What is the integral of 2zxe^{-x^2} ?

The integral of 2zxe^{-x^2} is -ze^{-x^2}+C

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integral of 2zxe^{-x^2}

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

- z e^{- x^{2}} + C

Solution steps

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

Take the constant out:

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

= 2 z \cdot \int e^{- x^{2}} x d x

Apply u-substitution

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

= 2 z \cdot \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

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

Use the common integral:

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

= 2 z (- \frac{1}{2} e^{u})

Substitute back

u = - x^{2}

= 2 z (- \frac{1}{2} e^{- x^{2}})

Simplify

2 z (- \frac{1}{2} e^{- x^{2}}) : - z e^{- x^{2}}

= - z e^{- x^{2}}

Add a constant to the solution

= - z e^{- x^{2}} + C

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What is the integral of 2zxe^{-x^2} ?
The integral of 2zxe^{-x^2} is -ze^{-x^2}+C