Pointwise convergence on a compact set to a continuous function AND integration I am looking for a counter-example to this statement:

If a sequence $f_n$ of continuous functions converges point-wise to the continuous function $f$ on a compact set $K$; then, the integral $\int_Kf_n$ converges to $\int_Kf$.

 A: Expanding on Kelenner's answer in the comments, using $f(x) = n^2 (1-x) x^n$ on $[0, 1]$. 
First we need to check pointwise convergence. We'd like to guess $f = 0$ works, and sketching a graph makes this seem sensible. For $x = 0$ or $1$, $f_n(0) = f_n(1) = 0$ for any $n$, so immediately done. For $x \in (0, 1)$, $n^2 (1-x) x^n < n^2 x^n$. Observe that as a function of $n$, $x^n$ is roughly exponential decay while $n^2$ is polynomial growth, so $n^2 x^n \to 0$ for any fixed $x$, giving us pointwise convergence.
Clearly $f(x) = 0$ is a continuous function, so we have pointwise convergence on $[0, 1]$. Finally, we need to check integrals:
$$\begin{align} \int_0^1 f_n(x) \, dx &= \int_0^1 n^2 (1-x) x^n \, dx \\ 
&= n^2 \int_0^1 x^n - x^{n+1} \, dx \\
&= n^2 \left[\frac{1}{n+1} x^{n+1} - \frac{1}{n+2} x^{n+2}\right]_0^1 \\
&= n^2 \times \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \\
&= \frac{n^2}{(n+1)(n+2)}\\ 
&\to 1 \text{ as } n \to \infty \end{align}$$
while $\int_0^1 f(x) \, dx = 0$, giving the required counterexample.
