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

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

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

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

- \frac{e ^{- s t} - 1}{s}

Solution steps

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

Apply u-substitution

: \int_{0}^{- s t} - \frac{e ^{u}}{s} d u

= \int_{0}^{- s t} - \frac{e ^{u}}{s} d u

Take the constant out:

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

= - \frac{1}{s} \cdot \int_{0}^{- s t} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{1}{s} [e^{u}]_{0}^{- s t}

Simplify

- \frac{1}{s} [e^{u}]_{0}^{- s t} : - \frac{[ e ^{u} ] _{0}^{- s t}}{s}

= - \frac{[ e ^{u} ] _{0}^{- s t}}{s}

Compute the boundaries:

e^{- s t} - 1

= - \frac{e ^{- s t} - 1}{s}

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