Converting partial derivative to ordinary in an integral

[phys_student1]

Hi,

I find my professor doing this a lot of times, here is it:

∫{ ∂(f[x])/∂x } dx = ∫d(f[x])

How is that possible?

 

[bigfooted]

Integration is the inverse of differentiation and vice versa.
Suppose you have a function f(x).
Take the derivative of the function: d/dx(f(x)) = df(x)/dx.
By taking the 'antiderivative' or the integral, you end up with f(x) again, which is exactly what your professor is doing.

For the rest, there is not really a difference between a partial and an ordinary differentiation operator, except that the ordinary differentiation just means that there is only one independent variable x, and with partial differentiation you can have more.

 

[phys_student1]

Thanks, but...

That's exactly my question. You see, the idea of having a partial derivative in itself means
that f is not only a function of (x) but also of another variable.

That's why I am wondering how that operator was changed to ordinary derivative.

Is it possibly because only in this integral we are interested in f as a function of x only, and
for that reason we can, only here, change the partial to ordinary derivative ?

 

[HallsofIvy]

The partial derivative of a function of several variables treats the other variables as constants so
$$\int \frac{\partial f(x,y,z,...)}{\partial x}dx= \int df(x,y,z,...)$$
follows from
$$\int \frac{df}{dx}dx= \int df$$
for functions of a single variable.

 

[phys_student1]

The partial derivative of a function of several variables treats the other variables as constants so
$$\int \frac{\partial f(x,y,z,...)}{\partial x}dx= \int df(x,y,z,...)$$
follows from
$$\int \frac{df}{dx}dx= \int df$$
for functions of a single variable.

Thanks,

So, because we are interested only in x, then the partial, in this case, is equivalent to the ordinary derivative, hence the change.

Thanks everyone for your help, appreciated :)