Let $f_n \to f$ pointwise on $[0,1]$. If $f_n, f$ are continuous, is it true that $\int_0^1 f_n(x) dx \to \int_0^1 f(x)dx$? 
Let $\{f_n\}$ be a sequence of continuous functions on on $[0,1]$. Let $f_n \to f$ pointwise. If $f$ is continuous on $[0,1]$, is it true that $$\int_0^1 f_n(x) dx  \to \int_0^1 f(x)dx?$$

I couldn't think of a counter-example, so my inclination is that it is true. If I can show that $f_n \to f$ uniformly, then I would be done, since I can choose an $N$ such that for all $n > N$ and $x\in [0,1]$ we have $|f_n(x) - f(x)| < \varepsilon$, which gives
$$ \left| \int_0^1 f_n(x) - f(x) dx \right| \leq \int_0^1 |f_n(x) - f(x)|dx < \varepsilon$$ 
Can it be shown that $f_n \to f$ uniformly since we're working on a compact set and $f$ is continuous?  Or can a counter-example be constructed from here?

Let $f_n(x) = n^2x^n(1-x).$ Then $f_n\to 0$ pointwise everywhere in $[0,1],$ but $\int_0^1 f_n(x)\,dx \to 1.$
