What is the integral of 1/4 e^{-2x} ?

The integral of 1/4 e^{-2x} is -1/8 e^{-2x}+C

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integral of 1/4 e^{-2x}

\int \frac{1}{4} e^{- 2 x} d x

- \frac{1}{8} e^{- 2 x} + C

Solution steps

\int \frac{1}{4} e^{- 2 x} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{- 2 x} d x

Apply u-substitution

= \frac{1}{4} \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

= \frac{1}{4} (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{4} (- \frac{1}{2} e^{u})

Substitute back

u = - 2 x

= \frac{1}{4} (- \frac{1}{2} e^{- 2 x})

Simplify

\frac{1}{4} (- \frac{1}{2} e^{- 2 x}) : - \frac{1}{8} e^{- 2 x}

= - \frac{1}{8} e^{- 2 x}

Add a constant to the solution

= - \frac{1}{8} e^{- 2 x} + C

d er i v a t i v e f (x) = 4 x^{2} - 5 x + 2

t a y l or ln (\frac{2}{2 - x})

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

ma c l a u r in \frac{1}{1 + x ^{2}}, x = 2

x \to - 9 - lim (\frac{x + 10}{x + 9})

What is the integral of 1/4 e^{-2x} ?
The integral of 1/4 e^{-2x} is -1/8 e^{-2x}+C