What is the integral of 1/(sqrt(2pi))e^{(-x^2)/2} ?

The integral of 1/(sqrt(2pi))e^{(-x^2)/2} is 1/2 \erf(x/(sqrt(2)))+C

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integral of 1/(sqrt(2pi))e^{(-x^2)/2}

\int \frac{1}{2 π} e^{\frac{- x ^{2}}{2}} d x

\frac{1}{2} erf (\frac{x}{2}) + C

Solution steps

\int \frac{1}{2 π} e^{\frac{- x ^{2}}{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2 π} \cdot \int e^{\frac{- x ^{2}}{2}} d x

2 π = 2^{\frac{1}{2}} π^{\frac{1}{2}}

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} \cdot \int e^{\frac{- x ^{2}}{2}} d x

Simplify

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} \cdot \int e^{- \frac{x ^{2}}{2}} d x

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{x ^{2}}{2} = (\frac{x}{2})^{2}

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} \cdot \int e^{- (\frac{x}{2})^{2}} d x

Apply u-substitution

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} \cdot \int e^{- u^{2}} 2 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} 2 \cdot \int e^{- u^{2}} d u

Use the nonelementary integral:

\int e^{- u^{2}} d u = \frac{π}{2} erf (u)

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} 2 \frac{π}{2} erf (u)

Substitute back

u = \frac{x}{2}

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} 2 \frac{π}{2} erf (\frac{x}{2})

Simplify

\frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} 2 \frac{π}{2} erf (\frac{x}{2}) : \frac{1}{2} erf (\frac{x}{2})

= \frac{1}{2} erf (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2} erf (\frac{x}{2}) + C

What is the integral of 1/(sqrt(2pi))e^{(-x^2)/2} ?
The integral of 1/(sqrt(2pi))e^{(-x^2)/2} is 1/2 \erf(x/(sqrt(2)))+C