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I've tried to prove the theorem in advance at the level that satisfies me. The notation used might not be correct, but I hope all major steps are correct.

DEFINITION (from Apostol's Calculus I):

Let $f$ be a function that is integrable on $[a,x]$, for each $x$ in $[a,b]$. Let $c$ be such that $a \le c \le b$ and define a new function $A$ as follows: $$A(x) = \int_c^xf(t)dt, a \le x \le b$$ Then the derivative $A'(x)$ exists at each point $x$ in the open interval $(a,b)$ where $f$ is continuous, and for such $x$ we have $$A'(x) = f(x).$$

PROOF:

I will show that the theorem is a consequence of the property of continuity of the function at some point.

First we make some derivations. For simplicity lets assume that $h$ is positive. The proof for negative $h$ is no harder.

$$A'(x) = \lim_{h\to0} \frac{A(x+h) - A(x)}{h} = \lim_{h\to0}\frac{\int_c^{x+h}f(t)dt - \int_c^{x}f(t)dt}{h} = \lim_{h\to0}\frac{\int_x^{x+h}f(t)dt}{h}$$

Since $f$ is continuous at $x$, for any number $\epsilon$ there is some $\delta$, such that when $t \in [x-\delta, x+\delta]$,

$$f(x) - \epsilon \le f(t) \le f(x) + \epsilon$$

Then for any $h < \delta$,

$$\int_x^{x+h} [f(x) - \epsilon]dt \le \int_x^{x+h} f(t)dt \le \int_x^{x+h} [f(x) + \epsilon]dt$$

$$[f(x) - \epsilon]h \le \int_x^{x+h} f(t)dt \le [f(x) + \epsilon]h$$

$$f(x) - \epsilon \le \frac {\int_x^{x+h} f(t)dt}{h} \le f(x) + \epsilon$$

I can make $\epsilon$ as small as possible, hence

$$\lim_{h\to0}\frac{\int_x^{x+h}f(t)dt}{h} = A'(x) = f(x)$$

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    $\begingroup$ Looks good to me. $\endgroup$ – Marcin Łoś Feb 24 '14 at 8:20
  • $\begingroup$ Isn't this same as the proof in Apostol?. $\endgroup$ – jdoicj May 14 '14 at 6:30
  • $\begingroup$ @boywholived: It was very close as I remember, but I wrote it before reading the one in the book. $\endgroup$ – Graduate May 14 '14 at 6:32
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    $\begingroup$ Notice that you are both trying to show that the limit defining $A'(x)$ exists and to determine its value. In particular, since you do not know that it does exist until the very last line of your proof, you should avoid doing manipulations with that limit until the very end —in general, you can only manipulate things that exist! $\endgroup$ – Mariano Suárez-Álvarez May 19 '14 at 4:47
  • $\begingroup$ In a smiliar vein, it is much better than «Since $f$ is continuous at $x$, for any number $\epsilon$ there is some $\delta$, such that [...]» to write somethign along the lines of «Let $\epsilon$ be a positive number. Since $f$ is continuous, there exists a $\delta>0$ such that [...]» and so on. Fixing $\epsilon$ allows you to avoid saying things like «since I can make $\epsilon$ as small as possible», which do not make any sense: a number cannot be made smaller than it is (you wrote «as small as possible» but you probably meeant «as small as we may wish» or some other such turn) $\endgroup$ – Mariano Suárez-Álvarez May 19 '14 at 4:49
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The proof you have shown is correct.

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