Integral limit of function on unit interval

By continuity, exists $M>0$ such that $|f(x)|<M$ for all $x$. Then $\int_0^1|f(x^n)|\,\mathrm{dx}\le \int_0^1M\,\mathrm{d}x$, so, by Dominated Convergence Theorem, $\lim_{n\to\infty}\int_0^1|f(x^n)|\,\mathrm{dx}=\int_0^1\lim_{n\to\infty}|f(x^n)|\,\mathrm{dx}=f(0)=1$

Therefore $\int_0^1f(x^n)\,\mathrm{dx}$ exists and is a number in $[-1,1]$ (you can't calculate its value cause $f$ can take negative values).

In both answers by sinbadh and Xander Henderson, I'm a little annoyed by the statement $\lim\limits_{n\to\infty} x^n=0$, this is only true for $x\in[0,1)$.

In fact I would prefer to introduce $f_n(x)=f(x^n)$ as a proper sequence of functions.

$\|f_n\|_{\infty}\le M$ since $f$ is continuous on $[0,1]$.

And now $f_n\to \ell:\begin{cases} f(0)=1& x\in[0,1)\\ f(1)=3&x=1\end{cases}$

With $\displaystyle \int_{[0,1]} \ell(x)\mathop{dx}=\int_{[0,1)} \ell(x)\mathop{dx}=1$ since $\ell(1)=3$ is finite.