Is the Integral Syntax Correct for dx = v dt and x = \int v dt?

[konik]

Is the following syntax correct?

dx = v\ dt
x = \int v\ dt

or should it be:

dx = v \ dt
dx = \int v \ dt

 

[Doc Al]

Is the following syntax correct?

dx = v\ dt
x = \int v\ dt

This is OK. Realize that this is just saying that:
\int \ dx = x

or should it be:

dx = v \ dt
dx = \int v \ dt

That's no good--you must integrate both sides.

 

[HallsofIvy]

The first. Obviously, the two right sides of the second are not the same and cannot both be equal to dx.

What you are doing is starting with dx= v dt and integrating both sides:
\int x= \int v dt. Since \int dx= x the result is x= \int v dt.

(The integral is not "well defined" so that should be x= \int v dt+ C.)
(Once again, Doc Al comes in 2 seconds ahead of me!)

 

[Gib Z]

The integral is not "well defined" so that should be x= \int v dt+ C

We don't really need to include the additional C, since indefinite integrals are only unique up to a additive constant anyway.

 

[HallsofIvy]

Yes, of course. The anti-derivative of x^2 is \int x^2 dx which is, itself, equal to (1/3)x^3+ C. It is only in the last that we need the "C".

 

[CompuChip]

Actually, I prefer to think of
\int v \, dt
as notation where the \int dt is a single symbol.
The equation
dx = v dt
wouldn't make sense then but can be written
dx/dt = v
or considered as a limit.

(Of course, I also use them as mnemonics and manipulate them as ordinary fractions, but sometimes it's good to keep things clear and separate legal operations from convenient notation).