# $f:[0,1] \to \mathbb R$ be a differentiable function with bounded derivative, then is $\int_0^1 f'(x)=f(1)-f(0)$?

Let $$f:[0,1] \to \mathbb R$$ be a differentiable function such that $$\sup_{x\in [0,1]} |f'(x)|$$ is finite. Then since $$f'(x)=\lim_{n\to \infty} \dfrac{f(x+1/n)-f(x)}{1/n}$$, so $$f'(x)$$ is measurable and also the Lebesgue integral $$\int_0^1|f'(x)|dx$$ is finite, thus $$f' \in L^1([0,1])$$.

My question is, is it true that $$\int_0^1 f'(x)=f(1)-f(0)$$ ?

Note that fundamental theorem of calculus does not apply here since $$f'(x)$$ is not continuous (not even known to be Riemann integrable)

• Use Barrow rule – Tito Eliatron Nov 4 '18 at 22:55
• @TitoEliatron: what is that ? – user521337 Nov 4 '18 at 22:57
• Fundamental Theorem of Calculus en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Corollary In Spanish it is known as Barrow's Rule. – Tito Eliatron Nov 4 '18 at 22:58
• @TitoEliatron: see my question again – user521337 Nov 4 '18 at 22:59
• @TitoEliatron: again ... we do not know that $f'$ is Riemann integrable ... we only know it is Lebesgue integrable ... – user521337 Nov 4 '18 at 23:02

Let $${[a,b]}$$ be a compact interval of positive length, let $${F: [a,b] \rightarrow {\bf R}}$$ be a differentiable function, such that $${F'}$$ is absolutely integrable. Then the Lebesgue integral $${\int_{[a,b]} F'(x)\ dx}$$ of $${F'}$$ is equal to $${F(b) - F(a)}$$.