How do I integrate $\int _0 ^1 \frac{d}{dx} \frac{1}{3} u^3 dx$, where $u = u(x,t)$? I have a note saying that this equals $\frac{1}{3} u^3(1,t) - u^3(0,t)$, but I thought it might be different from how you usually take the integral of a derivative since $u=u(x,t)$.

As is being mentioned, make the irrelevance of $t$ inside the interval clear by parametrizing $u$ in terms of $t$ and writing $u(t,x)=u_t(x)$. Then your integral is
$$
1/3\int_0^1\frac{d}{dx}[u_t(x)^3]dx\stackrel{\text{chain rule}}{=}\int_0^1u_t'(x)[u_t(x)]^2dx
$$
Now enforce the substitution $s=u_t(x)\Rightarrow ds=u_t'(x)dx$....
