What is the difference between a parametric equation and a mathematic law?

First of all sorry for my English, I'm not used to communicate with this language. I want to ask something about a thing that I discovered while studying physics (AKA applied mathematics).

There is this strange operator $\int$ that integrates, this not so curly "S" that means "infinite sum of infinitesimal quantity/numbers", but not only it does "a sum", but it also takes advantage of the "infinitesimal environment" to "linearize" every kind of figure. Ok? (I don't even know if I'm explaining this in the right way...)

So this guy is kind of an analytic and a geometric operators mixed together, and here comes my question: What is the result of an integration?

Is the result of an integration still a mathematical law? Is it a parametric equation? What's the analytical difference between the two, omitting all the rest we can say about integrals and the entire math?

Usually I keep hearing that "the derivative lowers the grade and the integral raises", but this is not at all about the grade. Also, the grade is only part of an analytical dissertation about a law or a resolution of a problem.

I hope that I'm explaining my question in the right way, thanks for all the replies.

PS. I also think that if I understood this better, I could really understand what the integration constant means.

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• This is better asked at math, not physics, site. It is purely about the notion of integration. I would recommend Lang's calculus book, or perhaps a 19th century book on calculus (these are always easy and interesting historically, and give you a different perspective). – Ron Maimon Nov 6 '11 at 21:40
• @ron-maimon can you be more specific? I'm really looking for a good answer that explain me this, and not one that just say this symbol stay for "bla bla bla" and does "bla bla bla". What is the title of the book you are suggesting? – Micro Nov 6 '11 at 21:54
• What do you mean by "mathematical law"? Can you give any example? – Zev Chonoles Nov 6 '11 at 21:59
• @zev-chonoles i.e. law=log; function=(y=log(x)) – Micro Nov 6 '11 at 22:15
• @Micro: I'm sorry, but that answer only confuses me more than I was before. – Arturo Magidin Nov 6 '11 at 22:23

Given the definition of mathematical law and parametric equation listed by @user10389, the result of an integration is a parametric equation. The definite integration operator essentially gives you the measure bounded by the function you're integrating, so it is redefining the measure in terms of the function.

As far as the integration constant, you may find it helpful to think of the indefinite integration operator as a halfway step- it tells you something like what the potential energy of the thing you're integrating is. The integration constant just says that the only thing that's important is the difference between potentials, not their actual value.

• First of all thanks, now i want to show you an example, let's consider $sin(x)$ if i integrate on $x$ in an indefinite interval i obtain $-cos(x)$, this mean that if i consider a variation of the x in the function $sin(x)$ along an indefinite interval ( the infinite ) $-cos(x)$ is the difference between who or what? And the costant of integration stand for? – Micro Nov 7 '11 at 0:16
• I wont to dedicate a separate post for a similar example but with an integration in a definite interval, let's always consider $sin(x)$, i integrate between 3 and 4 on $x$, and i obtain $cos(3)-cos(4)$ so what does that really mean? $cos(3)-cos(4)$ is almost $1$ in numeric terms, like the difference between my 2 terms of the interval of integration. at this link i have a picture of this 2 example that i can't really decode. – Micro Nov 7 '11 at 0:17
• Look at what role measure theory place in defining Lebesgue_integration. Measure (weight) functions play a role of a (real) value for solving proper integrals. Also consider objects like "high order" functions, which are mappings between different sets of functions or functions to values. Integrals can be seen as such functions as well. – Yauhen Yakimovich Nov 7 '11 at 9:04

I am unaware of any formal mathematical way to define a concept of "law", though it may be successfully used in the reference of some other science as physics or philosophy.

Informally, one can model or describe such laws of "nature" by using equations. For example, the process of finding the solution to differential equation, which would be a function, is called integration. This is similar to a more simple situation, when some functions that can be expressed polynomially $(a_n x^n + a_{n-1} x^{n-1} + ..)$ or analytically, once assigned to some value on the right hand - form respective equations, which can be solved as well.

The axiomatics of formal languages is another example.

• honestly when you talk about "functions that can be expressed polynomially" i recognize some theorems about approximation rather than the concept of a matematic law, for what i know the only reason to express a function polynomially is the approximation of this function and not the determination of a law. Can you explain this point in another way? – Micro Nov 7 '11 at 0:36
• Maybe you should take a look at foundations, e.g. at least two definitions of a function: (1) set theoretical and (2) category theoretical. Another good exercise could be to think about how one can formalize such notion in terms of a formal language or logics. Again, concepts as mathematical laws or patterns are not specified in any rigor. – Yauhen Yakimovich Nov 7 '11 at 8:52

The integral of $f$ from $a$ to $b$ gives the area between the graph of the function $f$ and the horizontal axis in the interval from $a$ to $b$.

It is an important theorem that this integral can be expressed by $F(b)-F(a)$ for any function that happens to have the property $F'=f$.

Notice that these properties do not change when you add a constant to $F$.

If $b$ is a variable $t$ and $a$ is a constant, say 0, then the constant of integration keeps track of the fact that you should subtract $F(a)$, that is, you have to make sure that you choose your constant so that your integral gives 0 for $t=0$, as it clearly should.