What is the integral of-3e^{-2t} ?

The integral of-3e^{-2t} is 3/2 e^{-2t}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of-3e^{-2t}

\int - 3 e^{- 2 t} d t

\frac{3}{2} e^{- 2 t} + C

Solution steps

\int - 3 e^{- 2 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^{- 2 t} d t

Apply u-substitution

= - 3 \cdot \int - \frac{1}{2} 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}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= - 3 (- \frac{1}{2} e^{u})

Substitute back

u = - 2 t

= - 3 (- \frac{1}{2} e^{- 2 t})

Simplify

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

= \frac{3}{2} e^{- 2 t}

Add a constant to the solution

= \frac{3}{2} e^{- 2 t} + C