= Solution
<Uniform convergence> on an interval $I$ means that for every $\epsilon>0$ there is $N$ such that $|f_n(x)-f(x)|<\epsilon$ for every $x\in I$ and every $n\ge N$. Equivalently, the supremum of this difference tends to zero. A <uniform limit> of <continuous functions> is continuous: at a point $x_0$, choose one $f_N$ uniformly within $\epsilon/3$ of $f$, then use <continuity> of $f_N$ to make $|f_N(x)-f_N(x_0)|<\epsilon/3$. The <triangle inequality> gives $|f(x)-f(x_0)|<\epsilon$. Thus $f$ is integrable on the compact interval, and
$$
\left|\int_a^b f_n-\int_a^b f\right|\le (b-a)\sup_{[a,b]}|f_n-f|\longrightarrow0.
$$
For the half-line example, differentiating gives $f_n'(x)=n^{-2}e^{-x/n}(1-x/n)$. The maximum is attained at $x=n$, where $f_n(n)=1/(en)$. Therefore \b[$f_n\to0$ uniformly on $[0,\infty)$]. Nevertheless, the substitution $x=ny$ gives
$$
\boxed{\int_0^\infty f_n(x)\,dx=\int_0^\infty y e^{-y}\,dy=1}.
$$
The integral does not tend to zero. This is the phenomenon that <uniformly vanishing functions can retain a nonzero integral>: the finite-length estimate above has no finite analogue for the entire half-line.
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