Don't worry, I have a really easy approach for you. Using this, your double integrals will become funny and nonsense, instead of troubling you.
Ok, forget everything, clear your mind, have fresh air..
Now, we have a double integral over a region, which is shown as
. You probably already know that
can be
or
, depending on your solution.
Your region is a triangle, which is not a rectangle. Integrals on rectangular regions can be expressed as
or
where a b c d are constant and limits of the rectangle.
Since your region is not a rectangle, your integral can't be expressed using only constants like a b c d. We need functions here.
In order to use my approach, the region has to be bounded by two lines parallel to one of the axes. For example, if a region is bounded by "x=a and x=b" or "y=a and y=b", then we can go on. If it's not bounded, you should split the region. You may also need to split the region if the bottom and top functions are piecewise.
When we use
and
as bounds, the top and bottom functions are top and bottom bounds of the region. Also dA = dy dx. If the bottom bound is
and top bound is
, our integral is:
When we use
and
as bounds, the top and bottom functions are left and right bounds of the region. Also dA = dx dy. If the bottom bound is
and top bound is
, our integral is:
Code:


2.oo
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1. o

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Aha! This triangle is bounded by two lines parallel to x axis:
and
. It also has nonpiecewise top and bottom functions.
(red: top, blue: bottom)
Code:


2.oo
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1. o

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Now, you find f(y) and g(y), top and bottom functions.
Then, the integral will be