What is the integral of-1/2 e^{-3x} ?

The integral of-1/2 e^{-3x} is 1/6 e^{-3x}+C

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integral of-1/2 e^{-3x}

\int - \frac{1}{2} e^{- 3 x} d x

\frac{1}{6} e^{- 3 x} + C

Solution steps

\int - \frac{1}{2} e^{- 3 x} d x

Take the constant out:

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

= - \frac{1}{2} \cdot \int e^{- 3 x} d x

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

= - \frac{1}{2} \cdot \int - \frac{1}{3} 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} (- \frac{1}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= - \frac{1}{2} (- \frac{1}{3} e^{u})

Substitute back

u = - 3 x

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

Simplify

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

= \frac{1}{6} e^{- 3 x}

Add a constant to the solution

= \frac{1}{6} e^{- 3 x} + C

x \to 0 lim (\frac{- cos ( x )}{x})

\int - x + 6 d x

\int (\frac{e ^{3 x}}{3} + 3 x \cdot 2^{x^{2}} - x^{4}) d x

\frac{d}{d x} (cos (6 x + 11))

\frac{d}{d x} (tan (x) + 1)

What is the integral of-1/2 e^{-3x} ?
The integral of-1/2 e^{-3x} is 1/6 e^{-3x}+C