Finding the area of trapezoid area question

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The discussion focuses on calculating the area of a trapezoid using a double integral of the function cos((y-x)/(y+x)) over a specified region defined by the points (1, 0), (2, 0), (0, 2), and (0, 1). The trapezoid is bounded by the lines y+x = 1, y+x = 2, y=0, and x=0. A substitution method involving u=y-x and v=y+x is recommended, along with computing the Jacobian to express the integral correctly. The final result of the integral is (3/2)*sin(1).

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(double integral) cos((y-x)/(y+x))dA
Where the double integral is over the region with points (1, 0), (2, 0), (0, 2), and (0, 1).
I think the trapezoid is enclosed by
y+x = 1
y+x = 2
y=0
x=0
How can I use this? Thanks
 
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Hello nemesest, for this problem, you can make a substitution u=y-x, v=y+x, compute the Jacobian, and express the integral in terms of u and v (don't forget that the final representation of the integral must involve the Jacobian). The result is (3/2)*sin1. Hope that helps :)
 

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