What is the integral of ce^{-θy} ?

The integral of ce^{-θy} is -ce^{-θy} 1/θ+C

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integral of ce^{-θy}

\int c e^{- θ y} d y

- c e^{- θ y} \frac{1}{θ} + C

Solution steps

\int c e^{- θ y} d y

Take the constant out:

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

= c \cdot \int e^{- θ y} d y

Apply u-substitution

: \int - \frac{e ^{u}}{θ} d u

= c \cdot \int - \frac{e ^{u}}{θ} d u

Take the constant out:

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

= c (- \frac{1}{θ} \cdot \int e^{u} d u)

Use the common integral:

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

= c (- \frac{1}{θ} e^{u})

Substitute back

u = - θ y

= c (- \frac{1}{θ} e^{- θ y})

Simplify

= - c e^{- θ y} \frac{1}{θ}

Add a constant to the solution

= - c e^{- θ y} \frac{1}{θ} + C

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