approximate +4 r4 0 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Approximate the double integral using the midpoint rule with m = n = 2:**
\[ \int_{0}^{4} \int_{0}^{4} \frac{1}{(x^3 + 1)((y^3 + 1)} \, dy \, dx \]
To carry out this approximation, follow these steps:
1. **Divide the region of integration [0,4] x [0,4] into subregions:**
Use the given m = 2 and n = 2 to divide this region into a 2x2 grid. Therefore, each subregion will be a square with sides of length 2.
2. **Determine the midpoints of each subregion:**
Calculate the midpoints of the 2x2 grid.
- For subregion 1: Midpoint is (1,1)
- For subregion 2: Midpoint is (1,3)
- For subregion 3: Midpoint is (3,1)
- For subregion 4: Midpoint is (3,3)
3. **Evaluate the integrand at each midpoint:**
Substitute each midpoint into the integrand \( \frac{1}{(x^3 + 1)((y^3 + 1)} \) and compute the values.
4. **Sum the values and multiply by the area of each subregion:**
- Area of each subregion = (width) x (height) = 2 x 2 = 4
- Approximate value of the integral by summing up the weighted function values at the midpoints.
\[ \text{Approximate Integral} \approx 4 \times \left[ \frac{1}{((1)^3 + 1)((1)^3 + 1)} + \frac{1}{((1)^3 + 1)((3)^3 + 1)} + \frac{1}{((3)^3 + 1)((1)^3 + 1)} + \frac{1}{((3)^3 + 1)((3)^3 + 1)} \right] \]
Compute individually and sum up the individual terms to find the result.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F62ca1c58-ac8a-4380-bd93-744ecdfbeb4a%2F027d8a02-0b08-4053-8ece-47ca9e8bbe66%2Fy9zrmi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Approximate the double integral using the midpoint rule with m = n = 2:**
\[ \int_{0}^{4} \int_{0}^{4} \frac{1}{(x^3 + 1)((y^3 + 1)} \, dy \, dx \]
To carry out this approximation, follow these steps:
1. **Divide the region of integration [0,4] x [0,4] into subregions:**
Use the given m = 2 and n = 2 to divide this region into a 2x2 grid. Therefore, each subregion will be a square with sides of length 2.
2. **Determine the midpoints of each subregion:**
Calculate the midpoints of the 2x2 grid.
- For subregion 1: Midpoint is (1,1)
- For subregion 2: Midpoint is (1,3)
- For subregion 3: Midpoint is (3,1)
- For subregion 4: Midpoint is (3,3)
3. **Evaluate the integrand at each midpoint:**
Substitute each midpoint into the integrand \( \frac{1}{(x^3 + 1)((y^3 + 1)} \) and compute the values.
4. **Sum the values and multiply by the area of each subregion:**
- Area of each subregion = (width) x (height) = 2 x 2 = 4
- Approximate value of the integral by summing up the weighted function values at the midpoints.
\[ \text{Approximate Integral} \approx 4 \times \left[ \frac{1}{((1)^3 + 1)((1)^3 + 1)} + \frac{1}{((1)^3 + 1)((3)^3 + 1)} + \frac{1}{((3)^3 + 1)((1)^3 + 1)} + \frac{1}{((3)^3 + 1)((3)^3 + 1)} \right] \]
Compute individually and sum up the individual terms to find the result.
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