# Fudamental Theorem of Calculus

1. Jul 14, 2005

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

2. Jul 14, 2005

### SteveRives

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 :rofl:, I reserve the right to review and edit this in the morning when I am thinking more clearly!

-SR

3. Jul 14, 2005

### Cyrus

But why not like this?

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

4. Jul 15, 2005

### 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.

5. Jul 15, 2005

### 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. )

6. Jul 15, 2005

### Icebreaker

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

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

7. Jul 15, 2005

### matt grime

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. Jul 15, 2005

### saltydog

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:

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

Last edited: Jul 15, 2005
9. Jul 15, 2005

### 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?

10. Jul 15, 2005

### 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.

11. Jul 15, 2005

### 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?

12. Jul 15, 2005

### 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.

13. Jul 15, 2005

### 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?

14. Jul 15, 2005

### quasi426

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. Jul 15, 2005

### 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.

16. Jul 15, 2005

### 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 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. Jul 15, 2005

### 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.

18. Jul 16, 2005

### Cyrus

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

19. Jul 16, 2005

### fourier jr

yes! to matt grime you listen!

20. Jul 16, 2005

### noslen

Now im confused, would this work...

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

:uhh: