TITLE: $-\iint_{A}(y+x)\,dA$ Evaluating
QUESTION [5 upvotes]: I am bit unsure about the following problem:
Evaluate the double integral:
$$-\iint_{A}(y+x)\,dA$$
over the triangle with vertices $(0,0), (1,1), (2,0)$
OK, so I figured here that I would do this by first evaluating the integral over the region bounded by the vertices $(0,0), (1,1), (1,0)$ and then evaluate the integral over the region bounded by the vertices $(1,0), (1,1), (2,0)$ before adding the two answers together, and then reversing the sign of this answer (since there is a minus sign in front of the original double integral).  Thus, I begin by finding:
$$\int_{0}^{1}dx \int_{0}^{x}(y+x)\,dy$$
When solved this gives me the answer $\frac{1}{2}$.
Next I solve:
$$\int_{1}^{2}dx \int_{1}^{2-x}(y+x)\,dy$$
When solved this gives me the answer $-\frac{7}{6}$.
I have verified both the integrals in Wolframalpha, and they give me the same answer.  I would therefore believe that the final answer should be:
$$-(\frac{1}{2} - \frac{7}{6}) = \frac{2}{3}$$
However, the final answer should, according to the book, be $-\frac{4}{3}$.
Thus, obviously I do something wrong here.  If anyone can help me out, I would greatly appreciate it.  Is it perhaps that it is not allowed to "split up" this into two separate integrals?  I couldn't find a way to solve this without doing this.

REPLY [2 votes]: Your second integral should be $$\int_{1}^{2}dx \int_{0}^{2-x}(y+x)dy.$$
Your lower $y$ limit was 1 instead of 0.
Draw the triangle to see the area you are integrating over.