Prove $ \lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a). $ 
Suppose $f(x)$ is continuous on $[a, b]$. Prove that $\forall x\in (a,b)$, we have
$$
\lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a).
$$

If $f$ id diffentiable, the conclution can be obtained by Newton-Leibniz formula easily. When $f$ is just continuous, I tried to use knowledge about integral with parameters. Write
$$
F(h,x):=\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t,
$$
but it seems we still need to compute $\lim_{h\rightarrow 0}\frac{f(t+h)-f(t)}{h}$, and prove uniform convergence. I have no idea how to go on.
Is my thoughts workable? If not, please suggest your way. Appreciate any help!
 A: This is a simple consequence of the Fundamental Theorem of Calculus. Also you need to have $h\to 0^{+}$ instead of $h\to 0$ (why? Explain yourself).
Clearly if $F$ is an anti-derivative of $f$ on $[a, b] $ (such an $F$ exists because of continuity of $f$ and Fundamental Theorem of Calculus) then we have $$\int_{a}^{x}f(t)\,dt=F(x)-F(a)$$ and $$\int_{a} ^{x} f(t+h) \, dt=F(x+h) - F(a+h) $$ and thus the desired limit equals the limit of expression $$\frac{F(x+h) - F(x)} {h} - \frac{F(a+h) - F(a)} {h} $$ and since $F$ is the anti-derivative of $f$ the desired limit equals $$F'(x) - F'(a) =f(x) - f(a) $$
A: Two things:

*

*This is only a partial answer (you'll see why at the end), but it might still help.

*It seems like the limit should be regarded as a right-hand limit.
Please let me know if I'm wrong.

The limit $\lim_{h\to 0^{+}}\frac{1}{h}\int_{a}^{x}\left[f(t+h)-f(t)\right]dt$ looks suspiciously similar to the limit that appears in the definition of the derivative. Given our knowledge of the first part of the Fundamental Theorem of calculus, this observation motivates defining a function $F$ with an integral like the one that FTC1 is concerned with.
$$F(x)=\int_{a}^{x}f(t)\text{ }dt$$
By FTC1, $F$ is continuous over $[a,b]$ and differentiable over $(a,b)$, and its derivative is $F'(x)=f(x)$ for all $x\in (a,b)$. Let's try to express $\frac{1}{h}\int_{a}^{x}\left[f(t+h)-f(t)\right]dt$ in terms of $F$.
If $h$ is small enough, then the integral $\int_{a}^{x}f(t+h)dt$ will be well defined. Applying the substitution rule and basic integral properties then gives
\begin{align*}
\int_{a}^{x}f(t+h)dt &= \int_{a+h}^{x+h}f(t)dt\\
&= \int_{a}^{x+h}f(t)dt-\int_{a}^{a+h}f(t)dt\\
&= F(x+h)-F(a+h)
\end{align*}
Perhaps you see why I think the limit should be a right-hand limit. If $h$ were allowed to be negative, then $a+h<a$, so the domain of integration $[a+h,x+h]$ would contain points outside the “domain” of $f$. Anyway, we may now write
\begin{align*}
\frac{1}{h}\int_{a}^{x}\left[f(t+h)-f(t)\right]dt &= \frac{1}{h}\left(\int_{a}^{x}f(t+h)\text{ }dt-\int_{a}^{x}f(t)\text{ }dt\right)\\
&= \frac{F(x+h)-F(a+h)-F(x)}{h}\\
&= \frac{F(x+h)-F(x)}{h}-\frac{F(a+h)-0}{h}\\
&= \frac{F(x+h)-F(x)}{h}-\frac{F(a+h)-F(a)}{h}
\end{align*}
because $F(a)=\int_{a}^{a}f(t)dt=0$
If one can prove that $\lim_{h\to 0^{+}}\frac{F(a+h)-F(a)}{h}=f(a)$ (this seems intuitively obvious to me, but I’m not sure how to prove it), then the desired result immediately follows from taking the limit and applying FTC1 to the first term. Hope this is helpful!
A: Define $F:\Bbb{R}^2\to \Bbb{R}$ as $F(\alpha,\beta):=\int_{\alpha}^{\beta}f(t)\, dt$. Then, by a simple change of variables $u=t+h$ in $\int_a^xf(t+h)\, dt$, it is easy to see that (for $h\neq 0$)
\begin{align}
\dfrac{1}{h}\int_a^xf(t+h)-f(t)\, dt = \dfrac{F(a+h,x+h)-F(a,x)}{h}
\end{align}
In other words, if we fix $a$ and $x$, and define $\gamma:\Bbb{R}\to\Bbb{R}^2$ as $\gamma(h)=(a+h,x+h)$, then the above expression equals $\dfrac{(F\circ\gamma)(h)-(F\circ\gamma)(0)}{h}$, which is precisely the difference quotient for $F\circ \gamma$. Now, since $f$ is continuous, it is easily seen that $F$ is continuously differentiable (the FTC tells you how to calculate the partial derivatives of $F$ in terms of $f$), and $\gamma$ is clearly differentiable. So, by the chain rule, the composition is also differentiable. Now, calculating your limit is a simple exercise in chain rule:
\begin{align}
\lim_{h\to 0}\dfrac{1}{h}\int_a^xf(t+h)-f(t)\, dt &= (F\circ \gamma)'(0)\\
&=\dfrac{\partial F}{\partial \alpha}(\gamma(0))\cdot \gamma_1'(0)+\dfrac{\partial F}{\partial \beta}(\gamma(0))\cdot \gamma_2'(0)\\
&= \left[-f(a)\right]\cdot \left[1\right] + \left[f(x)\right]\cdot \left[1\right]\\
&= f(x)-f(a)
\end{align}
Hopefully the evaluation of partial derivatives of $F$ is clear (this is just the first/second idk which fundamental theorem of calculus).
