Does $nx^n(1-x^2)$ converges uniformly to $0$ on $[0,1]$? 
Does the sequence of functions
$$f_n(x) = nx^n(1-x^2)$$
converges uniformly to $0$ on $[0,1]$?

I took the derivative, set it zero to find the critical points, which are $\{-\sqrt{\frac{n}{n+2}}, +\sqrt{\frac{n}{n+2}}\}$. So I picked the positive root which makes my function maximum. Then
$$ \lim_{n \to \infty} n\cdot \left(\sqrt{\frac{n}{n+2}}^n\cdot (1-\frac{n}{n+2}) \right) \neq 0$$
so can I conclude that $\{f_n\}$ does not converge to $0$ uniformly since this limit is not $0$? and hence $\int_0^1 f_n(x) dx$ does not converge to $\int_0^1 f(x) dx$ for non uniform continuity the intergrands?
 A: Actually, you have to take the positive root, since you are working on $[0,1]$. And, yes$$\lim_{n\to\infty}f_n\left(\sqrt{\frac n{n+2}}\right)\ne0$$(actually, it's equal to $24$), and so, since $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function, the convergence is not uniform.
However, from the fact that the convergence is not uniform does not follow that you don't have $\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm dx=\int_0^10\,\mathrm dx$. For instance, if you define$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}\sin(n\pi x)&\text{ if }x\in\left[0,\frac1n\right]\\0&\text{ otherwise,}\end{cases}\end{array}$$then $(f_n)_{n\in\Bbb N}$ also converges pointwise but not uniformly to the null function. However,$$(\forall n\in\Bbb N):\int_0^1f_n(x)\,\mathrm dx=\frac2{n\pi},$$and therefore$$\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm dx=0.$$
A: The first claim is right, $f_n$ doesn't converge uniformly to $f$.
The second isn't necessarily correct and in fact here it isn't. You can check it by an explicit calculation that $\int f_n\to \int f = 0$.
