= Solution
Ignoring the small radiative back-reaction, <conservation of energy> gives $\dot x=\sqrt{2(E-f(x))/m}$ and Newton's equation gives $\ddot x=-f'(x)/m$. Integrating the Larmor formula with $dt=dx/\dot x$ yields, for $E>V_0$,
$$
ΔE_{\rm Rad}={q^2μ_0\over6πm^2c}\sqrt{m/2}
\int_0^L{[f'(x)]^2\over\sqrt{E-f(x)}}\,dx.
$$
If $E<V_0$, the particle turns at $x_*$ defined by $f(x_*)=E$ and traverses the accelerated region twice, so the same formula holds with twice the integral from $0$ to $x_*$.
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