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