TITLE: Prove $ \lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a). $
QUESTION [11 upvotes]: 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!

REPLY [4 votes]: 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 $$F'(x) - F'(a) =f(x) - f(a) $$