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

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

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

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

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

Decimal

0.30189 \dots

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

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