What is the integral of 100e^{-0.01t} ?

The integral of 100e^{-0.01t} is -10000e^{-0.01t}+C

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integral of 100e^{-0.01t}

\int 100 e^{- 0.01 t} d t

- 10000 e^{- 0.01 t} + C

Solution steps

\int 100 e^{- 0.01 t} d t

Take the constant out:

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

= 100 \cdot \int e^{- 0.01 t} d t

Apply u-substitution

= 100 \cdot \int - \frac{1}{0.01} e^{u} d u

Take the constant out:

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

= 100 (- \frac{1}{0.01} \cdot \int e^{u} d u)

Simplify

= 100 (- 100 \cdot \int e^{u} d u)

Use the common integral:

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

= 100 (- 100 e^{u})

Substitute back

u = - 0.01 t

= 100 (- 100 e^{- 0.01 t})

Simplify

= - 10000 e^{- 0.01 t}

Add a constant to the solution

= - 10000 e^{- 0.01 t} + C

What is the integral of 100e^{-0.01t} ?
The integral of 100e^{-0.01t} is -10000e^{-0.01t}+C