Is It Correct to Express a Limit of an Integral as a Function Value Times dx?

[mnb96]

Hello,
How would you formally express the result of:

$$\lim_{\Delta \to 0}\int_{a}^{a+\Delta}f(x)\cdot dx$$

Is it correct to say that it is equal to $$f(a)\cdot dx$$ ?

Thanks!

 

[Matthollyw00d]

Hello,
How would you formally express the result of:

$$\lim_{\Delta \to 0}\int_{a}^{a+\Delta}f(x)\cdot dx$$

Is it correct to say that it is equal to $$f(a)\cdot dx$$ ?

Thanks!

Why do you think it's equal to that?
Try writing out a few elementary examples.

 

[mnb96]

$$\lim_{\Delta \to 0}\int_{a}^{a+\Delta}f(x)\cdot dx$$

$$= \lim_{\Delta \to 0}(F(a+\Delta) - F(a))$$

$$= \lim_{\Delta \to 0}(F(a+\Delta) - F(a)) \cdot \frac{\Delta}{\Delta}$$

$$= \frac{dF}{dx}(a) \cdot dx$$

$$= f(a) \cdot dx$$

There must be a mistake...where is it?

 

[Matthollyw00d]

$$\lim_{\Delta \to 0}\int_{a}^{a+\Delta}f(x)\cdot dx$$

$$= \lim_{\Delta \to 0}(F(a+\Delta) - F(a))$$

$$= \lim_{\Delta \to 0}(F(a+\Delta) - F(a)) \cdot \frac{\Delta}{\Delta}$$

$$= \frac{dF}{dx}(a) \cdot dx$$

$$= f(a) \cdot dx$$

There must be a mistake...where is it?

The Equality from line 3 to 4.
Line 3 $$=F'(a) \cdot 0 = 0$$

Consider
$$\lim_{\Delta \to 0}\int_{a}^{a+\Delta}e^xdx=\lim_{\Delta \to 0}(e^{a+\Delta}-e^a)=e^a-e^a=0$$

 

[mnb96]

Ok...so you are suggesting that changing a vanishing quantity $\Delta$ into dx is permitted only in ratios?
In that case the answer to my original post would be zero (for continuous and integrable functions), isn´t it?

 

[Matthollyw00d]

Ok...so you are suggesting that changing a vanishing quantity $\Delta$ into dx is permitted only in ratios?
In that case the answer to my original post would be zero (for continuous functions), isn´t it?

The reason you got the dx from the $$\Delta$$ in the quotient isn't because the $$\Delta$$ turned into it, it's because you gave the definition of a derivative and the substituted the two. So yes the answer is 0.

 

[mnb96]

Ok thanks.

So when I see in textbooks identities of the kind df=dx+dy+dz, they make sense only by accepting the fact that dx,dy,dz,df were originally linked by the definition of derivative (e.g.: f was a function f(x,y,z)) ?

 

[Matthollyw00d]

Normally it would be written like this
If
$$f(t)=f(x(t),y(t),z(t))$$
then
$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}$$

I feel that just "df=dx+dy+dz" is quite ambiguous.

 

[mnb96]

Thanks a lot for your help!
Since the original answer has been completely answered, I continued expressing my doubts in a https://www.physicsforums.com/showthread.php?t=416836".

Thanks again.

 

[HallsofIvy]

Normally it would be written like this
If
$$f(t)=f(x(t),y(t),z(t))$$
then
$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}$$

I feel that just "df=dx+dy+dz" is quite ambiguous.

Well, I would say not "ambiguous" but simply wrong for anything other than f(x,y,z)= x+ y+ z+ constant!

From
$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}$$
we can get, in differential form,
$$df=\left(\frac{\partial f}{\partial x}\right)dx+\left(\frac{\partial f}{\partial y}\right)dy+\left(\frac{\partial f}{\partial z}\right)dz$$

 

[yossell]

I feel that just "df=dx+dy+dz" is quite ambiguous.

I too have to say that I hate and fear these kinds of equations and differentials in general wherever they crop up. It drives me particularly mad in physics textbooks which seem to use them so often. It's one of the few things in life that's brought me to tears and I'm normally a happy-go-lucky kind of guy. I'm sure it's a personal block and I should try and get over my fears, but I still mentally try and recast arguments in a way that doesn't use them.