Why is the fundamental theorem of calculus astounding?

[appplejack]

Homework Statement

I just learned this idea from my lecture in calculus. I think I understand it at a surface level but don't know much about why it is an astounding discovery as my lecturer suggested.

Homework Equations

So, it states that let (f) be a continuous on an interval [a,b]. Let (g) be an anti-derivative of (f). Then ∫(b up) (a down) f(x) dx = g(b)-g(a)

The Attempt at a Solution

Your take on this?

 

[Char. Limit]

The Fundamental Theorem of Calculus, or as it's normally abbreviated, the FTC, is astounding because it connects the two branches of calculus together. The two branches of calculus are differential calculus and integral calculus. Differential calculus is the study of finding the slope of a function, and integral calculus is the study of finding the area between a function and the x-axis. At first glance, these two don't seem related, and indeed there exist many ways to find the area under a function that don't involve the derivative at all!

However, by the FTC, we can see that differential calculus and integral calculus ARE related, and that their fundamental operations, the derivative and integral, are inverses of each other. This incidentally also makes a lot more integrals easily calculable. Finding the integral of x^2 is not easy when you don't have the FTC handy! (Although it is doable; I've done it.)

 

[Deveno]

to elaborate, what one normally does to find the integral over [a,b] of f(x) is the following:

first, one partitions [a,b] into "pieces" (sub-intervals).

next, one picks a value of f(x) in each "piece".

then one calculates the area of the rectangle with height f(x_i*) (where the star means we picked x_i in the i-th "piece" or subinterval "in some way" (yes, this is hand-waving, but often we make a standard choice, like the left-hand side, the right-hand side, or the mid-point), and width the length of our "piece" (sub-interval).

finally, we add all the rectangle areas together.

then, we repeat with a "finer chopping up" of our interval (more subintervals...perhaps these should be sub-sub-intervals? i dunno).

then, we take the limit as the "size" of our pieces goes to 0 (technically, as the "mesh" of our partition goes to 0), and we pray to the gods above that such a limit indeed exists, or else we're fugazi.

this is a labor-intensive process, fraught with gnashing of teeth, and pulling of hair (the cosmetic degradation is perhaps one reason why women are not as numerous in the mathematical sciences as they should be).

AND...to be sure we have "the" integral, we need to verify that we get the same answer (yes, the same limit), no matter HOW we "chop up the interval" OR choose the "representative height" (the x_i* business). and this is a tall order, because there's just infinitely many ways to go about it (to be fair, the usual partitions are REGULAR partitions, where we just divide it into n equal pieces).

as you might suspect, continuous functions are our first choice for integrating, because as the sub-intervals get smaller, the different choices for f(x_i*) matter less and less (they all "converge" to a common value, once the (sub-)intervals get "small enough").

well, it would be nice if there was an easier way.

and (for certain functions, which, fortunately for YOU, continuous functions fall into) that way is to "anti-differentiate": that is, guess a function F(x) that f(x) is the derivative OF:

f(x) = F'(x).

this is why differentiation is taught FIRST...to be any good at all with guessing what function f might be a derivative of, you ought to have some experience finding derivatives (you suspected all along that differentiation was going to be used later, right?).

it's not immediately clear that finding the slope of a curve, is somehow related to finding the area under a different curve. after all, slope is a "local" thing, and area is a "global" thing. but this is GREAT news, because slopes are relatively easy to compute, while areas are very difficult to compute (unless a region is very simple, like a square, for example).

 

[HallsofIvy]

From an historical point of view, there were many people, including Fermat, who worked on the problem of finding slopes of graphs and tangent lines to graphs and using limit processes to find area under graphs goes back to Archimedes. But Newton and Leibniz are considered the founders of Calculus because the first realized the connection between the two types of problem.

 

[djrcalculus]

Applejack,

I've studied and taught the Fundamental Theorem for 53 years. I also fail to see what's "astounding".

Good for you!

DJRCALCULUS

 

[SteveL27]

From an historical point of view, there were many people, including Fermat, who worked on the problem of finding slopes of graphs and tangent lines to graphs and using limit processes to find area under graphs goes back to Archimedes. But Newton and Leibniz are considered the founders of Calculus because the first realized the connection between the two types of problem.

For the record, Newton's teacher Isaac Barrow is credited with the first proof of FTC.

http://en.wikipedia.org/wiki/Isaac_Barrow

What's also interesting is that Leibniz was well aware of Barrow's work.

Leibniz bought a copy of Barrow’s book during his stay in London that winter, and in that book is to be found (albeit in geometrical garb) most of the basic rules for differentiation and integration, as well as indications of the reciprocal nature of these operations. It is often said that both Newton and Leibniz were both greatly indebted to Barrow for the fundamental ideas involved in calculus.

http://www.mathpages.com/home/kmath335/kmath335.htm

Finding the integral of x^2 is not easy when you don't have the FTC handy! (Although it is doable; I've done it.)

I strongly recommend this exercise to everyone who hasn't done it. You'll discover a connection to a well-known formula of discrete math. And you can generalize to xⁿ. Specifically, see if you can calculate the integral of x² from 0 to 1 directly from the definition of the Riemann integral.

 

[1MileCrash]

Because it's basically what shows that integration and differentiation are opposites. It seems common sense now, but look at it in terms of what types of problems differentiation and integration came from.

"I want to know the rate of change at a single point, not over an interval. I'll do so by taking the limit as the size of the interval approaches zero."

"I want to know the area under this curve, I will do so by cutting it into pieces with easily defined area and adding all of them up."

These things are opposite.

 

[1MileCrash]

Applejack,

I've studied and taught the Fundamental Theorem for 53 years. I also fail to see what's "astounding".

Good for you!

DJRCALCULUS

Try a field you can appreciate?

 

[micromass]

Applejack,

I've studied and taught the Fundamental Theorem for 53 years. I also fail to see what's "astounding".

Good for you!

DJRCALCULUS

The fundamental theorem is truly one of the nicest results in lower level mathematics. If you don't appreciate it, then I pitty your students.

 

[jackmell]

Hi guys,

Lemme' tell you something I think is not well-known about it: it works for multi-valued functions too. Dorothy looked at the sign-post up ahead:

\mathop\oint\limits_{|z|=1} \log(z)dz=\left(z\log(z)-z\right)\biggr|_1^1

then thought, "well ain't that zero?" but then remembered she wasn't in Kansas anymore.

 

[micromass]

Hi guys,

Lemme' tell you something I think is not well-known about it: it works for multi-valued functions too. Dorothy looked at the sign-post up ahead:

\mathop\oint\limits_{|z|=1} \log(z)dz=\left(z\log(z)-z\right)\biggr|_1^1

then thought, "well ain't that zero?" but then remembered she wasn't in Kansas anymore.

I wonder if that means that we can generalize the fundamental theorem to Riemann surfaces (which is the natural setting for multivalued functions)

 

[jackmell]

I wonder if that means that we can generalize the fundamental theorem to Riemann surfaces (which is the natural setting for multivalued functions)

Absolutely. That makes it amazing I think. Consider the worst possible case, an algebraic function:

f(z,w)=a_0(z)+a_1(z)w+\cdots+a_n(z)w^n=0

w(z) even for moderate n like n=5 is basically swiss-cheeze in the complex plane. Now consider a contour which traverses a very complex path over this surface which avoids singular points. We can still compute the integral of w(z) using antiderivatives:

\mathop\int\limits_{C} wdz=W(z)\biggr|_{z_0}^{z_1}

but of course, just like Dorothy's integral, we need to use an antiderivative which is analytically continuous over the entire path. Also, w(z) over it's normal Riemann surface is a single-valued holomorphic function (of the "places" on the surface) except at the singular points so naturally one would think an integral of w(z) over it's Riemann surface which avoids the singular points would have an antiderivative.

 

[cjc0117]

How can you find the indefinite integral of x² without the FTC? Aren't Riemann sums only good for definite integrals?

 

[1MileCrash]

An indefinite integral is really just another word for antiderivative, which we know to be connected to integration via FTC.

 

[cjc0117]

Yeah, but Char. Limit and SteveL27 are implying that there's a way to do it without using the FTC.

 

[micromass]

Yeah, but Char. Limit and SteveL27 are implying that there's a way to do it without using the FTC.

They're saying that finding the definite integral of x² without the FTC is possible.

 

[1MileCrash]

I see no such implication.

 

[cjc0117]

Yeah, I guess there isn't. I just thought they meant indefinite integral when they said integral

 

[cjc0117]

So the general formula for the definite integral from a to b of x² that Char. Limit was speaking of is:

lim_{n\rightarrow∞}\sum^{n}_{i=1}(a+\frac{(b-a)i}{n})^{2}(\frac{b-a}{n})

=lim_{n\rightarrow∞}\frac{b-a}{n}\sum^{n}_{i=1}(a^{2}+\frac{2a(b-a)}{n}i+\frac{(b-a)^{2}}{n^{2}}i^{2})

=lim_{n\rightarrow∞}(\frac{a^{2}(b-a)}{n}(n)+\frac{2a(b-a)^{2}}{n^{2}}(\frac{n^{2}+n}{2})+\frac{(b-a)^{3}}{n^{3}}(\frac{2n^{3}+3n^{2}+n}{6}))

=a^{2}(b-a)+a(b-a)^{2}+\frac{1}{3}(b-a)^{3}

I've checked with different values of a and b and I think it's right. Now I just need to generalize to xⁿ.

EDIT: Of course it's right; it simplifies to \frac{1}{3}(b^{3}-a^{3})=\int^{b}_{a}x^{2}dx. So the general formula for the definite integral of xⁿ from a to b is just \frac{1}{n+1}(b^{n+1}-a^{n+1}). But then, SteveL27, why do you say that finding the integral of x² without the FTC will help find a well-known formula for discrete math, when the formula can just be found using the FTC? Or is it just the way that the formula is written before being simplified?

 

[djrcalculus]

I seem to have critics who confuse "understanding the fundamental" with "not appreciating the fundamental theorem".

The fundamental theorem is, or should be, very obvious.
You should pity the those who don't find it so.

DJRCALCULUS

 

[1MileCrash]

I seem to have critics who confuse "understanding the fundamental" with "not appreciating the fundamental theorem".

The fundamental theorem is, or should be, very obvious.
You should pity the those who don't find it so.

DJRCALCULUS

No, don't worry, we all figured your post was meant to be taken as a boast of some type.

It surely took humanity a good while to develop the FTC (not to mention calculus as a whole), considering how so very obvious it is!

 

[djrcalculus]

You pity students because they understand the FTC is obvious?
A synonym for "astounding" is "bewildering".

The FTC is beautiful, simple and obvious; not bewildering.

signing off,
DJR

 

[1MileCrash]

You pity students because they understand the FTC is obvious?
A synonym for "astounding" is "bewildering".

The FTC is beautiful, simple and obvious; not bewildering.

signing off,
DJR

Astounding: Surprisingly impressive or notable: "the summit offers astounding views".

Bewildering: Cause (someone) to become perplexed and confused: "she seemed frightened and bewildered"; "a bewildering array of desserts".


I agree that if you change the definition of words, your post could be considered reasonable.