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.