Double Integral Evaluation - Solution to Exercise 2

$$I = \int_{y=0}^{2} \Bigg( \frac{x^2}{2} - yx \Bigg) \Bigg|_{x=y}^{3} dy = \int_{0}^{2} \Bigg[ \Bigg( \frac{3^2}{2} - 3y \Bigg) - \Bigg( \frac{y^2}{2} - y^2 \Bigg) \Bigg] dy$$

$$= \int_{0}^{2} \Bigg( \frac{9}{2} - 3y - \frac{y^2}{2} + y^2 \Bigg) dy = \int_{0}^{2} \Bigg( \frac{9}{2} - 3y + \frac{y^2}{2} \Bigg) dy$$

$$= \Bigg( \frac{9}{2}y - \frac{3}{2}y^2 + \frac{1}{6}y^3 \Bigg) \Bigg|_{0}^{2} = \Bigg[ \frac{9}{2}(2) - \frac{3}{2}(2^2) + \frac{1}{6}(2^3) \Bigg] - 0$$

$$= 9 - 6 + \frac{8}{6} = 9 - 6 + \frac{4}{3} \approx 4.333$$