The other tutor's solution is correct but I think you should
calculate the vertices rather that assume they are exactly as 
they look.

Let's graph it:



The region looks like a parallelogram.

We need to find the corner points, which
are the two vertices and the two points of intersection

To find the vertex of  we set the part
in the absolute value = 0



and solve for x



And we substitute this in








So the vertex of  is (3,-2).

To find the vertex of y=4-2abs(x-2)}}} we also set the part
in the absolute value = 0



and solve for x



And we substitute this in








So the vertex of  is (2,4). 

Now we find the other two vertices of the figure

We solve the system



Since the right sides both equal to y, set them equal:



We can divide every term through by 2 without getting fractions,
so we do so:



Add 1 to both sides:



There are four cases to consider:

1.   and 
which is the same as 
     and 
which is the same as
    


becomes






Substitute in

y=2abs(x-3)-2
y=2abs(4-3)-2
y=2abs(1)-2
y=2(1)-2
y=0

So one point of intersection is (4,0)

2.   and 
which is the same as 
     and 

That's a contradiction, so we ignore this case.

3.   and 
which is the same as 
     and 
which is the same as
    


becomes




Thats a contradiction too. So we ignore this case

4.   and 
which is the same as 
     and 
which is the same as
    


becomes








Substitute in

y=2abs(x-3)-2
y=2abs(1-3)-2
y=2abs(-2)-2
y=2(2)-2
y=4-2
y=2

So the other point of intersection is (1,2)

So the area we want to find is the area of the polygon whose vertices
are (1,2), (3,-2), (4,0), and (2,4)  figure:



Since the polygon is convex (doesn't "sink in"

 anywhere). we use the determinant formula, whose
rows are the coordinates in counter-clockwise order, with the first
row repeated at the bottom:



To expand it add the sum of the products of the diagonals going down
to the right and subtract the sum of the products of the diagonals
going up to the right:

[]  

Edwin