What is the integral of (e^{-16x}) ?

The integral of (e^{-16x}) is -1/16 e^{-16x}+C

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integral of (e^{-16x})

\int (e^{- 16 x}) d x

- \frac{1}{16} e^{- 16 x} + C

Solution steps

\int e^{- 16 x} d x

Apply u-substitution

: \int - \frac{1}{16} e^{u} d u

= \int - \frac{1}{16} e^{u} d u

Take the constant out:

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

= - \frac{1}{16} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{16} e^{u}

Substitute back

u = - 16 x

= - \frac{1}{16} e^{- 16 x}

Add a constant to the solution

= - \frac{1}{16} e^{- 16 x} + C

\int t^{2} ln (t) d t

x \to π lim (csc (x + π))

\frac{\partial}{\partial x} (sec^{2} (x y^{2}) y^{2})

\frac{3 s - 5}{( s - 1 ) ^{2}}

f (x) = arctan (2 x)

What is the integral of (e^{-16x}) ?
The integral of (e^{-16x}) is -1/16 e^{-16x}+C