SOLUTION In using the Midpoint Rule with \(m = n = 2\), we evaluate \(f(x, y) = x - 3y^2\) at the centers of the
four subrectangles shown in Figure 10. So \(\bar{x}_1 = 1/2, \bar{x}_2 = 3/2, \bar{y}_1 =
5/4,\) and \(\bar{y}_2 = 7/4\).
The area of each subrectangle is \(\Delta A =
1/2\). Thus \[
\iint_R (x - 3y^2) dA \approx \sum_{i=1}^{2} \sum_{j=1}^{2} f(\bar{x}_i,
\bar{y}_j) \Delta A
\] \[
= f(\bar{x}_1, \bar{y}_1)\Delta A + f(\bar{x}_1, \bar{y}_2)\Delta A +
f(\bar{x}_2, \bar{y}_1)\Delta A + f(\bar{x}_2, \bar{y}_2)\Delta A
\] \[
= f(1/2, 5/4)\Delta A + f(1/2, 7/4)\Delta A + f(3/2, 5/4)\Delta A +
f(3/2, 7/4)\Delta A
\] \[
= \left(-\frac{67}{16}\right)\frac{1}{2} +
\left(-\frac{139}{16}\right)\frac{1}{2} +
\left(-\frac{51}{16}\right)\frac{1}{2} +
\left(-\frac{123}{16}\right)\frac{1}{2} = -\frac{1}{32}(67+139+51+123) =
-\frac{380}{32} = -11.875
\] Thus we have \[
\iint_R (x - 3y^2) dA \approx -11.875
\]