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

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

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integral from 0 to 1 of e^{-st}(t-1)

\int_{0}^{1} e^{- s t} (t - 1) d t

\frac{- e ^{- s} + 1}{s ^{2}} - \frac{1}{s}

Solution steps

\int_{0}^{1} e^{- s t} (t - 1) d t

Apply Integration By Parts

: [- e^{- s t} \frac{1}{s} (- 1 + t) - \int - e^{- s t} \frac{1}{s} d t]_{0}^{1}

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

\int - e^{- s t} \frac{1}{s} d t = e^{- s t} \frac{1}{s ^{2}}

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

Compute the boundaries:

- e^{- s} \frac{1}{s ^{2}} - \frac{1}{s} + \frac{1}{s ^{2}}

= - e^{- s} \frac{1}{s ^{2}} - \frac{1}{s} + \frac{1}{s ^{2}}

Simplify

= \frac{- e ^{- s} + 1}{s ^{2}} - \frac{1}{s}

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