What is the integral from 3 to 4 of 2^{-x} ?

The integral from 3 to 4 of 2^{-x} is 1/(16ln(2))

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integral from 3 to 4 of 2^{-x}

\int_{3}^{4} 2^{- x} d x

\frac{1}{16 ln ( 2 )}

Decimal

0.09016 \dots

Solution steps

\int_{3}^{4} 2^{- x} d x

Apply u-substitution

= \int_{- 3}^{- 4} - 2^{u} d u

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

= - \int_{- 4}^{- 3} - 2^{u} d u

Take the constant out:

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

= - (- \int_{- 4}^{- 3} 2^{u} d u)

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

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{- 4}^{- 3})

Simplify

= [\frac{2 ^{u}}{ln ( 2 )}]_{- 4}^{- 3}

Compute the boundaries:

\frac{1}{16 ln ( 2 )}

= \frac{1}{16 ln ( 2 )}

What is the integral from 3 to 4 of 2^{-x} ?
The integral from 3 to 4 of 2^{-x} is 1/(16ln(2))