Pointwise convergence to zero, with integrals converging to a nonzero value For $n\in{\mathbb{N}}$ let $$f_n(x)=nx(1-x^2)^n\qquad(0\le x\le 1).$$
Show that $\{f_n\}_{n=1}^\infty$ converges pointwise to $0$ on $[0,1]$.
Show that $\{\int_0^1f_n\}_{n=1}^\infty$ converges to $\frac12$.
I've already shown both of these statements to be true. What I don't understand is this: how can $f_n$ converge pointwise to $0$, yet the sequence $\{\int_0^1f_n\}_{n=1}^\infty$ converges to $\frac12$? Isn't that almost like saying $f(x)=0\qquad(a\le x\le b)$, but $\int_a^bf(x)=\frac12$?
Clearly that would be false. I know this has something to do with that this is a sequence of functions but it still baffles my mind. Thanks in advance.

Have you tried sketching the first few functions? 
As $n$ increases, the area under the function graph "flows" further and further to the right end like a wave while remaining constant.
As $f_n(1)=0$ for all $n$ and every point $x<1$ falls behind the wave sooner or later, the pointwise convergence follows.
The underlying problem is that swapping limits (and integration is a limit) is not always allowed; the basic example for sequences is $\lim_{n\to\infty}\lim_{m\to\infty}\frac{n}{n+m}=0\ne1= \lim_{m\to\infty}\lim_{n\to\infty}\frac{n}{n+m}$.
