= Solution
Put $\varepsilon_n=\sup_{[-1,1]}|F_n-F|$. <Uniform convergence> means $\varepsilon_n\to0$, and continuity ensures that the functions have <Riemann integrals>. The <integral triangle inequality> gives
$$
\left|\int_{-1}^1F_n(x)\,dx-\int_{-1}^1F(x)\,dx\right|
\leq\int_{-1}^1|F_n(x)-F(x)|\,dx\leq2\varepsilon_n\longrightarrow0.
$$
This proves that <uniform convergence preserves integrals on a compact interval>.
For the <derivative> assertion, take $F_n(x)=\sin(n^2x)/n$ and $F(x)=0$. These functions are differentiable, with $\sup|F_n|\leq1/n$, so their convergence is uniform. However, \b[$F_n'(0)=n$ whereas $F'(0)=0$]. The <derivatives> therefore need not converge even at one interior point.
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