What is the integral from 0 to ln(4) of e^{-x} ?

The integral from 0 to ln(4) of e^{-x} is 3/4

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integral from 0 to ln(4) of e^{-x}

\int_{0}^{l n (4)} e^{- x} d x

\frac{3}{4}

Decimal Notation

0.75

Solution steps

\int_{0}^{l n (4)} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 2 l n (2)} - e^{u} d u

= \int_{0}^{- 2 l n (2)} - e^{u} d u

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

= - \int_{- 2 l n (2)}^{0} - e^{u} d u

Take the constant out:

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

= - (- \int_{- 2 l n (2)}^{0} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- 2 l n (2)}^{0})

Simplify

= [e^{u}]_{- 2 l n (2)}^{0}

Compute the boundaries:

\frac{3}{4}

= \frac{3}{4}

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What is the integral from 0 to ln(4) of e^{-x} ?
The integral from 0 to ln(4) of e^{-x} is 3/4