Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

TLDR version:

Is it valid to say that $\int\mathrm{d}A\mathrm{d}t = A + c$ ?

Extended version:

I am having trouble to understand exactly how the integration works when I integrate an expression that has differential inside.

Suppose I have this basic physical equation $\mathrm{d}v = a\mathrm{d}t$

Now in every material I've seen the next step is $\int_{v(t_0)}^{v(t)}\mathrm{d}v = \int_{t_0}^{t}a\mathrm{d}t$ But what are the intermediate steps here? Because I would imagine this as the first step: $\int_{t_0}^{t}\mathrm{d}v\mathrm{d}t = \int_{t_0}^{t}a\mathrm{d}t\mathrm{d}t$. Now look at the right side, the left side can use the same trick.

Suppose $A$ is a primitive function of $a$. Then this equation should be true $\frac{\mathrm{d}A}{\mathrm{d}t} = a$ from this we get $\mathrm{d}A = a\mathrm{d}t$ . The integral should be $\int_{t_0}^{t}\mathrm{d}A\mathrm{d}t$.

Now is it valid to say that $\int\mathrm{d}A\mathrm{d}t = A + c$ ?

share|improve this question
1  
You surely meant $\frac{\text{d}v}{\text{d}t} {\text{d}t}$ rather than ${\text{d}v}{\text{d}t}$. –  Hurkyl Mar 9 '13 at 18:04
    
Actually I didn't. I was just applying the rule that when I integrate function $f$ by $x$ I write $\int{f\mathrm{d}x}$ now just substitute $f$ with $\mathrm{d}v$ –  Honza Brabec Mar 9 '13 at 18:09
    
Ah, there's your problem then. When one says "integrate $f$ with respect to $x$", they secretly mean "multiply $f$ by the differential form $\text{d}x$ to get another differential form, and then integrate that"; that's why the $\text{d}x$ appears. –  Hurkyl Mar 9 '13 at 18:13
    
So that means that when the expression is already in the differential form I don't need to multiply right? Because that is exactly the root of the problem I didn't know about :) –  Honza Brabec Mar 9 '13 at 18:15
    
Exactly right.. –  Hurkyl Mar 9 '13 at 18:16
show 2 more comments

1 Answer 1

up vote 2 down vote accepted

There is no intermediate step: by the hypothesis, $\text{d}v = a \text{d}t$, and both integrals

$$ \int_{v_0}^{v_f} \text{d}v = \int_{t_0}^{t_f} a \, \text{d}t $$

indicate the same path from the point where $v = v_0$ and $t = t_0$ to the point where $v = v_f$ and $t = t_f$ and have the same integrand, and so therefore are the same integral. (because we're in one dimension, only the two endpoints of the path matter)


When working in the style of notation where you work with variables that depend functionally on one another rather than working with functions, the differential form becomes the correct way to formulate things. A vague, heuristic definition is that a differential form is simply something you integrate along a path ('real' definitions generally don't come until you start studying differential geometry, which is unfortunately much more complicated than you actually need to work with the issue at hand).

So if $\omega$ is a differential form, then $\int_a^b \omega$ is something meaningful (assuming there is enough context to infer what path $a$ and $b$ indicate). However, $\int_a^b v$ doesn't make sense!


Contrast the above with the way integrals are formulated in the alternative style where one works with functions, where writing $\int_a^b f$ does make sense.

The connection between the two approaches to integral calculus is through identity

$$ \int_a^b f = \int_a^b f(x) \, \text{d}x $$

There is a common abuse of notation $v=v(t)$ that conflates the two styles which can sometimes create some confusion here; when this abuse is present, you have to take some extra care to infer what is really meant.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.