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

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

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integral of (e^{-2x})/2

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

- \frac{1}{4} e^{- 2 x} + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{4} e^{- 2 x} + C

\int \frac{1}{2 + 5 x ^{6}} (30 x^{5}) d x

\int \frac{1}{25 - x ^{2}} d x

f (x) = 9 x^{2} + 6 x - 6

7 π x^{2}

x \to 0 + lim (12 e^{- \frac{2}{x}})

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