# Proof of first Fundamental theorem of calculus

Can you please, check if my proof is correct?

Suppose that $$f:[a,b]\to \Bbb{R}$$ is continuous and $$F(x)=\int^{x}_{a}f(t)dt$$, then $$F\in C^{1}[a,b]$$ and $$\dfrac{d}{dx}\int^{x}_{a}f(t)dt:=F'(x)=f(x)$$

MY PROOF: Credits to Aweygan for the correction

Let $$x_0\in[a,b]$$ and $$\epsilon>0$$ be given. Since $$f$$ is continuous at $$x_0$$ then, there exists $$\delta>0$$ such that $$|t-x_0|<\delta$$ implies $$|f(t)-f(x_0)|<\epsilon.$$ Thus, $$f(x_0)=\dfrac{1}{x-x_0}\int^{x}_{x_0}f(x_0)dt,\;\;\text{where}\;\;x\neq x_0.$$ For any $$x\in (a,b),$$ with $$0<|x-x_0|<\delta,$$ such that $$x_1=\min\{x,x_0\}$$ and $$x_2=\max\{x,x_0\}$$. So, we have \begin{align}\left| \dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) \right|&= \left| \dfrac{1}{x-x_0}\int^{x}_{x_0}(f(t)-f(x_0))dt \right| \\&\leq \dfrac{1}{|x-x_0|}\int^{x}_{x_0} \left|f(t)-f(x_0) \right|dt\\&\leq \dfrac{1}{|x-x_0|}\int^{x_2}_{x_1} \left|f(t)-f(x_0) \right|dt\\&< \dfrac{1}{|x-x_0|}\epsilon|x_1-x_2| \\&\leq \dfrac{1}{|x-x_0|}\epsilon|x-x_0| =\epsilon \end{align} Hence, $$F\in C^{1}[a,b]\;\;\text{and}\;\;\dfrac{d}{dx}\int^{x}_{a}f(t)dt:=F'(x)=f(x)$$

• It's not sufficient to prove the limit is $f(x)$ only from one side. Jan 14 '19 at 13:33
• You still only have differentiability from the right, you need to show differentiability from the left (i.e. when $-\delta<h<0$). Jan 14 '19 at 13:47
• You're still computing a one-sided limit. Jan 14 '19 at 13:49
• It's still not quite correct. I'm all but certain you'll need to handle the two cases $h>0$ and $h<0$ separately, or introduce some new variables to avoid this. Jan 14 '19 at 13:55
• @Aweygan: I want to rewrite it, now. Some moments, please! Jan 14 '19 at 13:56

It's essentially correct, but you should either split up the last part of the proof into the cases where $$x and $$x_0, or write $$x_1=\min\{x,x_0\}$$, $$x_2=\max\{x,x_0\}$$ and do the following:
\begin{align} \left| \dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) \right|&= \left| \dfrac{1}{x-x_0}\int^{x}_{x_0}(f(t)-f(x_0))dt \right| \\ &\leq \dfrac{1}{|x-x_0|}\int^{x_2}_{x_1} \left|f(t)-f(x_0) \right|dt \\&< \dfrac{1}{|x-x_0|}\epsilon|x-x_0| \\ &=\epsilon. \end{align}
You can also directly use the mean value theorem for integrals: $$\lim_{h\downarrow0}\frac{\int_{a}^{x}f(t)dt-\int_{a}^{x+h}f(t)dt}{h}=\lim_{h\downarrow0}\frac{\int_{x}^{x+h}f(t)dt}{h}=\lim_{h\downarrow0}\frac{f(c_{h})h}{h}$$ where $$c_{h}$$ is between $$x$$ and $$x+h$$. Therefore, $$\lim_{h\downarrow0}c_{h}=x$$. By continuity, $$\lim_{h\downarrow0}f(c_{h})=f(x)$$.