= Solution
<Uniform convergence> of $f_n$ to $f$ on an interval $I$ means that for every $\epsilon>0$ there is $N$ such that $|f_n(x)-f(x)|<\epsilon$ for all $n\geq N$ and every $x\in I$. Equivalently $\sup_{x\in I}|f_n(x)-f(x)|\to0$. The same $N$ must work throughout the interval, unlike <pointwise convergence>.
For completeness, <continuity> is preserved: choose $N$ with uniform error below $\epsilon/3$, and use <continuity> of $f_N$ at a given $x$ to make $|f_N(y)-f_N(x)|<\epsilon/3$. The <triangle inequality> then gives $|f(y)-f(x)|<\epsilon$. Thus each function involved in the <integral> is continuous on the closed interval and Riemann integrable. The estimate
$$
\left|\int_a^b f_n(x)dx-\int_a^b f(x)dx\right|\leq\int_a^b|f_n-f|dx\leq(b-a)\sup_{[a,b]}|f_n-f|
$$
proves \b[convergence of the <integrals>] by <uniform convergence>. When $a=b$ both <integrals> are zero.
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