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

The integral of 2ze^{-2yz} is -e^{-2zy}+C

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integral of 2ze^{-2yz}

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

- e^{- 2 zy} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = - 2 zy

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

Simplify

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

= - e^{- 2 zy}

Add a constant to the solution

= - e^{- 2 zy} + C

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