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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

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}

\frac{d y}{d x} = \frac{sec ( y )}{( x - 1 ) ^{2}}

\int (- x^{2} - 2 x) d x

y = x, y = \frac{1}{2} x, x = 9

\frac{d}{d x} (\frac{1}{x ^{2} + 2 x + 1})

y = (s - 4) (s^{2} - 3), (1, 6)

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