How prove this $F'(x_{0})=f(x_{0})$ if $F(x)=\int_{a}^{x}f(t)dt,x\in[a,b]$ Question:
let $f$ be Riemann integrable on $[a,b]$,Assmue that $x_{0}\in [a,b]$,and let $f(x)$ is continuous on point $x=x_{0}$,and define
$$F(x)=\int_{a}^{x}f(t)dt,x\in[a,b]$$
show that
$$F'(x_{0})=f(x_{0})$$
This problem is my friend (He is a teacher) ask me,and I post my answer,maybe have some wrong,

since
  $$F'(x_{0})=\lim_{x\to x_{0}}\dfrac{F(x)-F(x_{0})}{x-x_{0}}=\lim_{x\to x_{0}}\dfrac{\int_{x_{0}}^{x}f(t)dt}{x-x_{0}}$$
  and other hand,

case1: $x>x_{0}$ we have

$$I=\left|\dfrac{\int_{x_{0}}^{x}f(t)dt}{x-x_{0}}-f(x_{0})\right|=\dfrac{\left|\int_{x_{0}}^{x}f(t)dt-f(x_{0})(x-x_{0})\right|}{x-x_{0}}\le\dfrac{\int_{x_{0}}^{x}|f(t)-f(x_{0})|dt}{x-x_{0}}$$
  since
  $f(x)$ is continous on point $x=x_{0}$
  so
  $$I\le\dfrac{\int_{x_{0}}^{x}\varepsilon dt}{x-x_{0}}=\varepsilon$$
  so
  $$F'(x_{0})=f(x_{0})$$

case2:$x<x_{0}$

$$I=\left|\dfrac{\int_{x_{0}}^{x}f(t)dt}{x-x_{0}}-f(x_{0})\right|=\dfrac{\left|\int_{x_{0}}^{x}f(t)dt-f(x_{0})(x-x_{0})\right|}{|x-x_{0}|}\le\dfrac{\int_{x}^{x_{0}}|f(t)-f(x_{0})|dt}{x_{0}-x}$$
  since
  $f(x)$ is continous on point $x=x_{0}$
  so
  $$I\le\dfrac{\int_{x}^{x_{0}}\varepsilon dt}{x_{0}-x}=\varepsilon$$

My question:  someone  have other methods? Thank you  very much!
 A: Wikipedia
We can use Leibniz integral rule for variable limit 
$$\frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt$$ =
$$f(g(x)).h'(x)-f(h(x)).g'(x)$$
We have 
$$F(x)=\int_{a}^{x}f(t)dt,x\in[a,b]$$
Therefore  $$\frac{d}{dx}F(x) =f(x) $$
and $$ F'(x)=f(x) $$
so 
$$F'(x_0)=f(x_0)$$
A: I don't know that this amounts to a different approach, but:
Building on the fact that if $f:[c,d]\to\mathbb{R}$ is integrable (this is actually important; one must know that $f$ is integrable on $[a,b]$ to define $F$ as above) and non-negative (i.e., $f(x)\geq 0$) then $\int_c^d f(x)dx\geq 0$ as well, we note that for sufficiently small $\epsilon>0$
$$\inf_{|t-x_0|<\epsilon} f(t)\leq\frac{1}{\epsilon}\int_{x_0}^{x_0+\epsilon} f(t)dt\leq\sup_{|t-x_0|<\epsilon} f(t),$$
and similarly
$$\inf_{|t-x_0|<\epsilon} f(t)\leq\frac{1}{\epsilon}\int_{x_0-\epsilon}^{x_0} f(t)dt\leq\sup_{|t-x_0|<\epsilon} f(t).$$
Since $f$ is continuous at $x_0$, we know that
$$\lim_{\epsilon\to 0}\inf_{|t-x_0|<\epsilon} f(t) = f(x_0) = \lim_{\epsilon\to 0}\sup_{|t-x_0|<\epsilon} f(t),$$
implying that
$$\exists\lim_{\epsilon\to 0}\frac{F(x_0+\epsilon)-F(x_0)}{\epsilon} = f(x_0).$$
A: That's correct but assumes $a$ and $f$ are fixed.
Just to challenge your friend, you can prove with the same argument that if 
$$F(x)=\int_x^b f(t)dt,x∈[a,b]$$
then
$$F'(x_0)=-f(x_0), x \in [a,b].$$
