$\lim_{h\to 0}\int \frac{f(x+h)-f(x)}{h}\,\mathrm{d}x = f(b)-f(a)$ 
Let $f:[a,b]\to  R$ be integrable and continuous on $(a,b)$. Prove that:
$$\lim_{h\to 0} \int_a^b\frac{f(x+h)-f(x)}{h} \,\mathrm{d}x = f(b)-f(a).$$

What I've tried so far: make a change of variables, which didn't help. I also thought of using the intermediate value theorem inside the integral, but the function being differentiable is not specified.
Any help, tips? Thanks in advance!
 A: $\begin{align}\\
 \int_a^b\frac{f(x+h)-f(x)}{h} dx
&=\frac1{h}\left( \int_a^b f(x+h)dx-\int_a^bf(x)dx\right)\\
&=\frac1{h}\left( \int_{a+h}^{b+h} f(x)dx-\int_a^bf(x)dx\right)\\
&=\frac1{h}\left( \int_{a+h}^{b} f(x)dx+\int_{b}^{b+h} f(x)dx-\left(\int_{a}^{a+h}f(x)dx+\int_{a+h}^bf(x)dx\right)\right)\\
&=\frac1{h}\left( \int_{b}^{b+h} f(x)dx-\int_{a}^{a+h}f(x)dx\right)\\
&=\frac1{h}\int_{b}^{b+h} f(x)dx-\frac1{h}\int_{a}^{a+h}f(x)dx\\
&\underbrace{\longrightarrow}_{h \to 0} f(b)-f(a)
\qquad\text{if }f \text{ is continuous on }[a, b+h]\\
\end{align}
$ 
A: It is tacitly assumed that $f$ is integrable over a slightly larger interval. Let a small $h>0$ be given (the proof for $h<0$ is analogous). Then
$$\eqalign{\int_a^b\bigl(f(x+h)-f(x)\bigr)\>dx&=\int_{a+h}^{b+h} f(x)\>dx-\int_a^b f(x)\>dx\cr &=\int_b^{b+h}f(x)\>dx-\int_a^{a+h}f(x)\>dx\ ,\cr}$$
using elementary properties of the Riemann integral. Since $f$ is continuous the MVT  provides us with two points $\alpha\in[a,a+h]$, $\>\beta\in[b,b+h]$ such that
$$\int_a^b\bigl(f(x+h)-f(x)\bigr)\>dx=h\bigl(f(\beta)-f(\alpha)\bigr)\ .$$
It follows that
$$\lim_{h\to0+}{1\over h}\int_a^b\bigl(f(x+h)-f(x)\bigr)\>dx=f(b)-f(a)\ .$$
