Answer
$m \ A(D) \leq \iint_{D} f(x,y) d A \leq M \ A(D)$
Therefore, the proof has been proved.
Work Step by Step
When $m \leq f(x,y) \leq M$ on the region $S$, then we can write as follows:
$\implies m \cdot A \leq \iint_{S} f(x,y) d A \leq M \cdot A $
and $A$ represents the area of the region $S$.
Now, $m \cdot dA \leq \iint_{D} f(x,y) d A \leq M \cdot \ dA $
$\implies m \iint_{D} 1 \cdot dA \leq \iint_{D} f(x,y) d A \leq M \iint_{D} 1 \cdot dA $
$\implies m \ A(D) \leq \iint_{D} f(x,y) d A \leq M \ A(D)$
Therefore, the proof has been proved.
