Confusion about the Substitution rule

[JohnnyGui]

Given is a function $P(E)$ and its derivative $f(E)$. Writing $E$ in terms of $v$ according to $E=\frac{1}{2}mv^2$ gives the derivative $g(v)=f(E)mv$ and $dE=mvdv$.

My issue arises from the premise that I learned; Integrals and derivatives are based on steps of a fixed constant interval. To make my issue clear, let's add a subscript "C" if an interval is considered constant.

When multiplying $f(E)$ by a fixed $dE_C$ I understand that
$$f(E)dE_C=P(v+\frac{dE_C}{mv})-P(v)$$
Furthermore, based on my premise, the derivative $g(v)$ is based on a fixed interval $dv_C$.
$$g(v)=\frac{P(v+dv_C)-P(v)}{dv_C}$$
Because the interval $\frac{dE_C}{mv}$ varies with $v$, this means $\frac{dE_C}{mv}\neq dv_C$ since $dv_C$ is considered constant.
My issue is about the following equation that I learned
$$\frac{P(v+dv_C)-P(v)}{dv_C}=\frac{P(v+\frac{dE_C}{mv})-P(v)}{\frac{dE_C}{mv}}$$
I can only accept this equation in the following cases
1. $P(v)$ is proportional to $v$
2. The equation should be rewritten such that the interval $\frac{dE}{mv}$ is adjusted to equal $dv_C$ or vice versa. For this, $dE$ is actually not a constant but varies with $v$ so that $\frac{dE}{mv}=dv_C$ holds for every $v$.

I learned case 1. is not necessary to make this equation valid. So is it actually case 2. or is my premise wrong in the first place?

 

[anuttarasammyak]

My issue arises from the premise that I learned; Integrals and derivatives are based on steps of a fixed constant interval. To make my issue clear, let's add a subscript "C" if an interval is considered constant.

I am afraid that derivatives and integrals are not based on constant interval or steps.

 

[Mark44]

Given is a function $P(E)$ and its derivative $f(E)$. Writing $E$ in terms of $v$ according to $E=\frac{1}{2}mv^2$ gives the derivative $g(v)=f(E)mv$ and $dE=mvdv$.

Your notation is confusing. If the function is P(E), its derivative could be written as P'(E) or $\frac{dP(E)}{dE}$. What you're calling g(v) appears to be $\frac{dP(E)}{dv} = P'(E) \frac{dE}{dv}$, using the chain rule. My P'(E) is your f(E), and $\frac{dE}{dv}$ is mv.

My issue arises from the premise that I learned; Integrals and derivatives are based on steps of a fixed constant interval. To make my issue clear, let's add a subscript "C" if an interval is considered constant.

No, that's not true (steps of a fixed constant interval). Both are defined in terms of limits as the step size decreases to zero.

When multiplying $f(E)$ by a fixed $dE_C$ I understand that
$$f(E)dE_C=P(v+\frac{dE_C}{mv})-P(v)$$
Furthermore, based on my premise, the derivative $g(v)$ is based on a fixed interval $dv_C$.
$$g(v)=\frac{P(v+dv_C)-P(v)}{dv_C}$$
Because the interval $\frac{dE_C}{mv}$ varies with $v$, this means $\frac{dE_C}{mv}\neq dv_C$ since $dv_C$ is considered constant.
My issue is about the following equation that I learned
$$\frac{P(v+dv_C)-P(v)}{dv_C}=\frac{P(v+\frac{dE_C}{mv})-P(v)}{\frac{dE_C}{mv}}$$

That is not how the derivative is defined - fixed intervals don't play a role at all. Your notations of $dE_C$ and $dv_C$ are incorrect and just add clutter. Following your notation, P'(v) is defined as
$$\lim_{dv \to 0}\frac{P(v + dv) - P(v)}{dv} $$

The above is the usual definition of the derivative of a function P(v), using the limit of the difference quotient. Take a look at any calculus textbook for their proof of the chain rule, using the definition of the derivative.

I can only accept this equation in the following cases
1. $P(v)$ is proportional to $v$
2. The equation should be rewritten such that the interval $\frac{dE}{mv}$ is adjusted to equal $dv_C$ or vice versa. For this, $dE$ is actually not a constant but varies with $v$ so that $\frac{dE}{mv}=dv_C$ holds for every $v$.

1. No. If P(v) is proportional to v, then P(v) = kv. The derivative has no such limitations.
2. $\frac{dE}{mv}$ doesn't make much sense in the context of differentiation.

I learned case 1. is not necessary to make this equation valid. So is it actually case 2. or is my premise wrong in the first place?

Yes, premise is wrong, based on a misunderstanding of what derivatives are (and antiderivatives, as well).

 

[JohnnyGui]

No, that's not true (steps of a fixed constant interval). Both are defined in terms of limits as the step size decreases to zero.

I have 2 questions to see if I'm grasping this.
1. So for example:
$$\int_a^bf(E)dE$$
It is wrong to say that during this integration, the integrand goes in infinitesimally small constant steps of $dE$?

2. For the following equation
$$\int_a^bf(E)dE=\int_{\sqrt{2a/m}}^{\sqrt{2b/m}}f(E(v))mvdv$$
Is it correct to say that $dE\neq mvdv$?

1. No. If P(v) is proportional to v, then P(v) = kv. The derivative has no such limitations.

That's not what meant. If P(v)=kv, then the derivative is a constant k, making it independent of the interval size that you're choosing. So if the interval $\frac{dE}{mv}\neq dv$, it would still give the same derivative when using either interval, making me able to accept the mentioned equation.

2. dEmv doesn't make much sense in the context of differentiation.

Aside from differentiation, I'm trying to understand the relationship between 2 infinitesimally small intervals using the substitution rule. If $dv=\frac{dE}{mv}$, why can't I substitute it by each other for the derivative formula?

 

[Mark44]

I have 2 questions to see if I'm grasping this.
1. So for example:
$$\int_a^bf(E)dE$$
It is wrong to say that during this integration, the integrand goes in infinitesimally small constant steps of $dE$?

Yes, that's wrong. In the definite integral, the main purpose of dE (in this instance) is to indicate what the variable of integration is. It's definitely wrong to say that "the integrand goes in infinitesimally small constant steps of $dE$."

Where you're probably confused is in how the definite integral (the Riemann integral) is defined, which is like this:
$$\int_a^bf(x)dx = \lim_{max \Delta x_k \to 0} \sum_{i = 1}^n f(x^*) \Delta x_k$$
(from https://mathworld.wolfram.com/Riema...de the confines of advanced mathematics texts.)
The interval [a, b] is divided up into subintervals: $[x_0, x_1], [x_1, x_2], \dots, [x_{n-1}, x_n]$, not necessarily the same size. $\Delta x_k$ is the length of the the k-th subinterval. It is NOT constant. Likely the "dx" part of the definite integral is a carryover from how the integral is defined as a Riemann sum.

2. For the following equation
$$\int_a^bf(E)dE=\int_{\sqrt{2a/m}}^{\sqrt{2b/m}}f(E(v))mvdv$$
Is it correct to say that $dE\neq mvdv$?

No. Since you have E as a function of v, i.e., $E = 1/2 * mv^2$, then $dE = mv*dv$.
It looks like you have done the substitution correctly, though.

That's not what meant. If P(v)=kv, then the derivative is a constant k, making it independent of the interval size that you're choosing. So if the interval $\frac{dE}{mv}\neq dv$, it would still give the same derivative when using either interval, making me able to accept the mentioned equation.

What you wrote in point 1 was this:

I can only accept this equation in the following cases
1. P(v) is proportional to v

Based on this, P(v) is necessarily equal to kv, for some number k. That's what "proportional to" means.
I couldn't make any sense out of your 2nd point, mostly because your notation was unclear due to a lot of extra variables and subscripts on differentials.

Aside from differentiation, I'm trying to understand the relationship between 2 infinitesimally small intervals using the substitution rule. If $dv=\frac{dE}{mv}$, why can't I substitute it by each other for the derivative formula?

What you're calling "intervals" are really differentials.
In the paragraph above, you wrote "If $dv=\frac{dE}{mv}$." This is the other relationship between dv and dE. (The other is $dE = mv dv$.)

 

[JohnnyGui]

(from https://mathworld.wolfram.com/Riema...de the confines of advanced mathematics texts.)
The interval [a, b] is divided up into subintervals: [x0,x1],[x1,x2],…,[xn−1,xn], not necessarily the same size. Δxk is the length of the the k-th subinterval. It is NOT constant. Likely the "dx" part of the definite integral is a carryover from how the integral is defined as a Riemann sum.

This explains it, thanks. One thing though; your link states that for the boundaries $(0,a)$ then $\Delta x_k=a/n=h$ which shows that $\Delta x_k$ only depends on the boundaries of the sum but is a constant $h$ during the Riemann summation. It even shows an example with $\Delta x_k$ staying constant at $f(x_1), f(x_2), f(x_3)$ (it goes in integer steps of $h$). How is $\Delta x_k$ then considered to vary during the summation?

No. Since you have E as a function of v, i.e., E=1/2∗mv2, then dE=mv∗dv.
It looks like you have done the substitution correctly, though.

Ok, this has brought me to my root question about the substitution rule:
If $dE=mvdv$, and $dE$ varies during integration, does that mean that $mvdv$ varies exactly the same way $dE$ does after each step of $dE$?

I couldn't make any sense out of your 2nd point, mostly because your notation was unclear due to a lot of extra variables and subscripts on differentials.

So when removing the subscripts, and knowing that $dv=\frac{dE}{mv}$, I can say that
$$\frac{P(v+dv)-P(v)}{dv}=\frac{P(v+\frac{dE}{mv})-P(v)}{\frac{dE}{mv}}$$
The varying $\frac{dE}{mv}$ is therefore always equal to the varying $dv$ at any value of $v$?

 

[vela]

This explains it, thanks. One thing though; your link states that for the boundaries $(0,a)$ then $\Delta x_k=a/n=h$ which shows that $\Delta x_k$ only depends on the boundaries of the sum but is a constant $h$ during the Riemann summation. It even shows an example with $\Delta x_k$ staying constant at $f(x_1), f(x_2), f(x_3)$ (it goes in integer steps of $h$). How is $\Delta x_k$ then considered to vary during the summation?

When the Riemann integral is first taught, the interval is divided into $n$ equal size pieces to keep thing simple. In the more general definition of a Riemann integral, the interval is divided up into $n$ pieces with the requirement being that the size of each piece goes to 0 in the limit $n \to \infty$. The size of the pieces don't have to be equal.

Note that the size of $\Delta x_k$ is fixed when doing the summation, but to get to the integral, you still have to take a limit. Obviously, as the number of pieces increases, the size of each piece gets smaller.

Ok, this has brought me to my root question about the substitution rule:
If $dE=mvdv$, and $dE$ varies during integration, does that mean that $mvdv$ varies exactly the same way $dE$ does after each step of $dE$?

Yes, that's what the equal sign tells you.

So when removing the subscripts, and knowing that $dv=\frac{dE}{mv}$, I can say that
$$\frac{P(v+dv)-P(v)}{dv}=\frac{P(v+\frac{dE}{mv})-P(v)}{\frac{dE}{mv}}$$
The varying $\frac{dE}{mv}$ is therefore always equal to the varying $dv$ at any value of $v$?

 

[Mark44]

So when removing the subscripts, and knowing that $dv=\frac{dE}{mv}$, I can say that
$$\frac{P(v+dv)-P(v)}{dv}=\frac{P(v+\frac{dE}{mv})-P(v)}{\frac{dE}{mv}}$$
The varying $\frac{dE}{mv}$ is therefore always equal to the varying $dv$ at any value of $v$?

I think @vela already answered your question here, because dE = mv dv. The bigger question is what good does it do to replace $dv$ by $\frac{dE}{mv}$ on the right side of the equation above?

Another point that should be emphasized is that for a given value of $\Delta x$ or $\Delta x_k$ in the Riemann sum, you get an approximation of the value of the integral. It's only when you take the limit of the Riemann sum (which forces $\Delta x$ or $\Delta x_k$ to zero), that you get the exact value of the integral.

 

[JohnnyGui]

When the Riemann integral is first taught, the interval is divided into n equal size pieces to keep thing simple. In the more general definition of a Riemann integral, the interval is divided up into n pieces with the requirement being that the size of each piece goes to 0 in the limit n→∞. The size of the pieces don't have to be equal.

Note that the size of Δxk is fixed when doing the summation, but to get to the integral, you still have to take a limit. Obviously, as the number of pieces increases, the size of each piece gets smaller.

Thanks, but if the interval is divided up into n equal pieces and in the general definition each piece goes to 0 in the limit $n \to \infty$, how does that make each piece different in size? Doesn't that merely mean that all equal pieces go to 0 to the same extent, making them still equal?

I think @vela already answered your question here, because dE = mv dv. The bigger question is why you would want to replace dv on the right side of the equation above.

Not that I want to replace it, I merely stumbled upon this equation and had trouble with it because I erroneously thought $\frac{dE}{mv}$ may not be always equal to $dv$ because constant intervals were playing a role. With my new understanding I wanted to make sure this equation can still be valid.

 

[vela]

Thanks, but if the interval is divided up into n equal pieces and in the general definition each piece goes to 0 in the limit $n \to \infty$, how does that make each piece different in size? Doesn't that merely mean that all equal pieces go to 0 to the same extent, making them still equal?

You missed the point. The interval has to be divided up into $n$ pieces, but those pieces don't have the be the same size. It's been a while, but if I recall correctly, the integral only exists if the Riemann sum doesn't depend on how you divide up the interval (other than the requirement that the size of all of the pieces goes to 0 with increasing $n$).

Not that I want to replace it, I merely stumbled upon this equation and had trouble with it because I erroneously thought $\frac{dE}{mv}$ may not be always equal to $dv$ because constant intervals were playing a role. With my new understanding I wanted to make sure this equation can still be valid.

I think I finally understand your question. If you have $dE = mv\,dv$, then equally sized chunks $\Delta v$ wouldn't translate to equal size chunks $\Delta E$. If you assume the Riemann sum depends on the interval being broken up into equally sized pieces, then it looks like changing variables will cause a problem when integrating with respect to $E$. Is that what you were thinking?

 

[JohnnyGui]

You missed the point. The interval has to be divided up into n pieces, but those pieces don't have the be the same size. It's been a while, but if I recall correctly, the integral only exists if the Riemann sum doesn't depend on how you divide up the interval (other than the requirement that the size of all of the pieces goes to 0 with increasing n).

Okay, so the integral emerges if the pieces go to 0, whether the pieces are equal or not? (based on your quote "doesn't depend on how you divide up the interval")
This confuses me a bit because in the first replies I was told that the pieces are not constant in any way, but now it is said that it doesn't matter whether they're constant or not, as long as they approach 0. This makes me think I am allowed to think that an integral goes in constant fixed steps because it doesn't matter for the outcome.

I think I finally understand your question. If you have dE=mvdv, then equally sized chunks Δv wouldn't translate to equal size chunks ΔE. If you assume the Riemann sum depends on the interval being broken up into equally sized pieces, then it looks like changing variables will cause a problem when integrating with respect to E. Is that what you were thinking?

Yes! This caused me a huge confusion about how substituting variables in integrals would still give the same outcome, since equal sized $dE$ chunks would not equal $mvdv$ if the $dv$'s are also considered equal sized.

 

[Mark44]

Okay, so the integral emerges if the pieces go to 0, whether the pieces are equal or not? (based on your quote "doesn't depend on how you divide up the interval")

Yes. In the definition I wrote back in post #5, the limit is as the maximum of the $x_k$ goes to zero. By implication, this forces all smaller subintervals to go to zero as well.

This confuses me a bit because in the first replies I was told that the pieces are not constant in any way, but now it is said that it doesn't matter whether they're constant or not, as long as they approach 0.

In the summation (before the limit is taken) the subintervals can be equal or unequal in size. As far as the summation (only) is concerned, the subintervals are fixed in size (they can be equal but don't have to be), but in the limit (which is the actual definition of the definite integral) none of the subintervals is constant.

This makes me think I am allowed to think that an integral goes in constant fixed steps because it doesn't matter for the outcome.

You're continuing to ignore the fact that the integral is a limit. The limit causes all the subintervals, again equal in length or not, to get smaller. This means that what you're saying about constant fixed steps is incorrect.

 

[vela]

Okay, so the integral emerges if the pieces go to 0, whether the pieces are equal or not? (based on your quote "doesn't depend on how you divide up the interval")
This confuses me a bit because in the first replies I was told that the pieces are not constant in any way, but now it is said that it doesn't matter whether they're constant or not, as long as they approach 0. This makes me think I am allowed to think that an integral goes in constant fixed steps because it doesn't matter for the outcome.

You do need to differentiate between the Riemann sum, which has a finite number of terms, and the Riemann integral, which is a limit. It doesn't really make sense to say the integral "goes in constant fixed steps." For the sum, you can assume the $\Delta x$'s all have the same size as that's a perfectly good division of the interval, but I fail to see the usefulness of this assumption when it comes to the integral. Perhaps you can elaborate on your thinking regarding this.

Yes! This caused me a huge confusion about how substituting variables in integrals would still give the same outcome, since equal sized $dE$ chunks would not equal $mvdv$ if the $dv$'s are also considered equal sized.

You just need to jettison the idea that $dv$ or $dE$ has to be equally sized.

You might want to consult a book on real analysis where Riemann integrals are defined more generally than in an introductory calculus book.

 

[JohnnyGui]

Thanks @Mark44 and @vela

You do need to differentiate between the Riemann sum, which has a finite number of terms, and the Riemann integral, which is a limit. It doesn't really make sense to say the integral "goes in constant fixed steps."

1. So based on the above replies, should I also say that it doesn't make sense either to say an integration goes in varying steps of dE (=mvdv) since integration is all about a limit where all steps go to 0?

2. If yes, does that mean that mvdv=dE stays close to zero to the same extent, even when $v$ goes very large?

 

[vela]

1. So based on the above replies, should I also say that it doesn't make sense either to say an integration goes in varying steps of dE (=mvdv) since integration is all about a limit where all steps go to 0?

Right.

2. If yes, does that mean that mvdv=dE stays close to zero to the same extent, even when $v$ goes very large?

Yes. $v$ is always finite, so because $dv$ is infinitesimal, $dE$ will be as well.

 

[JohnnyGui]

Yes. v is always finite, so because dv is infinitesimal, dE will be as well.

Thank you. I think there's a question left until I can fully grasp this.

1. If dE and dv are both infinitesimally small, and yet $dE\neq dv$, does that mean there are different "gradations" of being infinitesimally small?

2. If yes, I reckon the factor by which they differ is finite because $\frac{dE}{dv}=mv$. Is such a finite emergence actually possible in the realm of infinitesimally small values? Or would that mean that dE and dv are still finite to some extent?

 

[Mark44]

Thank you. I think there's a question left until I can fully grasp this.

1. If dE and dv are both infinitesimally small, and yet $dE\neq dv$, does that mean there are different "gradations" of being infinitesimally small?

2. If yes, I reckon the factor by which they differ is finite because $\frac{dE}{dv}=mv$. Is such a finite emergence actually possible in the realm of infinitesimally small values? Or would that mean that dE and dv are still finite to some extent?

Yes to both.

 

[JohnnyGui]

@vela

Yes to both.

Thanks for confirming.

Sorry, I have one final question about how boundaries relates to an infinitesimal interval for the Riemann sum.

Suppose a lower boundary $a$, which for the Riemann sum means the starting point with $k=1$ is $x_k=x_1=a$ And suppose making the upper boundary $n$ so small that $n=1$, simplifying the Riemann sum to just one 1 step, thus:
$$\lim_{max \Delta x_k \to 0} \sum_{k=1}^1 f(x_k^*) \Delta x_k$$
Would this Riemann sum equal $f(a)dx$ or $f(a)da$ if $\Delta x$ approaches 0? Is it $f(a)da$ purely because the step size $\Delta x_k$ is also denoted by $k$ such that $\Delta x_1=\Delta a$, thus giving $da$ when it approaches 0?

 

[Mark44]

@vela

Thanks for confirming.

Sorry, I have one final question about how boundaries relates to an infinitesimal interval for the Riemann sum.

Suppose a lower boundary $a$, which for the Riemann sum means the starting point with $k=1$ is $x_k=x_1=a$ And suppose making the upper boundary $n$ so small that $n=1$, simplifying the Riemann sum to just one 1 step, thus:
$$\lim_{max \Delta x_k \to 0} \sum_{k=1}^1 f(x_k^*) \Delta x_k$$
Would this Riemann sum equal $f(a)dx$ or $f(a)da$ if $\Delta x$ approaches 0? Is it $f(a)da$ purely because the step size $\Delta x_k$ is also denoted by $k$ such that $\Delta x_1=\Delta a$, thus giving $da$ when it approaches 0?

While a is the lower boundary, n is not a boundary at all -- it's the number of subintervals that make up the interval [a, b]. If the Riemann sum consists of just a single subinterval, $\Delta x = b - a$. So $\Delta x$ can't decrease in size, and taking the limit doesn't make any sense.
Essentially what you're doing is approximating $\int_a^b f(x) ~dx$ by $f(x^*)\Delta x$, where $f(x^*)$ is the function value at some number $x^* \in [a, b]$. Probably not a very good approximation unless the function is a constant function.
Also $f(a)da$ doesn't make sense because you're treating a (and therefore da) as variables.

 

[JohnnyGui]

While a is the lower boundary, n is not a boundary at all -- it's the number of subintervals that make up the interval [a, b]. If the Riemann sum consists of just a single subinterval, $\Delta x = b - a$. So $\Delta x$ can't decrease in size, and taking the limit doesn't make any sense.
Essentially what you're doing is approximating $\int_a^b f(x) ~dx$ by $f(x^*)\Delta x$, where $f(x^*)$ is the function value at some number $x^* \in [a, b]$. Probably not a very good approximation unless the function is a constant function.
Also $f(a)da$ doesn't make sense because you're treating a (and therefore da) as variables.

Ah, I forgot about the $n$ definition.

What if $a$ is actually a variable (I'm deliberately not choosing the variable $x$ in the boundaries because it's already defined in the integrand), would then..
$$\int_a^{a+da}f(x)dx=f(a)da$$
..make any sense?

It's a bit of an analogue to the known equation
$$\frac{d}{dx}\int_a^xf(t)dt=f(x)$$
Rewriting the numerator gives
$$d\bigg(\int_a^xf(t)dt \bigg) = \int_x^{x+dx}f(t)dt=f(x)dx$$

 

[Mark44]

Ah, I forgot about the $n$ definition.

What if $a$ is actually a variable (I'm deliberately not choosing the variable $x$ in the boundaries because it's already defined in the integrand), would then..
$$\int_a^{a+da}f(x)dx=f(a)da$$
..make any sense?

Not really. The da term is presumably infinitesimally small, so the interval of integration [a, a + da] would be of length zero.

EDIT

Thinking about it, perhaps it should be:
$$\int_a^{a+da}f(x)dx=g(a(x))dx$$
Where $g(a(x))=f(a(x))\cdot a'(x)$.

No, I don't see how that makes any sense, either.

 

[JohnnyGui]

No, I don't see how that makes any sense, either.

Sorry I edited my post again right after you replied. If it doesn't make sense, how is the following known equation possible?
$$\frac{d}{dx}\int_0^xf(t)dt=f(x)$$
Because rewriting the numerator gives
$$d\bigg(\int_c^xf(t)dt \bigg) = \int_c^xf(t)dt - \int_c^{x+dx}f(t)dt= \int_x^{x+dx}f(t)dt=f(x)dx$$
Which is an analogue to what I wrote (substituting $x$ by $a$ and $t$ by $x$)

 

[Mark44]

Sorry I edited my post again right after you replied. If it doesn't make sense, how is the following known equation possible?
$$\frac{d}{dx}\int_0^xf(t)dt=f(x)$$

This is part of the Fundamental Theorem of Calculus, for which a proof is given in most calculus textbooks.

Because rewriting the numerator gives
$$d\bigg(\int_c^xf(t)dt \bigg) = \int_c^xf(t)dt - \int_c^{x+dx}f(t)dt= \int_x^{x+dx}f(t)dt=f(x)dx$$
Which is an analogue to what I wrote (substituting $x$ by $a$ and $t$ by $x$)

I agree with the outer two expressions - $$d\bigg(\int_c^xf(t)dt \bigg) =f(x)dx$$

but not the two in the middle. In the left-most integral you're taking the differential of a definite integral, but you don't show that differential operating on either of the middle two integrals, nor does dx appear in either of the two middle integrals.

The middle two integrals should be like this:
$$\int_c^{x + dx}f(t)dt - \int_c^xf(t)dt= \int_x^{x+dx}f(t)dt$$
and have no relation to $d\bigg(\int_c^xf(t)dt \bigg)$

Also, I don't recall seeing a differential term as part of one of the limits of integration. If it made sense to do this, your third integral, $\int_x^{x+dx}f(t)dt$, would be zero.

 

[JohnnyGui]

The middle two integrals should be like this:
$$\int_c^{x + dx}f(t)dt - \int_c^xf(t)dt= \int_x^{x+dx}f(t)dt$$

My bad, I intended to write it like you did but I accidentally switched the two middle integrals.

but you don't show that differential operating on either of the middle two integrals, nor does dx appear in either of the two middle integrals

and have no relation to $d\bigg(\int_c^xf(t)dt \bigg)$

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 where the $x$ stands?

Also, I don't recall seeing a differential term as part of one of the limits of integration. If it made sense to do this, your third integral, ∫xx+dxf(t)dt, would be zero.

Wouldn't that be correct because it equals $f(x)dx$ which is also zero? (we've agreed that $dx$ is infinitesimally small).

 

[Mark44]

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 where the $x$ stands?

The definition of the differential of a function f, as I recall it is this:
$df(x) = f'(x)dx$
So $d\left(\int_a^x f(t)dt\right) = \left(\frac d{dx} \int_a^x f(t)dt \right) dx = f(x) dx$
But like I said, I don't see that integrating between x and x + dx makes any sense, which is what the middle integrals you wrote are doing.

Wouldn't that be correct because it equals $f(x)dx$ which is also zero? (we've agreed that $dx$ is infinitesimally small).

Sure. I've already agreed that $d\int_a^x f(t)dt = f(x)dx$, but my quarrel is with the integrals $\int_c^{x + dx}f(t)dt - \int_c^xf(t)dt$ and $\int_x^{x+dx}f(t)dt$.
I don't believe that having dx in the integration limits makes sense.

 

[vela]

Also, I don't recall seeing a differential term as part of one of the limits of integration. If it made sense to do this, your third integral, $\int_x^{x+dx}f(t)dt$, would be zero.

I've seen it before, for example, in the context of probability.
$$P(x<X<x+dx) = \int_x^{x+dx} f_X(x)\,dx=f_X(x)dx$$

 

[JohnnyGui]

I've seen it before, for example, in the context of probability.
$$P(x<X<x+dx) = \int_x^{x+dx} f_X(x)\,dx=f_X(x)dx$$

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 this corresponds to 1 step of a Riemann summation of $f(t_k)$ where $t_1=x$, thus
$$\lim_{max \Delta t_k \to 0} \sum_{k=1}^1 f(t_k^*) \Delta t_k =\lim_{max \Delta t_k \to 0}f(x)\Delta t$$
But I don't see how the $\Delta t$ should transform into $\Delta x$.

 

[vela]

Is it actually possible to correspond a Riemann summation for an integral like this?

I don't think so, but why would you want to?

 

[JohnnyGui]

I don't think so, but why would you want to?

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 $dx$?