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

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

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

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

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

Solution steps

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

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}

\int (x^{x}) d x

\frac{d y}{d x} = x^{5} (x - x y)

e^{x} 1

n = 0 \sum \infty \frac{1}{9}

\int \frac{5 x}{x - 1} d x

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