= Solution
The <mean value theorem> states that, if $f$ is <continuous> on $[a,b]$ and <differentiable> on $(a,b)$, with $a<b$, there is $c\in(a,b)$ such that
$$
f'(c)=\frac{f(b)-f(a)}{b-a}.
$$
To deduce it from <Rolle's theorem>, subtract the secant line: $k(t)=f(t)-f(a)-\frac{f(b)-f(a)}{b-a}(t-a)$. This <function> has the same <continuity> and <differentiability> properties and satisfies $k(a)=k(b)=0$. <Rolle's theorem> gives an interior point with $k'(c)=0$, which is precisely the required equality.
If $h'=0$ on $\mathbb R$, apply the <mean value theorem> on any $[x,y]$ with $x<y$. It gives $h(y)-h(x)=(y-x)h'(c)=0$. Hence \b[$h$ is constant].
For the <differential equation>, the <product rule> gives
$$
\frac{d}{dx}\bigl(e^{-ax}f(x)\bigr)=e^{-ax}\bigl(f'(x)-af(x)\bigr)=0.
$$
The preceding result makes this product constant. Thus \b[all solutions are $f(x)=Ce^{ax}$], and differentiation verifies every such solution.
For the <integral> assertion, <continuity> ensures that $f$ is <Riemann integrable> on every finite <closed interval>. Use the usual orientation for an <integral> with negative upper endpoint. For $h\ne0$, additivity of the <integral> gives
$$
\frac{F(x+h)-F(x)}h-f(x)
=\frac1h\int_x^{x+h}\bigl(f(t)-f(x)\bigr)\,dt.
$$
The <absolute value> of this expression is at most the <supremum> of $|f(t)-f(x)|$ on the segment joining $x$ and $x+h$. <Continuity> at $x$ makes this bound tend to zero for either sign of $h$. Consequently \b[$F'(x)=f(x)$], proving this form of the <fundamental theorem of calculus>.
Finally, a <differentiable> solution of the <integral equation> is <continuous>, so the <fundamental theorem of calculus> yields $g'=g$; setting $x=0$ gives $g(0)=A$. The differential-equation calculation above now forces
$$
\boxed{g(x)=Ae^x.}
$$
Conversely, $A+\int_0^xAe^t\,dt=A+A(e^x-1)=Ae^x$, so this <function> satisfies the original <integral equation>. It is unique because every <differentiable> solution has already been forced to this form and this <initial condition>.
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