What is the integral of ze^{-z^2} ?

The integral of ze^{-z^2} is -1/2 e^{-z^2}+C

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integral of ze^{-z^2}

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

- \frac{1}{2} e^{- z^{2}} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{1}{2} e^{u}

Substitute back

u = - z^{2}

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

Add a constant to the solution

= - \frac{1}{2} e^{- z^{2}} + C

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What is the integral of ze^{-z^2} ?
The integral of ze^{-z^2} is -1/2 e^{-z^2}+C