Section 15.1: Double Integrals over rectangles

SOLUTION 1 Fubini’s Theorem gives \[ \iint_R (x - 3y^2) dA = \int_0^2 \int_1^2 (x - 3y^2) dy dx = \int_0^2 [xy - y^3]_{y=1}^{y=2} dx \] \[ = \int_0^2 (2x - 8) - (x - 1) dx = \int_0^2 (x - 7) dx = \left[ \frac{x^2}{2} - 7x \right]_0^2 = 2 - 14 = -12 \] SOLUTION 2 Again applying Fubini’s Theorem, but this time integrating with respect to \(x\) first, we have \[ \iint_R (x - 3y^2) dA = \int_1^2 \int_0^2 (x - 3y^2) dx dy = \int_1^2 \left[ \frac{x^2}{2} - 3xy^2 \right]_{x=0}^{x=2} dy \] \[ = \int_1^2 (2 - 6y^2) dy = \left[ 2y - 2y^3 \right]_1^2 = (4 - 16) - (2 - 2) = -12 \]