Integral with triangle vertices $\int_Bxy +\frac1{(1 + x + y)^2}dλ^2$
$B$ denotes the interior of the triangle with vertices $(0,0)$, $(0, 1)$ and $(2,2)$.
I followed the same method got a double integral $$\int_0^2\int_0^yxy +\frac{1}{(1 + x + y)^2}\,dxdy$$ and the answer was $2,00954$
Is it correct?
or is it
$\int_0^2\int_{\frac12 y}^{2(y-1)}xy +\frac{1}{(1 + x + y)^2}\,dxdy$????
or $\int_0^2\int_{ 2(y-1) }^{y}xy +\frac{1}{(1 + x + y)^2}\,dxdy$
OR because the triangle is on the y axis we take 0<y<1
and the x is between 2(y-1) and y??
The more times I do it the more confused I get, so if anyone can tell me which one is correct I will be very grateful
 A: This is more or less a Multivariable Calculus problem. First recall the Lebesgue integrals (I am assume $\lambda$ stands for the Lebesgue measure) is equivalent to the Riemann integral when we are dealing with bounded domains with Riemann integrable functions.
Due to the fact the integrand is positive in the integration domain (which is bounded), we indeed have it being Riemann integrable with possibly being infinite without doing much analysis on it (it is actually a bounded function in a bounded domain).
In particular, we may evaluate the Lebesgue integral using Riemann techniques. This justifies why we are evaluating the Riemann integral instead. Now we refer to the diagram representing the integration domain given by @JohnWaylandBales:
Notice how when $x$ ranges in $(0, 2)$, the integration boundary is bounded below by $y = x$ and above by $y = \frac{x}{2} + 1$. This implies our integral is
$$
\int_0 ^2 \int_{x} ^{\frac{x}{2}} xy + \frac{1}{(1 + x + y)^2} \,dy \,dx.
$$
This then turns to a simple integration problem in Calculus: First integrate with respect to $y$ (the inner integral while treating $x$ as constant), then integrate with respect to $x$ (the outer integral). I was a bit too lazy to do the actual computation. Nevertheless, plugging the numbers into Wolframalpha gives an numerical approximation of 1.02719.
