Other guys beat me to it. The best thing to do in these programs is to draw out the total region, identify region as defined by our bounds, and then compute the double integral. In any case, the following is how I would solve it
This will be a parallelogram in the first quadrent. Notice how
and intersect at the top
and intersect at the bottem
So let us bound our function in terms of y and in terms of width (actual values here).
The width (x) can be found by equating the functions that intersect at the top and at the bottem and finding the difference.
We can now construct our double integral.
But wait! In this calculation we have included the sides to the left and to the right of our region. So we must find an expression for these areas so that we can subtract them.
On the left, we have the value and are extending to the intersection of which leads to
Thus the area THAT WE DONT WANT on the left side is
Similarly, on the right side we extend from the intersection of which leads to , to
Thus, total area is