Use the given transformation to evaluate the integral

double integral of R (4x+8y)dA, where R is the parallelogram with veritces (-1,3), (1,3), (3,-1) and (1,5); x=(1/4)(u+v), y=(1/4)(v-3u)
i did the jacobian and got (1/4) (although i cant promise it is correct
I have no idea what my bounds should be
thank you!!

Your vertices should be (-1,3), (1, -3), (3,-1) and (1,5) for the transformation to be meaningful in this question...
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The edges of the parallelogram "R" have equations (sketch the vertices to see which pairs of points produce the lines below...)
y = x + 4
y = x - 4
y = -3x
y = -3x + 8

Using the given transformations, these reduce to (respectively)
u = -4, u = 4, v = 0, and v = 8.

Next, the Jacobian ∂(x,y)/∂(u,v) equals
|1/4...1/4|
|-3/4..1/4| = 1/4.

So, change of variables yields
∫∫R (4x + 8y) dx dy
= ∫(v = 0 to 8) ∫(u = -4 to 4) [4 * (1/4)(u+v) + 8 * (1/4)(v-3u)] * |1/4| du dv
= (1/4) * ∫(v = 0 to 8) ∫(u = -4 to 4) (-5u + 3v) du dv
= (1/4) * ∫(v = 0 to 8) (-5u^2/2 + 3uv) {for u = -4 to 4} dv
= (1/4) * ∫(v = 0 to 8) (0 + 3 * 8v) dv
= ∫(v = 0 to 8) 6v dv
= 3v^2 {for v = 0 to 8}
= 192.

I hope this helps!