What is the integral from 0 to 30 of t^2e^{-t} ?

The integral from 0 to 30 of t^2e^{-t} is -(962)/(e^{30)}+2

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integral from 0 to 30 of t^2e^{-t}

\int_{0}^{30} t^{2} e^{- t} d t

- \frac{962}{e ^{30}} + 2

Decimal Notation

2

Solution steps

\int_{0}^{30} t^{2} e^{- t} d t

Apply Integration By Parts

: [- e^{- t} t^{2} - \int - 2 e^{- t} t d t]_{0}^{30}

= [- e^{- t} t^{2} - \int - 2 e^{- t} t d t]_{0}^{30}

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

= [- e^{- t} t^{2} - (- 2 (- e^{- t} t - e^{- t}))]_{0}^{30}

Simplify

= [- e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t})]_{0}^{30}

Compute the boundaries:

- \frac{962}{e ^{30}} + 2

= - \frac{962}{e ^{30}} + 2

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What is the integral from 0 to 30 of t^2e^{-t} ?
The integral from 0 to 30 of t^2e^{-t} is -(962)/(e^{30)}+2