What is the integral from 1 to 2 of e^{-st}(4-t) ?

The integral from 1 to 2 of e^{-st}(4-t) is (-2e^{-2s}+3e^{-s})/s+(e^{-2s}-e^{-s})/(s^2)

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integral from 1 to 2 of e^{-st}(4-t)

\int_{1}^{2} e^{- s t} (4 - t) d t

\frac{- 2 e ^{- 2 s} + 3 e ^{- s}}{s} + \frac{e ^{- 2 s} - e ^{- s}}{s ^{2}}

Solution steps

\int_{1}^{2} e^{- s t} (4 - t) d t

Apply Integration By Parts

: [- e^{- s t} \frac{1}{s} (4 - t) - \int e^{- s t} \frac{1}{s} d t]_{1}^{2}

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

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 1 to 2 of e^{-st}(4-t) ?
The integral from 1 to 2 of e^{-st}(4-t) is (-2e^{-2s}+3e^{-s})/s+(e^{-2s}-e^{-s})/(s^2)