What is the integral from 0 to t of e^{-3u} ?

The integral from 0 to t of e^{-3u} is -1/3 (e^{-3t}-1)

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integral from 0 to t of e^{-3u}

\int_{0}^{t} e^{- 3 u} d u

- \frac{1}{3} (e^{- 3 t} - 1)

Solution steps

\int_{0}^{t} e^{- 3 u} d u

Apply u-substitution

: \int_{0}^{- 3 t} - \frac{1}{3} e^{v} d v

= \int_{0}^{- 3 t} - \frac{1}{3} e^{v} d v

Take the constant out:

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

= - \frac{1}{3} \cdot \int_{0}^{- 3 t} e^{v} d v

Use the common integral:

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

= - \frac{1}{3} [e^{v}]_{0}^{- 3 t}

Compute the boundaries:

e^{- 3 t} - 1

= - \frac{1}{3} (e^{- 3 t} - 1)

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What is the integral from 0 to t of e^{-3u} ?
The integral from 0 to t of e^{-3u} is -1/3 (e^{-3t}-1)