Question

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.

Answer 1

Answer:

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

$R_{IO}=f(4/3,2/3)\times \Delta x +f(10/3,8/3)\times \Delta x + f(3,1)\times \Delta x + f(5,3)\times \Delta x + f(5,1)\times \Delta x+ f(7,3)\times \Delta x + f(7,1)\times \Delta x + f(26/3,8/3)\times \Delta x+f(9,1)\times \Delta x+f(32/3,2/3)\times \Delta x$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

$R_{IO}=(4/3*2/3)\times 2 +(10/3*8/3)\times 2 + (3*1)\times 2 + (5*3)\times 2+(5*1)\times 2+ (7*3)\times2 + (7*1)\times 2 + (26/3*8/3)\times2+(9*1)\times 2+(32/3*2/3)\times 2\\R_{IO}=200$

So the value of the Riemann sum is 200.