Given $f_n:= e^{-n(nx-1)^2} $, any suggested approaches for showing that $\lim_{n \to \infty}\int^1_0 f_n(x)dx = 0$? 
Given $f_n:= e^{-n(nx-1)^2} $, any suggested approaches for showing that $\displaystyle \lim_{n \to \infty}\int^1_0 f_n(x)dx = 0$? 

I have already shown that $f_n \to 0$ pointwise (and not uniformly) on $[0,1]$ and have been trying to show that $$ \lim_{n \to \infty}\int^1_0 f_n(x)dx = 0$$ I know that $ \forall  n \in \mathbb {N}, $ $\exists x_n = \frac {1}{n} \in [0,1]$ such that $f_n(x_n) = 1$, which has me thinking of doing something where I split the integral in such a way that isolates the interval to which the point $x_n$ belongs. Something along the lines of $$ \lim_{n \to \infty}\int^1_0 f_n(x)dx = \lim_{n \to \infty}\Big(\int^1_{x_n}f_n(x)dx + \int_0^{x_n} f_n(x)dx\Big)$$ However, this particular split doesn't help much because I can't really do anything with the first integral like I could with the second, bounding it with the sup norm and then observing that $x_n \to 0 $ as $ n \to \infty$ since $\sup_{[0,1]}|f_n| = 1$ (or so I think I can't). Any hints to other approaches I could use or to any modifications to my initial approach?
Thanks in advance.

Break the interval to $[0, 1/n]$ and $[1/n, 1]$ is a good idea, but may be still too stringent, so why don't try $[0, 2/n]$ and $[2/n, 1]$?
On the interval $[0, 2/n]$, the integrand is bounded by $1$.
On the interval $[2/n, 1]$, the integrand is bounded by 
$$\exp(-n(2 - 1)^2) = e^{-n}.$$
Therefore, 
$$\int_0^1 f_n(x) dx \leq 1 \times \frac{2}{n} + e^{-n} \times \left(1 - \frac{2}{n}\right) \to 0$$
as $n \to \infty$. Hence the result follows.
