What is the integral of-1/3 e^{-3t} ?

The integral of-1/3 e^{-3t} is 1/9 e^{-3t}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of-1/3 e^{-3t}

\int - \frac{1}{3} e^{- 3 t} d t

\frac{1}{9} e^{- 3 t} + C

Solution steps

\int - \frac{1}{3} e^{- 3 t} d t

Take the constant out:

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

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

Apply u-substitution

= - \frac{1}{3} \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

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

Use the common integral:

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

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

Substitute back

u = - 3 t

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

Simplify

- \frac{1}{3} (- \frac{1}{3} e^{- 3 t}) : \frac{1}{9} e^{- 3 t}

= \frac{1}{9} e^{- 3 t}

Add a constant to the solution

= \frac{1}{9} e^{- 3 t} + C

What is the integral of-1/3 e^{-3t} ?
The integral of-1/3 e^{-3t} is 1/9 e^{-3t}+C