Use D'Alembert's solution to find specific solution to wave equation Consider the PDE $$u_{tt}=c^2u_{xx}$$ where $u(x,0)=\phi(x)$ and $u_t(x,0)=\psi(x)$. I'm asked to derive the D'Alembert solution, which I have done and found to be $$u(x,t)=\frac{1}{2}(\phi(x-ct)+\phi(x+ct))+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s) \,  ds$$
I'm then asked to show that if $\phi(x)=0$ and $\psi(x)=\delta(x)$ then $$u(x,t)=\frac{1}{2c}\left(H(x+ct)-H(x-ct) \right)$$ where $H$ is the Heaviside function.
From my previous work I get that $$u(x,t)=\frac{1}{2c}\int_{x-ct}^{x+ct}\delta(s) \, ds$$
So in order to get the solution it seems as though it's the case that $\int_0^{x+ct}\delta(s) \, ds=H(x+ct)$.
Is this true? If so, why?

\begin{align}
   \int_{x-ct}^{x+ct}\delta(s)ds & = \int_{-\infty}^{x+ct}\delta(s)ds-\int_{-\infty}^{x-ct}\delta(s)ds \\
              & = H(x+ct)-H(x-ct)
\end{align}
