Double Integral Evaluation Solution to Exercise 1

$$I = \int_{y=-1}^{2} \Bigg( \frac{x^2}{2} - yx \Bigg) \Bigg|_{x=2}^{3} dy = \int_{y=-1}^{2} \Bigg[ \Bigg( \frac{3^2}{2} - 3y \Bigg) - \Bigg( \frac{2^2}{2} - 2y \Bigg) \Bigg] dy$$

$$= \int_{-1}^{2} \Bigg( \frac{9}{2} - 3y - \frac{4}{2} + 2y \Bigg) dy = \int_{-1}^{2} \Bigg( \frac{5}{2} - y \Bigg) dy$$

$$= \Bigg( \frac{5}{2}y - \frac{y^2}{2} \Bigg) \Bigg|_{-1}^{2} = \Bigg[ \frac{5}{2}(2) - \frac{(2)^2}{2} \Bigg] - \Bigg[ \frac{5}{2}(-1) - \frac{(-1)^2}{2} \Bigg]$$

$$= \Bigg( 5 - 2 \Bigg) - \Bigg( -\frac{5}{2} - \frac{1}{2} \Bigg) = 3 - (-3) = 6$$