# Is there a First Principles for integration?

1. May 27, 2013

### cmcraes

Is there a "First Principles" for integration?

More specifically, For indefinite integrals. Ive looked online and cant find anything.
Is there a method of finding indefinite integral analogous to finding the derivitive by:

Lim (f(c+h)-f(c))/h
h->0

So final question: Besides the 'inverse power rule', integration by parts or substitution, Is there a way to find the indefinite integral of a function? Thank you!

2. May 27, 2013

### Mandelbroth

For a definite Riemann integral, $\displaystyle \int_{a}^{b}f(x) \, dx = \lim_{n\rightarrow\infty^-}\sum_{i=1}^{n}\left[f(a+\frac{b-a}{n}i)(\frac{b-a}{n})\right]$. For indefinite integrals, I am unaware of such a limit form. I think it should just be seen as the integral of a function f is a function F for which the derivative of F is f.

3. May 27, 2013

### VantagePoint72

The "first principle" is the Fundamental Theorem of Calculus, which proves the definite integral / Riemann sum (which Mandelbroth gave) is equal to $F(b) - F(a)$ where $F'(x) = f(x)$. The indefinite integral of $f(x)$ is defined as the antiderivative of $f$ (plus a generic constant), by analogy with the Fundamental Theorem.

Perhaps this is the point of confusion. As a first principle thing, $\int_a^b f(x) dx$ is not defined as $F(b) - F(a)$, where $F'(x) = f(x)$. It's defined as the area under $f(x)$ between $x=a$ and $x=b$. It is the FToC that lets us connect this with the antiderivatives of $f(x)$. Having done this, the indefinite integral is just introduced as a shorthand for antidifferentiation.

4. May 27, 2013

### Mandelbroth

I disagree with parts of this.

The antiderivative of f is a function F such that the derivative of F is f. Thus, an arbitrary constant is implied by the nature of the derivative. However, what you said is, technically, correct, because a constant plus a constant is still a constant.

As to $\int_a^b f(x) \, dx$ being defined as the area under f(x), I completely disagree. It may have been originally defined as the area, but it is not, in general, going to give an area by modern definition.

Consider the integral $\int_{0}^{\pi}e^{ix} \, dx$. This will not be an area. If it was, good luck constructing any shape with an area of $2i$. :tongue:

5. May 27, 2013

### VantagePoint72

This was more a case of linguistic imprecision on my part than anything. Where I said "the antiderivative of $f$ (plus a generic constant)" I should have said "an antiderivative". As in, particular solution $F$ to $F'(x) = f(x)$.

And I, in turn, completely disagree with your approach. Considering it's pretty clear the OP is doing calculus of one real variable, I think it's much more sensible to consider the integral as an area so that the Riemann sum is suitably motivated. Then when we generalize to cases where the area doesn't make sense, we keep the sum and throw away the area interpretation. That's typically how it goes in mathematics: consider something concrete, derive some properties of it, and then abstract the whole thing so we forget about the original concrete thing and just keep the properties. I think it's a pedagogical mistake to jump directly to the abstract definitions. The definite integral as an area definition motivates the definition of the Riemann sum, and the former should be understood before the latter is taken a definition in its own right.

6. May 27, 2013

### micromass

While I agree that the area interpretation is a very nice one, it's not the most useful one. An integral is just a way to sum infinitely many elements which are infinitely small (very roughly). I don't think we shouldn't motivate the integral as an area, but I think we should leave the area interpretation very quickly and go to the more abstract "sum" interpretation. The way that integrals are used in mathematics and physics tend to have very little to do with areas anyway. Somebody who thinks of an integral as an area tends to have some difficulty with the fact that the arc length can also be given as an integral.

7. May 27, 2013

### VantagePoint72

Fair enough.

8. May 27, 2013

### cmcraes

My problem with the Riemann Integral is that you must know what F(x) is. Although usually do-able, its seems like something is missing from indefinite integration.

Also side question: what is the Indefinite integral of
y=0
Is it c, or 0?

9. May 27, 2013

### Office_Shredder

Staff Emeritus
It's c. And if you try to fine the area under the graph between x=a and x=b, the area is c-c=0 regardless of your choice of c

10. May 28, 2013

### epenguin

The method is - RECOGNITION!

I think the OP's question was that whereas we can almost infallibly differentiate any reasonable function (continuous derivatives etc.) we are given, it is much harder to do the inverse operation.

Isn't the answer (almost) that the only way to do it is to know the answer? Or at least the rough kind of shape of answer. By differentiation of a list of we can get a list of functions and their derivatives to be able to go backwards. Then the so called 'methods' of integration is basically a sort of handbook of recognising a number of types and when we have the rough general type we hammer it till it hopefully comes into the shape of some thing or things we recognise on our list.

I suspect that in years to come these 'methods' and excercises will disappear from curriculae, since the problems can be handled by calculators - it will go the way of doing long arithmetical calculations by hand.

11. May 28, 2013

### HallsofIvy

More generally, this is an example of the "inverse problem" in mathematics. It typically happens that we have some operation on a set that has a direct definition and we can determine a number of "rules" from the definition but then we have the problem of finding inverses for that operation- which typically requires knowing special examples of the direct operation.

12. May 28, 2013

### pwsnafu

It's worth pointing out, that is not the same thing as Riemann sums: according to what is written in the quote the characteristic function for the rationals would be integrable.