a contour map is shown for a function f on th...

# Explanation: ## Step1: Determine sub - rectangle dimensions The region $R=[0,6]\times[0,6]$. With $m = n=2$, $\Delta x=\frac{6 - 0}{2}=3$ and $\Delta y=\frac{6 - 0}{2}=3$. The sub - rectangles have dimensions $3\times3$. ## Step2: Identify mid - points of sub - rectangles The mid - points of the four sub - rectangles are $(1.5,1.5)$, $(4.5,1.5)$, $(1.5,4.5)$ and $(4.5,4.5)$. Reading from the contour map, $f(1.5,1.5)\approx5$, $f(4.5,1.5)\approx20$, $f(1.5,4.5)\approx15$, $f(4.5,4.5)\approx30$. ## Step3: Apply the Midpoint Rule The Midpoint Rule for a double integral $\iint_Rf(x,y)dA\approx\sum_{i = 1}^{m}\sum_{j = 1}^{n}f(\overline{x}_i,\overline{y}_j)\Delta A$, where $\Delta A=\Delta x\Delta y$. Here, $\Delta A = 3\times3=9$. So $\iint_Rf(x,y)dA\approx f(1.5,1.5)\times9+f(4.5,1.5)\times9+f(1.5,4.5)\times9+f(4.5,4.5)\times9=(5 + 20+15 + 30)\times9=70\times9 = 630$. ## Step4: Calculate the average value of $f$ The average value of a function $f$ over a region $R$ is $\bar{f}=\frac{1}{A(R)}\iint_Rf(x,y)dA$, where $A(R)$ is the area of $R$. Since $A(R)=6\times6 = 36$, $\bar{f}=\frac{630}{36}\approx17.5$. # Answer: (a) 630 (b) 17.5