What is the integral from 0 to 1 of 2^{-x} ?

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

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integral from 0 to 1 of 2^{-x}

\int_{0}^{1} 2^{- x} d x

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

Decimal Notation

0.72134 \dots

Solution steps

\int_{0}^{1} 2^{- x} d x

Apply u-substitution

: \int_{0}^{- 1} - 2^{u} d u

= \int_{0}^{- 1} - 2^{u} d u

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

= - \int_{- 1}^{0} - 2^{u} d u

Take the constant out:

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

= - (- \int_{- 1}^{0} 2^{u} d u)

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

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{- 1}^{0})

Simplify

= [\frac{2 ^{u}}{ln ( 2 )}]_{- 1}^{0}

Compute the boundaries:

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

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

\frac{d}{d x} (\frac{ln ( x )}{x})

\frac{d y}{d x} = (\frac{( 2 y + 7 )}{8 x + 9})^{2}

\int_{0}^{2} (6 t i - t^{3} j + 5 t^{9} k) d t

x \to 0 lim (\frac{x + 5}{3 x})

\frac{\partial}{\partial x} ((e^{\frac{x}{2}}) (2 x y))

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