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:
$$\frac{d}{dx} \int_{\sqrt{x}}^{x^3} f(x^2, x) d x$$

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

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?

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

because 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.

$$\frac{d}{dx} \int_{\sqrt{x}}^{x^3} f(x^2, t) dt$$

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, because 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.