general relativity - Why does the integral symbol disappear when applying a functional derivative?

it is known that variation is defined by following:

but could anyone tell me why the integral symbol disappears after following functional derivative?

Answer

Define functional $$ G[g]~:=~\int \!d^4x ~\sqrt{-g(x)}.\tag{0}$$ Method 1: $$ \int \!d^4x ~\color{red}{\frac{\delta G[g]}{\delta g_{\mu\nu}(x)}} \delta g_{\mu\nu}(x)~=~\delta G[g]~\stackrel{(0)}{=}~ \int \!d^4x ~\color{red}{\frac{\partial\sqrt{-g(x)}}{\partial g_{\mu\nu}(x)}} \delta g_{\mu\nu}(x). \tag{1}$$ Method 2: $$\color{red}{\frac{\delta G[g]}{\delta g_{\mu\nu}(y)}} ~\stackrel{(0)}{=}~ \int \!d^4x ~\frac{\delta\sqrt{-g(x)}}{\delta g_{\mu\nu}(y)} ~=~ \int \!d^4x ~\frac{\partial\sqrt{-g(x)}}{\partial g_{\kappa\lambda}(x)}\frac{\delta g_{\kappa\lambda}(x)}{\delta g_{\mu\nu}(y)} $$ $$~=~ \int \!d^4x ~\frac{\partial\sqrt{-g(x)}}{\partial g_{\mu\nu}(x)}\delta^4(x\!-\!y) ~=~\color{red}{\frac{\partial\sqrt{-g(y)}}{\partial g_{\mu\nu}(y)}}.\tag{2}$$