What is the integral from 0 to 1 of 5^{-θ} ?

The integral from 0 to 1 of 5^{-θ} is 4/(5ln(5))

integral from 0 to 1 of 5^{-θ}

\int_{0}^{1} 5^{- θ} d θ

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

Decimal

0.49706 \dots

Solution steps

\int_{0}^{1} 5^{- θ} d θ

Apply u-substitution

= \int_{0}^{- 1} - 5^{u} d u

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

= - \int_{- 1}^{0} - 5^{u} d u

Take the constant out:

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

= - (- \int_{- 1}^{0} 5^{u} d u)

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

= - (- [\frac{5 ^{u}}{ln ( 5 )}]_{- 1}^{0})

Simplify

= [\frac{5 ^{u}}{ln ( 5 )}]_{- 1}^{0}

Compute the boundaries:

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

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

\frac{d y}{d x} = - 2 x y^{2}

\frac{d y}{d x} = \frac{x ^{8} - 4 y}{x}

\int (1 - a x)^{\frac{1}{2}} d x

\int sin (2 t) e^{t} d t

\frac{d}{d x} (sin (4 1 - 2 x^{3}))

What is the integral from 0 to 1 of 5^{-θ} ?
The integral from 0 to 1 of 5^{-θ} is 4/(5ln(5))