The Fundamental Theorem of Calculus, Part I

[LucasGB]

According to Wikipedia, "the first fundamental theorem of calculus shows that an indefinite integration can be reversed by a differentiation."

Am I wrong or is this theorem very simple? Indefinite integrals are the same as antiderivatives. So isn't this theorem simply stating that the derivative of the antiderivative of a function is the function itself? But this already follows from the definition of antiderivatives itself! So isn't this theorem simply a re-statement of the definition of antiderivatives?

 

[Anonymous217]

As far as I know, it's a statement that connects differentials and integrals together as one concept, which is why this theorem is so important.

 

[pbandjay]

Integration isn't really defined as anti-differentiation. Consider the two as separate entities. For instance, derivatives calculate rate of change, and integrals calculate area under a function. But just how can an anti-derivative evaluate an area? The theorem proves the two are connected.

 

[LucasGB]

Integration isn't really defined as anti-differentiation.

Indefinite integration is, and the first part of the theorem concerns only indefinite integrals, since the derivative of any definite integral is zero.

 

[elibj123]

Integration is first defined via Riemann Sums over a defined interval. Then the Fundamental Theorem considers integration where the upper bound is x, that is, considers a function:

$$F(x)=\int^{x}_{a}f(s)ds$$

Where s is a dummy variable and a is an arbitrary constant. Then the Fundamental Theorem states that the derivative of F(x) is f(x), which proves that indefinite integration and differentiation can be considered as inverse operations. Which only then (!) brings in the notation of the indefinite integration $$\int$$ as a symbol of anti-differentiation.

 

[pbandjay]

Sorry, I misread. Ignore what was posted here before.

The fundamental theorem only deals with definite integrals, though, and the definition of the definite integral isn't the anti-derivative.

 

[LucasGB]

Integration is first defined via Riemann Sums over a defined interval. Then the Fundamental Theorem considers integration where the upper bound is x, that is, considers a function:

$$F(x)=\int^{x}_{a}f(s)ds$$

Where s is a dummy variable and a is an arbitrary constant. Then the Fundamental Theorem states that the derivative of F(x) is f(x), which proves that indefinite integration and differentiation can be considered as inverse operations. Which only then (!) brings in the notation of the indefinite integration $$\int$$ as a symbol of anti-differentiation.

But isn't the function F(x) you just wrote there a definite integral, instead of indefinite? It has bounds, and for every value of x it gives me a number, and not a function.

 

[espen180]

You should think about what a function is. A (single variable dependent) function is a defined operator which takes an input argument (a number, another function, a vector, a matrix, a tensor, you name it) and gives back another number, fuction, vector or whatever it was you stuffed into it. An function which gives you a function if you give it a number has to be dependent on multiple variables.

Here, x is a variable, and thus F(x) is a function. The fundamental theorem uses the definite integral which is already defined, defines F(x), then connects it with f(x).

 

[LucasGB]

You should think about what a function is. A (single variable dependent) function is a defined operator which takes an input argument (a number, another function, a vector, a matrix, a tensor, you name it) and gives back another number, fuction, vector or whatever it was you stuffed into it. An function which gives you a function if you give it a number has to be dependent on multiple variables.

Here, x is a variable, and thus F(x) is a function. The fundamental theorem uses the definite integral which is already defined, defines F(x), then connects it with f(x).

I think I get it then. Please correct me if I'm wrong.

The equation posted by elibj123 is the very definition of an antiderivative, and it relies on the definite integral, which is already defined. From what I can see, this is where the arbitrary constant of integration comes from, from the arbitrariness in the choice of the lower bound. Then, the theorem shows that the derivative of an antiderivative of a function is the function itself. Therefore, the antiderivative of a function isn't defined as another function which when differentiated yields the original function. It is simply defined by elibj123's equation, and then the theorem demonstrates that aforementioned property.

Is this correct?

 

[espen180]

I think I get it then. Please correct me if I'm wrong.

The equation posted by elibj123 is the very definition of an antiderivative, and it relies on the definite integral, which is already defined. From what I can see, this is where the arbitrary constant of integration comes from, from the arbitrariness in the choice of the lower bound. Then, the theorem shows that the derivative of an antiderivative of a function is the function itself. Therefore, the antiderivative of a function isn't defined as another function which when differentiated yields the original function. It is simply defined by elibj123's equation, and then the theorem demonstrates that aforementioned property.

Is this correct?

Correct. We call it the antiderivative because of the property, not the other way around. At least that is my understanding. In any case, it is the mathematics which is important, not the semantics.

The beauty of the theorem lies in that it takes two separate operations (differentiation and limits of Riemann sums) and connects them.

By the way, what I said before about functions has some exeptions. The only examples I can think of concern matrix and scalar multiplication for example

$$f(\vec{x})=\vec{x}\cdot \vec{a}$$

Which takes a vector and gives back a number.

 

[wofsy]

The fundamental theorem was taught to me this way.

starting with a function f(x) define a new function F(x) = Integral from a to x of f(t)dt
Then dF/dx = f(x).

 

[HallsofIvy]

Yes, where that integral can be defined in terms of Riemann sums. That is, actually, "part 1" of the fundamental theorem of calculus.

"Part 2" is the other say. Given a function F(x), the integral of F'(x) is F(x) plus, possibly, a constant.


Actually, prior to Newton and Leibniz, many people had worked on finding tangents to lines (notably Fermat and DesCartes) and finding areas of non-elementary figures goes back to Archimedes. Their development of the "fundamental theorem", tying those two problems together, is why Newton and Leigniz are recognized as the founders of Calculus.

 

[Landau]

Slightly offtopic:

You should think about what a function is. A (single variable dependent) function is a defined operator which takes an input argument (a number, another function, a vector, a matrix, a tensor, you name it) and gives back another number, fuction, vector or whatever it was you stuffed into it.

By the way, what I said before about functions has some exeptions. The only examples I can think of concern matrix and scalar multiplication for example

$$f(\vec{x})=\vec{x}\cdot \vec{a}$$

Which takes a vector and gives back a number.

It not only has some exceptions, it's just wrong and misleading to describe a function as an "operator" that takes some input and gives the same kind of output.

A function is nothing more than a relation from some set A to some set B, such that every element in the domain A is related to a unique element of the codomain B. That is, functions can be defined between arbritary sets, whatever (different) structure they might have. A can be a group, B a topological vector space, etc.

 

[wofsy]

Yes, where that integral can be defined in terms of Riemann sums. That is, actually, "part 1" of the fundamental theorem of calculus.

"Part 2" is the other say. Given a function F(x), the integral of F'(x) is F(x) plus, possibly, a constant.

I think Part 2 follows from part 1 if you know that the integral of zero is a constant.

In Newton's time Riemann sums had not been thought of yet. And I would guess that Riemann thought of these sums in a more general context - that of finding integrals that expressed physical quantities in terms of local averages. Physicists in Riemann's time assumed the average of a physical quantity - e.g. current or flux of a force- was well approximated by a random value of that quantity in a small enough volume. The smaller the volume (or area or length) the better the approximation. Multiplying these sample values by the small volumes and adding them all up gave a close approximation to the total.The limit - the Riemann integral then gives you the best possible approximation. The Riemann integral is really a grand average multiplied by the total volume.

This picture of the world - the ability to approximate the world accurately through local averages I think was a fundamental assumption about the nature of physical quantities.The Riemann integral - to Riemann and the other Physicists of his day - was an average multiplied by total volume.

Later, this idea of the integral proved inadequate for physics and other integrals were invented.

But this leaves the question of how the integral was thought of in Newton and Leibniz's day.

BTW: The 19'th century physicists view of the integral also made them look for a more general version of the fundamental theorem of calculus. To them the fundamental theorem really said that if two functions are related by taking a derivative then the integral of the derivative in the interior of a region - e.g. the interior of an interval - equals the integral of the function along the boundary of the region, F(b) - F(a) in the case of an interval. This general form of the fundamental theorem is Stoke's Theorem and the generalized idea of a derivative is today called the exterior derivative. it is still key in Physics and mathematics.