# Second fundamental theorem of calculus.

1. Nov 13, 2012

### 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:

$d\int_x^a f(s)ds$.

2. Nov 13, 2012

### 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.

Last edited by a moderator: Nov 14, 2012
3. Nov 13, 2012

### operationsres

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?

4. Nov 14, 2012

### 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.