I need help proving the following extension: Suppose $f$ is integrable on $[a,b]$ and let $F(x) := \int_a^x f(x) dx$. If f is continuous at $x_0$, then $F$ is differentiable at $x_0$ and $F'(x_0) = f(x_0)$.

This is an extension of the theorem that states: Let $f$ be a continuous function on $[a,b]$ and let $F(x) = \int_a^x f(x) dx$, for $x \in (a,b)$. Then $F$ is differentiable at $x$ and $F'(x) = f(x)$.

(This theorem assumes $f$ is continuous on $[a,b]$. the extension assumes $f$ is continuous at a point.)

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Hi Rocket12, Since this looks like a homework question, we want to know: what have you tried so far? – Johannes Kloos Aug 6 '12 at 19:26
here is what i have so far: – Rocket 12 Aug 6 '12 at 19:30
here is what i have so far: Let x0 be a point in (a,b). Since f is continuous at x0, we know that there is a number cn between xo and xn for which integral from xo to xn of f = f(cn)(xn-x0). So F(xn)-F(x0)/(xn-x0) = f(cn) where cn is bw xn and x0. Since cn is bw xn and x0 and since the lim as n->oo of xn= x0, it follows that lim as n->00of cn = x0. Because f is continuous at x0 this means that the lim as n->oo of F(xn)-F(x0)/(xn-x0) = lim as n->oo if f(cn) = f(x0). So F'(x0)- f(x0) – Rocket 12 Aug 6 '12 at 19:37

First, I think you should write $$F(x):=\int_a^xf(t)\,dt$$ otherwise the x's get messed.
Next, suppose $\,f\,$ is continuous at $\,x_0\,$ : $$(**)\,\,\frac{F(x)-F(x_0)}{x-x_0}=\frac{1}{x-x_0}\int_{x_0}^xf(t)dt=\frac{1} {x-x_0}\left[\int_{x_0}^x\left(f(t)-f(x_0)\right)dt\right]+f(x_0)$$
Thus, since for any $\epsilon > 0\,\,\exists\,\, \delta > 0\,\,s.t.\,\,|x-x_0|<\delta\Longrightarrow |f(x)-f(x_0)|<\epsilon\,$ , passing to the limit when $\,x\to x_0\,$ in (**) we get:
$$\left|\frac{F(x)-F(x_0)}{x-x_0}\right|\leq\left|\frac{1} {x-x_0}\left[\int_{x_0}^x\left(f(t)-f(x_0)\right)dt\right]\right|+|f(x_0)|<\epsilon+|f(x_0)|$$ whenever $\,|x-x_0|<\delta\,$ , and the claim follows.