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

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

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integral from 1 to 0 of e^{-x}

\int_{1}^{0} e^{- x} d x

- 1 + \frac{1}{e}

Solution steps

\int_{1}^{0} e^{- x} d x

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

= - \int_{0}^{1} e^{- x} d x

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

1 - \frac{1}{e}

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

Simplify

= - 1 + \frac{1}{e}

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