Uniform convergence of $f_n(x)=nx^n(1-x)$ for $x \in [0,1]$?

I want to decide whether or not $f_n(x)=nx^n(1-x)$ is uniformly convergent or not. I have shown that $\lim_{n\to\infty} f_n(x)=0$ for $x \in [0,1]$.

Now $f_n(0)=f_n(1)=0$. And in $(0,1)$, we have $f'_n(x)=n^2x^{n-1}-n(n+1)x^n$. Putting this to zero gives $x=\frac{n}{n+1}$. Also $f_n(\frac{n}{n+1})=(\frac{n}{n+1})^{n+1}$. But that tends to $\frac{1}{e}$ as $n$ tends to $\infty$ which is not $0$ so we don't have uniform convergence, right?

Nonetheless we have $\lim_{n\to\infty} \int_0^1 f_n(x)dx=\lim_{n\to\infty} (\frac{n}{n+2}-\frac{n}{n+1})=0=\int_0^1 \lim_{n\to\infty} f_n(x)dx$.

Is all this right? I have some doubts, I mostly doubt my intuition in analysis... Thank you!

Solution 1:

Yes you're right! Notice that this example shows that the uniform convergence is a sufficient condition to interchange limit and the integral sign $\int$ but not a necessary condition.