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

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

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integral of e^{-1x}

\int e^{- 1 x} d x

- e^{- x} + C

Solution steps

\int e^{- 1 \cdot x} d x

Simplify

= \int e^{- x} d x

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Use the common integral:

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

= - e^{u}

Substitute back

u = - x

= - e^{- x}

Add a constant to the solution

= - e^{- x} + C

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