Question

Approximate the integral integral integral integral f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2), and (0, 2) into eight equal squares and finding the sum Sigma 8 i = 1 f(xi, yi) where (xi, yi) is the center of the ith square. Evaluate the iterated integral and compare it with the approximation. Integral 4 0 Integral 2 0 (x + y) dy dx Sigma 8 i = 1 f(xi, yi) delta Ai = ___________ Integral 4 0 Integral 2 0 (x + y) dy dx = _____________

Answer 1

Answer:

Step-by-step explanation:

Approximate the integral $\int\int\limits_R {f(x,y)} \, dA$ by dividing the region $R$ with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.

Find the sum $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i$

Since all are equal squares, so $\delta A_i=1$ for every $i$

$\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=f(x_1,y_1)\delta A_1+f(x_2,y_2)\delta A_2+f(x_3,y_3)\delta A_3+f(x_4,y_4)\delta A_4+f(x_5,y_5)\delta A_5+f(x_6,y_6)\delta A_6+f(x_7,y_7)\delta A_7+f(x_8,y_8)\delta A_8\\\\=f(0.5,0.5)(1)+f(1.5,0.5)(1)+f(2.5,0.5)(1)+f(3.5,0.5)(1)+f(0.5,1.5)(1)+f(1.5,1.5)(1)+f(2.5,1.5)(1)+f(3.5,1.5)(1)\\\\=0.5+0.5+1.5+0.5+2.5+0.5+3.5+0.5+0.5+1.5+1.5+1.5+2.5+1.5+3.5+1.5\\\\=24$

Thus, $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=24$

Evaluating the iterate integral $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=\int\limits^4_0 {[xy+\frac{y^2}{2} ]}\limits^2_0 \, dx =\int\limits^4_0 {[2x+2]}dx\\\\=[x^2+2x]\limits^4_0=24.$

Thus, $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=24$