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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- 2 s} - 1

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

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

\int_{0}^{- a} x^{2} + a^{2} d x

\int_{- 3}^{0} (4 x^{\frac{1}{3}}) d x

\int_{1}^{2 e} 4 ln (x) d x

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

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