What is the integral of e^{-0.4t} ?

The integral of e^{-0.4t} is -2.5e^{-0.4t}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-0.4t}

\int e^{- 0.4 t} d t

- 2.5 e^{- 0.4 t} + C

Solution steps

\int e^{- 0.4 t} d t

Apply u-substitution

: \int - \frac{1}{0.4} e^{u} d u

= \int - \frac{1}{0.4} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.4} \cdot \int e^{u} d u

Simplify

= - 2.5 \cdot \int e^{u} d u

Use the common integral:

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

= - 2.5 e^{u}

Substitute back

u = - 0.4 t

= - 2.5 e^{- 0.4 t}

Add a constant to the solution

= - 2.5 e^{- 0.4 t} + C

\int \frac{3 cos ( x )}{1 + cos ( x )} d x

\int \frac{- 6}{1 - x ^{2}} d x

\int \frac{x - 1}{3} d x

\int \frac{tan ( x ) + sin ( x )}{cos ( x )} d x

\int \frac{x}{( x + 10 ) ( x ^{2} + 12 )} d x

What is the integral of e^{-0.4t} ?
The integral of e^{-0.4t} is -2.5e^{-0.4t}+C