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

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

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

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

e - \frac{1}{e}

Decimal Notation

2.35040 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

e - \frac{1}{e}

= e - \frac{1}{e}

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