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

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

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integral of-1/2 e^{-2v}

\int - \frac{1}{2} e^{- 2 v} d v

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

Solution steps

\int - \frac{1}{2} e^{- 2 v} d v

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 v} d v

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 v

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = - \frac{1}{2} y + x

\int \frac{ln ( x )}{2} d x

cos^{2} (3 x)

\int 6 x^{\frac{2}{3}} - 8 x^{\frac{3}{4}} d x

x \to 1 + lim (ln (x))

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