(1) ∫<0, 1> dx ∫<0, 2> (3x^2+y) dy = ∫<0, 1> dx [3x^2y + y^2/2]= ∫<0, 1> (6x^2 + 2)dx = [2x^3 + 2x]<0, 1> = 4(2) ∫<0, π/2> dy ∫<0, y> sin(x+y) dx = ∫<0, π/2> dy[-cos(x+y)] = ∫<0, π/2> (cosy - cos2y)dy = [siny - (1/2)sin2y]<0, π/2> = 1