What is the integral from 1 to infinity of 2^{-x} ?

The integral from 1 to infinity of 2^{-x} is 1/(2ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to infinity of 2^{-x}

\int_{1}^{\infty} 2^{- x} d x

\frac{1}{2 ln ( 2 )}

Decimal

0.72134 \dots

Solution steps

\int_{1}^{\infty} 2^{- x} d x

Compute the indefinite integral:

\int 2^{- x} d x = - \frac{2 ^{- x}}{ln ( 2 )} + C

Compute the boundaries:

\int_{1}^{\infty} 2^{- x} d x = 0 - (- \frac{1}{2 ln ( 2 )})

= 0 - (- \frac{1}{2 ln ( 2 )})

Simplify

= \frac{1}{2 ln ( 2 )}

What is the integral from 1 to infinity of 2^{-x} ?
The integral from 1 to infinity of 2^{-x} is 1/(2ln(2))