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

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

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

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

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

Decimal Notation

0.36114 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

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