What is the integral of 90e^{-0.12t} ?

The integral of 90e^{-0.12t} is -749.99999…e^{-0.12t}+C

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integral of 90e^{-0.12t}

\int 90 e^{- 0.12 t} d t

- 749.99999 \dots e^{- 0.12 t} + C

Solution steps

\int 90 e^{- 0.12 t} d t

Take the constant out:

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

= 90 \cdot \int e^{- 0.12 t} d t

Apply u-substitution

: \int - \frac{1}{0.12} e^{u} d u

= 90 \cdot \int - \frac{1}{0.12} e^{u} d u

Take the constant out:

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

= 90 (- \frac{1}{0.12} \cdot \int e^{u} d u)

Simplify

= 90 (- 8.33333 \dots \cdot \int e^{u} d u)

Use the common integral:

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

= 90 (- 8.33333 \dots e^{u})

Substitute back

u = - 0.12 t

= 90 (- 8.33333 \dots e^{- 0.12 t})

Simplify

= - 749.99999 \dots e^{- 0.12 t}

Add a constant to the solution

= - 749.99999 \dots e^{- 0.12 t} + C

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What is the integral of 90e^{-0.12t} ?
The integral of 90e^{-0.12t} is -749.99999…e^{-0.12t}+C