What is the integral from 0 to 3 of e^{-st}t^2 ?

The integral from 0 to 3 of e^{-st}t^2 is -(9e^{-3s})/s+(2(-3se^{-3s}-e^{-3s})+2)/(s^3)

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integral from 0 to 3 of e^{-st}t^2

\int_{0}^{3} e^{- s t} t^{2} d t

- \frac{9 e ^{- 3 s}}{s} + \frac{2 ( - 3 s e ^{- 3 s} - e ^{- 3 s} ) + 2}{s ^{3}}

Solution steps

\int_{0}^{3} e^{- s t} t^{2} d t

Apply Integration By Parts

: [- e^{- s t} \frac{1}{s} t^{2} - \int - e^{- s t} \frac{2}{s} t d t]_{0}^{3}

= [- e^{- s t} \frac{1}{s} t^{2} - \int - e^{- s t} \frac{2}{s} t d t]_{0}^{3}

\int - e^{- s t} \frac{2}{s} t d t = - \frac{2}{s ^{3}} (- s e^{- s t} t - e^{- s t})

= [- e^{- s t} \frac{1}{s} t^{2} - (- \frac{2}{s ^{3}} (- s e^{- s t} t - e^{- s t}))]_{0}^{3}

Simplify

= [- e^{- s t} \frac{1}{s} t^{2} + \frac{2}{s ^{3}} (- s e^{- s t} t - e^{- s t})]_{0}^{3}

Compute the boundaries:

- e^{- 3 s} \frac{9}{s} + \frac{2}{s ^{3}} (- 3 s e^{- 3 s} - e^{- 3 s}) + \frac{2}{s ^{3}}

= - e^{- 3 s} \frac{9}{s} + \frac{2}{s ^{3}} (- 3 s e^{- 3 s} - e^{- 3 s}) + \frac{2}{s ^{3}}

Simplify

= - \frac{9 e ^{- 3 s}}{s} + \frac{2 ( - 3 s e ^{- 3 s} - e ^{- 3 s} ) + 2}{s ^{3}}

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What is the integral from 0 to 3 of e^{-st}t^2 ?
The integral from 0 to 3 of e^{-st}t^2 is -(9e^{-3s})/s+(2(-3se^{-3s}-e^{-3s})+2)/(s^3)