What is the integral of 18e^{-3x} ?

The integral of 18e^{-3x} is -6e^{-3x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 18e^{-3x}

\int 18 e^{- 3 x} d x

- 6 e^{- 3 x} + C

Solution steps

\int 18 e^{- 3 x} d x

Take the constant out:

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

= 18 \cdot \int e^{- 3 x} d x

Apply u-substitution

= 18 \cdot \int - \frac{1}{3} e^{u} d u

Take the constant out:

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

= 18 (- \frac{1}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= 18 (- \frac{1}{3} e^{u})

Substitute back

u = - 3 x

= 18 (- \frac{1}{3} e^{- 3 x})

Simplify

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

= - 6 e^{- 3 x}

Add a constant to the solution

= - 6 e^{- 3 x} + C