What is the integral from 0 to 1/2 of e^{x-1} ?

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

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integral from 0 to 1/2 of e^{x-1}

\int_{0}^{\frac{1}{2}} e^{x - 1} d x

\frac{e - 1}{e}

Decimal

0.23865 \dots

Solution steps

\int_{0}^{\frac{1}{2}} e^{x - 1} d x

e^{x - 1} = e^{- 1} e^{x}

= \int_{0}^{\frac{1}{2}} e^{- 1} e^{x} d x

Take the constant out:

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

= e^{- 1} \cdot \int_{0}^{\frac{1}{2}} e^{x} d x

Apply exponent rule:

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

= \frac{1}{e} \cdot \int_{0}^{\frac{1}{2}} e^{x} d x

Use the common integral:

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

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

Compute the boundaries:

e - 1

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

Simplify

= \frac{e - 1}{e}

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