What is the integral of e^{-18x} ?

The integral of e^{-18x} is -1/18 e^{-18x}+C

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integral of e^{-18x}

\int e^{- 18 x} d x

- \frac{1}{18} e^{- 18 x} + C

Solution steps

\int e^{- 18 x} d x

Apply u-substitution

: \int - \frac{1}{18} e^{u} d u

= \int - \frac{1}{18} e^{u} d u

Take the constant out:

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

= - \frac{1}{18} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{18} e^{u}

Substitute back

u = - 18 x

= - \frac{1}{18} e^{- 18 x}

Add a constant to the solution

= - \frac{1}{18} e^{- 18 x} + C

\frac{\partial}{\partial t} (s - t)

f (x) = - 8 x^{3} + 24 x + 9

\frac{d}{d x} ((9 x^{6} + 2 x^{3})^{4})

x - 5

y^{^{''}} - 11 y^{^{'}} + 30 y = 0