What is the integral from 1 to 2 of 3^{-θ} ?

The integral from 1 to 2 of 3^{-θ} is 2/(9ln(3))

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integral from 1 to 2 of 3^{-θ}

\int_{1}^{2} 3^{- θ} d θ

\frac{2}{9 ln ( 3 )}

Decimal Notation

0.20227 \dots

Solution steps

\int_{1}^{2} 3^{- θ} d θ

Apply u-substitution

: \int_{- 1}^{- 2} - 3^{u} d u

= \int_{- 1}^{- 2} - 3^{u} d u

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

= - \int_{- 2}^{- 1} - 3^{u} d u

Take the constant out:

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

= - (- \int_{- 2}^{- 1} 3^{u} d u)

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

= - (- [\frac{3 ^{u}}{ln ( 3 )}]_{- 2}^{- 1})

Simplify

= [\frac{3 ^{u}}{ln ( 3 )}]_{- 2}^{- 1}

Compute the boundaries:

\frac{2}{9 ln ( 3 )}

= \frac{2}{9 ln ( 3 )}

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What is the integral from 1 to 2 of 3^{-θ} ?
The integral from 1 to 2 of 3^{-θ} is 2/(9ln(3))