Show that $x_n \rightarrow 0$ Let $f:[0,1] \rightarrow \mathbb{R}$ continuous, such that $f(0)=0$
We set $x_n=\int_0^1{f(x^n)}dx$
Show that $x_n \rightarrow 0$
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The function $f$ is continuous at a closed interval $\Rightarrow $ $f$ is bounded $\Rightarrow \exists M>0: |f(x)| \leq M, \forall x \in [0,1]$
The function $f$ is continuous and $f(0)=0$ $\Rightarrow \lim_{x \rightarrow 0}{f(x)}=0 \Rightarrow \forall \epsilon >0 \text{ } \exists \delta >0: \forall x \in [0,1], |x-0| < \delta \Rightarrow |f(x)-f(0)|< \epsilon$
So $0 \leq x< \delta \Rightarrow |f(x)| < \epsilon$
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How can I continue?

Hints using the notation and stuff you already did:
For all $\;\epsilon>0\;$ there exists $\;K\in\Bbb N\;\;s.t.\;\;n>K\implies |f(x^n)|<\epsilon\;$ (why?) , so:
$$\left|\int\limits_0^1f(x^n)\,dx\right|\le\int\limits_0^1|f(x^n)|dx\le\int\limits_0^1\epsilon\,dx=\epsilon$$
