What is the integral from 0 to 1 of e^{-sx} ?

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

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integral from 0 to 1 of e^{-sx}

\int_{0}^{1} e^{- s x} d x

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

Solution steps

\int_{0}^{1} e^{- s x} d x

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- s} - 1

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

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