Second fundamental theorem of calculus.

[operationsres]

Let f(x) be a non-stochastic mapping f: \mathbb{R} \to \mathbb{R}. The second fundamental theorem of calculus states that:

\frac{d}{dx} \int_a^x f(s)ds = f(x).
*QUESTION 1* Is the following true?

\frac{d}{dx} \int_x^a f(s)ds = f(x).

*QUESTION 2* Related to this, how can I evaluate/simplify/express:

d\int_x^a f(s)ds.

 

[Kreizhn]

For the first question, what happens to the integral when you reverse the bounds on integration? This should tell you that your answer is not quite correct, modulo only a sign error.

The next answer depends on your context and there are a few ways to do this. The first is to simply use the definition of the exterior derivative. Assuming you are just working on \mathbb R you have that for any function (0-form) F, then in a coordinate chart you have
dF = \frac{\partial F}{\partial x} dx.
Alternatively, if you let \gamma: [0,1] \to \mathbb R be a smooth path such that \gamma(0) = a, \gamma'(0) = v then
d_a F(v) = \left. \frac{d}{dt} \right|_{t=0} (F \circ \gamma).
Both will give you the same solution, the the latter is coordinate independent.

 

[operationsres]

For the first question, what happens to the integral when you reverse the bounds on integration? This should tell you that your answer is not quite correct, modulo only a sign error.

The next answer depends on your context and there are a few ways to do this. The first is to simply use the definition of the exterior derivative. Assuming you are just working on \mathbb R you have that for any function (0-form) F, then in a coordinate chart you have
dF = \frac{\partial F}{\partial x} dx.
Alternatively, if you let \gamma: [0,1] \to \mathbb R be a smooth path such that \gamma(0) = a, \gamma'(0) = v then
d_a F(v) = \left. \frac d{dt} \right|_{t=0} (F\circ \gamma).
Both will give you the same solution, the the latter is coordinate independent.

Hi,

I'm not sure what "modulo" means?

It would make sense if the solution to QUESTION1 is actually -f(x), is that what you were saying?

 

[HallsofIvy]

"modulo a sign error" meant "correct except possibly a sign error".

Yes, \int_x^a f(t)dt= -\int_a^x f(t)dt and then apply the fundamental theorem.