Why Use a Dummy Variable in the Fundamental Theorem of Calculus?

[Cyrus]

A quick question. The fundamental theorem of calclus states that:

\frac{d}{dx} \int^x_a f(t)dt= f(x)

I was wondering why the use of the dummy variable t, and not just x. Is it to distinguish that the function varies with the value t, and the limit of integration varies with a different variable x. I don't see what problem it would pose to call it f(x)dx.

 

[SteveRives]

A quick question. The fundamental theorem of calclus states that:

\frac{d}{dx} \int^x_a f(t)dt= f(x)

I was wondering why the use of the dummy variable t, and not just x. Is it to distinguish that the function varies with the value t, and the limit of integration varies with a different variable x. I don't see what problem it would pose to call it f(x)dx.

It is standard to express relation of change as change in y with respect to change in x. And so the use of x is established (by practice). It is really not more complicated than one word: tradition. If you wanted, we could put it this way:

\frac{d}{dt} \int^t_a f(x)dx= f(t)

Having just had two glasses of wine , I reserve the right to review and edit this in the morning when I am thinking more clearly!

-SR

 

[Cyrus]

But why not like this?

\frac{d}{dx} \int^x_a f(x)dx= f(x)

 

[abercrombiems02]

what you have is fine, pretty much all u need to worry about with the theory is that if you differentiate an expression that you just integrated, you'll get the same thing.

 

[qbert]

it is a convenient notation to keep things straight.

compare:
<br /> \frac{d}{dx} \int_{\sqrt{x}}^{x^3} f(x^2, x) d x<br />

with :
<br /> \frac{d}{dx} \int_{\sqrt{x}}^{x^3} f(x^2, t) dt<br />

in cases like this where you need to know
explicitly what's the variable being integrated
it's good to have the habit of "proper" notation.

(for simple cases, of course, one notation is as good as another. )

 

[Icebreaker]

Because to obtain the form that you've written you must first write

F(x) = \int_{a}^{x}f(t)dt

 

[matt grime]

But why not like this?

\frac{d}{dx} \int^x_a f(x)dx= f(x)

becuase you cannot have the x as both a dummy variable of the integral and the variable of the limit. it just makes no sense. they are different things. using the same letter for different things is 'not allowed' in mathematics.

 

[saltydog]

it is a convenient notation to keep things straight.

<br /> \frac{d}{dx} \int_{\sqrt{x}}^{x^3} f(x^2, t) dt<br />

You know, I've never looked at Leibnitz's rule with that type of integrand, that is:

f(g(x),t)

I assume it would be:

3x^2f(g(x),x^3)-\frac{1}{2}x^{-1/2}f(g(x),\sqrt{x})+\int_{\sqrt{x}}^{x^3}\frac{\partial f}{\partial g}\frac{dg}{dx}f(g(x),t)dt

 

[Cyrus]

But why do we even need a dummy variable matt? Could we not read it as, f is a function that varies on the value of x, and that we integrate from a to x. Then we take the derivative with resepct to x?

 

[matt grime]

because that's what it is. it is the end point of the interval that is the variable, not the subject of the integral. if you change the meaning of the symbol then the FTC no longer applies since you aren't dealing with the same object.

 

[matt grime]

perhaps it would help to think of sums

\sum_{r=1} ^n r= \frac{n(n+1)}{2}

r is the dummy variable. what happens if you replace r with n in that sum?

 

[Cyrus]

Are you saying that if i use f(x)dx, then instead of having f(x)dx vary between a and x, f(x)dx ALWAYS takes on the value of the upper limit, and is just added to itself x-a times? so f(x)dx is never changing once we pick a value for x, thus the need for the dummy variable t.

 

[matt grime]

i'm saying that it makes no sense to speak of adding (and I'm happy to use that abuse of notation) f(x)dx to itself as x varies from a to x. surely you can see that?

 

[quasi426]

But why not like this?

\frac{d}{dx} \int^x_a f(x)dx= f(x)

you can write it like this, but you have to know that the dummy variable x is different then the x in the function being integrated. So basically the reason it doesn't make any sense is that you are not communicating your idea to everyone else but simply yourself (since you know that the two variables represent different things.) So in order to communicate the idea that the two variables are different then you should use different characters.

If you assume that the dummy variable and the variable getting integrated are the same, then you get this sort of never ending loop.

 

[Crosson]

Let me try and define a function F(x) this way:

F(x)=\int_a ^x \frac{Sin (x)}{x} dx

Now let me evaluate F(3).

F(3)=\int_a ^3 \frac{Sin (3)}{3} d3

Is there a problem with those threes? There shouldn't be, because to evaluate a function at x = 3 we simply replace x by 3 everywhere it appears. Maybe you would say that I should evaluate F(3) this way:

F(3)=\int_a ^3\frac{Sin (x)}{x} dx

But then I would say that we are breaking the rule above, that to evaluate a function at x = 3 we replace x everywhere by 3. The only way out of this dilema is to use a dummy variable.

 

[Cyrus]

Yep yep, I see what it is used for now. I always wondered the use of that notation, but now it is clear. The only thing I don't see crosson is your notation of d3. Would that not be zero, since 3 is a constant? If not, does d3 really mean anything?

 

[Castilla]

Saltydog, I have not checked your Lebniz rule aplication, but it is easy to see if it was correct: f(g(x), t) is a function of x and t, so put (say)
f(g(x),t) = h(x,t) and work the leibniz rule with this instead of that.

Castilla.

 

[Cyrus]

It seems that you would not change dx to d3. It would stay as dx, no?

 

[fourier jr]

perhaps it would help to think of sums

\sum_{r=1} ^n r= \frac{n(n+1)}{2}

r is the dummy variable. what happens if you replace r with n in that sum?

yes! to matt grime you listen!

 

[noslen]

Now I am confused, would this work...

\frac{d}{dx} \int f(x)dx= f(x)

 

[matt grime]

that is true, since you have an indefinite integral there, and the notation

\int f(x)dx


means do the definite integral from a to x of f(t)dt where a is some arbitrary constant (remember indefinite integrals are only defined up to constant.

 

[mathwonk]

and please remember, when stating theorems, to give the hypothesis, and not just the conclusion. otherwise it makes no sense. in this case the correct hypothesis is that f is integrable and continuous at the point x where the derivative is taken.

i.e. the version of the FTC you are using is roughly like claiming that x+3 = 8, without saying what x is.

 

[Cyrus]

Going back to my question, would it be stay as dx, or d3, in which case if it is d3, that is phyiscally meaningless, because d3=0, since 3 is a constant, and just further shows the need for the use of a dummy variable.

 

[Cyrus]

that is true, since you have an indefinite integral there, and the notation

\int f(x)dx


means do the definite integral from a to x of f(t)dt where a is some arbitrary constant (remember indefinite integrals are only defined up to constant.

I believe you ment to say, \int f(t)dt, no?

 

[matt grime]

read the words out loud, what do they say?

 

[Cyrus]

you said the notation says to do the definite integral, but you have an indefinite integral right above it. Was the bottom part of your text where you said from a to x, referring to the equation above?

Actually, I provided an indefinite integral as well, i should have put:

\int^x_a f(t)dt


I re-read what you wrote, are you saying that:

\int f(x)dx= \int^x_a f(t)dt

I have not seen it expressed like this before, so it threw me off, sorry.

Makes sense, becuase the first integral would give you F(X) + C, and the second would give you F(X) - F(a), and so C= -F(a). Is that what you meant?

 

[matt grime]

yes, that is the relationship between indefinite and definte integrals.

 

[Cyrus]

As far as Crosson's post goes, would it actually be d3, or would it remain dx inside the integral? If it is infact, d3, then this is meaningless, and should give more reason as to why we need the dummy variable.

 

[Cyrus]

Any thoughts on that? The more I think about it the more it seems that crosson is right that it indeed would be d3, and that would be zero, since x takes on a constant value, and there is no change, so no "dx".

 

[matt grime]

Errm, no one's commented on it 'cos d3 is meaningless.

 

[Cyrus]

For the case of a regular definite integral,

\int^b_a f(x)dx

we have x changing from a to b. where it is a, a+deltax, a+2deltax ... b

but dx is always a constant, small incremental change, even though x takes on values between a and b. In other words, we would not plug in, f(a)d(a) + f(a+deltax)d(a+delta x).

as x takes on its values.

This is why I was wondering if it would take on the value d3, or remain dx.

See, once you pick a value for x, let's say 3, wouldent dx be,

(3-a)/n as n goes to infinity. Which is not the same as d3, which would be zero. Actually, d3 itself seems meaningless, it would have to be something like d/dx (3) to be equal to zero.

So could I revise what cosson wrote to say that:

F(3)=\int_a ^3 \frac{Sin (3)}{3} dx??

 

[matt grime]

it's notation. as you're using it, it is meaningless. that's all.

 

[Cyrus]

But given what crosson wrote, would you plug in 3 into dx, or would you have left it as dx, reguardless of its meaning?

 

[matt grime]

What he wrote was deliberarely incoorect: he was trying to show you an erroneous assumption in your reasoning. either using dx or d3 would be equally good at doing this. you cannot disregard its meaning, it is its meaning that was the important thing to consider.

 

[Cyrus]

I see, so this would be the most "correct form" to write the equation?

F(3)=\int_a ^3 \frac{Sin (3)}{3} dx

If so, that makes things ALOT clearer. Sorry to waste so much of your time on such nonsense matt.

 

[mathwonk]

look back at post 20 and ask yourself what it means if say f is the "greatest integer" function. i.e. hypotheses really do matter. do some thinking!

 

[Cyrus]

Im sorry, I don't understand what you mean mathwonk.

\frac{d}{dx} \int f(x)dx= f(x)


That is all that is in post number 20.

See, Crosson said to evalute:

F(x)=\int_a ^x \frac{Sin (x)}{x} dx

You would replace all the x's by 3's if you choose to do the function at the point 3, giving you:

F(3)=\int_a ^3 \frac{Sin (3)}{3} d3

But what I was asking matt about was the fact that crosson wrote d3, which does not make sense, it should have been left as dx.

I don't see how post 20 comes into play here.

 

[cepheid]

Correct me if I'm wrong (I didn't read everything), but I think that it has been mentioned already that Crosson was just giving an example to show the pitfalls of *failing* to distinguish between the variable that is the limit of integration and the dummy variable.

See, Crosson said to evalute:

F(x)=\int_a ^x \frac{Sin (x)}{x} dx

You would replace all the x's by 3's if you choose to do the function at the point 3

NO! You wouldn't! Or, if you did, you would be making a mistake. You wouldn't replace all the x's. What should be replaced with 3? Whatever variable the function F depends on. What variable is that? Well, F has been defined as an integral, so F changes when the integral changes. The integral changes as its upper limit varies. So the upper limit, x, is the variable on which F depends, an x that has NOTHING to do whatsoever with the other x's that are inside the integral. That suggests that the x's inside the integral must not be called x's at all, for if they are, you might naively be tempted to do this when you set x=3 to evaluate F(3):

F(3)=\int_a ^3 \frac{Sin (3)}{3} d3

resulting in utter nonsense. Yes, the above is utter nonsense because you replaced variables with 3 that were not supposed to be replaced. Yes, d3 is nonsense, and the integrand should still be a function. That is the whole point that Crosson was making in the first place, by showing what sort of error could potentially be made if these two very distsinct variables are not represented using two different symbols. So to prevent any chance of this error occurring, you would write it like this:

F(x) = \int_a^x \frac{\sin t}{t}dt

F(3) = \int_a^3 \frac{\sin t}{t}dt

So, Crosson nicely answered the question of why we do the above, which, if I'm not mistaken, was the whole point of the thread.

 

[Cyrus]

YES! Thank you Cephid, your the man. Thanks for helping me. Same goes to mathwonk, matt, steve, and the others! For some reason I picture you guys shaking your heads as you stand in a circle beating me with your shoe in your hand becuase I've asked the same question so many times.

 

[mathwonk]

take the integral of an integrable function f from a to x. ask yourself if that new function has a derivative. in fact it does not always have one. but it has one wherever the integrand is continuous.

so what does it mean to say that the derivative of the integral is the original function when the original function may not be continuous?

of course i realize now that in post 20 your notation may not refer to the integral, but to the antiderivative. still, what does it eman to write "antiderivative of f" when maybe f does not have an antiderivative?

i.e. there are many f's. like the greatest integer function, that have integrals everywhere, but not antiderivatives everywhere.


i.e. integration and differentiation are not really inverse operations unless you understand what the exceptions are and how much they matter.


the answer is, for riemann integration,l that a function which is rieman integrable, is also continuous almost everywhere. its indefinite integral, i.e. from a to x, is then lipschitz continuous, and differentiable almost everywhere. and the derivative doeas equal the original function everywhere the originjal function is continuous.

but how do you recognize an "antiderivative: in this sense? i.e. when does F have the properety that F(b) - F(a) equals the integral?

it happens if F is any lipschitz continuous function with derivative equal to the original function f wherever f is continuous. But it is not enough to say merely that F is contin with dertivative equal to f where f is continuous. so you cannot use this method of "antiderivatives" to evaluate all integrals, unless you know about lipschitz continuity.

i./e. it is not always true that the integral equals F(b)-F(a), just because f is integrable and F is an antiderivative wherever the integral HAS a derivative.

oh, and concdrning d3 and so on, there is a notion in logic of constants and variables. constants are symbols that are reserved as having an assigned meaning and variables are other symbols. so the symbol after d is supposed to be a variable. now it can be a variable that has already been used, so strictly speaking you could write the integral from a to x of f(x)dx, because in the context of f(x)dx, x is what is called a "bound" variable, i.e. having meaning only within the confines of the expression f(x)dx. but it is better to use not only a variable but a new variable, like t. and write integral from a to x of f(t)dt, as it is less confusing. but it is not allowed to use d3, since 3 already is reserved for the natural number, so d3 would mean zero.

i.e. a variable x is supposed to be able to stand for any number, but 3 can only stand for 3, unless you say otherwise, but that would really be coinfusing, such as my saying that every letter i just typed actually stands for the letter that follows it in the alphabet, (and z stands for a).

 

[Cyrus]

so d3 would mean zero.

Are you sure, I thought d3 is meaningless. For example, when you evaluate a definite integral, you would have the limits from a to b. and a function, f(x)dx. And x would vary between a and b. You would NOT plug in, f(a)d(a) +...f(b)d(b) as x takes on those values, becasue d(a) or d(b) makes no sense. That is one of the things that I thought about, because you guys said that it would be f(3)d3, but when you do the definite integral I just wrote above, you clearly would NOT write f(a)d(a). You would write
f(a)dx+...f(b)dx , like a rieman sum approximation. You would leave it alone as dx. Where dx~= b-a/n. Dont the rules of integration say you leave the dx alone?

As for post 20, that was not my post mathwonk, someone else wrote that one.

 

[mathwonk]

my remarks about post 20 apply even more accurately to your post 1.

crosson does not seem to understand the concept of "bound variable". i.e. there are certain places where a symbol appears, i.e. where it occurs as a bound variable, that it is not to be replaced by the value.

for instance in the statement: " (for every x, x = x) implies that (for some x, 2x = 3)" the two occurrences of x are bound separately, hence this statement is exactly the same as the statement "(for every y, y = y) implies that (for some x, 2x = 3)".

i.e. there is no need at all, in the original statement, for the first symbol x to represent the same number as the last one. as i said earlier, this is not normally done, because many people find it confusing, even if correct.

however one often sees the notation used by crosson and in fact no one ever substitutes in for all the x's occurring there, because it would be obviously wrong. thus indeed many people have the right intutition but are mistaken in thinking it cannot also be justified rigorously.

 

[Cyrus]

But would you agree, that it should have been written as: F(3) = \int^3_a f(3) dx ?

Where the dx stayed an x, and not a 3. I can see where cepheid is comming from in saying that you MIGHT be TEMPTED to put a 3 in there, but then I could argue that you might be tempted to put in f(a)d(a) +...f(b)d(b) when doing a riemann sum. It is just something you do not do, period. So I don't really see why you would be tempted to do such a thing in the first place.

 

[mathwonk]

no, the last two occurrences are both x's. the only place the three goes is at the top. to see this notice that another way to write the same thign is to put simply f under the integral sign. no x no dx.

 

[mathwonk]

one variation that is possible is to put f(x)dg(x) under the integral sign.

 

[mathwonk]

note too the most precise notation, which does not allow x everywhere, would have an x out at the bottom and top of the integral sign, i.e. the limits of integration, using t instead of x, would then run from t=a to t=x, and there would be f(t)dt under the integral sign.

this is the one i use in my classes as it leaves the least possibilioty for confusion.'

it also makes it clear that the avriable of F is the x at the top of the integral sign, and that the t is merely an index, i.e. "dummy variable", for describing the range of integration.

 

[Cyrus]

Right, I agree that the 3's have no business inside the integral. Thats fine. I am no longer in dispute about that fact; however, what I am saying is that let's assume you mistakenly left the variable inside the integral as an x and dx. When you came back an tried to evaluate.

F(x) = \int^x_a f(x) dx

at the value, x=3, wouldent you get a number equal to,

F(3) = \int^3_a f(3) dx = f(3)*(3-a)

Because once x takes on a value, the function f(x) now becomes a constant function.
It looks like there is a built in contradition, because x is a single value, so dx should be zero. At the same time, you could argue that dx should equal to 3-a/n, because those are the limits of the integral. Does that make any sense?

 

[matt grime]

For the love of God cyrus, crosson said the integral from a to 3 of f(3) d3 and that is what he meant!

\int_a^3 f(3)dx

actually makes sense and is exactly f(3)(3-a)

but almost the entire thread has been wasting bandwidth discussing nonsense! you realize that you are asking which of these incorrect notations is more correct? who cares? it is just wrong. why debate something that is false and therefore may be made whatever we feel like?

 

[Cyrus]

You are right, I appologize matt. From talking with you, I got the impression that:

F(x)=\int^x_af(x)dx = f(x)(x-a)

And it is a mathematically correct integral,

the only problem is that it does not do what we are after, and so we have to use a dummy variable, since f is constant function at a particular value for x. But apparently it is not true that F(x)=\int^x_af(x)dx = f(x)(x-a).

 

[matt grime]

You are right, I appologize matt. From talking with you, I got the impression that:

F(x)=\int^x_af(x)dx = f(x)(x-a)

And it is a mathematically correct integral,

no it isn't. this is the exact thing you first posted and it was explained to you then.

the only problem is that it does not do what we are after, and so we have to use a dummy variable, since f is constant function at a particular value for x. But apparently it is not true that F(x)=\int^x_af(x)dx = f(x)(x-a).

i am now even more confused as to what you want to talk about, and we're at 50 posts now.