What is the integral of e^{-sx} ?

The integral of e^{-sx} is -1/s e^{-sx}+C

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integral of e^{-sx}

\int e^{- s x} d x

- \frac{1}{s} e^{- s x} + C

Solution steps

\int e^{- s x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{s} d u

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

Use the common integral:

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

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

Substitute back

u = - s x

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

Add a constant to the solution

= - \frac{1}{s} e^{- s x} + C

\frac{d}{d x} (\frac{x ^{4}}{8} + \frac{x ^{- 2}}{4})

\frac{d}{d x} (\frac{x ^{2} + x + 1}{2 x})

\frac{d y}{d x} = 2 x + y, y (0) = 4

\frac{d}{d x} (6 e^{- 8 x})

f (x) = 2 x + 1, (4, 3)