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

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

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

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

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

Decimal Notation

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 )}

f (x) = (x^{3} + 2)^{5}, at x = - 1

x, 1

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

y = x^{3} - 6 x^{2} + 8 x, y = x^{2} - 4 x

ln (x e^{x} + 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))