What is the integral from 0 to 2 of 3e^{-st} ?

The integral from 0 to 2 of 3e^{-st} is -(3(e^{-2s}-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 2 of 3e^{-st}

\int_{0}^{2} 3 e^{- s t} d t

- \frac{3 ( e ^{- 2 s} - 1 )}{s}

Solution steps

\int_{0}^{2} 3 e^{- s t} d t

Take the constant out:

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

= 3 \cdot \int_{0}^{2} e^{- s t} d t

Apply u-substitution

: \int_{0}^{- s \cdot 2} - \frac{e ^{u}}{s} d u

= 3 \cdot \int_{0}^{- s \cdot 2} - \frac{e ^{u}}{s} d u

Take the constant out:

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

= 3 (- \frac{1}{s} \cdot \int_{0}^{- s \cdot 2} e^{u} d u)

Use the common integral:

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

= 3 (- \frac{1}{s} [e^{u}]_{0}^{- s \cdot 2})

Simplify

3 (- \frac{1}{s} [e^{u}]_{0}^{- s \cdot 2}) : - \frac{3 [ e ^{u} ] _{0}^{- 2 s}}{s}

= - \frac{3 [ e ^{u} ] _{0}^{- 2 s}}{s}

Compute the boundaries:

e^{- 2 s} - 1

= - \frac{3 ( e ^{- 2 s} - 1 )}{s}

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

\int_{16}^{36} x d x

\int_{0}^{x} sin (y) d y

\int_{0}^{x} \frac{8}{81} x y d y

\int_{- e - 1}^{- 2} \frac{1}{x + 1} d x

What is the integral from 0 to 2 of 3e^{-st} ?
The integral from 0 to 2 of 3e^{-st} is -(3(e^{-2s}-1))/s