What is the integral from 0 to 2 of 5^{-x} ?

The integral from 0 to 2 of 5^{-x} is (24)/(25ln(5))

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integral from 0 to 2 of 5^{-x}

\int_{0}^{2} 5^{- x} d x

\frac{24}{25 ln ( 5 )}

Decimal

0.59648 \dots

Solution steps

\int_{0}^{2} 5^{- x} d x

Apply u-substitution

: \int_{0}^{- 2} - 5^{u} d u

= \int_{0}^{- 2} - 5^{u} d u

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

= - \int_{- 2}^{0} - 5^{u} d u

Take the constant out:

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

= - (- \int_{- 2}^{0} 5^{u} d u)

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

= - (- [\frac{5 ^{u}}{ln ( 5 )}]_{- 2}^{0})

Simplify

= [\frac{5 ^{u}}{ln ( 5 )}]_{- 2}^{0}

Compute the boundaries:

\frac{24}{25 ln ( 5 )}

= \frac{24}{25 ln ( 5 )}

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\frac{d}{d x} (sin (3 x^{3}))

\int \frac{2 x - 13}{x ^{2} - 13 x + 50} d x

y^{^{'}} + 0.564 y^{2} = 0

\frac{\partial}{\partial x} (4 x z sin (y))

What is the integral from 0 to 2 of 5^{-x} ?
The integral from 0 to 2 of 5^{-x} is (24)/(25ln(5))