Answer
${S_{2,2}} = 0.983$
Work Step by Step
We have $N=2$, $M=2$ and $f\left( {x,y} \right) = \sqrt {x + y} $ over ${\cal R} = \left[ {0,1} \right] \times \left[ {0,1} \right]$.
Using the regular partition, we get the dimensions of the subrectangles:
$\Delta x = \frac{{1 - 0}}{2} = \frac{1}{2}$, ${\ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{2} = \frac{1}{2}$
The Riemann sum ${S_{2,2}}$ to estimate the double integral of $f\left( {x,y} \right) = \sqrt {x + y} $ is given by
${S_{2,2}} = \mathop \sum \limits_{i = 1}^2 \mathop \sum \limits_{j = 1}^2 f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = \frac{1}{4}\mathop \sum \limits_{i = 1}^2 \mathop \sum \limits_{j = 1}^2 f\left( {{P_{ij}}} \right)$
Using midpoints of the subrectangles as sample points, we get
${S_{2,2}} = \frac{1}{4}\left( {f\left( {\frac{1}{4},\frac{1}{4}} \right) + f\left( {\frac{3}{4},\frac{1}{4}} \right) + f\left( {\frac{1}{4},\frac{3}{4}} \right) + f\left( {\frac{3}{4},\frac{3}{4}} \right)} \right)$
$ = \frac{1}{4}\left( {\frac{1}{{\sqrt 2 }} + 1 + 1 + \sqrt {\frac{3}{2}} } \right)$
${S_{2,2}} = 0.983$
