Answer
We verify the following by evaluating directly:
$\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}} = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 2}^3 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 + x - y}}$
Work Step by Step
From the integral $\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}}$, we see that the order of integration is $dydx$ and the domain of integration is a rectangle: $2 \le x \le 3$, $0 \le y \le 2$.
Evaluate the integral in the order $dydx$:
$\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}} = - \mathop \smallint \limits_{x = 2}^3 \left( {\ln \left( {1 + x - y} \right)|_0^2} \right){\rm{d}}x$
(1) ${\ \ \ \ }$ $\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}} = - \mathop \smallint \limits_{x = 2}^3 \left( {\ln \left( {x - 1} \right) - \ln \left( {x + 1} \right)} \right){\rm{d}}x$
The antiderivative of $\ln \left( {x + a} \right)$ is $\left( {x + a} \right)\ln \left( {x + a} \right) - x$, because
$\frac{d}{{dx}}\left( {\left( {x + a} \right)\ln \left( {x + a} \right) - x} \right) = \frac{{x + a}}{{x + a}} + \ln \left( {x + a} \right) - 1 = \ln \left( {x + a} \right)$
Therefore, integral (1) becomes
$\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}} = - \left( {\left( {x - 1} \right)\ln \left( {x - 1} \right) - x - \left( {x + 1} \right)\ln \left( {x + 1} \right) + x} \right)|_2^3$
$ = - \left( {2\ln 2 - 4\ln 4 + 3\ln 3} \right) = 4\ln 4 - 3\ln 3 - 2\ln 2$
So, $\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}} = 4\ln 4 - 3\ln 3 - 2\ln 2$.
Using Fubini's Theorem (Theorem 3 in Section 16.1), we can change the order of integration to $dxdy$. Hence,
$\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}} = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 2}^3 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 + x - y}}$
Evaluate the integral in the order $dxdy$:
$\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 2}^3 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 + x - y}} = \mathop \smallint \limits_{y = 0}^2 \left( {\ln \left( {1 + x - y} \right)|_2^3} \right){\rm{d}}y$
(2) ${\ \ \ \ }$ $\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 2}^3 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 + x - y}} = \mathop \smallint \limits_{y = 0}^2 \left( {\ln \left( {4 - y} \right) - \ln \left( {3 - y} \right)} \right){\rm{d}}y$
The antiderivative of $\ln \left( {b - y} \right)$ is $ - \left( {b - y} \right)\ln \left( {b - y} \right) - y$, because
$\frac{d}{{dy}}\left( { - \left( {b - y} \right)\ln \left( {b - y} \right) - y} \right) = \frac{{b - y}}{{b - y}} + \ln \left( {b - y} \right) - 1 = \ln \left( {b - y} \right)$
Therefore, integral (2) becomes
$\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 2}^3 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 + x - y}} = \left( { - \left( {4 - y} \right)\ln \left( {4 - y} \right) - y + \left( {3 - y} \right)\ln \left( {3 - y} \right) + y} \right)|_0^2$
$ = - 2\ln 2 + 4\ln 4 - 3\ln 3$
So, $\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 2}^3 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 + x - y}} = 4\ln 4 - 3\ln 3 - 2\ln 2$.
Hence,
$\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}} = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 2}^3 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 + x - y}} = 4\ln 4 - 3\ln 3 - 2\ln 2$
Thus, we verified Fubini's Theorem (Theorem 3 in Section 16.1):
$\mathop \smallint \limits_{x = 2}^3 \mathop \smallint \limits_{y = 0}^2 \frac{{{\rm{d}}y{\rm{d}}x}}{{1 + x - y}} = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 2}^3 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 + x - y}}$
