L. Evans book on PDE shows the solution for the nonhomogeneous transport equation in the following way:
\begin{cases} u_t+b\cdot Du&=f & \text{in } \mathbb{R}^n\times(0,\infty)\\ u&=g &\text{on } \mathbb{R}^n\times\{t=0\} \end{cases} Fix $(x,t)\in\mathbb{R}^n\times(0,\infty)$ and set $z(s)=u(x+sb,t+s)$ for $s\in\mathbb{R}$. Then, $\dot{z}(s)=Du(x+sb,t+s)\cdot b+u_t(x+sb,t+s)=f(x+sb,s+t)$. Consequently $u(x,t)-g(x-tb)=z(0)-z(-t)=\int_{-t}^0\dot{z}\text{ds}=\int_{-t}^0f(x+sb,t+s)\text{ds}\stackrel{(!)}{=}\int_0^tf(x+(s-t)b,s)\text{ds}$
and so $u(x,t)=g(x-tb)+\int_o^tf(x+(s-t),s)\text{ds}$ solves the initital value problem.
I'm having trouble with the equality (!), why does this hold? I am probably missing come caclulas knowledge here. I think I do know $\int_{-x}^0=-\int_0^x$, but what happens in this case exactly?
(I'm studying PDE on my own with the help of L. Evans' book, W. Craig's book and some lecture notes from my university for those curious why I need help)