What is the integral of-e^{-2x} ?

The integral of-e^{-2x} is 1/2 e^{-2x}+C

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integral of-e^{-2x}

\int - e^{- 2 x} d x

\frac{1}{2} e^{- 2 x} + C

Solution steps

\int - e^{- 2 x} d x

Take the constant out:

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

= - \int e^{- 2 x} d x

Apply u-substitution

: \int - \frac{1}{2} e^{u} d u

= - \int - \frac{1}{2} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Simplify

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

Add a constant to the solution

= \frac{1}{2} e^{- 2 x} + C

\frac{64}{x ^{2} + 16}

f (x) = x^{2} - 4 x, at x = 1

\frac{d}{d x} (- 5 sin (x))

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

\frac{d}{d x} (- e^{- 4 x})