What is the integral of 6e^{-2x} ?

The integral of 6e^{-2x} is -3e^{-2x}+C

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integral of 6e^{-2x}

\int 6 e^{- 2 x} d x

- 3 e^{- 2 x} + C

Solution steps

\int 6 e^{- 2 x} d x

Take the constant out:

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

= 6 \cdot \int e^{- 2 x} d x

Apply u-substitution

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

= 6 \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

= 6 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= 6 (- \frac{1}{2} e^{u})

Substitute back

u = - 2 x

= 6 (- \frac{1}{2} e^{- 2 x})

Simplify

6 (- \frac{1}{2} e^{- 2 x}) : - 3 e^{- 2 x}

= - 3 e^{- 2 x}

Add a constant to the solution

= - 3 e^{- 2 x} + C

\frac{d}{d x} ((1 - x)^{2} + 10 (y - x^{2})^{2})

\frac{d}{d x} (3 x - sin^{2} (4 x))

x \to 0 lim (\frac{x}{e ^{7 x} - 1})

y^{^{'}} + 4 y = 32

x \to 3 lim (3)