What is the integral of 9e^{-x} ?

The integral of 9e^{-x} is -9e^{-x}+C

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integral of 9e^{-x}

\int 9 e^{- x} d x

- 9 e^{- x} + C

Solution steps

\int 9 e^{- x} d x

Take the constant out:

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

= 9 \cdot \int e^{- x} d x

Apply u-substitution

= 9 \cdot \int - e^{u} d u

Take the constant out:

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

= 9 (- \int e^{u} d u)

Use the common integral:

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

= 9 (- e^{u})

Substitute back

u = - x

= 9 (- e^{- x})

Simplify

= - 9 e^{- x}

Add a constant to the solution

= - 9 e^{- x} + C