What is the integral from 1 to ln(2) of e^{-x} ?

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

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integral from 1 to ln(2) of e^{-x}

\int_{1}^{l n (2)} e^{- x} d x

- \frac{1}{2} + \frac{1}{e}

Decimal Notation

- 0.13212 \dots

Solution steps

\int_{1}^{l n (2)} e^{- x} d x

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

= - \int_{l n (2)}^{1} e^{- x} d x

Apply u-substitution

: \int_{- l n (2)}^{- 1} - e^{u} d u

= - \int_{- l n (2)}^{- 1} - e^{u} d u

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

= - (- \int_{- 1}^{- l n (2)} - e^{u} d u)

Take the constant out:

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

= - (- (- \int_{- 1}^{- l n (2)} e^{u} d u))

Use the common integral:

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

= - (- (- [e^{u}]_{- 1}^{- l n (2)}))

Simplify

= - [e^{u}]_{- 1}^{- l n (2)}

Compute the boundaries:

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

= - (\frac{1}{2} - \frac{1}{e})

Simplify

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

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