A Variation of Scalar Field Action: Polchinski's AdS/CFT Review

[craigthone]

I am reading Polchinski's review on AdS/CFT https://arxiv.org/abs/1010.6134.
I have a very simple question, and please help me out. Thanks in advanced.

The question abou formula (3.19)
The scalar effective bulk action is given by
$$ S_0=\frac{\eta}{2}\epsilon^{1-D}\int d^Dx \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}$$
The variation of $S_0$ is given by
$$ \delta S_0={\eta}\epsilon^{1-D}\int d^Dx \delta \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}$$

My question is why the two terms from Leibnitz are equal?
The variation of $S_0$ is given by
$$\int d^Dx \delta \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}=\int d^Dx \phi_{\rm cl} \partial_\epsilon \delta \phi_{\rm cl}$$

 

[stevendaryl]

Which two terms are equal? Are you talking about the equality at the bottom of your post?

 

[craigthone]

Which two terms are equal? Are you talking about the equality at the bottom of your post?

yes, sorry for my expression.
$$ \delta S_0=\frac{\eta}{2}\epsilon^{1-D}\int d^Dx \delta \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}+\frac{\eta}{2}\epsilon^{1-D}\int d^Dx \phi_{\rm cl} \partial_\epsilon \delta \phi_{\rm cl} ={\eta}\epsilon^{1-D}\int d^Dx \delta \phi_{\rm cl} \partial_\epsilon \phi_{\rm cl}$$
why does the 2nd equality hold?