Double Integral Evaluation - Solution to Exercise 5

$$I = \int_{x=0}^{1} \Bigg( \int_{y=1}^{2} \frac{x}{y^2} e^{\frac{x}{y}} \, dy \Bigg) dx$$

$$\text{Let } u = \frac{1}{y}, \; du = -\frac{dy}{y^2} \; \Rightarrow \; dy = -y^2 du$$ $$\text{Limits: } y=1 \to u=1, \; y=2 \to u=\frac{1}{2}$$

$$I = \int_{0}^{1} x \int_{1}^{1/2} e^{xu} (-du) \, dx = \int_{0}^{1} x \Bigg[ -\frac{1}{x} e^{xu} \Bigg]_{1}^{1/2} dx = \int_{0}^{1} \Bigg( e^{x} - e^{x/2} \Bigg) dx$$

$$= \Bigg[ e^{x} - 2e^{x/2} \Bigg]_{0}^{1} = \Bigg( e - 2e^{1/2} \Bigg) - \Bigg( 1 - 2 \Bigg) = e - 2\sqrt{e} + 1 \approx 0.4208$$