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

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

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integral of e^{(-z^2)/2}

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

(\frac{π}{2})^{\frac{1}{2}} erf (\frac{z}{2}) + C

Solution steps

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

Simplify

= \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})^{\frac{1}{2}} erf (\frac{z}{2})

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

Add a constant to the solution

= (\frac{π}{2})^{\frac{1}{2}} erf (\frac{z}{2}) + C

\int - (9 x)^{\frac{1}{3}} d x

\int \frac{9}{x ^{2} 1 + x ^{2}} d x

\int (9 x + 2) d x

\int (4 x^{2} + 3 x^{- 2}) d x

sin^{2} (2 x)

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