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 $u=u(x,t)$.
How do I integrate $\int _0 ^1 \frac{d}{dx} \frac{1}{3} u^3 dx$, where $u = u(x,t)$?
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$\begingroup$
integration
derivatives
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3No different, because the integral is with respect to $x$, so it doesn't "care" about the $t$-dependency. So, you can think about $t$ just as some constant you can set (for the purposes of taking the integral). – 2017-01-07
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3However, they really should write partial derivative $\partial/\partial x$, and not $d/dx$. – 2017-01-07
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0For your purposes, $t$ is a constant. – 2017-01-07
1 Answers
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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 $s=u_t(x)\Rightarrow ds=u_t'(x)dx$....