What Does Integrating the Derivative \(\int \frac{d}{dx} dx\) Equal?

[autobot.d]

Just wondering what this is

$\int\frac{d}{dx}dx$

What does this equal? Is it even allowed. I was thinking it is equal to identity,
which in my case is 1.
Is it equivalent to

$\int\frac{dx}{dx}$



??

Thanks!

 

[micromass]

It makes no sense. The notation $\frac{d}{dx}$ is just the notation for taking the derivative of something. What you're writing

$$\int\frac{d}{dx}dx$$

is taking the integral of a notation. It's not defined.

However,

$$\int \frac{df}{dx}dx$$

IS defined: it is the integral of a function. The integral is equal to f (plus a constant).

 

[autobot.d]

What if I did something like this
$\int\left(\frac{d\left(\int^{a\left(x\right)}_{0}f\left(x,y\right)dy\right)}{dx}\right)dx$
does this make sense?

 

[autobot.d]

I am saying that this will give me

$\int^{a\left(x\right)}_{0}f\left(x,y\right)dy$

 

[HallsofIvy]

What if I did something like this
$\int\left(\frac{d\left(\int^{a\left(x\right)}_{0}f\left(x,y\right)dy\right)}{dx}\right)dx$
does this make sense?

I am saying that this will give me

$\int^{a\left(x\right)}_{0}f\left(x,y\right)dy$

No, it wouldn't. For one thing, that integral is a definite integral, a number, while your first integral will be a function of x.

Leibniz's rule:
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,y)dy= f(x, \beta(x))\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x} dy$$

In this particular example, the integral of the derivative would be
$$\int \left(f(x, a(x))\frac{da}{dx}\right)dx+ \int \int_a^{a(x)} \frac{\partial f}{\partial x} dy dx$$

 

[autobot.d]

Thanks for the help!

 

[paulfr]

Integration of a Derivative and
Differentiation of an Integral
both result in the original function
AND
both allude to
The Fundamental Theorem of Calculus
That is that Integration and Differentiation are Inverse Functions

 

[Bacle]

Actually, paulfr, I think it may be a good idea to be more careful:

I don't know how you would define integrals and derivatives as functions;

are you referring to definite, or indefinite integrals? If your integral is indefinite,

then the two processes cannot be inverses of each other, because the indefinite

integral of f' is f+C, for C real.

You also need to state that

f must be a.e continuous (or , having at-most countably-many discontinuities ) for

f' to be defined. The conditions for the FT Calculus for Lebesgue Integration is

a little different; I think f being absolutely-continuous is sufficient, but I think it

can be weakened.

 

[paulfr]

Bacle
My statements are not meant to be taken in a strict Mathematical sense with all conditions stated.
I was trying to simplify for the student.

But it is true that ... informally ...
For a function to be Differentiable, continuity is necessary but NOT sufficient.
For a function to be Integrable, continuity IS sufficient but not necessary.

In general though, as I said, The Fundamental Rule of Calculus is
Differentiation and Integration are Inverse Operations/Functions/Processes
You can find this in any Calculus text.

 

[Bacle]

Yes, Paulfr, I guess it is difficult for me to take off my Mathematical hat

and not address every possible case at times. Still, it is not too clear to me the level

of rigor that autobot.d wanted, so I just tried to complement/expand a bit , just

in case the OP wanted some more.