# Does an integral of a partial derivative make the partial derivative disappear?

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

• Sure it does. You get $A(x)=\phi(T,x)-\phi(0,x)$ – GReyes Feb 21 at 7:24
• 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. – GReyes Feb 21 at 7:28
• 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? – jaja Mar 3 at 0:15
• Absolutely right! – GReyes Mar 3 at 8:12