Provide examples that satisfy the following cases
1) $f_n: [0, ∞)$ → R that converges uniformly to the function $f (x) = 0$ on [0, ∞) but such that $\lim_{n→∞} \int_{0}^{∞} f_n(x) dx \neq \int_{0}^{∞} f (x) dx$
2) $f_n: [0, ∞)$ → R and $f_n$ is differentiable such that the sequence ($f_n(0)$) is a convergent sequence and $f'_n$ converges uniformly to a function $g(x)$ on [0, ∞) , while $f_n$ does not converge uniformly on [0, ∞).
For the first one, you can define $$f_n(x)=\begin{cases}\frac{2n-x}{2n^2};&x\in[0,2n],\\0;&x\in[2n,\infty).\end{cases}$$ Then $\lim_{n\to\infty}f_n(x)=0$, uniformly on $[0,\infty)$ and $\int_0^{\infty}f_n(x){\rm d}x=1\neq0$.
For the second one, you can use a primitive function of the first one, i.e. $g_n(x)=\int_{0}^xf_n(x){\rm d}x$. Then, $g_n$ is differentiable on $[0,\infty)$, $g_n(0)=0$ for all $n$, and $g_n$ does not converge uniformly on $[0,\infty)$ (you can see this by noticing that each $g_n$ is increasing and $\lim_{n\to\infty}g_n(\sqrt{n})=0$, which implies that $\lim_{n\to\infty}g_n(x)=0$ for all $x$, and $g_n(2n)=1$ which then shows that the convergence is not uniform) but the derivatives $g_n'$ do converge uniformly.