I am considering the reaction-diffusion equation \begin{equation}\tag{1} u_{t}=u_{xx}+f(u) \end{equation} with non-linearity \begin{equation}\tag{2} f(u)=-u+H(u-\alpha),\quad0<\alpha<1, \end{equation} where $H(u)$ is the Heaviside function.
The values $u=0$ and $u=1$ are stable spatially homogeneous equilibria.
Find speed $c$ such that $v(\xi), \xi:=x+ct$, is a traveling front with $\lim_{\xi\to-\infty}v(\xi)=0$ and $\lim_{\xi\to\infty}v(\xi)=1$.
Substituting $v(\xi)$ into (1) gives \begin{equation}\tag{3} v_{\xi\xi}-cv_{\xi}+f(v)=0 \end{equation} or, equivalently, the system \begin{align}\tag{3'} &v_{\xi}=w\\ &w_{\xi}=cw-f(v). \end{align}
If a such a front exists,
\begin{equation}\tag{4} c\int_{-\infty}^{\infty}w^2\, d\xi=\int_{0}^1f(v)\, du. \end{equation} This follows by multiplying (3) by $v_{\xi}$ and integrating from $\xi=-\infty$ to $\xi=\infty$.
For most functions $f(u)$, it is necessary to calculate $c$ numerically. However, in the special case of (2), it is said, the speed $c$ can be calculated directly to be \begin{equation}\tag{5} c=\frac{1-2\alpha}{\sqrt{\alpha-\alpha^2}}. \end{equation}
I tried to verify (5):
I think what I can compute is $$ \int_{0}^1f(v)\, dv=-\underbrace{\int_{0}^1 v\, dv}_{=\frac{1}{2}}+\underbrace{\int_0^1H(v-\alpha)\, dv}_{=\int_{-\alpha}^{1-\alpha}H(y)\, dy}=-\frac{1}{2}+[H(y)y]_{-\alpha}^{1-\alpha}=-\frac{1}{2}+1-\alpha=\frac{1-2\alpha}{2} $$
So what I got up to now is only $$ c=\frac{1-2\alpha}{2\int_{-\infty}^{\infty}w^2\, d\xi}. $$
But I do not know how to compute $$ \int_{-\infty}^{\infty}w^2\, d\xi=\int_{-\infty}^{\infty}(v_{\xi})^2\, d\xi. $$ Maybe it helps to use (3'), i.e. that $$ w=\frac{w_{\xi}+f(v)}{c}, $$ giving \begin{align} w^2&=\frac{(w_{\xi}+f(v))^2}{c^2}\\ &=\frac{1}{c^2}(f(v)^2+2w_{\xi}f(v)+(w_{\xi})^2)\\ &=\frac{1}{c^2}\left[v^2+(w_{\xi})^2-2w_{\xi}v+H(v-\alpha)\cdot (-2v+H(v-\alpha)+2w_{\xi})\right] \end{align}
In this case, you can find a traveling wave solution explicity. Look for a solution $v$ that satisfies $v(0) = \alpha$ and is continuously differentiable there. Then for $\xi < 0$ the equation becomes $v'' - cv' - v = 0$ and for $\xi > 0$ the equation becomes $v'' - cv' - v + 1 = 0$.
These are just ordinary differential equations with constant coefficients, so we can write down the solutions. For general $c$, the separate solutions with the correct limiting behavior at $ \pm \infty$ are $v_1(\xi) = \alpha e^{z_1 \xi}$ for $\xi < 0$ and $v_2(\xi) = 1 + (\alpha - 1) e^{z_2 \xi}$ for $\xi > 0$, where $z_1 = \frac{c}{2} + \sqrt{\frac{c^2}{4} +1}$ and $z_1 = \frac{c}{2} - \sqrt{\frac{c^2}{4} +1}$.
Then $v_1(0) = v_2(0)$ by construction. You also want $v_1'(0) = v_2'(0)$, which leads to the additional condition $\alpha z_1 = (\alpha - 1)z_2 $.
Now solve this equation for $c$ and the formula for the wave speed follows.