Let R be the region bounded by the trapezoid with vertices (0, 0), (4, 4), (8, 4), and (12, 0). Let P be the partition of R determined by vertical lines with x-intercepts 0, 2, 4, 6, 8, 10, 12 and by horizontal lines with y-intercepts 0, 2, 4. If f(x, y) = xy, find the Riemann sum of f for P if (ui, vi) is the centroid of Ri.

The value of the Riemann sum is 200.

Step-by-step explanation:

Let the region bounded by the trapezoid is shown as attached figure. Thus the different regions are formed for which the centroid points are calculated as follows

R-1 is the region formed as right angle triangle between points (x1,y1)=(0,0), (x2,y2)=(2,0) and (x3,y3)=(2,2). The centroid is given as

Centroid for this region is given as

C-R1 is (x1+x2+x3)/3,(y1+y2+y3)/3 which is (0+2+2)/3, (0+0+2)/3 given as ()

Similary the Centroids of other regions are given as

C-R2=(10/3,8/3), C-R3=(3,1), C-R4=(5,3), C-R5=(5,1), C-R6=(7,3), C-R7=(7,1),

C-R8=(26/3,8/3), C-R9=(9,1) and C-R10=(32/3,2/3)

Now using the following equation, the Riemann sum is given as

Here value of f is simple product of the points and the value of Δx is 2. Plugging in the values give the value of Riemann Sum as

So the value of the Riemann sum is 200.

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation: