Let R=[0,4]×[−1,2]. Create a Riemann sum by subdividing [0,4] into m=2 intervals, and [−1,2] into n=3 subintervals, then use it to estimate the value of∬R(1−xy2)dA.Take the sample points to be the upper left corner of each rectangle.

Answer: 

To estimate the value of the double integral \(\iint_R (1 - xy^2) \, dA\) over the region \(R = [0, 4] \times [-1, 2]\) using a Riemann sum, we will subdivide the intervals as specified.

### Step 1: Subdivision of the intervals

1. **For the interval \([0, 4]\)**:
   - Divide into \(m = 2\) intervals:
     - \([0, 2]\) and \([2, 4]\).
   - Width of each interval: 
     \[
     \Delta x = \frac{4 - 0}{2} = 2.
     \]

2. **For the interval \([-1, 2]\)**:
   - Divide into \(n = 3\) intervals:
     - \([-1, 0]\), \([0, 1]\), and \([1, 2]\).
   - Width of each interval:
     \[
     \Delta y = \frac{2 - (-1)}{3} = 1.
     \]

### Step 2: Sample points

We will use the upper-left corner of each rectangle as the sample point:

- For the first interval in \(x\) and \(y\):
  - \((0, -1)\)
  
- For the second interval in \(x\) (first \(y\) interval):
  - \((0, 0)\)
  
- For the third interval in \(x\) (first \(y\) interval):
  - \((0, 1)\)

- For the first interval in \(x\) (second \(y\) interval):
  - \((2, -1)\)
  
- For the second interval in \(x\) (second \(y\) interval):
  - \((2, 0)\)

- For the third interval in \(x\) (second \(y\) interval):
  - \((2, 1)\)

- For the first interval in \(x\) (third \(y\) interval):
  - \((4, -1)\)
  
- For the second interval in \(x\) (third \(y\) interval):
  - \((4, 0)\)

- For the third interval in \(x\) (third \(y\) interval):
  - \((4, 1)\)

### Step 3: Riemann sum calculation

The area of each rectangle is given by:
\[
\Delta A = \Delta x \cdot \Delta y = 2 \cdot 1 = 2.
\]

Now, we compute the value of the function \(f(x, y) = 1 - xy^2\) at each sample point and multiply by the area of each rectangle:

1. \(f(0, -1) = 1 - 0 \cdot (-1)^2 = 1\)
2. \(f(0, 0) = 1 - 0 \cdot 0^2 = 1\)
3. \(f(0, 1) = 1 - 0 \cdot 1^2 = 1\)

4. \(f(2, -1) = 1 - 2 \cdot (-1)^2 = 1 - 2 = -1\)
5. \(f(2, 0) = 1 - 2 \cdot 0^2 = 1\)
6. \(f(2, 1) = 1 - 2 \cdot 1^2 = 1 - 2 = -1\)

7. \(f(4, -1) = 1 - 4 \cdot (-1)^2 = 1 - 4 = -3\)
8. \(f(4, 0) = 1 - 4 \cdot 0^2 = 1\)
9. \(f(4, 1) = 1 - 4 \cdot 1^2 = 1 - 4 = -3\)

### Total Riemann sum:

Now, we sum the contributions from each rectangle:

\[
\text{Riemann sum} = 2 \left( f(0, -1) + f(0, 0) + f(0, 1) + f(2, -1) + f(2, 0) + f(2, 1) + f(4, -1) + f(4, 0) + f(4, 1) \right)
\]

Substituting the values:

\[
= 2 \left( 1 + 1 + 1 - 1 + 1 - 1 - 3 + 1 - 3 \right)
\]
\[
= 2 \left( 1 + 1 + 1 - 1 + 1 - 1 - 3 + 1 - 3 \right)
\]
\[
= 2 \left( -4 \right) = -8.
\]

### Final Answer:

Thus, the estimated value of the double integral using the Riemann sum is:
\[
\boxed{-8}.
\]

Do you have any questions or want further details? Here are some related questions you might consider:

1. How would the Riemann sum change if we used the lower right corner as the sample point?
2. What is the effect of increasing the number of subdivisions on the accuracy of the estimate?
3. Can you explain how to interpret the result of the double integral geometrically?
4. What other methods could be used to evaluate the double integral?
5. How would you compute the double integral exactly using analytical methods? 

**Tip:** Increasing the number of intervals can lead to a more accurate estimate of the integral.

Let R=[0,4]×[−1,2]. Create a Riemann sum by subdividing [0,4] into m=2 intervals, and [−1,2] into n=3 subintervals, then use it to estimate the value of ∬R(1−xy2)dA. Take the sample points to be the upper left corner of each rectangle.

Explore how to estimate the double integral ∬R(1−xy2)dA over the region R=[0,4]×[−1,2] using a Riemann sum with specific subdivisions. This step-by-step solution demonstrates the concepts of double integrals and Riemann sums, suitable for high school students in Grades 11-12.

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