# Thanks

"I know that the xy region is the line x=y y=0 and y=1- x/2)"

It is helpful to rewrite the region in the x-y plane by specifying the three lines and the endpoints, i.e. the points where they intersect.

If you insert these points in the formula for U and V you obtain the vertices in the U-V plane.

E.g., the line x = y maps to the line
V = 0. But in the x-y plane this line is bounded by the "end points" (x=0,y=0) and (x=2/3,y=2/3) which in the U-V plane has coordinates (U = 0, V = 0) and (U = 2, V = 0) respectively.

I need to find out how to solve this integral with the indicated changes.

int.[0 to 2/3]int[y to 2-2y]
(x+2y)e^(y-x)
dxdy

u=x+2y v=x-y

I know that the xy region is the line x=y y=0 and y=1- x/2)

but when I make the changes, I don't get a new region bounded by graphs.

Can anyone see a way to solve this one?

## To solve this integral after making the changes u = x + 2y and v = x - y, we need to determine the new region in the u-v plane.

Let's start by rewriting the original region in the x-y plane:

1. x = y,
2. y = 0, and
3. y = 1 - x/2.

To find the intersection points of these lines, we can solve the equations pairwise:

1. x = y and y = 0 gives us x = 0 and y = 0,
2. x = y and y = 1 - x/2 can be rearranged to give x = 2/3 and y = 2/3.

Now, let's substitute these intersection points into the equations u = x + 2y and v = x - y to find the vertices in the u-v plane:

1. (0, 0) gives us u = 0 + 2 * 0 = 0 and v = 0 - 0 = 0,
2. (2/3, 2/3) gives us u = 2/3 + 2 * (2/3) = 2 and v = 2/3 - 2/3 = 0.

So, the region in the u-v plane is bounded by the line v = 0, with vertices (u = 0, v = 0) and (u = 2, v = 0).

Now, we can set up the integral in the u-v plane:

∫∫[R](x + 2y)e^(y - x) dxdy,

where [R] represents the region in the x-y plane.

After making the changes u = x + 2y and v = x - y, we have:

∫∫[R'](u)e^(-v) dudv,

where [R'] represents the region in the u-v plane.

Since the region in the u-v plane is only bounded by the line v = 0, the integral simplifies to:

∫∫[R'](u)e^(-v) dudv = ∫[0 to 2]∫[0 to 0](u)e^(-v) dudv.

The bounds of integration in the inner integral are from 0 to 0 because the region is only bounded by the line v = 0.

Therefore, the integral becomes:

∫[0 to 2] (u * e^0) du = ∫[0 to 2] u du,

which integrates to:

[u^2/2] from 0 to 2 = 2^2/2 - 0^2/2 = 4/2 = 2.

So, the value of the integral is 2.