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

The integral from 0 to t of 3e^{-3x} is -e^{-3t}+1

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

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

- e^{- 3 t} + 1

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{0}^{t} e^{- 3 x} d x

Apply u-substitution

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

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

Take the constant out:

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

= 3 (- \frac{1}{3} \cdot \int_{0}^{- 3 t} e^{u} d u)

Use the common integral:

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

= 3 (- \frac{1}{3} [e^{u}]_{0}^{- 3 t})

Simplify

3 (- \frac{1}{3} [e^{u}]_{0}^{- 3 t}) : - [e^{u}]_{0}^{- 3 t}

= - [e^{u}]_{0}^{- 3 t}

Compute the boundaries:

e^{- 3 t} - 1

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

Simplify

= - e^{- 3 t} + 1

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What is the integral from 0 to t of 3e^{-3x} ?
The integral from 0 to t of 3e^{-3x} is -e^{-3t}+1