= Solution
The <fundamental theorem of calculus> says first that if $h$ is continuous on $[a,b]$ and $H(x)=\int_a^xh(t)\,dt$, then $H'(x)=h(x)$. Indeed,
$$
\frac{H(x+s)-H(x)}s-h(x)=\frac1s\int_x^{x+s}(h(t)-h(x))\,dt,
$$
whose absolute value tends to zero by continuity. Consequently, if $F$ is differentiable with continuous derivative, applying the first part to $h=F'$ shows that $F(x)-\int_a^xF'(t)\,dt$ has zero derivative and is constant. Hence
$$
\boxed{\int_a^bF'(t)\,dt=F(b)-F(a).}
$$
A derivative need not be <Riemann integrable>. Define $f(0)=0$ and $f(x)=x^2\sin(x^{-2})$ for $x\ne0$. Then $f$ is differentiable at zero with $f'(0)=0$, while for $x\ne0$,
$$
f'(x)=2x\sin(x^{-2})-\frac2x\cos(x^{-2}),
$$
which is unbounded near zero. Since every Riemann-integrable function on a compact interval is bounded, \b[$g=f'|_{[0,1]}$ need not be Riemann integrable].
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