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

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

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integral of ke^{-2x}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= k \cdot \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

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

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Simplify

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

Add a constant to the solution

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

y = \frac{x - 1}{x + 1}

x \to 0 - lim (arccos (e^{- x^{2}}))

(x + 2)^{2} + (y - 3)^{2} = 37, (- 3, - 3)

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