What is the integral from 2 to 3 of 2^{-x} ?

The integral from 2 to 3 of 2^{-x} is 1/(8ln(2))

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integral from 2 to 3 of 2^{-x}

\int_{2}^{3} 2^{- x} d x

\frac{1}{8 ln ( 2 )}

Decimal

0.18033 \dots

Solution steps

\int_{2}^{3} 2^{- x} d x

Apply u-substitution

: \int_{- 2}^{- 3} - 2^{u} d u

= \int_{- 2}^{- 3} - 2^{u} d u

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

= - \int_{- 3}^{- 2} - 2^{u} d u

Take the constant out:

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

= - (- \int_{- 3}^{- 2} 2^{u} d u)

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

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{- 3}^{- 2})

Simplify

= [\frac{2 ^{u}}{ln ( 2 )}]_{- 3}^{- 2}

Compute the boundaries:

\frac{1}{8 ln ( 2 )}

= \frac{1}{8 ln ( 2 )}

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What is the integral from 2 to 3 of 2^{-x} ?
The integral from 2 to 3 of 2^{-x} is 1/(8ln(2))