Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A \simeq 2.9375$
Work Step by Step
Write $f\left( {x,y} \right) = xy$.
From Figure 1, we see that $\Delta x = \Delta y = 0.5$.
Referring to Figure 1, there are $4 \times 4$ subrectangles, so using the midpoints as sample points, the Riemann sum to estimate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A$ is ${S_{4,4}}$ given by
${S_{4,4}} = \mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^4 f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$
$ = 0.25(f\left( {1.25,0.25} \right) + f\left( {1.75,0.25} \right)$
${\ \ \ }$ $ + f\left( {0.75,0.75} \right) + f\left( {1.25,0.75} \right) + f\left( {1.75,0.75} \right)$
${\ \ \ }$ $ + f\left( {0.75,1.25} \right) + f\left( {1.25,1.25} \right) + f\left( {1.75,1.25} \right)$
${\ \ \ }$ $ + f\left( {0.75,1.75} \right) + f\left( {1.25,1.75} \right))$
$ = 0.25(0.3125 + 0.4375 + 0.5625 + 0.9375 + 1.3125 + 0.9375$
${\ \ \ }$ $ + 1.5625 + 2.1875 + 1.3125 + 2.1875)$
$ = 2.9375$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A \simeq 2.9375$.
