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

The integral from 0 to 1 of t^2e^{-2t} is -5/(4e^2)+1/4

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

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

- \frac{5}{4 e ^{2}} + \frac{1}{4}

Decimal Notation

0.08083 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int - e^{- 2 t} t d t = - \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t})

= [- \frac{1}{2} e^{- 2 t} t^{2} - (- \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t}))]_{0}^{1}

Simplify

= [- \frac{1}{2} e^{- 2 t} t^{2} + \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t})]_{0}^{1}

Compute the boundaries:

- \frac{5}{4 e ^{2}} + \frac{1}{4}

= - \frac{5}{4 e ^{2}} + \frac{1}{4}

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What is the integral from 0 to 1 of t^2e^{-2t} ?
The integral from 0 to 1 of t^2e^{-2t} is -5/(4e^2)+1/4