Answer
$$1 - {e^{ - 1/4}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_{y/2}^{1/2} {{e^{ - {x^2}}}} dxdy} \cr
& = \int_{y = 0}^{y = 1} {\int_{x = y/2}^{x = 1/2} {{e^{ - {x^2}}}} dxdy} \cr
& x = \frac{y}{2} \to y = 2x \cr
& \left( {{\text{Graph below}}} \right) \cr
& {\text{Switching the order of integration, the new region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant 2x,{\text{ }}0 \leqslant x \leqslant \frac{1}{2}} \right\} \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\int_{y/2}^{1/2} {{e^{ - {x^2}}}} dxdy} = \int_0^{1/2} {\int_0^{2x} {{e^{ - {x^2}}}} dydx} \cr
& = \int_0^{1/2} {\left[ {\int_0^{2x} {{e^{ - {x^2}}}} dy} \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^{1/2} {\left[ {{e^{ - {x^2}}}y} \right]_0^{2x}dx} \cr
& = \int_0^{1/2} {2x{e^{ - {x^2}}}dx} \cr
& = - \int_0^{1/2} {{e^{ - {x^2}}}\left( { - 2x} \right)dx} \cr
& {\text{Integrate}} \cr
& = - \left[ {{e^{ - {x^2}}}} \right]_0^{1/2} \cr
& = - \left( {{e^{ - {{\left( {1/2} \right)}^2}}} - {e^{ - {{\left( 0 \right)}^2}}}} \right) \cr
& = - {e^{ - 1/4}} - 1 \cr
& = 1 - {e^{ - 1/4}} \approx 0.2211 \cr} $$
