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6.20 Theorem Let $f$ be Riemann integrable on $[a,b]$. For $a \le x \le b$, put $$F(x)=\int_a^x f(t) \, dt.$$ Then $F$ is continuous on $[a,b]$; furthermore, if $f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is differentiable at $x_0$, and $$F'(x_0)=f(x_0).$$

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In the last line, can I use MVT to conclude $\frac{F(t)-F(s)}{t-s}=F'(x_0)$ ?
I think it is possible since F is continuous and differentiable on [a,b], so there exists a point $x_0$ belongs to (a,b) such that $\frac{F(t)-F(s)}{t-s}=F'(x_0)$ where $a \le s <t \le b$.
Is that right?

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To use the MVT you need to know that $F'$ exists, which is what you are trying to prove. So, you can't use the MVT here. The second last line is basically the definition of the derivative of $F$ at $x_0$ (set $s=x_0$), so no further estimates are needed. –  copper.hat Nov 23 '12 at 15:55
    
Hmm.. I thought that if f is continuous on $x_0$ where $a \le x_0 \le b$ then F is differentiable at $x_0$, so it means that F is differentiable on [a,b]. But now I'm a bit suspicious of my logic. In here, I can't guarantee that F is differentiable in [a,b] whole range so it is impossible to apply MVT? –  pascl Nov 23 '12 at 16:02
    
Well, the fact is true, so you can of course use the MVT. However, you are trying to prove the fact is true, do you can't use the fact that it is true to prove that it is true without having something circular going on. Besides, the second last line basically gives it to you on a plate... –  copper.hat Nov 23 '12 at 16:08

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