each rectangle having a width of Δx. The height of each rectangle is then determined by evaluating the function at a specific x-value within the width of the rectangle. The sum of the areas of all the rectangles is then taken as the approximation of the integral.

3. What is the relationship between an integral and summation?

An integral and summation are both mathematical methods used to find the total value of a function over a specific interval. However, an integral is a continuous process that considers all the values within the interval, while summation is a discrete process that only considers specific values within the interval.

4. Can an integral in terms of summation be used for any type of function?

Yes, an integral in terms of summation can be used for any type of function as long as it is defined and continuous over the given interval. This includes both polynomial and non-polynomial functions.

5. What is the significance of using an integral in terms of summation?

An integral in terms of summation allows for the calculation of the area under a curve, which has many practical applications in fields such as physics, economics, and engineering. It also allows for the calculation of other important quantities such as volume, distance, and work.

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