What is the integral from 0 to 1 of t^2e^{-t} ?

The integral from 0 to 1 of t^2e^{-t} is -5/e+2

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integral from 0 to 1 of t^2e^{-t}

\int_{0}^{1} t^{2} e^{- t} d t

- \frac{5}{e} + 2

Decimal

0.16060 \dots

Solution steps

\int_{0}^{1} t^{2} e^{- t} d t

Apply Integration By Parts

: [- e^{- t} t^{2} - \int - 2 e^{- t} t d t]_{0}^{1}

= [- e^{- t} t^{2} - \int - 2 e^{- t} t d t]_{0}^{1}

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

= [- e^{- t} t^{2} - (- 2 (- e^{- t} t - e^{- t}))]_{0}^{1}

Simplify

= [- e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t})]_{0}^{1}

Compute the boundaries:

- \frac{5}{e} + 2

= - \frac{5}{e} + 2

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What is the integral from 0 to 1 of t^2e^{-t} ?
The integral from 0 to 1 of t^2e^{-t} is -5/e+2