What is the integral from 0 to t of e^{-6t} ?

The integral from 0 to t of e^{-6t} is -1/6 (e^{-6t}-1)

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integral from 0 to t of e^{-6t}

\int_{0}^{t} e^{- 6 t} d t

- \frac{1}{6} (e^{- 6 t} - 1)

Solution steps

\int_{0}^{t} e^{- 6 t} d t

Apply u-substitution

: \int_{0}^{- 6 t} - \frac{1}{6} e^{u} d u

= \int_{0}^{- 6 t} - \frac{1}{6} e^{u} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int_{0}^{- 6 t} e^{u} d u

Use the common integral:

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

= - \frac{1}{6} [e^{u}]_{0}^{- 6 t}

Compute the boundaries:

e^{- 6 t} - 1

= - \frac{1}{6} (e^{- 6 t} - 1)

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What is the integral from 0 to t of e^{-6t} ?
The integral from 0 to t of e^{-6t} is -1/6 (e^{-6t}-1)