Fudamental Theorem of Calculus

  • Thread starter Cyrus
  • Start date
  • #1
Cyrus
3,150
16
A quick question. The fundamental theorem of calclus states that:

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

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 dont see what problem it would pose to call it f(x)dx.
 

Answers and Replies

  • #2
SteveRives
56
0
cyrusabdollahi said:
A quick question. The fundamental theorem of calclus states that:

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

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 dont 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:

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

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

-SR
 
  • #3
Cyrus
3,150
16
But why not like this?

[tex] \frac{d}{dx} \int^x_a f(x)dx= f(x) [/tex]
 
  • #4
abercrombiems02
114
0
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.
 
  • #5
qbert
185
5
it is a convenient notation to keep things straight.

compare:
[tex]
\frac{d}{dx} \int_{\sqrt{x}}^{x^3} f(x^2, x) d x
[/tex]

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

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. )
 
  • #6
Because to obtain the form that you've written you must first write

[tex]F(x) = \int_{a}^{x}f(t)dt[/tex]
 
  • #7
matt grime
Science Advisor
Homework Helper
9,426
4
cyrusabdollahi said:
But why not like this?

[tex] \frac{d}{dx} \int^x_a f(x)dx= f(x) [/tex]


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.
 
  • #8
saltydog
Science Advisor
Homework Helper
1,591
3
qbert said:
it is a convenient notation to keep things straight.

[tex]
\frac{d}{dx} \int_{\sqrt{x}}^{x^3} f(x^2, t) dt
[/tex]

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

[tex]f(g(x),t)[/tex]

I assume it would be:

[tex]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[/tex]
 
Last edited:
  • #9
Cyrus
3,150
16
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?
 
  • #10
matt grime
Science Advisor
Homework Helper
9,426
4
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.
 
  • #11
matt grime
Science Advisor
Homework Helper
9,426
4
perhaps it would help to think of sums

[tex]\sum_{r=1} ^n r= \frac{n(n+1)}{2}[/tex]

r is the dummy variable. what happens if you replace r with n in that sum?
 
  • #12
Cyrus
3,150
16
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.
 
  • #13
matt grime
Science Advisor
Homework Helper
9,426
4
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?
 
  • #14
quasi426
208
0
cyrusabdollahi said:
But why not like this?

[tex] \frac{d}{dx} \int^x_a f(x)dx= f(x) [/tex]


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.
 
  • #15
Crosson
1,259
4
Let me try and define a function F(x) this way:

[tex]F(x)=\int_a ^x \frac{Sin (x)}{x} dx [/tex]

Now let me evaluate F(3).

[tex]F(3)=\int_a ^3 \frac{Sin (3)}{3} d3 [/tex]

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:

[tex]F(3)=\int_a ^3\frac{Sin (x)}{x} dx [/tex]

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.
 
  • #16
Cyrus
3,150
16
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 dont see crosson is your notation of d3. Would that not be zero, since 3 is a constant? If not, does d3 really mean anything?
 
  • #17
Castilla
241
0
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.
 
  • #18
Cyrus
3,150
16
It seems that you would not change dx to d3. It would stay as dx, no?
 
  • #19
fourier jr
757
13
matt grime said:
perhaps it would help to think of sums

[tex]\sum_{r=1} ^n r= \frac{n(n+1)}{2}[/tex]

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

yes! to matt grime you listen!

000015.gif
 
  • #20
noslen
26
0
Now im confused, would this work...

[tex] \frac{d}{dx} \int f(x)dx= f(x) [/tex]


:uhh:
 
  • #21
matt grime
Science Advisor
Homework Helper
9,426
4
that is true, since you have an indefinite integral there, and the notation

[tex]\int f(x)dx[/tex]


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.
 
  • #22
mathwonk
Science Advisor
Homework Helper
11,383
1,608
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.
 
Last edited:
  • #23
Cyrus
3,150
16
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.
 
  • #24
Cyrus
3,150
16
matt grime said:
that is true, since you have an indefinite integral there, and the notation

[tex]\int f(x)dx[/tex]


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, [tex]\int f(t)dt[/tex], no?
 
  • #25
matt grime
Science Advisor
Homework Helper
9,426
4
read the words out loud, what do they say?
 
  • #26
Cyrus
3,150
16
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:

[tex]\int^x_a f(t)dt[/tex]


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

[tex]\int f(x)dx= \int^x_a f(t)dt [/tex]

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?
 
Last edited:
  • #27
matt grime
Science Advisor
Homework Helper
9,426
4
yes, that is the relationship between indefinite and definte integrals.
 
  • #28
Cyrus
3,150
16
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.
 
  • #29
Cyrus
3,150
16
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".
 
  • #30
matt grime
Science Advisor
Homework Helper
9,426
4
Errm, no one's commented on it 'cos d3 is meaningless.
 
  • #31
Cyrus
3,150
16
For the case of a regular definite integral,

[tex] \int^b_a f(x)dx [/tex]

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, lets 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:

[tex]F(3)=\int_a ^3 \frac{Sin (3)}{3} dx [/tex]??
 
Last edited:
  • #32
matt grime
Science Advisor
Homework Helper
9,426
4
it's notation. as you're using it, it is meaningless. that's all.
 
  • #33
Cyrus
3,150
16
But given what crosson wrote, would you plug in 3 into dx, or would you have left it as dx, reguardless of its meaning?
 
  • #34
matt grime
Science Advisor
Homework Helper
9,426
4
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.
 
  • #35
Cyrus
3,150
16
I see, so this would be the most "correct form" to write the equation?

[tex]F(3)=\int_a ^3 \frac{Sin (3)}{3} dx [/tex]

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

Suggested for: Fudamental Theorem of Calculus

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
9
Views
8K
  • Last Post
Replies
2
Views
1K
Top