Integral Syntax Question

konik

Is the following syntax correct?

[tex] dx = v\ dt [/tex]
[tex] x = \int v\ dt [/tex]

or should it be:

[tex] dx = v \ dt[/tex]
[tex] dx = \int v \ dt [/tex]

Doc Al

This is OK. Realize that this is just saying that:
[tex] \int \ dx = x [/tex]

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:
[itex]\int x= \int v dt[/itex]. Since [itex]\int dx= x[/itex] the result is [itex]x= \int v dt[/itex].

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

Gib Z

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 [itex]x^2[/itex] is [itex]\int x^2 dx[/itex] which is, itself, equal to [itex](1/3)x^3+ C[/itex]. It is only in the last that we need the "C".

CompuChip

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