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

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

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

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

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

Solution steps

\int_{0}^{x} t^{2} e^{- s t} 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}^{x}

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

\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}^{x}

Simplify

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

Compute the boundaries:

- x^{2} e^{- x s} \frac{1}{s} + \frac{2}{s ^{3}} (- x s e^{- x s} - e^{- x s}) + \frac{2}{s ^{3}}

= - x^{2} e^{- x s} \frac{1}{s} + \frac{2}{s ^{3}} (- x s e^{- x s} - e^{- x s}) + \frac{2}{s ^{3}}

Simplify

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

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