What is the integral from 0 to infinity of 1/(2^x) ?

The integral from 0 to infinity of 1/(2^x) is 1/(ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of 1/(2^x)

\int_{0}^{\infty} \frac{1}{2 ^{x}} d x

\frac{1}{ln ( 2 )}

Decimal Notation

1.44269 \dots

Solution steps

\int_{0}^{\infty} \frac{1}{2 ^{x}} d x

Compute the indefinite integral:

\int \frac{1}{2 ^{x}} d x = - \frac{1}{ln ( 2 ) \cdot 2 ^{x}} + C

Compute the boundaries:

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

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

Simplify

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

x \to 0 lim (x^{6} cos (\frac{5}{x}))

n = 1 \sum \infty (\frac{9}{10})^{n}

\frac{d}{d x} (3 x^{2} + 1 (3 x^{4} + 1)^{3})

\int x^{3} cos (x^{4} - 5) d x

t \to \infty lim (e^{- t})

What is the integral from 0 to infinity of 1/(2^x) ?
The integral from 0 to infinity of 1/(2^x) is 1/(ln(2))