Sequence of functions; uniform convergence and integrating

Appa

1. The problem statement, all variables and given/known data
The sequence f_n: [-1,1] -> R, f_n(x)= nxe^-nx2 converges pointwise to f(x)= 0, x in [-1,1]. Can you verify the following:

lim_{n->[tex]\infty[/tex]} ([tex]\int[/tex][tex]^{1}_{0}[/tex]f_n(x)dx) = [tex]\int[/tex][tex]^{1}_{0}[/tex] (lim_{n->[tex]\infty[/tex]} f_n(x))dx

2. Relevant equations
Theorem: If f_n is continuous on the interval D for every n and f_n converges uniformly to f on D=[a,b], then

lim_{n->[tex]\infty[/tex]} ([tex]\int[/tex][tex]^{a}_{x}[/tex]f_n(t)dt) = [tex]\int[/tex][tex]^{a}_{x}[/tex] (lim_{n->[tex]\infty[/tex]} f_n(t))dt = [tex]\int[/tex][tex]^{a}_{x}[/tex]f(t)dt) for every x in D.

3. The attempt at a solution
The main idea is to find out whether the sequence is uniformly convergent on D= [0,1]. f_n = nxe^-nx2 is continuous for every n because e^t is always continuous. I tried something like this to verify uniform convergence:
f_n(x) is uniformly convergent on D if there is a positive number _{[tex]\epsilon[/tex]} >0 and an index N, so that
|f_n(x) -f(x)|<_{[tex]\epsilon[/tex]} for every n[tex]\geq[/tex]N and every x in [0,1]
[tex]\Leftrightarrow[/tex] |nxe^-nx2-0|<_{[tex]\epsilon[/tex]}
[tex]\Leftrightarrow[/tex] nxe^-nx2 <_{[tex]\epsilon[/tex]}

But I don't know how to go forward. How can I show that the greater the index n gets, the closer f_n gets to 0? I tried comparison but couldn't find any sequence, the term of which would be greater than f_n so that the sequence would still converge.

Dick

Finding a uniform limit for f_n isn't that hard. You just need to find the extrema of the functions as a function of n. Take the derivative, set it equal to zero etc. But I'm going to guess that what you are really expected to do is actually calculate the integral of f_n and verify that the limit of the integrals is zero.