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

The integral from 1 to 0 of ze^{-1} is -1/(2e)

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integral from 1 to 0 of ze^{-1}

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

- \frac{1}{2 e}

Decimal Notation

- 0.18393 \dots

Solution steps

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

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

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

Take the constant out:

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

= - e^{- 1} \cdot \int_{0}^{1} z d z

Apply exponent rule:

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

= - \frac{1}{e} \cdot \int_{0}^{1} z d z

Apply the Power Rule

: [\frac{z ^{2}}{2}]_{0}^{1}

= - \frac{1}{e} [\frac{z ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= - \frac{1}{e} \cdot \frac{1}{2}

Simplify

= - \frac{1}{2 e}

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What is the integral from 1 to 0 of ze^{-1} ?
The integral from 1 to 0 of ze^{-1} is -1/(2e)