I'm stuck on a double integration problem:
$$\iint ⌊x⌋*⌊y⌋ \,dA $$ over the region bounded by $x = -2$, $x = 1$, $y = 0$, $y = 2$.
I know that the region formed is a rectangle and the integral should by from -2 to 1, 0 to 2 $\,dy\,dx$, but I'm not sure how to approach integrating the multiplication of two floor functions.
On
Recall that when integrating with respect to one variable, we may treat the other as a constant. Hence,
$$\iint\lfloor x\rfloor\lfloor y\rfloor dxdy = \int\lfloor y\rfloor\left(\int\lfloor x\rfloor dx\right)dy.$$
Now, consider your bound, and split the discussion into regions where each floor function takes a constant value. Once you have that, sum all the values up to get your answer.
As an example for a single-variable case: Say we want to evaluate $\int_0^3\lfloor x\rfloor ^2dx$. We can do this: $$\int_0^3\lfloor x\rfloor ^2dx = \int_0^10\,dx+\int_1^21\,dx+\int_2^34\,dx=0+1+4=5.$$ Can you adapt this to your case?
Hint:
Let me reduce the problem from handling double integral to single integrals. $$\int_{-2}^1 \int_0^2 \lfloor x\rfloor\lfloor y\rfloor\,\, dydx= \left(\int_{-2}^1\lfloor x\rfloor \, dx \right)\left(\int_0^2 \lfloor y\rfloor\, dy\right)$$