Answer
$$\dfrac{e-2}{2}$$
Work Step by Step
Write the region of integration $R$ as follows:
$R=${$ (x,y) | 0 \leq x \leq x^3, 0 \leq y \leq 1$}
$I=\int_{0}^{1} \int_{0}^{x^3} e^{y/x} \ dy \ dx \\=\int_{0}^{1} [x \times e^{(y/x)}]_{0}^{x^3} \ dx \\=[\dfrac{(e^{x^2}-x^2}{2} ]_{0}^{1}\\=\dfrac{[(e^{1}-e^{0})-(1^2-0)]}{2} \\=\dfrac{e-2}{2}$
