What is the integral from 0 to ln(2) of e^{-t} ?

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

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integral from 0 to ln(2) of e^{-t}

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Apply u-substitution

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

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

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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