What is the integral of 1/10 e^{-x/(10)} ?

The integral of 1/10 e^{-x/(10)} is -e^{-x/(10)}+C

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integral of 1/10 e^{-x/(10)}

\int \frac{1}{10} e^{- \frac{x}{10}} d x

- e^{- \frac{x}{10}} + C

Solution steps

\int \frac{1}{10} e^{- \frac{x}{10}} d x

Take the constant out:

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

= \frac{1}{10} \cdot \int e^{- \frac{x}{10}} d x

Apply u-substitution

: \int - 10 e^{u} d u

= \frac{1}{10} \cdot \int - 10 e^{u} d u

Take the constant out:

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

= \frac{1}{10} (- 10 \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{10} (- 10 e^{u})

Substitute back

u = - \frac{x}{10}

= \frac{1}{10} (- 10 e^{- \frac{x}{10}})

Simplify

\frac{1}{10} (- 10 e^{- \frac{x}{10}}) : - e^{- \frac{x}{10}}

= - e^{- \frac{x}{10}}

Add a constant to the solution

= - e^{- \frac{x}{10}} + C

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What is the integral of 1/10 e^{-x/(10)} ?
The integral of 1/10 e^{-x/(10)} is -e^{-x/(10)}+C