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

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

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integral from 0 to 2 of 2e^{-st}

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

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

Solution steps

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

Apply u-substitution

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

= 2 \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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- 2 s} - 1

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

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