Questions: A contour map is shown for a function f on the square R=[0,4] × [0,4] Use the Midpoint Rule with m=n=4 to estimate the value for ∫∫R f(x, y) d A

A contour map is shown for a function f on the square R=[0,4] × [0,4]

Use the Midpoint Rule with m=n=4 to estimate the value for ∫∫R f(x, y) d A
Transcript text: A contour map is shown for a function $f$ on the square $R=[0,4] \times[0,4]$ Use the Midpoint Rule with $m=n=4$ to estimate the value for $\iint_{R} f(x, y) d A$
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Solution

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Solution Steps

Step 1: Divide the Region into Subrectangles

The region R=[0,4]×[0,4] R = [0, 4] \times [0, 4] is divided into m=n=4 m = n = 4 subrectangles. Each subrectangle will have dimensions Δx=Δy=1 \Delta x = \Delta y = 1 .

Step 2: Identify the Midpoints of Each Subrectangle

The midpoints of each subrectangle are:

  • (0.5, 0.5), (1.5, 0.5), (2.5, 0.5), (3.5, 0.5)
  • (0.5, 1.5), (1.5, 1.5), (2.5, 1.5), (3.5, 1.5)
  • (0.5, 2.5), (1.5, 2.5), (2.5, 2.5), (3.5, 2.5)
  • (0.5, 3.5), (1.5, 3.5), (2.5, 3.5), (3.5, 3.5)
Step 3: Evaluate the Function at Each Midpoint

Using the contour map, approximate the function values at the midpoints:

  • f(0.5,0.5)23 f(0.5, 0.5) \approx 23
  • f(1.5,0.5)15 f(1.5, 0.5) \approx 15
  • f(2.5,0.5)15 f(2.5, 0.5) \approx 15
  • f(3.5,0.5)23 f(3.5, 0.5) \approx 23
  • f(0.5,1.5)15 f(0.5, 1.5) \approx 15
  • f(1.5,1.5)5 f(1.5, 1.5) \approx 5
  • f(2.5,1.5)5 f(2.5, 1.5) \approx 5
  • f(3.5,1.5)15 f(3.5, 1.5) \approx 15
  • f(0.5,2.5)15 f(0.5, 2.5) \approx 15
  • f(1.5,2.5)5 f(1.5, 2.5) \approx 5
  • f(2.5,2.5)5 f(2.5, 2.5) \approx 5
  • f(3.5,2.5)15 f(3.5, 2.5) \approx 15
  • f(0.5,3.5)23 f(0.5, 3.5) \approx 23
  • f(1.5,3.5)15 f(1.5, 3.5) \approx 15
  • f(2.5,3.5)15 f(2.5, 3.5) \approx 15
  • f(3.5,3.5)23 f(3.5, 3.5) \approx 23
Step 4: Apply the Midpoint Rule

The Midpoint Rule for double integrals is given by: Rf(x,y)dAΔxΔyi=1mj=1nf(xi,yj) \iint_R f(x, y) \, dA \approx \Delta x \Delta y \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_i, y_j) Here, Δx=Δy=1 \Delta x = \Delta y = 1 , and the sum of the function values at the midpoints is: 23+15+15+23+15+5+5+15+15+5+5+15+23+15+15+23=237 23 + 15 + 15 + 23 + 15 + 5 + 5 + 15 + 15 + 5 + 5 + 15 + 23 + 15 + 15 + 23 = 237

Final Answer

Rf(x,y)dA11237=237 \iint_R f(x, y) \, dA \approx 1 \cdot 1 \cdot 237 = 237

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