Let $X$ be the vector field given by $X = b \cdot \nabla_x + \partial_t$ where $b \in \mathbb{R}^n$ is fixed. Let $f \in C^1(\mathbb{R}^{n+1})$. Assume that $u \in C^1(\mathbb{R}^n \times [0,\infty))$ and $u|_{t=0} = g \in C^1(\mathbb{R}^n)$. Then $Xu = f$ if and only if $u(x,t) = g(x-tb) + \int_0^t f(x+(s-t)b,s) \ ds$.
I have shown that if $u \in C^1(\mathbb{R}^n \times [0, \infty))$ solves $Xu = f$, then $u(x,t) = g(x-tb) + \int_0^t f(x+(s-t),s) \ ds$, where $u|_{t=0} = g \in C^1(\mathbb{R}^n)$. What I am having some difficulty doing is showing that such a $u$ is actually a solution to the given partial differential equation $Xu = f$.
It is clear to me that $Xg = 0$. However, I don't know exactly how to deal with the integral term with applying the operator to it. I have tried using difference quotients to show that when the operator is applied to the integral, I obtain $f(x,t)$, but it seems to get messy.
Can anyone point me in the correct direction?
Check this link to see how to deal with differentiating an integral over an evolving domain: https://en.wikipedia.org/wiki/Time_evolution_of_integrals.
We use this to see that $$\partial_t\left(\int_0^tf(x+(s-t)b,s)\,ds\right)=\int_0^t\frac{\partial f(x+(s-t)b,s)}{\partial t}\,ds+f(x,t),$$ now notice that $$\int_0^t\frac{\partial f(x+(s-t)b,s)}{\partial t}\,ds=-\int_0^tb\cdot\nabla_xf(x+(s-t)b,s)\,ds.$$ It is also more clear that $$b\cdot\nabla_x\int_0^tf(x+(s-t)b,s)\,ds=\int_0^tb\cdot\nabla_xf(x+(s-t)b,s)\,ds.$$ Thus $$(b\cdot\nabla_x+\partial_t)\left(\int_0^tf(x+(s-t)b,s)\,ds\right)=f(x,t).$$
Let's try to prove the extension of the FTOC, by the FTOC (let $F$ be the antiderivative of $f$) $$\int_{a(t)}^{b(t)}f(t,x)\,dx=F(t,b(t))-F(t,a(t)),$$ then $$\frac{d}{dt}\left(\int_{a(t)}^{b(t)}f(t,x)\,dx\right)=\frac{d}{dt}\left(F(t,b(t))-F(t,a(t))\right)$$ $$=\frac{\partial F}{\partial t}(t,b(t))-\frac{\partial F}{\partial t}(t,a(t))+\frac{\partial F}{\partial x}(t,b(t))b'(t)-\frac{\partial F}{\partial x}(t,a(t))a'(t)$$ $$=\int_{a(t)}^{b(t)}\frac{\partial f(t,x)}{\partial t}+f(t,b(t))b'(t)-f(t,a(t))a'(t).$$