What is the integral of-25e^{-0.5t} ?

The integral of-25e^{-0.5t} is 50e^{-0.5t}+C

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integral of-25e^{-0.5t}

\int - 25 e^{- 0.5 t} d t

50 e^{- 0.5 t} + C

Solution steps

\int - 25 e^{- 0.5 t} d t

Take the constant out:

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

= - 25 \cdot \int e^{- 0.5 t} d t

Apply u-substitution

: \int - \frac{1}{0.5} e^{u} d u

= - 25 \cdot \int - \frac{1}{0.5} e^{u} d u

Take the constant out:

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

= - 25 (- \frac{1}{0.5} \cdot \int e^{u} d u)

Simplify

= - 25 (- 2 \cdot \int e^{u} d u)

Use the common integral:

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

= - 25 (- 2 e^{u})

Substitute back

u = - 0.5 t

= - 25 (- 2 e^{- 0.5 t})

Simplify

= 50 e^{- 0.5 t}

Add a constant to the solution

= 50 e^{- 0.5 t} + C

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What is the integral of-25e^{-0.5t} ?
The integral of-25e^{-0.5t} is 50e^{-0.5t}+C