Answer
We choose the point $P = \left( {0.8622,0.5} \right) \in {\cal D}$ such that:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = f\left( P \right)\cdot Area\left( {\cal D} \right)$
Thus, the Mean Value Theorem for Double Integrals is verified.
Work Step by Step
We have $f\left( {x,y} \right) = {{\rm{e}}^{x - y}}$ and the triangle bounded by $y=0$, $x=1$, and $y=x$. We can consider the triangle as a vertically simple region whose domain description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le x} \right\}$
To verify the Mean Value Theorem for Double Integrals for $f\left( {x,y} \right)$, we first evaluate the double integral of $f\left( {x,y} \right)$ as an iterated integral over ${\cal D}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = 0}^x {{\rm{e}}^{x - y}}{\rm{d}}y} \right){\rm{d}}x$
$ = - \mathop \smallint \limits_{x = 0}^1 \left( {{{\rm{e}}^{x - y}}|_0^x} \right){\rm{d}}x$
$ = - \mathop \smallint \limits_{x = 0}^1 \left( {1 - {{\rm{e}}^x}} \right){\rm{d}}x$
$ = - \left( {\left( {x - {{\rm{e}}^x}} \right)|_0^1} \right)$
$ = - \left( {1 - {\rm{e}} + 1} \right)$
$ = {\rm{e}} - 2$
Notice that the area of ${\cal D}$ is the area of a triangle with base $1$ and height $1$. So, $Area\left( {\cal D} \right) = \frac{1}{2}$.
By Theorem 5, the Mean Value Theorem for Double Integrals, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = f\left( P \right)\cdot Area\left( {\cal D} \right)$,
where $f\left( P \right) = \bar f$.
Thus,
${\rm{e}} - 2 = \frac{1}{2}f\left( P \right)$
$f\left( P \right) = 2{\rm{e}} - 4$
Now, Theorem 5 guarantees that there exist a point $P = \left( {a,b} \right) \in {\cal D}$ such that $f\left( P \right) = {{\rm{e}}^{a - b}} = 2{\rm{e}} - 4$. Taking the logarithm on both sides, we get
$a - b = \ln \left( {2{\rm{e}} - 4} \right)$
$a = \ln \left( {2{\rm{e}} - 4} \right) + b$
We choose $b = \frac{1}{2}$, so then $a = \ln \left( {2{\rm{e}} - 4} \right) + \frac{1}{2} \simeq 0.8622$.
Since the point $P = \left( {0.8622,0.5} \right) \in {\cal D}$, the following is satisfied:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = f\left( P \right)\cdot Area\left( {\cal D} \right)$
Thus, the Mean Value Theorem for Double Integrals is verified.
