Q:

Where is the $\int_0^t\Delta F$?

$Rm$ is curvature tensor , and the $g_{ij}$ meet Ricci flow.
For getting the inequality 1 in the below paper's 191th page , I compute that:
$$
\because ~~~\frac{\partial F}{\partial t}\leq \Delta F+\overline{C}M^3\\
\therefore ~~~
\int_0^t\frac{\partial F}{\partial t}\leq 
\int_0^t\Delta F+\int_0^t\overline{C}M^3\\
\therefore ~~~
F(t)-F(0)\leq \overline{C}M^3t+\int_0^t\Delta F  \\
\therefore  ~~~~
F(t)\leq F(0)+ \overline{C}M^3t+\int_0^t\Delta F
$$
But in the below picture ,there is not $\int_0^t\Delta F$, Why ?
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A:

Since the goal is to get an estimate $|\nabla Rm|^2 \le C/t$ which is in particular independent of spatial position, I think that what's going on here is a maximum principle argument. As written the inequality you have highlighted does not hold pointwise - you're correct that it should be $$F(x,t) \le F(x,0) + \bar C M^3 t + \int_0^t \Delta F(x,s)ds.$$
However, if we replace $F(x,0)$ with the maximum value of $F$ at time zero it is true: $$ \tag{1} F(x,t) \le \sup_x F(x,0) + \bar C M^3 t. $$
Since the second inequality was estimating $F(0) = A|Rm|^2 \le AM^2$ anyway, this weaker version is still bounded above by $F \le (A+\bar C)M^2$.
You can deduce $(1)$ by noticing that $G(x,t) = F(x,t) - \bar C M^3 t$ satisfies $\partial_t G \le \Delta G$ and applying the weak parabolic maximum principle to $G$: since $G$ is bounded above by the constant $\sup_x F(x,0)$ at the initial time, this bound persists for all time.

