What is the double integral for a triangle with vertices (1,1),(2,4),(5,2)?

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SUMMARY

The discussion focuses on computing the double integral of the function (16-2x-3y)/11 over a triangular region defined by the vertices (1,1), (2,4), and (5,2). The user successfully derived the equations of the triangle's sides: y = x(1/4) + 3/4, y = 3x - 2, and y = -x(2/3) + 16/3. They established the limits for the double integral as 1 < x < 2 with y ranging from x(1/4) + 3/4 to 3x - 2, and 2 < x < 5 with y ranging from (1/4)x + 3/4 to -x(2/3) + 16/3. Despite correct setup, the user encountered difficulties in obtaining the correct integral result.

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Homework Statement



Compute the double integral where the region is a triangle with vertices (1,1),(2,4),(5,2).(please see the attachment)

Homework Equations


double integral((16-2x-3y)1/11)da




The Attempt at a Solution


first i found the equation of the three lines :

(1,1)-->(5,2) y=x(1/4)+3/4
(1,1)-->(2,4) y=3x-2
(2,4)--->(5,2) y=-x(2/3)+16/3

and then i set up the double integral by defining the range of x and y limits.
1<x<2 , x(1/4)+3/4<y<3x-2
2<x<5, (1/4)x+3/4<y<-x(2/3)+16/3

i integrated these 2 integrals but I am not getting the right answer!
 

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Your set-up sounds correct. Show us what you did when you integrated. Maybe the problem is there.
 
if the setup sounds good then i will manage to get the right answer by myself. Thanks
 

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