What is the integral of Ae^{-kt} ?

The integral of Ae^{-kt} is -Ae^{-kt} 1/k+C

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integral of Ae^{-kt}

\int A e^{- k t} d t

- A e^{- k t} \frac{1}{k} + C

Solution steps

\int A e^{- k t} d t

Take the constant out:

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

= A \cdot \int e^{- k t} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - k t

= A (- \frac{1}{k} e^{- k t})

Simplify

= - A e^{- k t} \frac{1}{k}

Add a constant to the solution

= - A e^{- k t} \frac{1}{k} + C

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