Answer
True
Work Step by Step
The statement is true as shown below:
$0=\iint_{R} f(x, y) d A$
$0=\iint_{R_{1}+R_{2}} f(x, y) d A$
$0=\iint_{R_{1}} f(x, y) d A+\iint_{R_{2}} f(x, y) d A$
$-\iint_{R_{2}} f(x, y) d A=\iint_{R_{1}} f(x, y) d A$