What is the integral of-2e^{-6x} ?

The integral of-2e^{-6x} is 1/3 e^{-6x}+C

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integral of-2e^{-6x}

\int - 2 e^{- 6 x} d x

\frac{1}{3} e^{- 6 x} + C

Solution steps

\int - 2 e^{- 6 x} d x

Take the constant out:

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

= - 2 \cdot \int e^{- 6 x} d x

Apply u-substitution

: \int - \frac{1}{6} e^{u} d u

= - 2 \cdot \int - \frac{1}{6} e^{u} d u

Take the constant out:

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

= - 2 (- \frac{1}{6} \cdot \int e^{u} d u)

Use the common integral:

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

= - 2 (- \frac{1}{6} e^{u})

Substitute back

u = - 6 x

= - 2 (- \frac{1}{6} e^{- 6 x})

Simplify

- 2 (- \frac{1}{6} e^{- 6 x}) : \frac{1}{3} e^{- 6 x}

= \frac{1}{3} e^{- 6 x}

Add a constant to the solution

= \frac{1}{3} e^{- 6 x} + C

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