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Let f be continuous on [a,b] and differentiable almost everywhere on (a,b). Show that $ \int_a^b f'(x)\,\mathrm{d}x=f(b)-f(a) $ if and only if $ \int_a^b\lim_{n\to\infty}\text{Diff}_{1/n}f=\lim_{n\to\infty}\int_a^b\text{Diff}_{1/n}f $

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    martini:Assume f is continuous. Extend f to take the value f(b) on (b,b+1] and for 0Diff_h f(x)=[f(x+h)-f(x)]/h for all x in [a,b]. – 2012-11-23

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$\def\D{\operatorname{Diff}}$If I should guess, I'd say $\D_h$ is the diffrence quotient operator. We have \begin{align*} \int_a^b \D_{1/n}f(x) \,dx &= \int_a^b \frac{f(x+1/n) - f(x)}{1/n} \, dx\\ &= n \cdot \left(\int_a^b f(x+1/n)\, dx - \int_a^b f(x)\, dx \right)\\ &= n \cdot \left(\int_{a+1/n}^{b+1/n} f(x)\, dx - \int_a^b f(x)\, dx\right)\\ &= n \cdot\int_b^{b+1/n} f(x)\, dx - n \cdot \int_a^{a+1/n} f(x)\, dx \end{align*} Now by continuity of $f$, for each $n$ there is an $a \le x_n \le a+1/n$ with $\int_a^{a+1/n} f(x)\, dx = \frac 1n f(x_n)$, analogously for the $b$-integral, again by continuity $n\int_a^{a+1/n} f(x)\, dx \to f(a)$, hence we have $ \int_a^b \D_{1/n}f(x) \, dx \to f(b) - f(a) $ The limit on the left hand side is almost every where equal to $f'$ by definition. So your second equation is equivalent to $ \int_a^b f'(x)\, dx = f(b) - f(a) $ which is what we wanted to show.

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    As $f$ is continuous, there are $y_i, y_s \in [a, a+1/n]$ with $f(y_i) \le f(x) \le f(y_s)$ for all $x \in [a,a+1/n]$. Integrating gives $f(y_i) \le n\int_a^{a+1/n}f(x)\, dx \le f(y_s)$. Now $x_n$ exists by the intermediate value theorem. Note that $x_n \to a$ for $n \to \infty$, hence $f(x_n) \to f(a)$ by continuity.2012-11-24