If $f_n$ and $f$ are integrable and $f_n$ tends to $f$ uniformly on $[a,b]$ then $\displaystyle \int_a^b (f_n(t) - f(t))\,\mathrm d t \to 0 $, since $$\int_a^b |f_n(t) - f(t)| \, \mathrm d t \le \sup_{t \in (a,b)} |f_n(t)-f(t)|(b-a) \to 0 $$ This is not true in general if the convergence is pointwise which can be shown by constructing a sequence of triangles with height $n$ and width $2/n$.
It is also true that if we have $f_n \to f$ uniformly then $\displaystyle \int_a^b (f_n(t) - f(t))^2 \, \mathrm d t \to 0$ by essentially the same argument. But is it true in general if the convergence is only pointwise? I can't seem to find a counterexample. Is it still true if $f$ is uniformly bounded?
Try the triangles you mention in your question with width $2/n$, but with height $\sqrt{n}$.