First Fundamental Theorem of Calculus

Main Question or Discussion Point

Hi,

I just learned about the First Fundamental theorem of calculus. From my understanding, it talks specifically about definite integrals. I was wondering if there is any sort of theorem that proves that the derivative of the indefinite integral of a function is equal to the function itself.

Me too,

I have learned Differential Calculus recently, And then integral Calculus.

I am not sure if there is a theorem because we derive other equations from calculus itself.

I am not sure Peter, but i will get back to you after i figure it out.

WannabeNewton
That is the very definition of the indefinite integral.

lurflurf
Homework Helper
The definition of indefinite integral is the derivative of the indefinite integral of a function is equal to the function itself. The fundamental theorem establishes that definite and indefinite integrals are almost the same thing. Which is why they have very similar notation. this causes confusion at times.

Thank you very much guys! I hope this is useful to you too, utkarshraj!

Bacle2
But there are some subtleties to consider too: Not every (differentiable) function can be recovered from its pointwise derivative . The Cantor function f is not the integral of its derivative, since f is a.e. constant , so you do not recover the integral from its pointwise derivative --since f' is a.e. 0, so that:

∫f' =∫0 * =0 ≠ f , but f is not constant.

* f' is a.e. 0 but not actually 0 .

http://en.wikipedia.org/wiki/Cantor_function

But there are some subtleties to consider too: Not every (differentiable) function can be recovered from its pointwise derivative . The Cantor function f is not the integral of its derivative, since f is a.e. constant , so you do not recover the integral from its pointwise derivative --since f' is a.e. 0, so that:

∫f' =∫0 * =0 ≠ f , but f is not constant.

* f' is a.e. 0 but not actually 0 .

http://en.wikipedia.org/wiki/Cantor_function
OK, but the Cantor function is not even differentiable. It's only a.e. differentiable.

Bacle2
Right, but the point is that it is Riemann-integrable but does not have an antiderivative, so you
work a.e.

Right, but the point is that it is Riemann-integrable but does not have an antiderivative, so you
work a.e.
Why wouldn't it have an antiderivative?

Bacle2
By post #6 and the fact that integrals of a.e functions are equal, there is no F with F'=f , where f is the Cantor function.

By post #6 and the fact that integrals of a.e functions are equal, there is no F with F'=f , where f is the Cantor function.
Since $f$ is continuous, it seems to me that $\int_0^x f$ would do the trick...

Bacle2
∫f'=∫0=0 , where the integral is over [0,1]

∫f'=∫0=0 , where the integral is over [0,1]
That's true. So why does that imply that $f$ has no antiderivative?

Bacle2
I absolutely agree that the cantor function isn't absolutely continuous.

But why does that imply that it doesn't have an antiderivative. Since the Cantor function $f$ is continuous, doesn't the fundamental theorem of calculus imply that $\int_0^x f$ is an antiderviative?

pwsnafu
But why does that imply that it doesn't have an antiderivative. Since the Cantor function $f$ is continuous, doesn't the fundamental theorem of calculus imply that $\int_0^x f$ is an antiderviative?
Further, it's a standard exercise in measure theory to show that for any $f \in L^1[0,1]$, the antiderivative $\int_{[0,x]} f$ is absolutely continuous. This of course is just a part of Fundamental Theorem of Lebesgue integration.