The Riemannian metric is $g_{ij}$,its inverse is $g^{ij}$,and the induced measure is $du=u(x)du$ where $u(x)=\sqrt{det(g_{ij})}$.The scalar curvature is $R=g^{ij}R_{ij}$ . $r=\frac{\int R du}{\int du}$ is the average scalar.Consider the equation $\frac{\partial }{\partial t}g_{ij}=\frac{2}{n}rg_{ij}-2R_{ij}$ . I understand that $\frac{\partial }{\partial t}\log u=\frac{1}{2}g^{ij}\frac{\partial}{\partial t}g_{ij}=r-R$ . How to show that :
$$ \frac{\partial }{\partial t}\int du=\int (r-R)du=0 $$
Hint:
$$\frac{\partial}{\partial t} \int_M du = \int_M \frac{\partial}{\partial t} du = \int_M \left(\frac{\partial}{\partial t}u(x)\right) \,dx^1\cdots dx^n$$
From here you should be able to derive that
$$\frac{\partial}{\partial t} \int_M du = \int_M (r-R) du.$$
That the last term is zero has nothing to do with Ricci flow: For any function $f$, if $A$ is the average of $f$, then
$$\int_M (f-A) du = 0.$$