What is the integral from 0 to 2 of 20t^2e^{-t} ?

The integral from 0 to 2 of 20t^2e^{-t} is 20(-(10)/(e^2)+2)

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integral from 0 to 2 of 20t^2e^{-t}

\int_{0}^{2} 20 t^{2} e^{- t} d t

20 (- \frac{10}{e ^{2}} + 2)

Decimal Notation

12.93294 \dots

Solution steps

\int_{0}^{2} 20 t^{2} e^{- t} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 \cdot \int_{0}^{2} t^{2} e^{- t} d t

Apply Integration By Parts

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

= 20 [- e^{- t} t^{2} - \int - 2 e^{- t} t d t]_{0}^{2}

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

= 20 [- e^{- t} t^{2} - (- 2 (- e^{- t} t - e^{- t}))]_{0}^{2}

Simplify

= 20 [- e^{- t} t^{2} + 2 (- e^{- t} t - e^{- t})]_{0}^{2}

Compute the boundaries:

- \frac{10}{e ^{2}} + 2

= 20 (- \frac{10}{e ^{2}} + 2)

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What is the integral from 0 to 2 of 20t^2e^{-t} ?
The integral from 0 to 2 of 20t^2e^{-t} is 20(-(10)/(e^2)+2)