Convergence of sequence of functions, $f_n(x) = n^2 x(1-nx) \dots $

I'd strongly encourage you to draw a picture of the graph of $f_{n}$.

Your argument that $f_{n} \to 0$ is not quite correct. I'd argue as follows:

We have $f_{n}(x) \to 0$ as $(n \to \infty)$ for all $x \in [0,1]$. This is clear for $x = 0$ and for $x > 0$ we have $f_{n}(x) = 0$ for all $n$ so large that $\frac{1}{n} < x$.

If $f_{n} \to f$ uniformly on $[0,1]$ then $f_{n} \to f$ pointwise, hence we must have $f = 0$.

On the other hand, the function $f_{n}$ has a maximum at $\frac{1}{2n}$ (not $\frac{2}{n}$ as you've written in your question) as can be found by differentiation, for example. Evaluation gives \[ f_{n}(\frac{1}{2n}) = n^{2} \frac{1}{2n} ( 1 - n\frac{1}{2n}) = \frac{n}{4} \] Therefore \[ \sup_{x \in [0,1]} |f_{n}(x) - f(x)| = \sup_{x \in [0,1]} |f_{n}(x)| = \frac{n}{4} \xrightarrow{n \to \infty} \infty \] and hence $f_{n}$ does not converge uniformly to $f = 0$.

Another way to proceed is to look at the area of the function in your domain $[0,1]$. (I usually use this as a first step to check if the function is not uniformly convergent, since it is relatively easy.)

The area of the function $f_n(x)$ in the domain is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$ irrespective of $n$.

Note that $f_n(0) = 0, \forall n$ and $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = 0$, $\forall x \in (0,1]$.

This can be seen since for any $x$, $\exists N \in \mathbb{N}$ such that $\forall n > N$, $\frac{1}{n} < x \Rightarrow f_n(x) = 0$.

Hence, $f(x) = \displaystyle \lim_{n \rightarrow \infty} f_n(x) = 0$, $\forall x \in [0,1]$.

So we have $\displaystyle \int_{0}^{1} f_n(x) dx = \frac{1}{6}$, $\forall n \in \mathbb{N}$ and $\displaystyle \int_{0}^{1} f(x) dx = 0$. So we have $$\displaystyle \lim_{n \rightarrow \infty} \int_{0}^{1} f_n(x) dx = \frac{1}{6} \neq 0 = \displaystyle \int_{0}^{1} f(x) dx$$

Hence, we have $\displaystyle \lim_{n \rightarrow \infty} \int_{0}^{1} f_n(x) dx \neq \displaystyle \int_{0}^{1} \lim_{n \rightarrow \infty} f_n(x) dx$

And we know that if a sequence of functions converge uniformly, we can swap the limit and the integrals to get the same integral.

Hence, the function is not uniformly convergent.

The limit function is $f(x) = 0 \quad \forall x \in [0,1]$ because, as $n \rightarrow \infty$ you have that the region where the sequence is $n^2 x(1-nx)$ is always smaller and smaller (this is a nice way of approaching sequences with boundary conditions that involve $n$).

As for uniform convergence, you should take the supremum for $x \in [0,1]$ and since the function has a maxmimum inside the interval, you can say that $\sup_{x \in [0,1]}\left| n^2 x(1- n x)\right| = \frac{n}{4}$, and if $n \rightarrow \infty$ it doesn't approach zero, so the convergence is not uniform.