Answer
$2$
Work Step by Step
Consdier $A=(\dfrac{1}{2}) \int_m^n (r^2) d \theta=\int_{0}^{\pi/2} (\dfrac{1}{2}) 4 \sin 2 \theta d \theta$
$\implies A= \int_{0}^{\pi/2} (\dfrac{1}{2}) 4 \sin 2 \theta d \theta$
This implies that
$A=(2)[- \cos (2\theta) ]_{0}^{(\pi/2)} =2$