What is the integral from 0 to 1 of 2e^{-st} ?

The integral from 0 to 1 of 2e^{-st} is -(2(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 2e^{-st}

\int_{0}^{1} 2 e^{- s t} d t

- \frac{2 ( e ^{- s} - 1 )}{s}

Solution steps

\int_{0}^{1} 2 e^{- s t} d t

Take the constant out:

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

= 2 \cdot \int_{0}^{1} e^{- s t} d t

Apply u-substitution

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

= 2 \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

= 2 (- \frac{1}{s} \cdot \int_{0}^{- s} e^{u} d u)

Use the common integral:

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

= 2 (- \frac{1}{s} [e^{u}]_{0}^{- s})

Simplify

2 (- \frac{1}{s} [e^{u}]_{0}^{- s}) : - \frac{2 [ e ^{u} ] _{0}^{- s}}{s}

= - \frac{2 [ e ^{u} ] _{0}^{- s}}{s}

Compute the boundaries:

e^{- s} - 1

= - \frac{2 ( e ^{- s} - 1 )}{s}

\int_{0}^{1} (9 x - 9^{x}) d x

\int_{- 1}^{1} x^{7} + x^{3} + x d x

\int_{2}^{0} 1 d x

\int_{- 1}^{3} (x + 4) d x

\int_{0}^{2} (4 x - 8) d x

What is the integral from 0 to 1 of 2e^{-st} ?
The integral from 0 to 1 of 2e^{-st} is -(2(e^{-s}-1))/s