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.

Here is a somewhat.. cheap way of doing the problem that you might feel a bit unsatisfied with but consider the following instead:
$$\int_{-\infty}^{\infty} e^{-n(nx-1)^2}\,dx.$$
Clearly
$$\int_0^1 e^{-n(nx-1)^2}\,dx \le \int_{-\infty}^{\infty} e^{-n(nx-1)^2}\,dx$$
since the integrand is positive. Making a change of variable of $y = nx-1$, we see that $dy = n\,dx$ and we get
$$\int_{-\infty}^{\infty} e^{-n(nx-1)^2}\,dx \Longrightarrow \frac{1}{n} \int_{-\infty}^{\infty} e^{-ny^2}\,dy.$$
Making another change of variables of $z= \sqrt{n}y$, we get $dz = \sqrt{n}\,dy$ and so
$$\frac{1}{n} \int_{-\infty}^{\infty} e^{-ny^2}\,dy \Longrightarrow \frac{1}{n^{\frac{3}{2}}} \int_{-\infty}^{\infty} e^{-z^2}\,dz.$$
What happens to this integral as you let $n$ tend to infinity? Do you see how this answers the question at hand?
