What is the integral from 0 to 1 of 10e^{-st} ?

The integral from 0 to 1 of 10e^{-st} is -(10(e^{-s}-1))/s

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integral from 0 to 1 of 10e^{-st}

\int_{0}^{1} 10 e^{- s t} d t

- \frac{10 ( e ^{- s} - 1 )}{s}

Solution steps

\int_{0}^{1} 10 e^{- s t} d t

Take the constant out:

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

= 10 \cdot \int_{0}^{1} e^{- s t} d t

Apply u-substitution

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

= 10 \cdot \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

= 10 (- \frac{1}{s} \cdot \int_{0}^{- s} e^{u} d u)

Use the common integral:

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

= 10 (- \frac{1}{s} [e^{u}]_{0}^{- s})

Simplify

10 (- \frac{1}{s} [e^{u}]_{0}^{- s}) : - \frac{10 [ e ^{u} ] _{0}^{- s}}{s}

= - \frac{10 [ e ^{u} ] _{0}^{- s}}{s}

Compute the boundaries:

e^{- s} - 1

= - \frac{10 ( e ^{- s} - 1 )}{s}

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What is the integral from 0 to 1 of 10e^{-st} ?
The integral from 0 to 1 of 10e^{-st} is -(10(e^{-s}-1))/s