What is the integral from 0 to 4 of e^{-3t} ?

The integral from 0 to 4 of e^{-3t} is (e^{12}-1)/(3e^{12)}

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integral from 0 to 4 of e^{-3t}

\int_{0}^{4} e^{- 3 t} d t

\frac{e ^{12} - 1}{3 e ^{12}}

Decimal Notation

0.33333 \dots

Solution steps

\int_{0}^{4} e^{- 3 t} d t

Apply u-substitution

: \int_{0}^{- 12} - \frac{1}{3} e^{u} d u

= \int_{0}^{- 12} - \frac{1}{3} e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 12}^{0} - \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

= - (- \frac{1}{3} \cdot \int_{- 12}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{3} [e^{u}]_{- 12}^{0})

Simplify

= \frac{1}{3} [e^{u}]_{- 12}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{12}}

= \frac{1}{3} (1 - \frac{1}{e ^{12}})

Simplify

= \frac{e ^{12} - 1}{3 e ^{12}}

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What is the integral from 0 to 4 of e^{-3t} ?
The integral from 0 to 4 of e^{-3t} is (e^{12}-1)/(3e^{12)}