I have often seen integrals such as

$$A = \int_{t=0}^{t=T} \frac{\partial}{\partial t} \phi(t,x) dt$$ and I'm wondering if the integral cancels the partial derivative when the variable that the $\frac{\partial}{\partial t}$ is the same as the $dt$ variable?

Does the integral cancel with the partial derivative or do both remain? And, does the answer to that question change based if the $dt$ term had been $dx$?

Similar question: Integral of a partial derivative

  • 5
    $\begingroup$ Sure it does. You get $A(x)=\phi(T,x)-\phi(0,x)$ $\endgroup$ – GReyes Feb 21 at 7:24
  • 2
    $\begingroup$ If you had $dx$ instead, then $\partial\phi/\partial t$ is just some function of $(t,x)$ with no special structure to be able to "cancel" it or express otherwise. $\endgroup$ – GReyes Feb 21 at 7:28
  • $\begingroup$ got it so in general $$A = \int_{t=0}^{t=T} \frac{\partial}{\partial x} \phi(t,x) dt \ne \phi(T,x)-\phi(0,x) $$ and also $$A = \int_{t=0}^{t=T} \frac{\partial}{\partial t} \phi(t,x) dx \ne \phi(T,x)-\phi(0,x) $$ right? $\endgroup$ – jaja Mar 3 at 0:15
  • $\begingroup$ Absolutely right! $\endgroup$ – GReyes Mar 3 at 8:12

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