Answer
Please see the figure attached.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln \left( {x + y} \right){\rm{d}}A \simeq 5$
Work Step by Step
We have $f\left( {x,y} \right) = \ln \left( {x + y} \right)$ and the domain ${\cal D} = \left\{ {1 \le y \le {\rm{e}},y \le x \le 2y} \right\}$.
Considering ${\cal D}$ as a horizontally simple region, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln \left( {x + y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 1}^{\rm{e}} \mathop \smallint \limits_{x = y}^{2y} \ln \left( {x + y} \right){\rm{d}}x{\rm{d}}y$
The antiderivative of $\ln \left( {x + y} \right)$ is $x\ln \left( {x + y} \right) + y\ln \left( {x + y} \right) - x$ because
$\frac{d}{{dx}}\left( {x\ln \left( {x + y} \right) + y\ln \left( {x + y} \right) - x} \right)$
$ = \ln \left( {x + y} \right) + \frac{x}{{x + y}} + \frac{y}{{x + y}} - 1$
$ = \ln \left( {x + y} \right)$
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln \left( {x + y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 1}^{\rm{e}} \left( {\left( {x\ln \left( {x + y} \right) + y\ln \left( {x + y} \right) - x} \right)|_y^{2y}} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 1}^{\rm{e}} \left( {3y\ln \left( {3y} \right) - 2y - 2y\ln \left( {2y} \right) + y} \right){\rm{d}}y$
(1) ${\ \ \ \ }$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln \left( {x + y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 1}^{\rm{e}} \left( {3y\ln \left( {3y} \right) - 2y\ln \left( {2y} \right) - y} \right){\rm{d}}y$
Write $u = \ln \left( {ay} \right)$ and $dv = ydy$. So, $du = \frac{1}{y}dy$ and $v = \frac{1}{2}{y^2}$. Using Integration by Parts Formula (Section 8.1),
$\smallint u{\rm{d}}v = uv - \smallint v{\rm{d}}u$
we get
$\smallint y\ln \left( {ay} \right){\rm{d}}y = \frac{1}{2}{y^2}\ln \left( {ay} \right) - \frac{1}{2}\smallint y{\rm{d}}y$
$\smallint y\ln \left( {ay} \right){\rm{d}}y = \frac{1}{2}{y^2}\ln \left( {ay} \right) - \frac{1}{4}{y^2}$
Thus, equation (1) becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln \left( {x + y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 1}^{\rm{e}} \left( {3y\ln \left( {3y} \right) - 2y\ln \left( {2y} \right) - y} \right){\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln \left( {x + y} \right){\rm{d}}A = \left( {\frac{3}{2}{y^2}\ln \left( {3y} \right) - \frac{3}{4}{y^2} - {y^2}\ln \left( {2y} \right) + \frac{1}{2}{y^2} - \frac{1}{2}{y^2}} \right)|_1^{\rm{e}}$
$ = \frac{3}{2}{{\rm{e}}^2}\ln \left( {3{\rm{e}}} \right) - \frac{3}{4}{{\rm{e}}^2} - {{\rm{e}}^2}\ln \left( {2{\rm{e}}} \right) - \frac{3}{2}\ln 3 + \frac{3}{4} + \ln 2 \simeq 5$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln \left( {x + y} \right){\rm{d}}A \simeq 5$.
