What is the integral from 1 to 2 of 5^{-θ} ?

The integral from 1 to 2 of 5^{-θ} is 4/(25ln(5))

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integral from 1 to 2 of 5^{-θ}

\int_{1}^{2} 5^{- θ} d θ

\frac{4}{25 ln ( 5 )}

Decimal Notation

0.09941 \dots

Solution steps

\int_{1}^{2} 5^{- θ} d θ

Apply u-substitution

: \int_{- 1}^{- 2} - 5^{u} d u

= \int_{- 1}^{- 2} - 5^{u} d u

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

= - \int_{- 2}^{- 1} - 5^{u} d u

Take the constant out:

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

= - (- \int_{- 2}^{- 1} 5^{u} d u)

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

= - (- [\frac{5 ^{u}}{ln ( 5 )}]_{- 2}^{- 1})

Simplify

= [\frac{5 ^{u}}{ln ( 5 )}]_{- 2}^{- 1}

Compute the boundaries:

\frac{4}{25 ln ( 5 )}

= \frac{4}{25 ln ( 5 )}

\frac{\partial}{\partial x} (x^{2} + y^{2} + 1)

\frac{d ^{2}}{d x ^{2}} (x e^{- x})

\frac{1 + s}{( s ( s ^{2} + s + 1 ) )}

\int \frac{cos ( x )}{( sin ^{2} ( x ) + 25 ) ^{2}} d x

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

What is the integral from 1 to 2 of 5^{-θ} ?
The integral from 1 to 2 of 5^{-θ} is 4/(25ln(5))