Evaluating the limit of an integral I am trying to solve this problem from a past exam.

Let $f:[0,1]\rightarrow\mathbb R$ be a continuous function such that $f(0)=0$ and $f(1)=1$. Evaluate the limit
  $$
\lim_{n\rightarrow\infty}n\int_0^1f(x)x^{2n}dx.
$$

Since $nf(x)x^{2n}$ is not uniformly convergent on [0,1] (even on [0,1)), I cannot swap $\lim$ and $\int$. I could use integration by parts, but I would get a horrendous alternating sum that involves repeated antiderivatives.
I would appreciate if you could give a clue to this problem.
EDIT:  While Peter Tamaroff's answer works and is elegant, I am also looking forward for a solution that one can easily come up with.
 A: I'd rather evaluate $$\left( {2n + 1} \right)\int_0^1 f (x){x^{2n}}dx$$ since $$\int_0^1 {{x^{2n}}dx}  = \frac{1}{{2n + 1}}$$
Note then that $$(2n+1)\int_0^1 {f\left( x \right){x^{2n}}dx}  - f\left( 1 \right) = \left( {2n + 1} \right)\int_0^1  ({f\left( x \right) - f\left( 1 \right)){x^{2n}}dx} $$
and that $(2n+1)x^{2n}$ converges uniformly to zero on $[0,\delta)$ for any $\delta >0$. Use that $f(x)-f(1)$ is continuous and is zero at $x=1$ to make the integral small in a neighborhood $[\delta,1]$. If the above isn't clear enough:
$$\begin{align}\left|\int_0^1(f(x)-f(1))x^{2n}dx\right|&\leqslant\int_0^1|f(x)-f(1)|x^{2n}dx\\&=\int_0^\delta|f(x)-f(1)|x^{2n}dx+\int_\delta^1|f(x)-f(1)|x^{2n}dx\\&\leqslant M\int_0^\delta x^{2n}dx+\varepsilon\int_\delta^1x^{2n}\\&\leqslant M\int_0^\delta x^{2n}dx+\varepsilon\cdot \int_0^1x^{2n}\\&=M \int_0^\delta x^{2n}dx+\frac{\epsilon}{2n+1}\end{align}$$
Upon multiplication by $2n+1$, you get the limit in question must indeed be $f(1)$. Thus your limit is $f(1)/2$.
A: \begin{eqnarray*}
\lim_{n \to \infty}n\int_{0}^{1}{\rm f}\left(x\right)x^{2n}\,{\rm d}x
& = &
\lim_{n \to \infty}\left\lbrack%
n\int_{0}^{1}{\rm f}\left(1 - \epsilon\right)\left(1 - \epsilon\right)^{2n}\,{\rm d}\epsilon
\right\rbrack
\\[3mm]
& = &
\lim_{n \to \infty}\left\lbrack n\int_{0}^{1}
{\rm e}^{\ln\left(\vphantom{\LARGE A}{\rm f}\left(1\ -\ \epsilon\right)\right)\
         +\
         2n\ln\left(1\ -\ \epsilon\right)}\,{\rm d}\epsilon\right\rbrack
\\[3mm]
& = &
\lim_{n \to \infty}\left\lbrace n\int_{0}^{1}
{\rm e}^{\ln\left(\vphantom{\LARGE A}{\rm f}\left(1\right)\right)\
-\
\left\lbrack\vphantom{\LARGE A}{\rm f}'\left(1\right)/{\rm f}\left(1\right)\right\rbrack\epsilon\ -\ 2n\epsilon}\,{\rm d}\epsilon\right\rbrace
\\[3mm]
& = &
{\rm f}\left(1\right)\lim_{n \to \infty}\left\lbrack n\ \times\ 
{{\rm e}^{-{\rm f}'\left(1\right)/{\rm f}\left(1\right)\ -\ 2n} - 1
 \over
-{\rm f}'\left(1\right)/{\rm f}\left(1\right) -2n}\right\rbrack
=
{\rm f}\left(1\right)\,{1 \over 2}
=
{1 \over 2}
\end{eqnarray*}
