What is the integral of e^{(-x^2)/4} ?

The integral of e^{(-x^2)/4} is sqrt(pi)\erf(x/2)+C

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integral of e^{(-x^2)/4}

integral

e^{\frac{- x ^{2}}{4}}

π erf (\frac{x}{2}) + C

Solution steps

\int e^{\frac{- x ^{2}}{4}} d x

Simplify

= \int e^{- \frac{x ^{2}}{4}} d x

Apply exponent rule:

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

\frac{x ^{2}}{4} = (\frac{x}{2})^{2}

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

Apply u-substitution

: \int e^{- u^{2}} \cdot 2 d u

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

Take the constant out:

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

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

Use the nonelementary integral:

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

= 2 \cdot \frac{π}{2} erf (u)

Substitute back

u = \frac{x}{2}

= 2 \cdot \frac{π}{2} erf (\frac{x}{2})

Simplify

2 \cdot \frac{π}{2} erf (\frac{x}{2}) : π erf (\frac{x}{2})

= π erf (\frac{x}{2})

Add a constant to the solution

= π erf (\frac{x}{2}) + C

\frac{( ( 1 + x ) ^{3} )}{( x )}

3 - ∣ x ∣

\frac{1}{( 4 + x ^{2} ) ^{2}}

x^{3} \cdot e^{x}

\frac{x}{( x ^{2} - 2 x + 2 )}

What is the integral of e^{(-x^2)/4} ?
The integral of e^{(-x^2)/4} is sqrt(pi)\erf(x/2)+C