What is the integral from 1 to infinity of te^{-t} ?

The integral from 1 to infinity of te^{-t} is 2/e

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integral from 1 to infinity of te^{-t}

\int_{1}^{\infty} t e^{- t} d t

\frac{2}{e}

Decimal

0.73575 \dots

Solution steps

\int_{1}^{\infty} t e^{- t} d t

Compute the indefinite integral:

\int t e^{- t} d t = - e^{- t} t - e^{- t} + C

Compute the boundaries:

\int_{1}^{\infty} t e^{- t} d t = 0 - (- \frac{2}{e})

= 0 - (- \frac{2}{e})

Simplify

= \frac{2}{e}

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What is the integral from 1 to infinity of te^{-t} ?
The integral from 1 to infinity of te^{-t} is 2/e