Answer
$2 \ln 2 -\dfrac{1}{2}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{2}_{1} \int^{\ln y}_{1-y} (1) \space dx \space dy = \int^{2}_{1}[x]^{\ln (x)}_{1-y} \space dy $
or, $=\int^{2}_{1}(\ln (y)-1+y)dy=[y \ln y-2y+\dfrac{y^2}{2}]_1^2$
So, $ Area=2 \ln 2 -\dfrac{1}{2}$
