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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3e^{-3t}

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

- e^{- 3 t} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

= 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

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

Use the common integral:

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

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

Substitute back

u = - 3 t

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

Simplify

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

= - e^{- 3 t}

Add a constant to the solution

= - e^{- 3 t} + C

\int x a r cos (x) d x

\int (3 x - 5)^{3} d x

\int (6^{2} y - 3 x y) d x

\int (2 - \frac{1}{x}) d x

\int x \frac{d}{1} (x) d x