Let $A=\{(x, y): x+y \leq 3\}$. Then the value of the integral $$ \iint_A f(x) g(y) \,d x\, d y $$ equals where $f(x)=\left\{\begin{array}{ll}x & \text { if } x \in[0,2] \\ 0 & \text { if } x \notin[0,2]\end{array}\right.$
$g(y)=\left\{\begin{array}{ll}1 & \text { if } x \in[0,2] \\ 0 & \text { if } x \notin[0,2]\end{array}\right.$
My approach:
by the form ($\frac{x}{a}+\frac{y}{b} = 1$) I am getting a triangle, whose area should be $$\frac{1}{2}\cdot3 \cdot3$$ $$= \frac{9}{2}$$ Am I correct?
Since $f(x) g(y)$ is non-zero only if $(x, y) \in [0,2]^2$, we can recast our domain of integration as $$A' = [0,2]^2 \setminus T$$ where $T \subset [0,2]^2$ is the triangle with vertices at $(1, 2), (2, 2)$, and $(2, 1)$. It's convenient to evaluate the integral over the square and subtract this triangle, writing $$\iint_A f(x) g(y) \; dx \, dy = \int_0^2 \int_0^2 f(x) g(y) \; dx \, dy - \int_1^2 \int_{3-x}^2 f(x) g(y) \; dy \, dx$$ The first integral on the RHS is straightforward, $$\int_0^2 \int_0^2 x \; dx \, dy = 4$$ The second requires only slightly more work, $$\int_1^2 x \int_{3-x}^2 \; dy \, dx = \int_1^2 x(x-1) \, dx = \int_0^1 (x^2 + x) \, dx = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}$$ and hence we obtain the final answer $$\iint_A f(x) g(y) \; dx \, dy = 4 - \frac{5}{6} = \frac{19}{6}$$