What is the integral of e^{-0.004t} ?

The integral of e^{-0.004t} is -250e^{-0.004t}+C

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integral of e^{-0.004t}

\int e^{- 0.004 t} d t

- 250 e^{- 0.004 t} + C

Solution steps

\int e^{- 0.004 t} d t

Apply u-substitution

: \int - \frac{1}{0.004} e^{u} d u

= \int - \frac{1}{0.004} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.004} \cdot \int e^{u} d u

Simplify

= - 250 \cdot \int e^{u} d u

Use the common integral:

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

= - 250 e^{u}

Substitute back

u = - 0.004 t

= - 250 e^{- 0.004 t}

Add a constant to the solution

= - 250 e^{- 0.004 t} + C

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\frac{\partial}{\partial y} (x^{5} + x^{2} y + x + 2)

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\int \frac{1}{4 x + 6 x} d x

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What is the integral of e^{-0.004t} ?
The integral of e^{-0.004t} is -250e^{-0.004t}+C