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))

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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 )}

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

\int_{0}^{2} 1 - ∣ 1 - x ∣ d x

\int_{0}^{x} cos (x) sin (x) d x

\int_{0}^{2} (x^{2} + 8 x) d x

\int_{0}^{s} 1 d s

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))