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.

In my way of understanding, you can say that at each point $x$, the behavior $f(x)$ has nothing to do with the rest of the values $f(y), y \neq x$. On the other hand, the integral $\int_{0}^{1} f(x) dx = \| f\|_{1,[0,1]}$ takes into account all of $f$. 
That is, pointwise convergence of $f_{n}$ (local behavior) does not tell you much about the convergence of $\int_{0}^{1} f_{n}(x) dx$ (global behavior). Which is why we need the idea of uniform convergence.
