Answer
$$\frac{1}{3}\left( {{e^2} - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^{{x^2}} {{e^{{x^3} + 1}}} dydx} \cr
& \int_0^1 {\left[ {\int_0^{{x^2}} {{e^{{x^3} + 1}}} dy} \right]dx} \cr
& \int_0^1 {{e^{{x^3} + 1}}\left[ {\int_0^{{x^2}} {dy} } \right]dx} \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^{{x^2}} {dy} = \left[ y \right]_{y = 0}^{y = {x^2}} \cr
& = {x^2} \cr
& {\text{Therefore,}} \cr
& \int_0^1 {{e^{{x^3} + 1}}\left[ {\int_0^{{x^2}} {dy} } \right]dx} = \int_0^1 {{e^{{x^3} + 1}}\left( {{x^2}} \right)dx} \cr
& = \frac{1}{3}\int_0^1 {{e^{{x^3} + 1}}\left( {3{x^2}} \right)dx} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{1}{3}\left[ {{e^{{x^3} + 1}}} \right]_0^1 \cr
& = \frac{1}{3}\left[ {{e^{{{\left( 1 \right)}^3} + 1}} - {e^{{{\left( 0 \right)}^3} + 1}}} \right] \cr
& = \frac{1}{3}\left( {{e^2} - 1} \right) \cr} $$
