Fundamental Theorem of Calculus

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agelessdrifter
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Fundamental Theorem of Calculus

Postby agelessdrifter » Wed Nov 18, 2009 12:29 am UTC

So we sort of touched on this in my calc one class today, in what seems to me, upon further inspection, to be a sort of ass-backwards way. Anyway, we're not going to go through the proof of the theorem in class at any point from what I understand, so I looked it up on wiki to see if I could make heads or tales of it instead of just accepting it.

The way this was all presented to my class, in order, was

a) we're introduced to anti-derivatives, and told they're synonymous with indefinite integrals
b) we go over Riemann sums
c) we're taught that the area under the graph of a function over an interval [a,b] is equivalent to the value of the anti-derivative at b less the value of the anti-derivative at a

so when I'm starting to look into this, in my mind, the integral of f is defined as the anti-derivative of f. So when I looked up the formal proof that the area was equal to the anti-derivative, and the first line was "Given: Image", I was like what the hell?

Then I realized after looking a little farther that the integral is apparently defined primarily as the area under the graph of a function, and the fact that this value is closely related to the anti-derivative was discovered later, and not the other way around, as I had originally thought. Am I understanding that correctly?

Anyway, is there a proof of the first part of the FToC that doesn't involve the Mean Value Thm for Integrals (which I don't know) and that I can understand at the level I'm at with calc? Or can someone kinda help me understand the MVThm for Integrals a little bit (ie what's the significance of it, what does it mean, roughly -- I don't really expect to develop a rigorous understanding of it from few posts on a msg board) so I can wrap my head around the proof a little better?

I guess basically what I'm looking for is a (very) rough idea of the order in which these mathematical concepts were developed initially, or and how one lead to another.

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Re: Fundamental Theorem of Calculus

Postby mmmcannibalism » Wed Nov 18, 2009 2:05 am UTC

I guess basically what I'm looking for is a (very) rough idea of the order in which these mathematical concepts were developed initially, or and how one lead to another.


The derivative is the change in y divided by the change in x

areas is the change in x times y(the average y)

If I recall correctly, they figured out derivatives, suspected that derivatives had something to do with area and then figured that out(they being newton who deserves a plural for being that awesome :D ).
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Re: Fundamental Theorem of Calculus

Postby BlackSails » Wed Nov 18, 2009 2:24 am UTC

Newton and leibnitz really

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Re: Fundamental Theorem of Calculus

Postby romulox » Wed Nov 18, 2009 2:35 am UTC

The way we proved the FToC required the mean value theorem. The MVT only depends on the continuity of f though and could easily be proven (uses continuity,min-max thm, and fixed pt thm/intermediate value thm).

I am not sure of the specifics of the historical development. Keep in mind, mathematics has a way of revising itself (thing infinitesimals to epsilon-delta or Riemann to Lebesgue) as time progresses and various ideas are better understood.

hope this helps.

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Re: Fundamental Theorem of Calculus

Postby MidsizeBlowfish » Wed Nov 18, 2009 4:13 am UTC

Here's a proof of the Fundamental Theorem of Calculus not requiring the Integral Mean Value Theorem (which, btw, I would use the FTC to prove). Not all details are shown; you should think about them yourself.

Theorem
i) If f is continuous on [a,b], then for any a < x < b, we have [math]\frac{d}{dx} \left ( \int_a^x f ( t ) dt \right ) = f ( x ).[/math]
ii) If f is continuous on [a,b] and F is any antiderivative of f, then [math]\int_a^b f ( t ) dt = F( b ) - F( a ).[/math]

Proof:
i) Define [math]g( x ) = \int_a^x f ( t ) dt.[/math] Let h > 0 (a similar argument works for h < 0) be such that a < x + h < b. We then have that
[math]\begin{align}
\frac{ g( x + h ) - g( x ) }{ h } & = \frac{ 1 }{ h } \left( \int_a^{ x + h } f ( t ) dt - \int_a^x f ( t ) dt \right ) = \frac{ 1 }{ h } \int_x^{ x + h } f( t ) dt.
\end{align}[/math]
Since f is continuous and the interval [x,x+h] is closed and bounded, there exist u and v in [x,x+h] such that [imath]f(u) \leq f(t) \leq f(v)[/imath] for all t in [x,x+h]. Comparison properties of the definite integral show that [math]f(u) \leq \frac{ g( x + h ) - g( x ) }{ h } \leq f(v).[/math] As [imath]h \to 0[/imath] we have [imath]u, v \to x[/imath], so since f is continuous, [imath]f(u), f(v) \to f(x)[/imath], and therefore by the Squeeze Theorem, g'(x)=f(x), which is what we wanted to show.

ii) As above, let [math]g( x ) = \int_a^x f ( t ) dt,[/math] and let F be any antiderivative of f. Then g'(x)=f(x)=F'(x), so [math]\frac{d}{dx} ( g( x ) - F( x ) ) = 0,[/math] which implies that there is a constant C so that g(x) = F(x) + C. We then get that [math]g(b) - g(a) = F(b) + C - ( F(a) + C ) = F(b) - F(a)[/math]. Now, [math]g(a) = \int_a^a f(t) dt = 0[/math], so the previous equation becomes [math]F(b) - F(a) = g(b) = \int_a^b f(t) dt,[/math] as desired.

QED.

I think it's awesome you want to understand the proof and not just accept the FTC as a given fact. Hopefully you'll be able to fill in some of the details I've left out. If you're confused, ask away! I'll do my best to give helpful answers.

EDIT: Fixed one of my images.

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Re: Fundamental Theorem of Calculus

Postby skeptical scientist » Wed Nov 18, 2009 7:36 am UTC

MidsizeBlowfish wrote:Here's a proof of the Fundamental Theorem of Calculus not requiring the Integral Mean Value Theorem.
*snip*
Then g'(x)=f(x)=F'(x), so [math]\frac{d}{dx} ( g( x ) - F( x ) ) = 0,[/math] which implies that there is a constant C so that g(x) = F(x) + C.

This is, of course, the mean value theorem for derivatives. So while you don't need to use the mean value theorem for integrals, you do need to use the one for derivatives. In fact, the proofs of the FToC that I know all use the mean value theorem for derivatives.

P.S. when using [math] tags followed by punctuation, you should always include the punctuation before the closing tag, or it will start the next line with a punctuation mark.
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Re: Fundamental Theorem of Calculus

Postby MidsizeBlowfish » Wed Nov 18, 2009 2:41 pm UTC

This is, of course, the mean value theorem for derivatives. So while you don't need to use the mean value theorem for integrals, you do need to use the one for derivatives. In fact, the proofs of the FToC that I know all use the mean value theorem for derivatives.


True enough. The OP asked for a proof that didn't use the Integral Mean Value Theorem because they hadn't yet seen it in his class. I presume if they're at the FTC they've covered the derivative MVT. I'm also unaware of any proofs of the FTC that avoid either MVT.

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Re: Fundamental Theorem of Calculus

Postby agelessdrifter » Wed Nov 18, 2009 7:57 pm UTC

Thanks for all the responses. I do know the MVT for derivatives, and that proof (blowfish's) makes much more sense to me than the first few I found. Piecing it together with some others I've looked at, I think I have a fairly solid understanding of it now.

MidsizeBlowfish wrote:I think it's awesome you want to understand the proof and not just accept the FTC as a given fact.


Thanks. It always worries me when I need to apply something without knowing how or why it works. Every once in a while I forget the operations I need to solve a problem, and I find that it helps to remember the reason for the process I'm trying to apply. And of course in this case I was so surprised by the simplicity and convenience of the theorem I just had to know how anyone managed to figure it out in the first place.

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Re: Fundamental Theorem of Calculus

Postby Yesila » Wed Nov 18, 2009 9:43 pm UTC

It's refreshing to have people want to know why a theorem works, especially when you want it enough to go and research it a bit and to ask others for some help. In the School where I've taught Calc I I've only had a very small handful of students who felt this way, most seem to just want the bare minimum out of the class -- learn how to solve the problems, learn the formula of the day, that sort of thing.

The first few times I taught the course I ignored the plea's of the masses and taught mainly at the theory and at the proofs, with some focus to how to do problems, but mainly leaving the problems to the students homework. This came back to haunt me since at the end of the term the majority of the students when filling out their end of term evaluations thought they learned nothing from me... all of the things they cared about (how to mechanically answer a series of procedural questions) was left mostly in their own hands and I just "wasted" their time during class time talking about "useless" things. Hence course evaluation suffered and "the powers that be" at my school had a talk with me saying "get those evaluations up!"

So I had to modify my approach to the class and try to appease the whims of the students. This meant more lessons like your FTC lesson where the result is presented and the proof is skimmed over or skipped altogether and most of the lesson time is spent doing multiple examples of very similar (to each other) problems. At the end of the term I have a class full of mostly happy students who can now do the same sort of math that a computer program can do, only with less accuracy and at a much slow speed. But the students give me good evaluations, and I get to keep my job.

Teaching is too much of a service industry, we need to keep the consumer happy. At the lower levels teachers have to keep parents happy and teach the students a little bit, while making sure they: have fun, don't have to much homework (that interferes with sports and other after school activities), get good grades, are able to pass standardized tests. In the college environment the parents are out of the picture but many students still have a very similar desires. They want to pass the tests, get decent grades and not have to work hard to get them. In either case if the customer is not happy they will sometimes complain to the school administrator and the teacher has a chance to lose their job.

I'd like to apologize to you, on behalf of your teacher, and to apologize to all my past and future students that want to learn, for the state of the system. The system that makes it hard for us to teach to the "good student" and encourages us to teach to the masses at a level that insults the intelligence of the people that really want and deserve to be in the class.

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Re: Fundamental Theorem of Calculus

Postby JayDee » Wed Nov 18, 2009 10:05 pm UTC

agelessdrifter wrote: And of course in this case I was so surprised by the simplicity and convenience of the theorem I just had to know how anyone managed to figure it out in the first place.
The history of calculus (and plenty of other mathematical and scientific developments) is fascinating stuff. With calculus, it may be useful to know that the original developments and ideas, the way people thought about these things back in the seventeenth century, is quite different to how we think about them today. At that time, geometric analysis and proofs were prefered, and the types of formal analytical or logical or whatever proofs we use these days would not have been accepted. So the way these things are taught doesn't correspond to the way they were developed. The rigour and theories often came later.

It's not a bad thing, but learning mathematics and learning the history of mathematics are two different things (or at least should be.)
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Re: Fundamental Theorem of Calculus

Postby Brunch » Wed Nov 18, 2009 10:06 pm UTC

I don't think anyone answered the OP's question about why it seems backwards.

Yes, integrals were useful way before the FTC made them super easy to evaluate. It made intuitive sense that the area under the curve had a meaningful value (especially for physics, when the unit on the x-axis times the unit on the y-axis yielded something meaningful) for quite a while. The indefinite integral is just an integral with a variable upper limit of integration which evaluates to a function instead of a number. Mathematicians had an intuitive sense for the FTC before it was proven just from looking at the units (if the integrating velocity with respect to time yielded position, and if differentiating position with respect to time yielded velocity, integration and differentiation must be somehow fundamentally related).

The mean value theorem for integrals just says that a function has to take on its average value for an interval somewhere on that interval.

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Re: Fundamental Theorem of Calculus

Postby skeptical scientist » Wed Nov 18, 2009 10:18 pm UTC

Brunch wrote:The mean value theorem for integrals just says that a continuous function has to take on its average value for an interval somewhere on that interval.

Fixed.
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Re: Fundamental Theorem of Calculus

Postby agelessdrifter » Thu Nov 19, 2009 2:12 pm UTC

Yesila wrote:The system that makes it hard for us to teach to the "good student" and encourages us to teach to the masses at a level that insults the intelligence of the people that really want and deserve to be in the class.


Yeah, I've often thought that this must be the case. I can't imagine anyone having gone through all the trouble to become the professor of a subject having as little interest in the material they teach as many professors I've had seem to when going through the rudimentary material they're forced to dwell on. I should just as soon apologize to you on behalf of the less inquisitive of your students that force you to dumb down the material. It's a shame so many people seem to see education as a hoop they have to jump through to make money. C'est la vie, I guess.

Brunch wrote: (if the integrating velocity with respect to time yielded position, and if differentiating position with respect to time yielded velocity, integration and differentiation must be somehow fundamentally related)

I hadn't thought of that (which makes me feel silly) -- the suspicion rising out of meaningful numbers with identifiable relationships to one another, I mean. That definitely takes away the sense of out-of-the-clear-blue I first had when I read the theorem.

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Re: Fundamental Theorem of Calculus

Postby RogerMurdock » Fri Nov 20, 2009 1:06 am UTC

As a senior in High School in Calc BC, I agree with the sentiments expressed here (though that's obvious at this point if I'm on an internet forum for mathematics).

I did exceptionally well (or at least think I learned a lot) in Calc AB due to a teacher that taught the theory first, and how it worked into equations and numbers second. He knew the derivations offhand for most formulas, and even if he didn't he would look them up and explain them to you later on if you wanted. In fact one of the most annoying things I found about calc is one of the most "memorizy", the inverse trig derivatives/integrals. It's simply given to you that the derivative of arctan(x)=1/1+x^2., with no explanation about how this was found or of what importance it signifies.

I also have a teacher this year that focuses on theory (I got lucky in my math teachers the last 2 years). He has expressed how he finds it silly how we are simply taught to find the answers mechanically to these problems, even though no real life mathematician or engineering would be caught dead doing "The integral of sin(2x) from 0 to pi" by hand and showing all their work. So most questions on tests are tricky ones really testing your knowledge of calculus. I have a TI-89 (which as you may or may not know can evaluate most derivatives and integrals very easily, and trivializes a lot of Calc I stuff), and so far it's use has been limited by the fact most problems simply do not call for "Find this derivative and circle the correct multiple choice answer". We've probably had about 2, all year.

So Yesila, take heart, your attempts to teach theory are not futile everywhere in the world. It is admirable that you try, despite what your students may think. We need more teachers like you in the world.

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Re: Fundamental Theorem of Calculus

Postby skeptical scientist » Fri Nov 20, 2009 2:26 am UTC

RogerMurdock wrote:It's simply given to you that the derivative of arctan(x)=1/1+x^2., with no explanation about how this was found or of what importance it signifies.

That depends a lot on your class. In fact, that rule is quite easy to derive, so I imagine the derivation is frequently done in class. We start with the rule dx/dy = 1/(dy/dx) for computing the derivatives of inverse functions. For y=tan x, we have dy/dx=(sec x)2 (you should either know this, or be able to work it out from the derivatives of sine and cosine and the quotient rule). Then (on the right interval), x=arctan y, and dx/dy=1/(sec x)2=1/(1 + (tan x)2)=1/(1+y2).

In fact, all trig and inverse trig derivatives and integrals can be easily worked out from three facts:
1) [imath]\lim_{x \to 0} \frac{\sin x}{x}=1[/imath]; [imath]\lim_{x \to 0} \frac{\cos x-1}{x}=0[/imath]
2) [imath]\cos^2 x + \sin^2 x = 1[/imath]
3) the angle sum identities: [math]\sin(\alpha + \beta) = \sin \alpha \cos \beta +\cos \alpha \sin \beta[/math][math]\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta[/math]
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Re: Fundamental Theorem of Calculus

Postby RogerMurdock » Fri Nov 20, 2009 8:14 am UTC

skeptical scientist wrote:
RogerMurdock wrote:It's simply given to you that the derivative of arctan(x)=1/1+x^2., with no explanation about how this was found or of what importance it signifies.

That depends a lot on your class. In fact, that rule is quite easy to derive, so I imagine the derivation is frequently done in class. We start with the rule dx/dy = 1/(dy/dx) for computing the derivatives of inverse functions. For y=tan x, we have dy/dx=(sec x)2 (you should either know this, or be able to work it out from the derivatives of sine and cosine and the quotient rule). Then (on the right interval), x=arctan y, and dx/dy=1/(sec x)2=1/(1 + (tan x)2)=1/(1+y2).

In fact, all trig and inverse trig derivatives and integrals can be easily worked out from three facts:
1) [imath]\lim_{x \to 0} \frac{\sin x}{x}=1[/imath]; [imath]\lim_{x \to 0} \frac{\cos x-1}{x}=0[/imath]
2) [imath]\cos^2 x + \sin^2 x = 1[/imath]
3) the angle sum identities: [math]\sin(\alpha + \beta) = \sin \alpha \cos \beta +\cos \alpha \sin \beta[/math][math]\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta[/math]


Ah, that was quite the simple explanation. I had always assumed since my original teacher glossed over the derivations of those derivatives, they were difficult to find and not worth my time (easier to just memorize). I'll remember this though, thanks.

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Re: Fundamental Theorem of Calculus

Postby EduardoLeon » Fri Nov 20, 2009 5:48 pm UTC

Let f(x) be a differentiable and (without loss of generality and just for the sake of better visualization) positive function for every x in the domain [a, b]. In the endpoints a and b, f(x) should be at least right and left differentiable respectively. Draw y=f(x) in the XY plane.

Now, for any t in the domain of f(x), let F(t) be the area of the surface bounded by y=f(x), y=0, x=a and x=t.

For any Δt such that t+Δt is the domain of f(x), let Fp(t,Δt) = [F(t+Δt)-F(t)] / Δt. The student should notice by simple inspection that the numerator is the area of the surface bounded by y=f(x), y=0, x=t and x=t+Δt; and that the denominator is the width of that surface.

Geometrically, as Δt tends to zero, F(t+Δt)-F(t) tends to become the area of a thin rectangle whose height is f(t). And Δt is the width of that rectangle. Thus, Fp(t,Δt) tends to become f(t).

Analytically, as Δt tends to zero, Fp(t,Δt) tends to F'(t). This is because of our definition of what a derivative is.

Thus, F'(t) = f(t). And, thus, the general antiderivative of f(t) is F(t) + C.

The student should draw deep conclusions about the geometrical interpretation of an antiderivative from this analysis.

---

NOTE: If the student sets t=a, what will be found is that the right derivative of F(t) at t=a is f(a). Likewise, if the student sets t=b, what will be found is that the left derivative of F(t) at t=b is f(b).

---

EXERCISE FOR THE STUDENT: Pick any new a* such that a < a* < t. Replace a with a* in the previous analysis. Notice that Fp(t,Δt), and, thus, F'(t) is not affected by this change. Explain why this happens. Relate this with the fact that two particular antiderivatives of a single function must differ by a constant. In this particular case, which is the constant?

---

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