What is the integral from 0 to 1 of e^{-5t} ?

The integral from 0 to 1 of e^{-5t} is (e^5-1)/(5e^5)

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integral from 0 to 1 of e^{-5t}

\int_{0}^{1} e^{- 5 t} d t

\frac{e ^{5} - 1}{5 e ^{5}}

Decimal

0.19865 \dots

Solution steps

\int_{0}^{1} e^{- 5 t} d t

Apply u-substitution

: \int_{0}^{- 5} - \frac{1}{5} e^{u} d u

= \int_{0}^{- 5} - \frac{1}{5} e^{u} d u

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

= - \int_{- 5}^{0} - \frac{1}{5} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{5} \cdot \int_{- 5}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{5} [e^{u}]_{- 5}^{0})

Simplify

= \frac{1}{5} [e^{u}]_{- 5}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{5}}

= \frac{1}{5} (1 - \frac{1}{e ^{5}})

Simplify

= \frac{e ^{5} - 1}{5 e ^{5}}

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What is the integral from 0 to 1 of e^{-5t} ?
The integral from 0 to 1 of e^{-5t} is (e^5-1)/(5e^5)