\[ \text{At } p = 0.5, \text{ solving for } x: \] \[ 0.5 = e^{-x^2} \] \[ \text{Taking the natural logarithm of both sides:} \] \[ \ln(0.5) = -x^2 \] \[ x = \sqrt{-\ln(0.5)} \] \[ = \sqrt{\ln(2)} \] \[ = \sqrt{0.6931} = 1.386 \] \[ \text{The consumer surplus is the area under the demand curve from } 0 \text{ to } 1.386: \] \[ \text{Consumer Surplus} = \int_0^{1.386} e^{-x^2} \,dx \] \[ \text{The result of the integral is } 0.30. \]

