Thank you.
How about the upper limit of the equation in pbuk's post of this thread if we substitute ##\delta x## by ##\Delta x## giving:
$$\lim_{\Delta x \to 0} \int_a^{a+\Delta x} f(x) dx \approx \lim_{\Delta x \to 0} f(a) \Delta x$$
I'd expect you won't agree with this upper limit, even if...
Yes but not insisted though. I only said it in 1 post because the source ultimately used ##\Delta t \rightarrow 0## for his conclusions, which I thought is the same as ##dt##. After your correction about this I kept using ##\Delta t## instead.
@PeroK @fresh_42
Thank you for the further...
Please look at the source carefully because you said I didn't understand the earlier posts and stated the source completely wrong. I did understand it which is why the source's ##\Delta t## actually confused me.
@fresh_42
Thanks, but I have no idea where you're seeing ##\Delta x## as an upper limit in the source. He is clearly stating ##\Delta t## in the upper limit and says:
$$G(x+\Delta t)-G(x)=\int_x^{x+\Delta t}f(t)dt$$
He is constantly using ##\Delta t## in his expressions afterwards. He might as...
@PeroK @Frabjous
Thank you, this cleared it up for me.
I have one last question because this source seems to say otherwise, or so I think.
It says that if
$$G(x)=\int_a^xf(t)dt$$
then one can deduce
$$G'(x)=f(x)$$
Which I can grasp. However, the poster also says that from... (note the ##dt##...
Thank you, I'm trying but I fail to see how this answers whether my conclusion in post #14 is correct or not about getting the same ##f(a)## regardless of using ##dx## or ##da##.
Thanks. I have one other question that came up when approaching this another way.
##f(x)dx## is also equal to ##dF(x)##. Now, if ##x## takes on a value ##a##, then ##dF(a)/dx=f(a)##. But we just saw that ##dF(a)/da=f(a)## as well.
Because ##da## is not equal to ##dx##, such a similarity is only...
So different variables that don't have a relation between them, since ##x## ultimately becomes ##a## after the integration?
Also, what is your stance on the upper limit being ##a+dx##?
Thank you. I'm more curious about the difference between ##dx## and ##da##. Your explanation made me think of the following.
Can I say that in the case of
$$\int_a^{a+da}f(x)dx \approx f(a)da$$
...there is no relationship between ##a## and ##x## whatsoever (##x## stays merely as an integrand...
Can I deduce from this that even if #a# is a variable for an abstract argument, then there's still no such thing as #da# because it must represent numbers? Or did I miss something here?
I don't get why it is stated differently here and here though. Can the limits be seen as a different variable...
Some sources state a similar format of the following
$$\int_a^{a+da}f(x)dx=f(a)da$$
Which had me thinking whether the following integration can exist
$$\int_a^{a+dx}f(x)dx=f(a)dx$$
I have difficulty grasping some aspects about these integrations
1. Regarding the 1st integration, shouldn't ##a##...
It was more of a curiosity to see how it changes the ##\Delta t ##
Since the integrand's variable ##t## takes on the limits of the integration, does that mean that for
$$\int_x^{x+dx}f(t)dt=f(x)dx$$
The ##dt## essentially becomes equal to the value ##dx##?
Yes, my original question actually comes from a probability function, namely the MB Distribution. Hence the antiderivative being denoted as ##P## and the derivative as ##f##
Is it actually possible to correspond a Riemann summation for an integral like this?
$$\int_x^{x+dx}f(t)dt$$
I'd assume...
My bad, I intended to write it like you did but I accidentally switched the two middle integrals.
I'm stumped. Isn't the whole definition of a differential (the "##d##" prefix) the difference between any function with variable ##(x+dx)## and the same function with variable ##x##, regardless of...