Double Integral Evaluation - Solution to Exercise 3

$$I = \int_{y=0}^{2} \int_{x=0}^{3} e^{x-y} \, dx \, dy = \int_{0}^{2} e^{-y} \Bigg( \int_{0}^{3} e^{x} \, dx \Bigg) dy = \int_{0}^{2} e^{-y} \Bigg[ e^{x} \Bigg]_{0}^{3} dy = \int_{0}^{2} e^{-y} (e^{3} - 1) \, dy$$

$$= (e^{3} - 1) \int_{0}^{2} e^{-y} \, dy = (e^{3} - 1) \Bigg[ -e^{-y} \Bigg]_{0}^{2} = (e^{3} - 1) (-e^{-2} + 1)$$

$$= (e^{3} - 1)(1 - e^{-2}) = e^{3} - e^{3}e^{-2} - 1 + e^{-2} = e^{3} - e + e^{-2} - 1$$

$$\approx 16.502$$