Answer:
Step-by-step explanation:
Approximate the integral [tex]\int\int\limits_R {f(x,y)} \, dA[/tex] by dividing the region [tex]R[/tex] with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.
Find the sum [tex]\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i[/tex]
Since all are equal squares, so [tex]\delta A_i=1[/tex] for every [tex]i[/tex]
[tex]\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[/tex]
Thus, [tex]\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=24[/tex]
Evaluating the iterate integral [tex]\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.[/tex]
Thus, [tex]\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=24[/tex]
