What is the integral of 1/8 e^{-x/8} ?

The integral of 1/8 e^{-x/8} is -e^{-x/8}+C

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integral of 1/8 e^{-x/8}

\int \frac{1}{8} e^{- \frac{x}{8}} d x

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

Solution steps

\int \frac{1}{8} e^{- \frac{x}{8}} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int e^{- \frac{x}{8}} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - \frac{x}{8}

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of 1/8 e^{-x/8} ?
The integral of 1/8 e^{-x/8} is -e^{-x/8}+C