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

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

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

\int_{1}^{\infty} (\frac{1}{2})^{x} d x

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

Decimal Notation

0.72134 \dots

Solution steps

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

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

Simplify

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

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