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.
 A: 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.
A: 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?
A: In the same spirit as Cameron Williams's answer, assuming you know the error function, $$ \int e^{-n(nx-1)^2}\,dx=\frac{\sqrt{\pi } }{2 n^{3/2}}\text{erf}\left(\sqrt{n} (n x-1)\right)$$ which makes  $$ \int_0^1 e^{-n(nx-1)^2}\,dx=\frac{\sqrt{\pi } }{2 n^{3/2}}\Big(\text{erf}\left(\sqrt{n}\right)+\text{erf}\left((n-1)
   \sqrt{n}\right)\Big)$$ For large values of the argument, a very fast convergence expansion of the error function is $$\text{erf}(x)=1-\frac{e^{-x^2}}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(2n-1)!!}{2^n}\frac 1{x^{n+1}}$$ So, limited to the first term $$ \int_0^1 e^{-n(nx-1)^2}\,dx \approx\frac{\sqrt{\pi } }{ n^{3/2}}$$ which is an extremely close approximation; for $n=5$, the exact value is $\approx \color{red}{0.158}4090079$ and the approximation is $\approx \color{red}{0.158}5330919$; for $n=10$, these values are respectively $\approx \color{red}{0.056049}69513$ and $\approx\color{red}{0.056049}91216$;for $n=15$, these values are respectively $\approx \color{red}{0.030509707}10$ and $\approx\color{red}{0.030509707}76$.
For sure, much better approximation could be obtained introducing the second term of the expansion.
