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Kazeer [188]
3 years ago
5

Please help check my answers I beg you!!!!!!!

Mathematics
1 answer:
Alchen [17]3 years ago
7 0
1. the answes for the graph ones is the 3rd and 4th graph because both graphs are 1 line and a line is infinite

2. (-3,5) x=(-) y=(+)that means that it must to the left and up part of graph, therefor it is the 4th graph

1. it would be the 4th graph that you didn't show because none of the included answers are correct
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Approximate the integral integral integral integral f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2),
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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

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