What is the integral of 8e^{-kx} ?

The integral of 8e^{-kx} is -e^{-kx} 8/k+C

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integral of 8e^{-kx}

\int 8 e^{- k x} d x

- e^{- k x} \frac{8}{k} + C

Solution steps

\int 8 e^{- k x} d x

Take the constant out:

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

= 8 \cdot \int e^{- k x} d x

Apply u-substitution

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

= 8 \cdot \int - \frac{e ^{u}}{k} d u

Take the constant out:

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

= 8 (- \frac{1}{k} \cdot \int e^{u} d u)

Use the common integral:

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

= 8 (- \frac{1}{k} e^{u})

Substitute back

u = - k x

= 8 (- \frac{1}{k} e^{- k x})

Simplify

8 (- \frac{1}{k} e^{- k x}) : - e^{- k x} \frac{8}{k}

= - e^{- k x} \frac{8}{k}

Add a constant to the solution

= - e^{- k x} \frac{8}{k} + C

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