The integral \[\frac{1}{{\sqrt {2\pi } }}\int\limits_{ – \infty }^\infty {{{\rm{e}}^{ – \frac{{{x^2}}}{2}}}} {\rm{dx}}\] is equal to A. \[\frac{1}{2}\] B. \[\frac{1}{{\sqrt 2 }}\] C. 1 D. \[\infty \]

” option2=”\[\frac{1}{{\sqrt 2 }}\]” option3=”1″ option4=”\[\infty \]” correct=”option1″]

The correct answer is $\boxed{\frac{1}{\sqrt{2\pi}}}$.

This integral is the probability density function of the standard normal distribution, which is a continuous probability distribution that is often used to model real-world data. The standard normal distribution is symmetric about the mean, with a variance of 1.

The integral can be evaluated using the following steps:

1. Substitute $u = -\frac{x^2}{2}$.
2. The integral becomes $\int_0^\infty e^{-u} du$.
3. This integral can be evaluated using the following formula:

$$\int_0^\infty e^{-u} du = \sqrt{2\pi}$$

1. Substitute back for $u$.
2. The integral becomes $\frac{1}{\sqrt{2\pi}}$.

Therefore, the integral $\frac{1}{\sqrt {2\pi } }}\int\limits_{ – \infty }^\infty {{{\rm{e}}^{ – \frac{{{x^2}}}{2}}}} {\rm{dx}}$ is equal to $\frac{1}{\sqrt{2\pi}}$.

The other options are incorrect because they do not represent the value of the integral.