
Double Integral Problem
If ∫−1to3 f(x) dx = −3 and ∫−4to−2 g(x) dx = −2, what is the value of ∫∫D f(x)g(y) dA where D is the square: −1 ≤ x ≤ 3, −4 ≤ y ≤ −2?
I am confused on where to start for this problem.
BTW, 1to3, and 4to2 are the integral bounds for their respected variables.
Thanks!

Your double integral is written as
$\displaystyle \displaystyle \int_{1}^3{\int_{4}^{2}{f(x)\,g(y)\,dy}\,dx}$
$\displaystyle \displaystyle = \int_{1}^3{f(x)\,\int_{4}^{2}{g(y)\,dy}\,dx}$
$\displaystyle \displaystyle = \int_{1}^3{2f(x)\,dx}$
$\displaystyle \displaystyle = 2\int_{1}^3{f(x)\,dx}$
$\displaystyle \displaystyle = 2(3)$
$\displaystyle \displaystyle = 6$.

Ok great!
They key I know understand is breaking the problem up, recognizing the integral bound and going from there.
Thanks a lot!