If $L_1\in R$ and $L_2$ is non-trivial language then $L_1\leq L_2$
We know there are two $x_1,x_2\in \Sigma^*$ such that $x_1\in L_2$ and $x_2\notin L_2$, since $L_2$ is not trivial. Now, we will build the following reduction:
$f(x)=\cases{x_1 &amp; $x\in L(M_1)=L_1$ \\ x_2 &amp; otherwise}$
Notice that $f$ is computable: calculate $M_1(x)$, and output either $x_1$ or $x_2$ according to the answer you got.