I would like to show that :
$$\int_{0}^1(3t^2 -2t^3)^n \mathrm{d}t \underset{n \rightarrow+\infty}{\sim} \frac{1}{\sqrt{n}} \int_0^{+\infty}e^{-3x^2}\mathrm{d}x$$
I thought of showing the equivalent result:
$$\sqrt{n}\int_{0}^1(3t^2 -2t^3)^n \mathrm{d}t \underset{n \rightarrow+\infty}{\longrightarrow} \int_0^{+\infty}e^{-3x^2}\mathrm{d}x$$
I tried using the binomial expansion and that for all $x\in\mathbb{R},$ $e^{-3x^2} = \displaystyle \sum_{k=0}^{+\infty}\frac{(-3x^2)^k}{k!}$, I also thought of changing the bounds of the integrals. But I don't really know yet how to proceed.
Edited based on comment by Travor Liu
This is going to be a bit messy, and I'm not going to be super precise with justifying why everything works, but we're essentially going to use Laplace's method for integral asymptotics.
We begin with rewriting the integral as follows: $$\int_{0}^1(3t^2 -2t^3)^n \, dt = \int_{0}^{1}e^{n\ln(3t^{2} - 2t^{3})}\,dt,$$ and we note that $\ln(3t^2 - 2t^3) < 0$ on $(0,1)$, and that as $n\to \infty$ the area on $(0,\varepsilon)$, for $\varepsilon > 0$ becomes negligible. With that in mind, we expand $\ln(3t^{2} - 2t^{3})$ around $t = 1$ (via Taylor series) to get $$\ln(3t^{2} - 2t^{3}) = -3(t-1)^{2} + O((t-1)^3),$$ and so $$\int_{0}^{1}e^{n\ln(3t^{2} - 2t^{3})}\,dt \sim \int_{0}^{1}e^{-3n(t-1)^{2}}\,dt, \quad\text{as }n\to\infty.$$ Now, we do a change of variables \begin{align} u &= 3n(t-1)^2 \implies t- 1 = \pm\sqrt{\frac{u}{3n}}\\ du &= 6n(t-1)\,dt = -6n\sqrt{\frac{u}{3n}}\,dt = -2\sqrt{3n}\sqrt{u}\,dt. \end{align} Note that, as we will see momentarily, we choose the negative root so that the integral remains positive. With this substitution, we get \begin{align} \int_{0}^{1}e^{-3n(t-1)^{2}}\,dt = -\frac{1}{2\sqrt{3n}}\int_{3n}^{0}u^{-1/2}e^{-u}\,du = \frac{1}{2\sqrt{3n}}\int_{0}^{3n}u^{-1/2}e^{-u}\,du. \end{align} Now, the because we have the factor of $e^{-u}$, the integrand becomes exponentially small as $u\to \infty$, and so we only introduce exponentially small error by extending the upper limit of the integral to infinity. This gives us $$ \int_{0}^1(3t^2 -2t^3)^n \, dt \sim \frac{1}{2\sqrt{3n}}\int_{0}^{\infty}u^{-1/2}e^{-u}\,du\quad\text{as }n\to\infty.$$
At this point we make us of the change of variables $u = 3x^2$, which gives us
$$\frac{1}{2\sqrt{3n}}\int_{0}^{\infty}u^{-1/2}e^{-u}\,du =\frac{1}{2\sqrt{3n}}\int_0^\infty(3x^2)^{-1/2}e^{-3x^2}(6x)\,dx = \frac{1}{\sqrt{n}}\int_{0}^{\infty} e^{-3x^2}\, dx.$$
So, we obtain
$$\boxed{\int_{0}^1(3t^2 -2t^3)^n \, dt\sim \frac{1}{\sqrt{n}}\int_{0}^{\infty} e^{-3x^2}\, dx\quad\text{as }n\to\infty}$$
Now, if we wanted to be a bit more precise, we could write \begin{align} \frac{1}{2\sqrt{3n}}\int_{0}^{\infty}u^{-1/2}e^{-u}\,du&= \frac{1}{2\sqrt{3n}}\Gamma\left(\frac{1}{2}\right)\\[5pt] &= \frac{1}{\sqrt{n}}\frac{1}{2}\sqrt{\frac{\pi}{3}}, \end{align}
where here $\Gamma$ is the gamma function, and so in reality we have $$\int_{0}^1(3t^2 -2t^3)^n \, dt\sim \frac{1}{\sqrt{n}}\frac{1}{2}\sqrt{\frac{\pi}{3}},\qquad\text{as }n\to\infty.$$