How do you do this double integral....2xe^(3xy) dxdy

evaluate the integral (2xe^(3xy) dxdy) over the rectangle A = {(x,y): 0≤x≤1, 0≤y≤2}.

∫∫A 2x e^(3xy) dA
= ∫(x = 0 to 1) ∫(y = 0 to 2) 2x e^(3xy) dy dx
= ∫(x = 0 to 1) (2/3) e^(3xy) {for y = 0 to 2} dx
= ∫(x = 0 to 1) (2/3) [e^(6x) - 1] dx
= (2/3) [(1/6) e^(6x) - x] {for x = 0 to 1}
= (2/3) [((1/6) e^6 - 1) - (1/6 - 0)]
= (1/9) (e^6 - 7).

I hope this helps!