Answer
${S_{4,3}} = 13.5$
Work Step by Step
We have $f\left( {x,y} \right) = xy$ over ${\cal R} = \left[ {1,3} \right] \times \left[ {1,2.5} \right]$.
The Riemann sum ${S_{4,3}}$ implies that $N=4$, $M=3$.
Using the regular partition, we get the dimensions of the subrectangles:
$\Delta x = \frac{{3 - 1}}{4} = \frac{1}{2}$, ${\ \ \ \ }$ $\Delta y = \frac{{2.5 - 1}}{3} = \frac{1}{2}$
The Riemann sum ${S_{4,3}}$ to estimate the double integral of $f\left( {x,y} \right) = xy$ is given by
${S_{4,3}} = \mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^3 f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = \frac{1}{4}\mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^3 f\left( {{P_{ij}}} \right)$
Using upper-right vertices of the subrectangles as sample points, we get
${S_{4,3}} = \frac{1}{4}(f\left( {\frac{3}{2},\frac{3}{2}} \right) + f\left( {2,\frac{3}{2}} \right) + f\left( {\frac{5}{2},\frac{3}{2}} \right) + f\left( {3,\frac{3}{2}} \right)$
$\ \ \ \ \ \ \ \ \ \ $ $ + f\left( {\frac{3}{2},2} \right) + f\left( {2,2} \right) + f\left( {\frac{5}{2},2} \right) + f\left( {3,2} \right)$
$\ \ \ \ \ \ \ \ \ \ $ $ + f\left( {\frac{3}{2},\frac{5}{2}} \right) + f\left( {2,\frac{5}{2}} \right) + f\left( {\frac{5}{2},\frac{5}{2}} \right) + f\left( {3,\frac{5}{2}} \right))$
$ = \frac{1}{4}\left( {\frac{9}{4} + 3 + \frac{{15}}{4} + \frac{9}{2} + 3 + 4 + 5 + 6 + \frac{{15}}{4} + 5 + \frac{{25}}{4} + \frac{{15}}{2}} \right)$
${S_{4,3}} = 13.5$
