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

The integral from 1 to 2 of 2^{-x} is 1/(4ln(2))

Popular >

integral from 1 to 2 of 2^{-x}

\int_{1}^{2} 2^{- x} d x

\frac{1}{4 ln ( 2 )}

Decimal Notation

0.36067 \dots

Solution steps

\int_{1}^{2} 2^{- x} d x

Apply u-substitution

: \int_{- 1}^{- 2} - 2^{u} d u

= \int_{- 1}^{- 2} - 2^{u} d u

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

= - \int_{- 2}^{- 1} - 2^{u} d u

Take the constant out:

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

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

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{- 2}^{- 1})

Simplify

= [\frac{2 ^{u}}{ln ( 2 )}]_{- 2}^{- 1}

Compute the boundaries:

\frac{1}{4 ln ( 2 )}

= \frac{1}{4 ln ( 2 )}

\int_{0}^{1} \frac{25}{4 y - 1} d y

x \to 0 + lim (x ln (x + x^{2}))

\frac{d P}{d x} = P - P^{2}

\frac{d}{d x} (3 x - 6)

\int e^{x} \frac{1}{x} d x

What is the integral from 1 to 2 of 2^{-x} ?
The integral from 1 to 2 of 2^{-x} is 1/(4ln(2))