Does $f_n(x) = n^2x^2(1-x)$ converge uniformly to $f$ on [0,1]?

Question

Does $f_n(x) = n^2x^2(1-x)$ converge uniformly to $f$ on [0,1]?

Attempt: If $f_n$ converges uniformly, then $\int f_n \rightarrow \int f$. However, $\int f_n = \frac{n^2}{12}$ which does not converge to 0 as n tends to infinity. Thus, contradiction. Is this process correct?

I was following the process for this question: Is $f_n(x)=n^2x(1-x^2)^n$ uniformly convergent on $[0,1]$?

Answer 1

Uniform convergence on $[a,b]$ implies pointwise convergence on $[a,b]$.

The sequence does not pointwise converge on $(0,1)$, therefore it does not uniformly converge on $(0,1) \subseteq [0,1]$.