What is the integral from 0 to infinity of 8e^{-8x} ?

The integral from 0 to infinity of 8e^{-8x} is 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to infinity of 8e^{-8x}

\int_{0}^{\infty} 8 e^{- 8 x} d x

1

Solution steps

\int_{0}^{\infty} 8 e^{- 8 x} d x

Compute the indefinite integral:

\int 8 e^{- 8 x} d x = - e^{- 8 x} + C

Compute the boundaries:

\int_{0}^{\infty} 8 e^{- 8 x} d x = 0 - (- 1)

= 0 - (- 1)

Simplify

= 1

What is the integral from 0 to infinity of 8e^{-8x} ?
The integral from 0 to infinity of 8e^{-8x} is 1