Let f(x,y)=x(1+y)
 and D
 be the rectangle determined by the inequalities −1≤x≤1
 and 2≤y≤4
. Choose a subdivision of D
 into squares with side length 1
. Use sample points of the Riemann Sum at the top left corner of each square of the subdivision to approximate

To approximate the double integral of the function \( f(x, y) = x(1 + y) \) over the region \( D \), we can use a Riemann sum. The region \( D \) is defined by the rectangle determined by the inequalities \( -1 \leq x \leq 1 \) and \( 2 \leq y \leq 4 \).

### Step-by-step procedure for the Riemann sum approximation:

1. **Subdivide the region \( D \):**
   - \( D \) is the rectangle defined by \( -1 \leq x \leq 1 \) and \( 2 \leq y \leq 4 \), with side lengths of 2 units in the \( x \)-direction and 2 units in the \( y \)-direction.
   - We will divide the rectangle into squares with side length 1. This results in 4 squares, since the rectangle has dimensions \( 2 \times 2 \).

   Specifically, the squares have the following coordinates (with their corners listed as \((x, y)\)):
   - Square 1: \( (-1, 2) \) to \( (0, 3) \)
   - Square 2: \( (0, 2) \) to \( (1, 3) \)
   - Square 3: \( (-1, 3) \) to \( (0, 4) \)
   - Square 4: \( (0, 3) \) to \( (1, 4) \)

2. **Determine the sample points:**
   - The Riemann sum uses the top-left corner of each square as the sample point.
   - The top-left corners of the squares are:
     1. Square 1: \( (-1, 3) \)
     2. Square 2: \( (0, 3) \)
     3. Square 3: \( (-1, 4) \)
     4. Square 4: \( (0, 4) \)

3. **Evaluate \( f(x, y) = x(1 + y) \) at each sample point:**
   - For \( (-1, 3) \): 
     \[
     f(-1, 3) = -1(1 + 3) = -1(4) = -4
     \]
   - For \( (0, 3) \): 
     \[
     f(0, 3) = 0(1 + 3) = 0
     \]
   - For \( (-1, 4) \): 
     \[
     f(-1, 4) = -1(1 + 4) = -1(5) = -5
     \]
   - For \( (0, 4) \): 
     \[
     f(0, 4) = 0(1 + 4) = 0
     \]

4. **Compute the Riemann sum:**
   - Since each square has an area of 1 (side length \( 1 \times 1 \)), the Riemann sum approximation of the double integral is:
     \[
     R = (-4 \times 1) + (0 \times 1) + (-5 \times 1) + (0 \times 1) = -4 + 0 - 5 + 0 = -9
     \]

Thus, the Riemann sum approximation of the double integral is \( \boxed{-9} \).

Would you like more details on any of these steps? 

Here are 5 questions that could further explore this concept:
1. How would the result change if we used the midpoint of each square instead of the top-left corner?
2. What happens if we increase the number of subdivisions in the region \( D \)?
3. How does this approximation compare with the exact value of the double integral?
4. What is the significance of the choice of sample points in a Riemann sum?
5. Can we approximate the integral using other methods, such as the trapezoidal rule?

**Tip**: Increasing the number of subdivisions in the region tends to improve the accuracy of the Riemann sum approximation for integrals.

Let f(x,y) = x(1+y) and D be the rectangle determined by the inequalities −1≤x≤1 and 2≤y≤4. Choose a subdivision of D into squares with side length 1. Use sample points of the Riemann Sum at the top left corner of each square of the subdivision to approximate the double integral.

Learn how to approximate the double integral of the function f(x, y) = x(1 + y) over the rectangle −1 ≤ x ≤ 1 and 2 ≤ y ≤ 4 using a Riemann sum. This problem involves subdividing the region into squares with side length 1 and using the top-left corner sample points. A step-by-step guide is provided, suitable for Calculus I students.

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