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

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

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integral of e^{-2u}

\int e^{- 2 u} d u

- \frac{1}{2} e^{- 2 u} + C

Solution steps

\int e^{- 2 u} d u

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

v = - 2 u

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

Add a constant to the solution

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

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