Answer
$$e - 1$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^x {2{e^{{x^2}}}dydx} } \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^1 {\left[ {2{e^{{x^2}}}y} \right]_0^xdx} \cr
& = \int_0^1 {\left[ {2{e^{{x^2}}}\left( x \right) - 2{e^{{x^2}}}\left( 0 \right)} \right]dx} \cr
& = \int_0^1 {2x{e^{{x^2}}}dx} \cr
& {\text{Integrate with respect to }}x \cr
& = \left[ {{e^{{x^2}}}} \right]_0^1 \cr
& {\text{Evaluating}} \cr
& = {e^{{{\left( 1 \right)}^2}}} - {e^{{{\left( 0 \right)}^2}}} \cr
& = e - 1 \cr} $$
