User:Dgdean/sandbox

First it is proved that F is differentiable on the open interval (a, b), which in turn proves that F is continuous on (a, b) since differentiability implies continuity.

Now, for x in (a, b) with a < c < b

$$ \frac{F(x)-F(c)}{x-c} = \frac{1}{x-c}(F(x)-F(c)) = \frac{1}{x-c}[\int_{a}^{x}f(t)dt - \int_{a}^{c}f(t)dt] $$ (substitution into F)

If a < x < c then

$$ \frac{F(x)-F(c)}{x-c} = \frac{1}{x-c}[\int_{a}^{x}f(t)dt - (\int_{a}^{x}f(t)dt + \int_{x}^{c}f(t)dt)] = \frac{1}{c-x}\int_{x}^{c}f(t)dt $$ ( additivity of integration on intervals)

If a < c < x then

$$ \frac{F(x)-F(c)}{x-c} = \frac{1}{x-c}[(\int_{a}^{c}f(t)dt + \int_{c}^{x}f(t)dt) - \int_{a}^{c}f(t)dt] = \frac{1}{x-c}\int_{c}^{x}f(t)dt $$

Since f is continuous on [a, b] it follows by