What is this formula in Calculus?

[MarcAlexander]

Hey guys. I'm trying to teach my self Calculus using the internet and I've got to the point were I need to learn what Integration is. One obstacle along my way is this equation:
\int_{a}^{b}f(x)dx=\sum_{i=0}^{n-1}f(xi)\Deltax

What is the value of d? And what is the use of integration in Mathematics?

NOTE: If you reply with "Just buy a book.", respond I will not.

 

[DeadOriginal]

d does not have a value. In f(x)dx, f(x) is the function and dx is the symbol for change in x which is delta x that you have on the other side of the equation.

Integration is basically used to calculate the area under a function in mathematics. I don't want to go too deep into what it is because I will start confusing you and myself. That is the jist of it though.

Maybe someone with a much deeper understanding can elaborate.

 

[MarcAlexander]

d does not have a value. In f(x)dx, f(x) is the function and dx is the symbol for change in x which is delta x that you have on the other side of the equation.

Integration is basically used to calculate the area under a function in mathematics. I don't want to go too deep into what it is because I will start confusing you and myself. That is the jist of it though.

Maybe someone with a much deeper understanding can elaborate.

May I ask for an example of a function that would create such curve worthy of calculating?

 

[DeadOriginal]

It can really be any function you are interested in. I guess the easiest function would be f(x)=x. If we for example want to find out what's the area underneath f(x)=x from x=0 to x=2 we can integrate the function which would give us \frac{x^2}{2}. From the fundamental theorem of calculus, we can find the area by plugging in the top limit and subtracting the substitution of the bottom limit which would give us something like \frac{2^2}{2} - \frac{0^2}{2} = 2 - 0 = 2. Thus the area underneath f(x)=x from 0 to 2 is equal to 2.

This can be easily verified by graphing y=x from 0 to 2. You will see a right triangle being formed by the x axis, the y=x, and an imaginary straight line you can draw in connecting the point (2,2) to the x axis. The area of a triangle is \frac{1}{2}(b)(h). Our triangle's base is 2 and its height is also 2. Plugging that into the equation for area of a triangle we will be left with 2 which verifies that our integration is correct.

This is an easy function so we were able to easily compute it with the triangle thing but with harder functions like \frac{5x^4-3x^7}{29x-209(x^3)} integrating will easily tell us what the area is.

 

[MarcAlexander]

It can really be any function you are interested in. I guess the easiest function would be f(x)=x. If we for example want to find out what's the area underneath f(x)=x from x=0 to x=2 we can integrate the function which would give us \frac{x^2}{2}. From the fundamental theorem of calculus, we can find the area by plugging in the top limit and subtracting the substitution of the bottom limit which would give us something like \frac{2^2}{2} - \frac{0^2}{2} = 2 - 0 = 2. Thus the area underneath f(x)=x from 0 to 2 is equal to 2.

This can be easily verified by graphing y=x from 0 to 2. You will see a right triangle being formed by the x axis, the y=x, and an imaginary straight line you can draw in connecting the point (2,2) to the x axis. The area of a triangle is \frac{1}{2}(b)(h). Our triangle's base is 2 and its height is also 2. Plugging that into the equation for area of a triangle we will be left with 2 which verifies that our integration is correct.

This is an easy function so we were able to easily compute it with the triangle thing but with harder functions like \frac{5x^4-3x^7}{29x-209(x^3)} integrating will easily tell us what the area is.

THANK YOU! I understand it now.
So what does \oint mean?

 

[DeadOriginal]

It is the integration sign. Notice the S like shape, its practically another way of saying sum aka Ʃ.

 

[MarcAlexander]

It is the integration sign. Notice the S like shape, its practically another way of saying sum aka Ʃ.

But I thought \int was the integration operator? Or do they mean the same thing?

 

[DeadOriginal]

Wait... Those are different? LOL I didn't notice.

I'm not sure actually.

 

[AlephZero]

So what does \oint mean?

It means "integration along a curve that is a closed loop". But if are only just starting to learn what integration is, don't worry about what that means. If you were taking formal math courses and just starting on integration, it would probably be about two years before you got to study path integrals.

One of the big problems with trying to learn from the internet rather than from a book is not studying topics in a "logical" order.

 

[MarcAlexander]

It means "integration along a curve that is a closed loop". But if are only just starting to learn what integration is, don't worry about what that means. If you were taking formal math courses and just starting on integration, it would probably be about two years before you got to study path integrals.

One of the big problems with trying to learn from the internet rather than from a book is not studying topics in a "logical" order.

Well my order is somewhat logical: function, derivative, integral...and now path integral.

Would you mind actually giving me a better definition of derivative than Wikipedia and an example please. My knowledge of it seems somewhat un-satisfied.

 

[bp_psy]

If you really want to learn calculus a great place to start is here:
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/

 

[paulfr]

Integration is just multiplication when one of the multiplicands is changing.

The formula in your post is misleading for a beginner.
It is saying that the integral is approximately the sum of n values of f(x) times the delta x (dx)
The actual exact value is the Limit as n=> infinity of this summation.

Think of f(x) as being y=3 from 0 to 4
The curve is a horizontal line.
So the area is 3 times [ 4-0 = 4] = 12
Now if y=f(x) is changing like y=sine x or y=3x+6 or y=x squared
then the integral will give the "multiplication" and thus the area under the curve.

Path integrals are for Calculus 3, not earlier.
There are too many topics you need to learn before moving to them.

And
Hint; Making threats on a public website is bad form.
You could have stated your concern much more politely.

 

[MarcAlexander]

Integration is just multiplication when one of the multiplicands is changing.

The formula in your post is misleading for a beginner.
It is saying that the integral is approximately the sum of n values of f(x) times the delta x (dx)
The actual exact value is the Limit as n=> infinity of this summation.

Think of f(x) as being y=3 from 0 to 4
The curve is a horizontal line.
So the area is 3 times [ 4-0 = 4] = 12
Now if y=f(x) is changing like y=sine x or y=3x+6 or y=x squared
then the integral will give the "multiplication" and thus the area under the curve.

Path integrals are for Calculus 3, not earlier.
There are too many topics you need to learn before moving to them.

And
Hint; Making threats on a public website is bad form.
You could have stated your concern much more politely.

I apologise if it startled you as 'bad form', but I didn't want to have to reply to people who make that silly suggestion, as I clearly stated: I'm trying to learn from the internet, without having to pay for a book.

NOTE: To put it politely: I have no idea what all that meant. I understand the 'lim a->b' sort of thing, but not the rest of it.

 

[JonF]

NOTE: To put it politely: I have no idea what all that meant. I understand the 'lim a->b' sort of thing, but not the rest of it.

Do you understand it formally? i.e. from a δ and ε point of view? Also do you understand differentiation from the point of view of limits and difference quotients? Before you move on to integrals I suggest you can do all of the following fluidly (and this is the typical order to learn them in also):

1. Formally define a limit, and be able to prove a function has a limit at a given point.
2. Understand and use: continuity, the mean value theorem, intermediate value theorem.
3. Formally define a derivative. Take a derivative by definition. Know the criteria for differentiability and be able to apply it.
4. Have an understanding of what a derivative is, and how it relates to change. Especially related to position, velocity, and acceleration.
5. Be able to use, know, and derive: the sum/difference rule, the product rule, the quotient rule, the power rule, and the chain rule
6. Know the derivative of six basic trig functions
7. Be able to take the nth derivative of any given algebraic or trig function.
8. Use derivatives to curve sketch, find max/mins, related rates, or optimize.
9. Understand and be able to compute Riemann Sums

You also should do many practice problems relating to each of these steps before moving on to the next.

 

[pwsnafu]

You also should do many practice problems relating to each of these steps before moving on to the next.

QFT.

You will only be able to start understanding calculus after you are familiar with it.
And the fastest way to become familiar to a concept in mathematics is to use it.

 

[micromass]

Learning things from sites like wikipedia is a really bad idea. Would you trust a surgeon who claimed he learned everything from wikipedia?? I understand you don't want to pay for expensive book, and the good thing is that you don't need to! There are many free sources on the internet that are much better than wikipedia. Here is a list of free calculus books: http://hbpms.blogspot.com/2008/05/stage-1-introductory-calculus.html

I also like the notes by Chen: http://rutherglen.science.mq.edu.au/wchen/ln.html
And Khan academy is also very good!

Be sure to make lots of problems. It's the only way to understand calculus...

 

[pari777]

@MarcAlexande
you told that u have done derivatives and functions. then we can define an integral as simply anti-derivative or primitive of a function f(x).that is,a function ∅(x) is called an integral of a function f(x) if ∅'(x) = f(x).
for eg:- x^4/4 is a primitive of x^3, because d/dx( x^4/4)= x^3.
but I agree with JonF and others. you have to do limits ,continuity, etc. thoroughly before u come to integration. there is no shortcut bro :).
Also do indefinite integrals before u come 2 definite ones. the example that u used in ur 1st post is a definite one.Indefinite ones r relatively easier than the definite ones and once u get the hang of things definite integrals and their properties will b easy as well.

 

[diligence]

Somebody posted it once already, but it appears that you overlooked it since i see no acknowledgment at all.

SERIOUSLY. The ABSOLUTELY BEST place to learn calculus for free on the internet is right here:

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/

This is a calculus class at MIT, videotaped and posted on the internet for your benefit. Not only are there videotapes of the lectures, but many resources (on the same website) that will help you learn the material. This is the page for Calculus 1. They have similar pages for other courses that I will label "Elementary Advanced Mathematics" (up to Linear Algebra). For higher mathematics, they have somewhat similar pages but without video lectures and tons of perks; basically just lesson plans, but still quite helpful to the independent student. Furthermore, they have similar online resources for ALL areas of science and technology.

If you don't understand the gravity of those statements, specifically why FREE online courses from MIT are probably the best thing since sliced bread, just google "MIT".