What is the integral of e^{1-8t} ?

The integral of e^{1-8t} is -1/8 e^{1-8t}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^{1-8t}

\int e^{1 - 8 t} d t

- \frac{1}{8} e^{1 - 8 t} + C

Solution steps

\int e^{1 - 8 t} d t

Apply u-substitution

= \int - \frac{1}{8} e^{u} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{8} e^{u}

Substitute back

u = 1 - 8 t

= - \frac{1}{8} e^{1 - 8 t}

Add a constant to the solution

= - \frac{1}{8} e^{1 - 8 t} + C