Question

Approximate the integral \(\int_{R} \int f(x, y) d A\) by dividing the rectangle \(R\) with vertices \((0,0),\) \((4,0),(4,2),\) and (0,2) into eight equal squares and finding the sum \(\sum_{i=1}^{8} f\left(x_{i}, y_{i}\right) \Delta A_{i}\) where \(\left(x_{i}, y_{i}\right)\) is the center of the \(i\) th square. Evaluate the iterated integral and compare it with the approximation. $$ \int_{0}^{4} \int_{0}^{2}\left(x^{2}+y^{2}\right) d y d x $$

Short answer

The Riemann sum approximation of the integral was 20, whereas the actual value of the iterated integral was 40. The approximation underestimated the integral value since the squares used for approximation were rather large.

Step-by-step solution

Divide the rectangle

First, we need to divide the rectangle R into 8 equal squares. Since R is a rectangle with vertices at (0,0), (4,0), (4,2) and (0,2), each square will have dimensions of 1x1. The centers of the squares will have coordinates \((0.5, 0.5), (1.5, 0.5), (2.5, 0.5), (3.5, 0.5), (0.5, 1.5), (1.5, 1.5), (2.5, 1.5), (3.5, 1.5)\).

Calculate the Riemann Sum Approximation

Now, we need to approximate the integral using the Riemann sum. According to Riemann sum approximation, we plug in the centers of each square into our function \(f\), then multiply each by the ΔA (which is 1 for each square, given the dimensions). After doing that for all 8 squares and summing them up, we have: \(1[(0.5)^2 + (0.5)^2] + 1[(1.5)^2 + (0.5)^2] + 1[(2.5)^2 + (0.5)^2] + 1[(3.5)^2 + (0.5)^2] + 1[(0.5)^2 + (1.5)^2] + 1[(1.5)^2 + (1.5)^2] + 1[(2.5)^2 + (1.5)^2] + 1[(3.5)^2 + (1.5)^2]\) which evaluates to 20.

Solve the iterated integral

To find the actual value of the integral, we integrate \(f\) over R using the limits of integration. The iterated integral \(\int_{0}^{4} \int_{0}^{2} (x^{2} + y^{2}) dy dx\) simplifies to \(\int_{0}^{4}[x^{2}y + y^{3}/3]_{0}^{2} dx = \int_{0}^{4} [2x^{2} + 8/3] dx = [2x^{3}/3 + 8x/3]_{0}^{4} = 40\).

Compare the approximation and the actual value

The Riemann sum approximation of the integral gave a value of 20 whereas the iterated integral calculation gave a value of 40. So, the approximation underestimates the integral value in this case. The discrepancy is due to the coarse granularity (large size) of the squares.