What is the integral from 0 to infinity of x2^{-x} ?

The integral from 0 to infinity of x2^{-x} is 1/(ln^2(2))

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integral from 0 to infinity of x2^{-x}

\int_{0}^{\infty} x 2^{- x} d x

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

Decimal

2.08136 \dots

Solution steps

\int_{0}^{\infty} x \cdot 2^{- x} d x

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 0 to infinity of x2^{-x} ?
The integral from 0 to infinity of x2^{-x} is 1/(ln^2(2))