What is the integral of e^{-1/2 z^2} ?

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

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integral of e^{-1/2 z^2}

\int e^{- \frac{1}{2} z^{2}}

\frac{π}{2} erf (\frac{z}{2}) + C

Solution steps

\int e^{- \frac{1}{2} z^{2}} d z

Simplify

- \frac{1}{2} z^{2} : - \frac{z ^{2}}{2}

= \int e^{- \frac{z ^{2}}{2}} d z

Apply exponent rule:

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

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

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

Apply u-substitution

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

= \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

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

Use the nonelementary integral:

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

= 2 \frac{π}{2} erf (u)

Substitute back

u = \frac{z}{2}

= 2 \frac{π}{2} erf (\frac{z}{2})

Simplify

2 \frac{π}{2} erf (\frac{z}{2}) : \frac{π}{2} erf (\frac{z}{2})

= \frac{π}{2} erf (\frac{z}{2})

Add a constant to the solution

= \frac{π}{2} erf (\frac{z}{2}) + C

\frac{d x}{d t} = 0.6 - \frac{5 x}{1000 + t}

y = ln (x + 4)

x \to 0 lim (∣ e^{x} - 1 ∣)

\int 1 + 9 x d x

x y^{^{''}} + 2 y^{^{'}} = 16 x^{2}

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