91Ó°ÊÓ

Given the integral\(\int\limits_0^\infty {{e^{ - 3{x^2}}}dx} \)

Except the constant factor, this integrand has the form of the pdf of a normal distribution for which\(\mu = 0\)and\({\sigma ^2} = \frac{1}{6}\)

Therefore if we multiply the integrand by\(\frac{1}{{{{\left( {2\pi } \right)}^{\frac{1}{{2\sigma }}}}}} = {\left( {\frac{3}{\pi }} \right)^{\frac{1}{2}}}\), ee obtain the pdf of normal distribution .

The integral of this pdf over the entire real line must equal to 1.therefore,

\(\int\limits_{ - \infty }^\infty {\exp \left( { - 3{x^2}} \right)dx = {{\left( {\frac{\pi }{3}} \right)}^{\frac{1}{2}}}} \)

Since the integrand is symmetric with respect to x=0, the integral over the positive half of the real line must be equal to the integral over the negative half of the real line.

Hence,\(\int\limits_0^\infty {\exp \left( { - 3{x^2}} \right)dx} = \frac{1}{2}{\left( {\frac{\pi }{3}} \right)^{\frac{1}{2}}}\)