Let T:R^2-->R2 be the map given by t(u,v) = (2u+v,u+2v)

let D u R2 be the region determined by y=2x, y=x/2, y=2x-3, y = (x+3)/2

(a) find the region D* such that T(D*) = D
(b) evaluate 3(2x/3 - y/3)^2 exp(-x/3 + 2y/3) over D by using (a)

(a) Note that the given transformation, T, is a linear transformation. We can write T as the following matrix:
[ T(u) T(v) ] = [ 2 1 ] [ u ]
. . . . . . . . . . [ 1 2 ] [ v ].

Then, to find D* such that T(D*) = D, we need to find T^-1(u, v), the inverse of T. To do this, note that the inverse of the transformation matrix is:
1/3 * [ 2 -1 ]
. . . . [ -1 2 ]
= [ 2/3 -1/3 ]
. .[ -1/3 2/3 ].

Thus:
T^-1(u, v) = ((2/3)u - (1/3)v, (-1/3)u + (2/3)v)
= ((2u - v)/3, (-u + 2v)/3).

Now, note that this transformation transforms the point (x, y) to ((2x - y)/3, (-x + 2y)/3). Also, D is a parallelogram, so D* must also be a parallelogram. Using the above inverse transformation, we can find the equations of the lines that make up D*.

Any point on the line y = 2x must take the form:
(t, 2t).

By the above inverse transformation, this corresponds to the point:
T(t, 2t) = ((2t - 2t)/3, (-t + 4t)/3) = (0, t),

which corresponds to the line x = 0.

In a similar fashion, the other three sides of D* are:
(i) y = x/2 ==> (2t, t) ==> T(2t, t) = ((4t - t)/3, (-2t + 2t)/3) = (t, 0) ==> y = 0
(ii) y = 2x - 3 ==> (t, 2t - 3) ==> T(t, 2t - 3) = (1, t - 2) ==> x = 1
(iii) y = (x + 3)/2 ==> (2t - 3, t) ==> T(2t - 3, t) = (t - 2, 1) ==> y = 1.

Thus, D* = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

(b) To compute this integral, apply the following change of coordinates:
u = (2x - y)/3 and v = (-x + 2y)/3 ==> x = 2u + v and y = u + 2v.

This transformation has Jacobian:
| 2 1 |
| 1 2 |
= (2)(2) - (1)(1)
= 3.

Therefore:
∫∫D 3[(2x - y)/3]^2*e^[(-x + 2y)/3] dA = ∫∫ 3u*e^v du dv (from u=0 to 1) (from v=0 to 1)
= 3 ∫ u du (from u=0 to 1) ∫ e^v dv (from v=0 to 1)
= (3)(1/2)(e - 1)
= 3(e - 1)/2.

I hope this helps!