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

The integral from 1 to infinity of t^2e^{-t} is 5/e

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integral from 1 to infinity of t^2e^{-t}

\int_{1}^{\infty} t^{2} e^{- t} d t

\frac{5}{e}

Decimal Notation

1.83939 \dots

Solution steps

\int_{1}^{\infty} t^{2} e^{- t} d t

Compute the indefinite integral:

\int t^{2} e^{- t} d t = - e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t}) + C

Compute the boundaries:

\int_{1}^{\infty} t^{2} e^{- t} d t = 0 - (- \frac{5}{e})

= 0 - (- \frac{5}{e})

Simplify

= \frac{5}{e}

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What is the integral from 1 to infinity of t^2e^{-t} ?
The integral from 1 to infinity of t^2e^{-t} is 5/e