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

The integral from 0 to 3 of e^{-st} is -(e^{-3s}-1)/s

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

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

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

Solution steps

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

Apply u-substitution

= \int_{0}^{- s \cdot 3} - \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 \cdot 3} e^{u} d u

Use the common integral:

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

= - \frac{1}{s} [e^{u}]_{0}^{- s \cdot 3}

Simplify

- \frac{1}{s} [e^{u}]_{0}^{- s \cdot 3} : - \frac{[ e ^{u} ] _{0}^{- 3 s}}{s}

= - \frac{[ e ^{u} ] _{0}^{- 3 s}}{s}

Compute the boundaries:

e^{- 3 s} - 1

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

What is the integral from 0 to 3 of e^{-st} ?
The integral from 0 to 3 of e^{-st} is -(e^{-3s}-1)/s