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)}

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

π erf (\frac{x}{2}) + C

Solution steps

\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{d y}{d t} = \frac{t}{y - 2}

\int cot (x) csc (x) d x

\frac{d}{d x} (x - 8)

\int 2 + x d x

y = - x^{3}, (- 1, 1)

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