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