Integral of the derivative function

Question

Let $ f:[a,b] \to \ \mathbb{R} $ function that continuous on $[a,b]$ and derivable on $(a,b)$. Show that if $f'$ (that isn't necessarily defined in the edges of $[a,b]$) is Riemann-integrable, then ${ \int_{a}^{b}{ f'\left(x\right) \,dx} = f(b) - f(a) }$

I was able to prove it by showing that for every $\epsilon > 0$ there exist $\delta > 0$ so that for every partition $p$ that satisfies $\lambda(p) < \delta $ and for every choose of $ x_{i-1} \le c_i \le x_i \Rightarrow \left| \sum_{i=1}^{n}f(c_i)\triangle {x_i} - I \right| < \epsilon $.

However, I would really like to prove it with Darboux sums. Is there an elegant way to prove it?

Answer 1

Well if you wish to talk about the symbol $\int_{a} ^{b} f'(x) \, dx$ then you must have $f'$ defined on $[a, b] $. What you probably wish to know is this:

Let $f:[a, b] \to \mathbb {R} $ be continuous on $[a, b] $ and differentiable on $(a, b) $. Further let $g:[a, b] \to\mathbb {R} $ be such that $g$ is Riemann integrable on $[a, b] $ and $g(x) =f'(x) \, \forall x\in (a, b) $ then $$\int_{a} ^{b} g(x) \, dx=f(b) - f(a) $$

This is true and usually called second part of Fundamental Theorem of Calculus and it is proved via mean value theorem. The proof uses Riemann sums and one can show that given any partition $P$ there is a Riemann sum over $P$ which equals $f(b) - f(a) $.

The proof using Darboux sums can be given but only by showing that Darboux sums can be approximated well enough by Riemann sums.