What is the integral of 4Ae^{-Ax} ?

The integral of 4Ae^{-Ax} is -4e^{-Ax}+C

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integral of 4Ae^{-Ax}

\int 4 A e^{- A x} d x

- 4 e^{- A x} + C

Solution steps

\int 4 A e^{- A x} d x

Take the constant out:

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

= 4 A \cdot \int e^{- A x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{A} d u

= 4 A \cdot \int - \frac{e ^{u}}{A} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - A x

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

Simplify

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

= - 4 e^{- A x}

Add a constant to the solution

= - 4 e^{- A x} + C

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