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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Simplify

= \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_{2}^{3} t^{2} + 3 d t

\int_{- 3}^{0} (x + 3) d x

\int_{- 7}^{2} (- x^{2} - 3 x) d x

\int_{- 3}^{0} x + 4 d x

\int_{0}^{y} x e^{y^{3}} d x

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