What is the integral of 2e^{-x/5} ?

The integral of 2e^{-x/5} is -10e^{-x/5}+C

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integral of 2e^{-x/5}

\int 2 e^{- \frac{x}{5}} d x

- 10 e^{- \frac{x}{5}} + C

Solution steps

\int 2 e^{- \frac{x}{5}} d x

Take the constant out:

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

= 2 \cdot \int e^{- \frac{x}{5}} d x

Apply u-substitution

: \int - 5 e^{u} d u

= 2 \cdot \int - 5 e^{u} d u

Take the constant out:

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

= 2 (- 5 \cdot \int e^{u} d u)

Use the common integral:

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

= 2 (- 5 e^{u})

Substitute back

u = - \frac{x}{5}

= 2 (- 5 e^{- \frac{x}{5}})

Simplify

= - 10 e^{- \frac{x}{5}}

Add a constant to the solution

= - 10 e^{- \frac{x}{5}} + C

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