Counter example for The Composition Theorem for Riemann Integrability

Question

Based on The Composition Theorem (found in the last Lemma here) we can say that if a function f is Riemann Integrable, then $f^n$ is Riemann Integrable as well.
The converse is not true, but I can't think of a counter example, any help would be appreciated.

Answer 1

Define $f:[0,1]\to\mathbb{R}$ by $f(x)=1$ if $x\in\mathbb{Q}$ and $f(x)=-1$ if $x\not\in\mathbb{Q}$. Then $f$ is not Riemann integrable on $[0,1]$, but $f(x)^2$ is the constant function $1$, hence is integrable on $[0,1]$.