Evaluate $\lim _{n\to \infty }\int _{0}^{1}nx^ne^{x^2}dx$ 
Evaluate $\lim _{n\to \infty }\int _{0}^{1}nx^ne^{x^2}dx.$

I applied the mean value thorem of integral to $\int _{0}^{1}nx^ne^{x^2}dx.$ We get $c\in (0,1):$
$$\int _{0}^{1}nx^ne^{x^2}dx=(1-0)nc^ne^{c^2}.$$ Taking limit ($\lim_{n\to \infty}$)on the both side,
We get, $$\lim_{n\to \infty}\int _{0}^{1}nx^ne^{x^2}dx=\lim_{n\to \infty} nc^ne^{c^2}=0.$$
My answer in the examination was wrong. I don't know the correct answer. Where is my mistake?

A bit late answer but maybe worth noting it.
Partial integration gives
$$I_n :=\int_0^1 \underbrace{nx^{n-1}}_{u'}\cdot\underbrace{xe^{x^2}}_{v}dx= \left.x^{n+1}e^{x^2}\right|_0^1- \underbrace{\int_0^1 x^n(1+2x^2)e^{x^2}dx}_{J_n=}=e-J_n$$
Now, $J_n$ can be easily estimated as follows
$$0\leq J_n \leq 3e\int_0^1x^ndx=\frac{3e}{n+1}\stackrel{n\to \infty}{\longrightarrow}0$$
Hence, $I_n \stackrel{n\to \infty}{\longrightarrow} e$.
