Calculate Double integral

Integral from D (2x+3y)dxdy with respect to D: limited line x+y>=-2,x<=0,y<=0

Probably the trickiest part with these equations is determining the limits of integration. The lines you describe give a triangular region with corners at the origin, (-2, 0), (0, -2) (you should draw this). If we make x the outer integral, then it can run from -2 to 0. Then at each x value, we need to know the y limits, that is, if we have a vertical line representing a particular x value, what are the y values representing the intersections with the triangle? The lower bound occurs for x + y = -2, y = -2 - x, and the upper bound occurs for y = 0. So the limits of our y integral, the inner integral, are -2 - x and 0.

I will use a text-based integral notation like so:

(a, b) integral (f(x)dx) means integrate f(x) from a to b.
and
(a, b)[g(x)] means g(b) - g(a)

Now your double integral is:

(-2, 0) integral ((-2 - x, 0) integral ((2x + 3y)dy)dx)

Let's look at the inner part first:
(-2 - x, 0) integral ((2x + 3y)dy)
= (-2 - x, 0)[2xy + (3/2)y^2]
= [(0 + 0) - (2x(-2 - x) + (3/2)(-2 - x)^2]
= [0 - (-4x - 2x^2 + (3/2)(x^2 + 4x + 4)]
= [0 - (-4x - 2x^2 + (3/2)x^2 + 6x + 6)]
= 4x + 2x^2 - (3/2)x^2 - 6x - 6
= (1/2)x^2 - 2x - 6

Now we substitute this into your double integral to get an ordinary integral:
(-2, 0) integral (((1/2)x^2 - 2x - 6)dx)
= (-2, 0)[(1/6)x^3 - x^2 - 6x]
= [(0 - 0 - 0) - ((1/6) * (-8) - 4 + 12)]
= [(0) - (-8/6 - 4 + 12)]
= 12 - 5 - 1/3
= 6 + 2/3 ... ... ... (6 and 2 thirds)